### Archive

Archive for the ‘Statistics’ Category

## Using gsem to combine estimation results

gsem is a very flexible command that allows us to fit very sophisticated models. However, it is also useful in situations that involve simple models.

For example, when we want to compare parameters among two or more models, we usually use suest, which combines the estimation results under one parameter vector and creates a simultaneous covariance matrix of the robust type. This covariance estimate is described in the Methods and formulas of [R] suest as the robust variance from a “stacked model”. Actually, gsem can estimate these kinds of “stacked models”, even if the estimation samples are not the same and eventually overlap. By using the option vce(robust), we can replicate the results from suest if the models are available for gsem. In addition, gsem allows us to combine results from some estimation commands that are not supported by suest, like models including random effects.

### Example: Comparing parameters from two models

Let’s consider the childweight dataset, described in [ME] mixed. Consider the following models, where weights of boys and girls are modeled using the age and the age-squared:

. webuse childweight, clear
(Weight data on Asian children)

. regress  weight age c.age#c.age if girl == 0, noheader
------------------------------------------------------------------------------
weight |      Coef.   Std. Err.      t    P>|t|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
age |   7.985022   .6343855    12.59   0.000     6.725942    9.244101
|
c.age#c.age |   -1.74346   .2374504    -7.34   0.000    -2.214733   -1.272187
|
_cons |   3.684363   .3217223    11.45   0.000     3.045833    4.322893
------------------------------------------------------------------------------

. regress  weight age c.age#c.age if girl == 1, noheader
------------------------------------------------------------------------------
weight |      Coef.   Std. Err.      t    P>|t|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
age |   7.008066   .5164687    13.57   0.000     5.982746    8.033386
|
c.age#c.age |  -1.450582   .1930318    -7.51   0.000    -1.833798   -1.067365
|
_cons |   3.480933   .2616616    13.30   0.000     2.961469    4.000397
------------------------------------------------------------------------------


To test whether birthweights are the same for the two groups, we need to test whether the intercepts in the two regressions are the same. Using suest, we would proceed as follows:

. quietly regress weight age c.age#c.age if girl == 0, noheader

. estimates store boys

. quietly regress weight age c.age#c.age if girl == 1, noheader

. estimates store girls

. suest boys girls

Simultaneous results for boys, girls

Number of obs   =        198

------------------------------------------------------------------------------
|               Robust
|      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
boys_mean    |
age |   7.985022   .4678417    17.07   0.000     7.068069    8.901975
|
c.age#c.age |   -1.74346   .2034352    -8.57   0.000    -2.142186   -1.344734
|
_cons |   3.684363   .1719028    21.43   0.000      3.34744    4.021286
-------------+----------------------------------------------------------------
boys_lnvar   |
_cons |   .4770289   .1870822     2.55   0.011     .1103546    .8437032
-------------+----------------------------------------------------------------
girls_mean   |
age |   7.008066   .4166916    16.82   0.000     6.191365    7.824766
|
c.age#c.age |  -1.450582   .1695722    -8.55   0.000    -1.782937   -1.118226
|
_cons |   3.480933   .1556014    22.37   0.000      3.17596    3.785906
-------------+----------------------------------------------------------------
girls_lnvar  |
_cons |   .0097127   .1351769     0.07   0.943    -.2552292    .2746545
------------------------------------------------------------------------------


Invoking an estimation command with the option coeflegend will give us a legend we can use to refer to the parameters when we use postestimation commands like test.

. suest, coeflegend

Simultaneous results for boys, girls

Number of obs   =        198

------------------------------------------------------------------------------
|      Coef.  Legend
-------------+----------------------------------------------------------------
boys_mean    |
age |   7.985022  _b[boys_mean:age]
|
c.age#c.age |   -1.74346  _b[boys_mean:c.age#c.age]
|
_cons |   3.684363  _b[boys_mean:_cons]
-------------+----------------------------------------------------------------
boys_lnvar   |
_cons |   .4770289  _b[boys_lnvar:_cons]
-------------+----------------------------------------------------------------
girls_mean   |
age |   7.008066  _b[girls_mean:age]
|
c.age#c.age |  -1.450582  _b[girls_mean:c.age#c.age]
|
_cons |   3.480933  _b[girls_mean:_cons]
-------------+----------------------------------------------------------------
girls_lnvar  |
_cons |   .0097127  _b[girls_lnvar:_cons]
------------------------------------------------------------------------------

. test  _b[boys_mean:_cons] = _b[girls_mean:_cons]

( 1)  [boys_mean]_cons - [girls_mean]_cons = 0

chi2(  1) =    0.77
Prob > chi2 =    0.3803


We find no evidence that the intercepts are different.

Now, let’s replicate those results by using the gsem command. We generate the variable weightboy, a copy of weight for boys and missing otherwise, and the variable weightgirl, a copy of weight for girls and missing otherwise.

. quietly generate weightboy = weight if girl == 0

. quietly generate weightgirl = weight if girl == 1

. gsem (weightboy <- age c.age#c.age) (weightgirl <- age c.age#c.age), ///
>      nolog vce(robust)

Generalized structural equation model             Number of obs   =        198
Log pseudolikelihood =  -302.2308

-------------------------------------------------------------------------------
|              Robust
|      Coef.  Std. Err.     z   P>|z|     [95% Conf. Interval]
-----------------+-------------------------------------------------------------
weightboy <-     |
age |   7.985022  .4678417   17.07  0.000     7.068069    8.901975
|
c.age#c.age |   -1.74346  .2034352   -8.57  0.000    -2.142186   -1.344734
|
_cons |   3.684363  .1719028   21.43  0.000      3.34744    4.021286
-----------------+-------------------------------------------------------------
weightgirl <-    |
age |   7.008066  .4166916   16.82  0.000     6.191365    7.824766
|
c.age#c.age |  -1.450582  .1695722   -8.55  0.000    -1.782937   -1.118226
|
_cons |   3.480933  .1556014   22.37  0.000      3.17596    3.785906
-----------------+-------------------------------------------------------------
var(e.weightboy)|   1.562942  .3014028                    1.071012    2.280821
var(e.weightgirl)|    .978849  .1364603                    .7448187    1.286414
-------------------------------------------------------------------------------

. gsem, coeflegend

Generalized structural equation model             Number of obs   =        198
Log pseudolikelihood =  -302.2308

-------------------------------------------------------------------------------
|      Coef.  Legend
-----------------+-------------------------------------------------------------
weightboy <-     |
age |   7.985022  _b[weightboy:age]
|
c.age#c.age |   -1.74346  _b[weightboy:c.age#c.age]
|
_cons |   3.684363  _b[weightboy:_cons]
-----------------+-------------------------------------------------------------
weightgirl <-    |
age |   7.008066  _b[weightgirl:age]
|
c.age#c.age |  -1.450582  _b[weightgirl:c.age#c.age]
|
_cons |   3.480933  _b[weightgirl:_cons]
-----------------+-------------------------------------------------------------
var(e.weightboy)|   1.562942  _b[var(e.weightboy):_cons]
var(e.weightgirl)|    .978849  _b[var(e.weightgirl):_cons]
-------------------------------------------------------------------------------

. test  _b[weightgirl:_cons]=  _b[weightboy:_cons]

( 1)  - [weightboy]_cons + [weightgirl]_cons = 0

chi2(  1) =    0.77
Prob > chi2 =    0.3803


gsem allowed us to fit models on different subsets simultaneously. By default, the model is assumed to be a linear regression, but several links and families are available; for example, you can combine two Poisson models or a multinomial logistic model with a regular logistic model. See [SEM] sem and gsem for details.

Here, I use the vce(robust) option to replicate the results for suest. However, when estimation samples don’t overlap, results from both estimations are assumed to be independent, and thus the option vce(robust) is not needed. When performing the estimation without the vce(robust) option, the joint covariance matrix will contain two blocks with the covariances from the original models and 0s outside those blocks.

### An example with random effects

The childweight dataset contains repeated measures, and it is, in the documentation, analyzed used the mixed command, which allows us to account for the intra-individual correlation via random effects.

Now, let’s use the techniques described above to combine results from two random-effects models. Here are the two separate models:

. mixed weight age c.age#c.age if girl == 0 || id:, nolog

Mixed-effects ML regression                     Number of obs      =       100
Group variable: id                              Number of groups   =        34

Obs per group: min =         1
avg =       2.9
max =         5

Wald chi2(2)       =   1070.28
Log likelihood = -149.05479                     Prob > chi2        =    0.0000

------------------------------------------------------------------------------
weight |      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
age |   8.328882   .4601093    18.10   0.000     7.427084    9.230679
|
c.age#c.age |  -1.859798   .1722784   -10.80   0.000    -2.197458   -1.522139
|
_cons |   3.525929   .2723617    12.95   0.000      2.99211    4.059749
------------------------------------------------------------------------------

------------------------------------------------------------------------------
Random-effects Parameters  |   Estimate   Std. Err.     [95% Conf. Interval]
-----------------------------+------------------------------------------------
id: Identity                 |
var(_cons) |   .7607779   .2439115      .4058409    1.426133
-----------------------------+------------------------------------------------
var(Residual) |   .7225673   .1236759      .5166365    1.010582
------------------------------------------------------------------------------
LR test vs. linear regression: chibar2(01) =    30.34 Prob >= chibar2 = 0.0000

. mixed weight age c.age#c.age if girl == 1 || id:, nolog

Mixed-effects ML regression                     Number of obs      =        98
Group variable: id                              Number of groups   =        34

Obs per group: min =         1
avg =       2.9
max =         5

Wald chi2(2)       =   2141.72
Log likelihood =  -114.3008                     Prob > chi2        =    0.0000

------------------------------------------------------------------------------
weight |      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
age |   7.273082   .3167266    22.96   0.000     6.652309    7.893854
|
c.age#c.age |  -1.538309    .118958   -12.93   0.000    -1.771462   -1.305156
|
_cons |   3.354834   .2111793    15.89   0.000      2.94093    3.768738
------------------------------------------------------------------------------

------------------------------------------------------------------------------
Random-effects Parameters  |   Estimate   Std. Err.     [95% Conf. Interval]
-----------------------------+------------------------------------------------
id: Identity                 |
var(_cons) |   .6925554   .1967582       .396848    1.208606
-----------------------------+------------------------------------------------
var(Residual) |   .3034231   .0535359      .2147152    .4287799
------------------------------------------------------------------------------
LR test vs. linear regression: chibar2(01) =    47.42 Prob >= chibar2 = 0.0000


Random effects can be included in a gsem model by incorporating latent variables at the group level; these are the latent variables M1[id] and M2[id] below. By default, gsem will try to estimate a covariance when it sees two latent variables at the same level. This can be easily solved by restricting this covariance term to 0. Option vce(robust) should be used whenever we want to produce the mechanism used by suest.

. gsem (weightboy <- age c.age#c.age M1[id])   ///
>      (weightgirl <- age c.age#c.age M2[id]), ///
>      cov(M1[id]*M2[id]@0) vce(robust) nolog

Generalized structural equation model             Number of obs   =        198
Log pseudolikelihood = -263.35559

( 1)  [weightboy]M1[id] = 1
( 2)  [weightgirl]M2[id] = 1
(Std. Err. adjusted for clustering on id)
-------------------------------------------------------------------------------
|              Robust
|      Coef.  Std. Err.     z   P>|z|     [95% Conf. Interval]
-----------------+-------------------------------------------------------------
weightboy <-     |
age |   8.328882  .4211157   19.78  0.000      7.50351    9.154253
|
c.age#c.age |  -1.859798  .1591742  -11.68  0.000    -2.171774   -1.547823
|
M1[id] |          1 (constrained)
|
_cons |   3.525929  .1526964   23.09  0.000      3.22665    3.825209
-----------------+-------------------------------------------------------------
weightgirl <-    |
age |   7.273082  .3067378   23.71  0.000     6.671887    7.874277
|
c.age#c.age |  -1.538309   .120155  -12.80  0.000    -1.773808    -1.30281
|
M2[id] |          1 (constrained)
|
_cons |   3.354834  .1482248   22.63  0.000     3.064319     3.64535
-----------------+-------------------------------------------------------------
var(M1[id])|   .7607774  .2255575                     .4254915    1.360268
var(M2[id])|   .6925553  .1850283                    .4102429    1.169144
-----------------+-------------------------------------------------------------
var(e.weightboy)|   .7225674  .1645983                     .4623572    1.129221
var(e.weightgirl)|   .3034231  .0667975                    .1970877    .4671298
-------------------------------------------------------------------------------


Above, we have the joint output from the two models, which would allow us to perform tests among parameters in both models. Notice that option vce(robust) implies that standard errors will be clustered on the groups determined by id.

gsem, when called with the vce(robust) option, will complain if there are inconsistencies among the groups in the models (for example, if the random effects in both models were crossed).

### Checking that you are fitting the same model

In the previous model, gsem‘s default covariance structure included a term that wasn’t in the original two models, so we needed to include an additional restriction. This can be easy to spot in a simple model, but if you don’t want to rely just on a visual inspection, you can write a small loop to make sure that all the estimates in the joint model are actually also in the original models.

Let’s see an example with random effects, this time with overlapping data.

. *fit first model and save the estimates
. gsem (weightboy <- age c.age#c.age M1[id]), nolog

Generalized structural equation model             Number of obs   =        100
Log likelihood = -149.05479

( 1)  [weightboy]M1[id] = 1
-------------------------------------------------------------------------------
|      Coef.  Std. Err.     z    P>|z|     [95% Conf. Interval]
----------------+--------------------------------------------------------------
weightboy <-    |
age |   8.328882  .4609841   18.07   0.000     7.425369    9.232394
|
c.age#c.age |  -1.859798  .1725233  -10.78   0.000    -2.197938   -1.521659
|
M1[id] |          1 (constrained)
|
_cons |   3.525929  .2726322   12.93   0.000      2.99158    4.060279
----------------+--------------------------------------------------------------
var(M1[id])|   .7607774  .2439114                     .4058407    1.426132
----------------+--------------------------------------------------------------
var(e.weightboy)|   .7225674  .1236759                     .5166366    1.010582
-------------------------------------------------------------------------------

. mat b1 = e(b)

. *fit second model and save the estimates
. gsem (weight <- age M2[id]), nolog

Generalized structural equation model             Number of obs   =        198
Log likelihood = -348.32402

( 1)  [weight]M2[id] = 1
------------------------------------------------------------------------------
|      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
weight <-    |
age |   3.389281   .1152211    29.42   0.000     3.163452    3.615111
|
M2[id] |          1  (constrained)
|
_cons |   5.156913   .1803059    28.60   0.000      4.80352    5.510306
-------------+----------------------------------------------------------------
var(M2[id])|   .6076662   .2040674                      .3146395    1.173591
-------------+----------------------------------------------------------------
var(e.weight)|   1.524052   .1866496                      1.198819    1.937518
------------------------------------------------------------------------------

. mat b2 = e(b)

. *stack estimates from first and second models
. mat stacked = b1, b2

. *estimate joint model and save results
. gsem (weightboy <- age c.age#c.age M1[id]) ///
>      (weight <- age M2[id]), cov(M1[id]*M2[id]@0) vce(robust) nolog

Generalized structural equation model             Number of obs   =        198
Log pseudolikelihood = -497.37881

( 1)  [weightboy]M1[id] = 1
( 2)  [weight]M2[id] = 1
(Std. Err. adjusted for clustering on id)
-------------------------------------------------------------------------------
|              Robust
|      Coef.  Std. Err.     z    P>|z|     [95% Conf. Interval]
----------------+--------------------------------------------------------------
weightboy <-    |
age |   8.328882  .4211157   19.78   0.000      7.50351    9.154253
|
c.age#c.age |  -1.859798  .1591742  -11.68   0.000    -2.171774   -1.547823
|
M1[id] |          1 (constrained)
|
_cons |   3.525929  .1526964   23.09   0.000      3.22665    3.825209
----------------+--------------------------------------------------------------
weight <-       |
age |   3.389281  .1157835   29.27   0.000      3.16235    3.616213
|
M2[id] |          1 (constrained)
|
_cons |   5.156913  .1345701   38.32   0.000      4.89316    5.420665
----------------+--------------------------------------------------------------
var(M1[id])|   .7607774  .2255575                     .4254915    1.360268
var(M2[id])|   .6076662     .1974                     .3214791    1.148623
----------------+--------------------------------------------------------------
var(e.weightboy)|   .7225674  .1645983                     .4623572    1.129221
var(e.weight)|   1.524052  .1705637                     1.223877    1.897849
-------------------------------------------------------------------------------

. mat b = e(b)

. *verify that estimates from the joint model are the same as
. *from models 1 and 2
. local stripes : colfullnames(b)

. foreach l of local stripes{
2.    matrix  r1 =  b[1,"l'"]
3.    matrix r2 = stacked[1,"l'"]
4.    assert reldif(el(r1,1,1), el(r2,1,1))<1e-5
5. }


The loop above verifies that all the labels in the second model correspond to estimates in the first and that the estimates are actually the same. If you omit the restriction for the variance in the joint model, the assert command will produce an error.

### Technical note

As documented in [U] 20.21.2 Correlated errors: Cluster-robust standard errors, the formula for the robust estimator of the variance is

$V_{robust} = \hat V(\sum_{j=1}^N u'_ju_j) \hat V$

where $$N$$ is the number of observations, $$\hat V$$ is the conventional estimator of the variance, and for each observation $$j$$, $$u_j$$ is a row vector (with as many columns as parameters), which represents the contribution of this observation to the gradient. (If we stack the rows $$u_j$$, the columns of this matrix are the scores.)

When we apply suest, the matrix $$\hat V$$ is constructed as the stacked block-diagonal conventional variance estimates from the original submodels; this is the variance you will see if you apply gsem to the joint model without the vce(robust) option. The $$u_j$$ values used by suest are now the values from both estimations, so we have as many $$u_j$$ values as the sum of observations in the two original models and each row contains as many columns as the total number of parameters in both models. This is the exact operation that gsem, vce(robust) does.

When random effects are present, standard errors will be clustered on groups. Instead of observation-level contributions to the gradient, we would use cluster-level contributions. This means that observations in the two models would need to be clustered in a consistent manner; observations that are common to the two estimations would need to be in the same cluster in the two estimations.

Categories: Statistics Tags:

## How to simulate multilevel/longitudinal data

I was recently talking with my friend Rebecca about simulating multilevel data, and she asked me if I would show her some examples. It occurred to me that many of you might also like to see some examples, so I decided to post them to the Stata Blog.

### Introduction

We simulate data all the time at StataCorp and for a variety of reasons.

One reason is that real datasets that include the features we would like are often difficult to find. We prefer to use real datasets in the manual examples, but sometimes that isn’t feasible and so we create simulated datasets.

We also simulate data to check the coverage probabilities of new estimators in Stata. Sometimes the formulae published in books and papers contain typographical errors. Sometimes the asymptotic properties of estimators don’t hold under certain conditions. And every once in a while, we make coding mistakes. We run simulations during development to verify that a 95% confidence interval really is a 95% confidence interval.

Simulated data can also come in handy for presentations, teaching purposes, and calculating statistical power using simulations for complex study designs.

And, simulating data is just plain fun once you get the hang of it.

Some of you will recall Vince Wiggins’s blog entry from 2011 entitled “Multilevel random effects in xtmixed and sem — the long and wide of it” in which he simulated a three-level dataset. I’m going to elaborate on how Vince simulated multilevel data, and then I’ll show you some useful variations. Specifically, I’m going to talk about:

1. How to simulate single-level data
2. How to simulate two- and three-level data
3. How to simulate three-level data with covariates
4. How to simulate longitudinal data with random slopes
5. How to simulate longitudinal data with structured errors

### How to simulate single-level data

Let’s begin by simulating a trivially simple, single-level dataset that has the form

$y_i = 70 + e_i$

We will assume that e is normally distributed with mean zero and variance $$\sigma^2$$.

We’d want to simulate 500 observations, so let’s begin by clearing Stata’s memory and setting the number of observations to 500.

. clear
. set obs 500


Next, let’s create a variable named e that contains pseudorandom normally distributed data with mean zero and standard deviation 5:

. generate e = rnormal(0,5)


The variable e is our error term, so we can create an outcome variable y by typing

. generate y = 70 + e

. list y e in 1/5

+----------------------+
|        y           e |
|----------------------|
1. | 78.83927     8.83927 |
2. | 69.97774   -.0222647 |
3. | 69.80065   -.1993514 |
4. | 68.11398    -1.88602 |
5. | 63.08952   -6.910483 |
+----------------------+


We can fit a linear regression for the variable y to determine whether our parameter estimates are reasonably close to the parameters we specified when we simulated our dataset:

. regress y

Source |       SS       df       MS              Number of obs =     500
-------------+------------------------------           F(  0,   499) =    0.00
Model |           0     0           .           Prob > F      =       .
Residual |  12188.8118   499  24.4264766           R-squared     =  0.0000
Total |  12188.8118   499  24.4264766           Root MSE      =  4.9423

------------------------------------------------------------------------------
y |      Coef.   Std. Err.      t    P>|t|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
_cons |   69.89768    .221027   316.24   0.000     69.46342    70.33194
------------------------------------------------------------------------------


The estimate of _cons is 69.9, which is very close to 70, and the Root MSE of 4.9 is equally close to the error’s standard deviation of 5. The parameter estimates will not be exactly equal to the underlying parameters we specified when we created the data because we introduced randomness with the rnormal() function.

This simple example is just to get us started before we work with multilevel data. For familiarity, let’s fit the same model with the mixed command that we will be using later:

. mixed y, stddev

Mixed-effects ML regression                     Number of obs      =       500

Wald chi2(0)       =         .
Log likelihood = -1507.8857                     Prob > chi2        =         .

------------------------------------------------------------------------------
y |      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
_cons |   69.89768   .2208059   316.56   0.000     69.46491    70.33045
------------------------------------------------------------------------------

------------------------------------------------------------------------------
Random-effects Parameters  |   Estimate   Std. Err.     [95% Conf. Interval]
-----------------------------+------------------------------------------------
sd(Residual) |    4.93737   .1561334      4.640645    5.253068
------------------------------------------------------------------------------


The output is organized with the parameter estimates for the fixed part in the top table and the estimated standard deviations for the random effects in the bottom table. Just as previously, the estimate of _cons is 69.9, and the estimate of the standard deviation of the residuals is 4.9.

Okay. That really was trivial, wasn’t it? Simulating two- and three-level data is almost as easy.

### How to simulate two- and three-level data

I posted a blog entry last year titled “Multilevel linear models in Stata, part 1: Components of variance“. In that posting, I showed a diagram for a residual of a three-level model.

The equation for the variance-components model I fit had the form

$y_{ijk} = mu + u_i.. + u_{ij.} + e_{ijk}$

This model had three residuals, whereas the one-level model we just fit above had only one.

This time, let’s start with a two-level model. Let’s simulate a two-level dataset, a model for children nested within classrooms. We’ll index classrooms by i and children by j. The model is
$y_{ij} = mu + u_{i.} + e_{ij}$

For this toy model, let’s assume two classrooms with two students per classroom, meaning that we want to create a four-observation dataset, where the observations are students.

To create this four-observation dataset, we start by creating a two-observation dataset, where the observations are classrooms. Because there are two classrooms, we type

. clear
. set obs 2
. generate classroom = _n


From now on, we’ll refer to classroom as i. It’s easier to remember what variables mean if they have meaningful names.

Next, we’ll create a variable that contains each classroom’s random effect $$u_i$$, which we’ll assume follows an N(0,3) distribution.

. generate u_i = rnormal(0,3)

. list

+----------------------+
| classr~m         u_i |
|----------------------|
1. |        1    .7491351 |
2. |        2   -.0031386 |
+----------------------+


We can now expand our data to include two children per classroom by typing

. expand 2

. list

+----------------------+
| classr~m         u_i |
|----------------------|
1. |        1    .7491351 |
2. |        2   -.0031386 |
3. |        1    .7491351 |
4. |        2   -.0031386 |
+----------------------+


Now, we can think of our observations as being students. We can create a child ID (we’ll call it child rather than j), and we can create each child’s residual $$e_{ij}$$, which we will assume has an N(0,5) distribution:

. bysort classroom: generate child = _n

. generate e_ij = rnormal(0,5)

. list

+------------------------------------------+
| classr~m         u_i   child        e_ij |
|------------------------------------------|
1. |        1    .7491351       1    2.832674 |
2. |        1    .7491351       2    1.487452 |
3. |        2   -.0031386       1    6.598946 |
4. |        2   -.0031386       2   -.3605778 |
+------------------------------------------+


We now have nearly all the ingredients to calculate $$y_{ij}$$:

$$y_{ij} = mu + u_{i.} + e_{ij}$$

We’ll assume mu is 70. We type

. generate y = 70 + u_i + e_ij

. list y classroom u_i child e_ij, sepby(classroom)

+-----------------------------------------------------+
|        y   classr~m         u_i   child        e_ij |
|-----------------------------------------------------|
1. | 73.58181          1    .7491351       1    2.832674 |
2. | 72.23659          1    .7491351       2    1.487452 |
|-----------------------------------------------------|
3. | 76.59581          2   -.0031386       1    6.598946 |
4. | 69.63628          2   -.0031386       2   -.3605778 |
+-----------------------------------------------------+


Note that the random effect u_i is the same within each school, and each child has a different value for e_ij.

Our strategy was simple:

2. Create variables for the level ID and its random effect.
3. Expand the data by the number of observations within that level.
4. Repeat steps 2 and 3 until the bottom level is reached.

Let’s try this recipe for three-level data where children are nested within classrooms which are nested within schools. This time, I will index schools with i, classrooms with j, and children with k so that my model is

$y_{ijk} = mu + u_{i..} + u_{ij.} + e_{ijk}$

where

$$u_{i..}$$ ~ N(0,2)
$$u_{ij.}$$ ~ N(0,3)
$$u_{ijk}$$ ~ N(0,5)

Let’s create data for

(level 3, i)   2 schools

(level 2, j)   2 classrooms in each school

(level 1, k)  2 students in most classrooms; 3 students in i==2 & j==2

Begin by creating the level-three data for the two schools:

. clear
. set obs 2
. generate school = _n
. generate u_i = rnormal(0,2)
. list school u_i

+--------------------+
| school         u_i |
|--------------------|
1. |      1    3.677312 |
2. |      2   -3.193004 |
+--------------------+


Next, we expand the data so that we have the three classrooms nested within each of the schools, and we create its random effect:

. expand 2
. bysort school: generate classroom = _n
. generate u_ij = rnormal(0,3)
. list school u_i classroom u_ij, sepby(school)

+-------------------------------------------+
| school         u_i   classr~m        u_ij |
|-------------------------------------------|
1. |      1    3.677312          1    .9811059 |
2. |      1    3.677312          2   -3.482453 |
|-------------------------------------------|
3. |      2   -3.193004          1   -4.107915 |
4. |      2   -3.193004          2   -2.450383 |
+-------------------------------------------+


Finally, we expand the data so that we have three students in school 2′s classroom 2, and two students in all the other classrooms. Sorry for that complication, but I wanted to show you how to create unbalanced data.

In the previous examples, we’ve been typing things like expand 2, meaning double the observations. In this case, we need to do something different for school 2, classroom 2, namely,

. expand 3 if school==2 & classroom==2


and then we can just expand the rest:

. expand 2 if !(school==2 & clasroom==2)


Obviously, in a real simulation, you would probably want 16 to 25 students in each classroom. You could do something like that by typing

. expand 16+int((25-16+1)*runiform())


In any case, we will type

. expand 3 if school==2 & classroom==2

. expand 2 if !(school==2 & classroom==2)

. bysort school classroom: generate child = _n

. generate e_ijk = rnormal(0,5)

. generate y = 70 + u_i + u_ij + e_ijk

. list y school u_i classroom u_ij child e_ijk, sepby(classroom)

+------------------------------------------------------------------------+
|        y   school       u_i   classr~m        u_ij   child       e_ijk |
|------------------------------------------------------------------------|
1. | 76.72794        1  3.677312          1    .9811059       1    2.069526 |
2. | 69.81315        1  3.677312          1    .9811059       2   -4.845268 |
|------------------------------------------------------------------------|
3. | 74.09565        1  3.677312          2   -3.482453       1    3.900788 |
4. | 71.50263        1  3.677312          2   -3.482453       2    1.307775 |
|------------------------------------------------------------------------|
5. | 64.86206        2 -3.193004          1   -4.107915       1    2.162977 |
6. | 61.80236        2 -3.193004          1   -4.107915       2   -.8967164 |
|------------------------------------------------------------------------|
7. | 66.65285        2 -3.193004          2   -2.450383       1    2.296242 |
8. | 49.96139        2 -3.193004          2   -2.450383       2   -14.39522 |
9. | 64.41605        2 -3.193004          2   -2.450383       3    .0594433 |
+------------------------------------------------------------------------+


Regardless of how we generate the data, we must ensure that the school-level random effects u_i are the same within school and the classroom-level random effects u_ij are the same within classroom.

Concerning data construction, the example above we concocted to produce a dataset that would be easy to list. Let’s now create a dataset that is more reasonable:

$y_{ijk} = mu + u_{i..} + u_{ij.} + e_{ijk}$

where

$$u_{i..}$$ ~ N(0,2)
$$u_{ij.}$$ ~ N(0,3)
$$u_{ijk}$$ ~ N(0,5)

Let’s create data for

(level 3, i)   6 schools

(level 2, j)   10 classrooms in each school

(level 1, k)   16-25 students

. clear
. set obs 6
. generate school = _n
. generate u_i = rnormal(0,2)
. expand 10
. bysort school: generate classroom = _n
. generate u_ij = rnormal(0,3)
. expand 16+int((25-16+1)*runiform())
. bysort school classroom: generate child = _n
. generate e_ijk = rnormal(0,5)
. generate y = 70 + u_i + u_ij + e_ijk


We can use the mixed command to fit the model with our simulated data.

. mixed y || school: || classroom: , stddev

Mixed-effects ML regression                     Number of obs      =      1217

-----------------------------------------------------------
|   No. of       Observations per Group
Group Variable |   Groups    Minimum    Average    Maximum
----------------+------------------------------------------
school |        6        197      202.8        213
classroom |       60         16       20.3         25
-----------------------------------------------------------

Wald chi2(0)       =         .
Log likelihood = -3710.0673                     Prob > chi2        =         .

------------------------------------------------------------------------------
y |      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
_cons |   70.25941   .9144719    76.83   0.000     68.46707    72.05174
------------------------------------------------------------------------------

------------------------------------------------------------------------------
Random-effects Parameters  |   Estimate   Std. Err.     [95% Conf. Interval]
-----------------------------+------------------------------------------------
school: Identity             |
sd(_cons) |   2.027064   .7159027      1.014487    4.050309
-----------------------------+------------------------------------------------
classroom: Identity          |
sd(_cons) |   2.814152   .3107647       2.26647    3.494178
-----------------------------+------------------------------------------------
sd(Residual) |   4.828923   .1003814      4.636133     5.02973
------------------------------------------------------------------------------
LR test vs. linear regression:       chi2(2) =   379.37   Prob > chi2 = 0.0000


The parameter estimates from our simulated data match the parameters used to create the data pretty well: the estimate for _cons is 70.3, which is near 70; the estimated standard deviation for the school-level random effects is 2.02, which is near 2; the estimated standard deviation for the classroom-level random effects is 2.8, which is near 3; and the estimated standard deviation for the individual-level residuals is 4.8, which is near 5.

We’ve just done one reasonable simulation.

If we wanted to do a full simulation, we would need to do the above 100, 1,000, 10,000, or more times. We would put our code in a loop. And in that loop, we would keep track of whatever parameter interested us.

### How to simulate three-level data with covariates

Usually, we’re more interested in estimating the effects of the covariates than in estimating the variance of the random effects. Covariates are typically binary (such as male/female), categorical (such as race), ordinal (such as education level), or continuous (such as age).

Let’s add some covariates to our simulated data. Our model is

$y_{ijk} = mu + u_{i..} + u_{ij.} + e_{ijk}$

where

$$u_{i..}$$ ~ N(0,2)
$$u_{ij.}$$ ~ N(0,3)
$$u_{ijk}$$ ~ N(0,5)

We create data for

(level 3, i)   6 schools

(level 2, j)   10 classrooms in each school

(level 1, k)   16-25 students

(level 3, school i)       whether the school is in an urban environment

(level 2, classroom j)  teacher’s experience (years)

(level 1, student k)    student’s mother’s education level

We can create a binary covariate called urban at the school level that equals 1 if the school is located in an urban area and equals 0 otherwise.

. clear
. set obs 6
. generate school = _n
. generate u_i = rnormal(0,2)
. generate urban = runiform()<0.50


Here we assigned schools to one of the two groups with equal probability (runiform()<0.50), but we could have assigned 70% of the schools to be urban by typing

. generate urban = runiform()<0.70


At the classroom level, we could add a continuous covariate for the teacher's years of experience. We could generate this variable by using any of Stata's random-number functions (see help random_number_functions. In the example below, I've generated teacher's years of experience with a uniform distribution ranging from 5-20 years.

. expand 10
. bysort school: generate classroom = _n
. generate u_ij = rnormal(0,3)
. bysort school: generate teach_exp = 5+int((20-5+1)*runiform())


When we summarize our data, we see that teaching experience ranges from 6-20 years with an average of 13 years.

. summarize teach_exp

Variable |       Obs        Mean    Std. Dev.       Min        Max
-------------+--------------------------------------------------------
teach_exp |        60    13.21667    4.075939          6         20


At the child level, we could add a categorical/ordinal covariate for mother's highest level of education completed. After we expand the data and create the child ID and error variables, we can generate a uniformly distributed random variable, temprand, on the interval [0,1].

. expand 16+int((25-16+1)*runiform())
. bysort school classroom: generate child = _n
. generate e_ijk = rnormal(0,5)
. generate temprand = runiform()


We can assign children to different groups by using the egen command with cutpoints. In the example below, children whose value of temprand is in the interval [0,0.5) will be assigned to mother_educ==0, children whose value of temprand is in the interval [0.5,0.9) will be assigned to mother_educ==1, and children whose value of temprand is in the interval [0.9,1) will be assigned to mother_educ==2.

. egen mother_educ = cut(temprand), at(0,0.5, 0.9, 1) icodes
. label define mother_educ 0 "HighSchool" 1 "College" 2 ">College"
. label values mother_educ mother_educ


The resulting frequencies of each category are very close to the frequencies we specified in our egen command.

. tabulate mother_educ, generate(meduc)

mother_educ |      Freq.     Percent        Cum.
------------+-----------------------------------
HighSchool |        602       50.17       50.17
College |        476       39.67       89.83
>College |        122       10.17      100.00
------------+-----------------------------------
Total |      1,200      100.00


We used the option generate(meduc) in the tabulate command above to create indicator variables for each category of mother_educ. This will allow us to specify an effect size for each category when we create our outcome variable.

. summarize meduc*

Variable |       Obs        Mean    Std. Dev.       Min        Max
-------------+--------------------------------------------------------
meduc1 |      1200    .5016667    .5002057          0          1
meduc2 |      1200    .3966667    .4894097          0          1
meduc3 |      1200    .1016667    .3023355          0          1


Now, we can create an outcome variable called score by adding all our fixed and random effects together. We can specify an effect size (regression coefficient) for each fixed effect in our model.

. generate score = 70
+ (-2)*urban
+ 1.5*teach_exp
+ 0*meduc1
+ 2*meduc2
+ 5*meduc3
+ u_i + u_ij + e_ijk


I have specified that the grand mean is 70, urban schools will have scores 2 points lower than nonurban schools, and each year of teacher's experience will add 1.5 points to the students score.

Mothers whose highest level of education was high school (meduc1==1) will serve as the referent category for mother_educ(mother_educ==0). The scores of children whose mother completed college (meduc2==1 and mother_educ==1) will be 2 points higher than the children in the referent group. And the scores of children whose mother completed more than college (meduc3==1 and mother_educ==2) will be 5 points higher than the children in the referent group. Now, we can use the mixed command to fit a model to our simulated data. We used the indicator variables meduc1-meduc3 to create the data, but we will use the factor variable i.mother_educ to fit the model.

. mixed score urban teach_exp i.mother_educ  || school: ||
classroom: , stddev baselevel

Mixed-effects ML regression                     Number of obs      =      1259

-----------------------------------------------------------
|   No. of       Observations per Group
Group Variable |   Groups    Minimum    Average    Maximum
----------------+------------------------------------------
school |        6        200      209.8        217
classroom |       60         16       21.0         25
-----------------------------------------------------------

Wald chi2(4)       =    387.64
Log likelihood = -3870.5395                     Prob > chi2        =    0.0000

------------------------------------------------------------------------------
score |      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
urban |  -2.606451    2.07896    -1.25   0.210    -6.681138    1.468237
teach_exp |   1.584759    .096492    16.42   0.000     1.395638     1.77388
|
mother_educ |
HighSchool  |          0  (base)
College  |   2.215281   .3007208     7.37   0.000     1.625879    2.804683
>College  |   5.065907   .5237817     9.67   0.000     4.039314      6.0925
|
_cons |   68.95018   2.060273    33.47   0.000     64.91212    72.98824
------------------------------------------------------------------------------

------------------------------------------------------------------------------
Random-effects Parameters  |   Estimate   Std. Err.     [95% Conf. Interval]
-----------------------------+------------------------------------------------
school: Identity             |
sd(_cons) |   2.168154   .7713944      1.079559    4.354457
-----------------------------+------------------------------------------------
classroom: Identity          |
sd(_cons) |    3.06871   .3320171      2.482336    3.793596
-----------------------------+------------------------------------------------
sd(Residual) |   4.947779   .1010263      4.753681    5.149802
------------------------------------------------------------------------------
LR test vs. linear regression:       chi2(2) =   441.25   Prob > chi2 = 0.0000


“Close” is in the eye of the beholder, but to my eyes, the parameter estimates look remarkably close to the parameters that were used to simulate the data. The parameter estimates for the fixed part of the model are -2.6 for urban (parameter = -2), 1.6 for teach_exp (parameter = 1.5), 2.2 for the College category of mother_educ (parameter = 2), 5.1 for the >College category of mother_educ (parameter = 5), and 69.0 for the intercept (parameter = 70). The estimated standard deviations for the random effects are also very close to the simulation parameters. The estimated standard deviation is 2.2 (parameter = 2) at the school level, 3.1 (parameter = 3) at the classroom level, and 4.9 (parameter = 5) at the child level.

Some of you may disagree that the parameter estimates are close. My reply is that it doesn’t matter unless you’re simulating a single dataset for demonstration purposes. If you are, simply simulate more datasets until you get one that looks close enough for you. If you are simulating data to check coverage probabilities or to estimate statistical power, you will be averaging over thousands of simulated datasets and the results of any one of those datasets won’t matter.

### How to simulate longitudinal data with random slopes

Longitudinal data are often conceptualized as multilevel data where the repeated observations are nested within individuals. The main difference between ordinary multilevel models and multilevel models for longitudinal data is the inclusion of a random slope. If you are not familiar with random slopes, you can learn more about them in a blog entry I wrote last year (Multilevel linear models in Stata, part 2: Longitudinal data).

Simulating longitudinal data with a random slope is much like simulating two-level data, with a couple of modifications. First, the bottom level will be observations within person. Second, there will be an interaction between time (age) and a person-level random effect. So we will generate data for the following model:

$weight_{ij} = mu + age_{ij} + u_{0i.} + age*u_{1i.} + e_{ij}$

where

$$u_{0i.}$$ ~ N(0,3)   $$u_{1i.}$$ ~ N(0,1)   $$e_{ij}$$ ~ N(0,2)

Let’s begin by simulating longitudinal data for 300 people.

. clear
. set obs 300
. gen person = _n


For longitudinal data, we must create two person-level random effects: the variable u_0i is analogous to the random effect we created earlier, and the variable u_1i is the random effect for the slope over time.

. generate u_0i = rnormal(0,3)
. generate u_1i = rnormal(0,1)


Let’s expand the data so that there are five observations nested within each person. Rather than create an observation-level identification number, let’s create a variable for age that ranges from 12 to 16 years,

. expand 5
. bysort person: generate age = _n + 11


and create an observation-level error term from an N(0,2) distribution:

. generate e_ij = rnormal(0,2)

. list person u_0i u_1i age e_ij if person==1

+-------------------------------------------------+
| person       u_0i        u_1i   age        e_ij |
|-------------------------------------------------|
1. |      1   .9338312   -.3097848    12    1.172153 |
2. |      1   .9338312   -.3097848    13    2.935366 |
3. |      1   .9338312   -.3097848    14   -2.306981 |
4. |      1   .9338312   -.3097848    15   -2.148335 |
5. |      1   .9338312   -.3097848    16   -.4276625 |
+-------------------------------------------------+


The person-level random effects u_0i and u_1i are the same at all ages, and the observation-level random effects e_ij are different at each age. Now we’re ready to generate an outcome variable called weight, measured in kilograms, based on the following model:

$weight_{ij} = 3 + 3.6*age_{ij} + u_{0i} + age*u_{1i} + e_{ij}$

. generate weight = 3 + 3.6*age + u_0i + age*u_1i + e_ij


The random effect u_1i is multiplied by age, which is why it is called a random slope. We could rewrite the model as

$weight_{ij} = 3 + age_{ij}*(3.6 + u_{1i}) + u_{01} + e_{ij}$

Note that for each year of age, a person’s weight will increase by 3.6 kilograms plus some random amount specified by u_1j. In other words,the slope for age will be slightly different for each person.

We can use the mixed command to fit a model to our data:

. mixed weight age || person: age , stddev

Mixed-effects ML regression                     Number of obs      =      1500
Group variable: person                          Number of groups   =       300

Obs per group: min =         5
avg =       5.0
max =         5

Wald chi2(1)       =   3035.03
Log likelihood = -3966.3842                     Prob > chi2        =    0.0000

------------------------------------------------------------------------------
weight |      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
age |   3.708161   .0673096    55.09   0.000     3.576237    3.840085
_cons |   2.147311   .5272368     4.07   0.000     1.113946    3.180676
------------------------------------------------------------------------------

------------------------------------------------------------------------------
Random-effects Parameters  |   Estimate   Std. Err.     [95% Conf. Interval]
-----------------------------+------------------------------------------------
person: Independent          |
sd(age) |   .9979648   .0444139      .9146037    1.088924
sd(_cons) |    3.38705   .8425298      2.080103    5.515161
-----------------------------+------------------------------------------------
sd(Residual) |   1.905885   .0422249      1.824897    1.990468
------------------------------------------------------------------------------
LR test vs. linear regression:       chi2(2) =  4366.32   Prob > chi2 = 0.0000


The estimate for the intercept _cons = 2.1 is not very close to the original parameter value of 3, but the estimate of 3.7 for age is very close (parameter = 3.6). The standard deviations of the random effects are also very close to the parameters used to simulate the data. The estimate for the person level _cons is 2.1 (parameter = 2), the person-level slope is 0.997 (parameter = 1), and the observation-level residual is 1.9 (parameter = 2).

### How to simulate longitudinal data with structured errors

Longitudinal data often have an autoregressive pattern to their errors because of the sequential collection of the observations. Measurements taken closer together in time will be more similar than measurements taken further apart in time. There are many patterns that can be used to descibe the correlation among the errors, including autoregressive, moving average, banded, exponential, Toeplitz, and others (see help mixed##rspec).

Let’s simulate a dataset where the errors have a Toeplitz structure, which I will define below.

We begin by creating a sample with 500 people with a person-level random effect having an N(0,2) distribution.

. clear
. set obs 500
. gen person = _n
. generate u_i = rnormal(0,2)


Next, we can use the drawnorm command to create error variables with a Toeplitz pattern.

A Toeplitz 1 correlation matrix has the following structure:

. matrix V = ( 1.0, 0.5, 0.0, 0.0, 0.0  \     ///
0.5, 1.0, 0.5, 0.0, 0.0  \     ///
0.0, 0.5, 1.0, 0.5, 0.0  \     ///
0.0, 0.0, 0.5, 1.0, 0.5  \     ///
0.0, 0.0, 0.0, 0.5, 1.0 )

. matrix list V

symmetric V[5,5]
c1  c2  c3  c4  c5
r1   1
r2  .5   1
r3   0  .5   1
r4   0   0  .5   1
r5   0   0   0  .5   1


The correlation matrix has 1s on the main diagonal, and each pair of contiguous observations will have a correlation of 0.5. Observations more than 1 unit of time away from each other are assumed to be uncorrelated.

We must also define a matrix of means to use the drawnorm command.

. matrix M = (0 \ 0 \ 0 \ 0 \ 0)

. matrix list M

M[5,1]
c1
r1   0
r2   0
r3   0
r4   0
r5   0


Now, we’re ready to use the drawnorm command to create five error variables that have a Toeplitz 1 structure.

. drawnorm e1 e2 e3 e4 e5, means(M) cov(V)

. list in 1/2

+---------------------------------------------------------------------------+
| person        u_i         e1         e2        e3          e4          e5 |
|---------------------------------------------------------------------------|
1. |      1   5.303562  -1.288265  -1.201399   .353249    .0495944   -1.472762 |
2. |      2  -.0133588   .6949759    2.82179  .7195075   -1.032395    .1995016 |
+---------------------------------------------------------------------------+


Let’s estimate the correlation matrix for our simulated data to verify that our simulation worked as we expected.

. correlate e1-e5
(obs=300)

|       e1       e2       e3       e4       e5
-------------+---------------------------------------------
e1 |   1.0000
e2 |   0.5542   1.0000
e3 |  -0.0149   0.4791   1.0000
e4 |  -0.0508  -0.0364   0.5107   1.0000
e5 |   0.0022  -0.0615   0.0248   0.4857   1.0000


The correlations are 1 along the main diagonal, near 0.5 for the contiguous observations, and near 0 otherwise.

Our data are currently in wide format, and we need them in long format to use the mixed command. We can use the reshape command to convert our data from wide to long format. If you are not familiar with the reshape command, you can learn more about it by typing help reshape.

. reshape long e, i(person) j(time)
(note: j = 1 2 3 4 5)

Data                               wide   ->   long
-----------------------------------------------------------------------------
Number of obs.                      300   ->    1500
Number of variables                   7   ->       4
j variable (5 values)                     ->   time
xij variables:
e1 e2 ... e5   ->   e
-----------------------------------------------------------------------------


Now, we are ready to create our age variable and the outcome variable weight.

. bysort person: generate age = _n + 11
. generate weight = 3 + 3.6*age + u_i + e

. list weight person u_i time age e if person==1

+-------------------------------------------------------+
|   weight   person        u_i   time   age           e |
|-------------------------------------------------------|
1. |  50.2153        1   5.303562      1    12   -1.288265 |
2. | 53.90216        1   5.303562      2    13   -1.201399 |
3. | 59.05681        1   5.303562      3    14     .353249 |
4. | 62.35316        1   5.303562      4    15    .0495944 |
5. |  64.4308        1   5.303562      5    16   -1.472762 |
+-------------------------------------------------------+


We can use the mixed command to fit a model to our simulated data.

. mixed weight age || person:, residual(toeplitz 1, t(time)) , stddev

Mixed-effects ML regression                     Number of obs      =      1500
Group variable: person                          Number of groups   =       300

Obs per group: min =         5
avg =       5.0
max =         5

Wald chi2(1)       =  33797.58
Log likelihood = -2323.9389                     Prob > chi2        =    0.0000

------------------------------------------------------------------------------
weight |      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
age |   3.576738   .0194556   183.84   0.000     3.538606     3.61487
_cons |   3.119974   .3244898     9.62   0.000     2.483985    3.755962
------------------------------------------------------------------------------

------------------------------------------------------------------------------
Random-effects Parameters  |   Estimate   Std. Err.     [95% Conf. Interval]
-----------------------------+------------------------------------------------
person: Identity             |
sd(_cons) |   3.004718   .1268162      2.766166    3.263843
-----------------------------+------------------------------------------------
Residual: Toeplitz(1)        |
rho1 |   .4977523   .0078807      .4821492    .5130398
sd(e) |   .9531284   .0230028      .9090933    .9992964
------------------------------------------------------------------------------
LR test vs. linear regression:       chi2(2) =  3063.87   Prob > chi2 = 0.0000


Again, our parameter estimates match the parameters that were used to simulate the data very closely.

The parameter estimate is 3.6 for age (parameter = 3.6) and 3.1 for _cons (parameter = 3). The estimated standard deviations of the person-level random effect is 3.0 (parameter = 3). The estimated standard deviation for the errors is 0.95 (parameter = 1), and the estimated correlation for the Toeplitz structure is 0.5 (parameter = 0.5).

### Conclusion

I hope I’ve convinced you that simulating multilevel/longitudinal data is easy and useful. The next time you find yourself teaching a class or giving a talk that requires multilevel examples, try simulating the data. And if you need to calculate statistical power for a multilevel or longitudinal model, consider simulations.

Categories: Statistics Tags:

## Using resampling methods to detect influential points

As stated in the documentation for jackknife, an often forgotten utility for this command is the detection of overly influential observations.

Some commands, like logit or stcox, come with their own set of prediction tools to detect influential points. However, these kinds of predictions can be computed for virtually any regression command. In particular, we will see that the dfbeta statistics can be easily computed for any command that accepts the jackknife prefix. dfbeta statistics allow us to visualize how influential some observations are compared with the rest, concerning a specific parameter.

We will also compute Cook’s likelihood displacement, which is an overall measure of influence, and it can also be compared with a specific threshold.

### Using jackknife to compute dfbeta

The main task of jackknife is to fit the model while suppressing one observation at a time, which allows us to see how much results change when each observation is suppressed; in other words, it allows us to see how much each observation influences the results. A very intuitive measure of influence is dfbeta, which is the amount that a particular parameter changes when an observation is suppressed. There will be one dfbeta variable for each parameter. If $$\hat\beta$$ is the estimate for parameter $$\beta$$ obtained from the full data and $$\hat\beta_{(i)}$$ is the corresponding estimate obtained when the $$i$$th observation is suppressed, then the $$i$$th element of variable dfbeta is obtained as

$dfbeta = \hat\beta - \hat\beta_{(i)}$

Parameters $$\hat\beta$$ are saved by the estimation commands in matrix e(b) and also can be obtained using the _b notation, as we will show below. The leave-one-out values $$\hat\beta_{(i)}$$ can be saved in a new file by using the option saving() with jackknife. With these two elements, we can compute the dfbeta values for each variable.

Let’s see an example with the probit command.

. sysuse auto, clear
(1978 Automobile Data)

. *preserve original dataset
. preserve

. *generate a variable with the original observation number
. gen obs =_n

. probit foreign mpg weight

Iteration 0:   log likelihood =  -45.03321
Iteration 1:   log likelihood = -27.914626
Iteration 2:   log likelihood = -26.858074
Iteration 3:   log likelihood = -26.844197
Iteration 4:   log likelihood = -26.844189
Iteration 5:   log likelihood = -26.844189

Probit regression                                 Number of obs   =         74
LR chi2(2)      =      36.38
Prob > chi2     =     0.0000
Log likelihood = -26.844189                       Pseudo R2       =     0.4039

------------------------------------------------------------------------------
foreign |      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
mpg |  -.1039503   .0515689    -2.02   0.044    -.2050235   -.0028772
weight |  -.0023355   .0005661    -4.13   0.000     -.003445   -.0012261
_cons |   8.275464   2.554142     3.24   0.001     3.269437    13.28149
------------------------------------------------------------------------------

. *keep the estimation sample so each observation will be matched
. *with the corresponding replication
. keep if e(sample)
(0 observations deleted)

. *use jackknife to generate the replications, and save the values in
. *file b_replic
. jackknife, saving(b_replic, replace):  probit foreign mpg weight
(running probit on estimation sample)

Jackknife replications (74)
----+--- 1 ---+--- 2 ---+--- 3 ---+--- 4 ---+--- 5
..................................................    50
........................

Probit regression                               Number of obs      =        74
Replications       =        74
F(   2,     73)    =     10.36
Prob > F           =    0.0001
Log likelihood = -26.844189                     Pseudo R2          =    0.4039

------------------------------------------------------------------------------
|              Jackknife
foreign |      Coef.   Std. Err.      t    P>|t|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
mpg |  -.1039503   .0831194    -1.25   0.215     -.269607    .0617063
weight |  -.0023355   .0006619    -3.53   0.001    -.0036547   -.0010164
_cons |   8.275464   3.506085     2.36   0.021     1.287847    15.26308
------------------------------------------------------------------------------

. *verify that all the replications were successful
. assert e(N_misreps) ==0

. merge 1:1 _n using b_replic

Result                           # of obs.
-----------------------------------------
not matched                             0
matched                                74  (_merge==3)
-----------------------------------------

. *see how values from replications are stored
. describe, fullnames

Contains data from .../auto.dta
obs:            74                          1978 Automobile Data
vars:            17                          13 Apr 2013 17:45
size:         4,440                          (_dta has notes)
--------------------------------------------------------------------------------
storage   display    value
variable name   type    format     label      variable label
--------------------------------------------------------------------------------
make            str18   %-18s                 Make and Model
price           int     %8.0gc                Price
mpg             int     %8.0g                 Mileage (mpg)
rep78           int     %8.0g                 Repair Record 1978
trunk           int     %8.0g                 Trunk space (cu. ft.)
weight          int     %8.0gc                Weight (lbs.)
length          int     %8.0g                 Length (in.)
turn            int     %8.0g                 Turn Circle (ft.)
displacement    int     %8.0g                 Displacement (cu. in.)
gear_ratio      float   %6.2f                 Gear Ratio
foreign         byte    %8.0g      origin     Car type
obs             float   %9.0g
foreign_b_mpg   float   %9.0g                 [foreign]_b[mpg]
foreign_b_weight
float   %9.0g                 [foreign]_b[weight]
foreign_b_cons  float   %9.0g                 [foreign]_b[_cons]
_merge          byte    %23.0g     _merge
--------------------------------------------------------------------------------
Sorted by:
Note:  dataset has changed since last saved

. *compute the dfbeta for each covariate
. foreach var in mpg weight {
2.  gen dfbeta_var' = (_b[var'] -foreign_b_var')
3. }

. gen dfbeta_cons = (_b[_cons] - foreign_b_cons)

. label var obs "observation number"
. label var dfbeta_mpg "dfbeta for mpg"
. label var dfbeta_weight "dfbeta for weight"
. label var dfbeta_cons "dfbeta for the constant"

. *plot dfbeta values for variable mpg
. scatter dfbeta_mpg obs, mlabel(obs) title("dfbeta values for variable mpg")

. *restore original dataset
. restore


Based on the impact on the coefficient for variable mpg, observation 71 seems to be the most influential. We could create a similar plot for each parameter.

jackknife prints a dot for each successful replication and an ‘x’ for each replication that ends with an error. By looking at the output immediately following the jackknife command, we can see that all the replications were successful. However, we added an assert line in the code to avoid relying on visual inspection. If some replications failed, we would need to explore the reasons.

### A computational shortcut to obtain the dfbeta values

The command jackknife allows us to save the leave-one-out values in a different file. To use these, we would need to do some data management and merge the two files. On the other hand, the same command called with the option keep saves pseudovalues, which are defined as follows:

$\hat{\beta}_i^* = N\hat\beta - (N-1)\hat\beta_{(i)}$

where $$N$$ is the number of observations involved in the computation, returned as e(N). Therefore, using the pseudovalues, $$\beta_{(i)}$$ values can be computed as $\hat\beta_{(i)} = \frac{ N \hat\beta - \hat\beta^*_i}{N-1}$

Also, dfbeta values can be computed directly from the pseudovalues as $\hat\beta - \hat\beta_{(i)} = \frac{\hat\beta_{i}^* -\hat\beta} {N-1}$

Using the pseudovalues instead of the leave-one-out values simplifies our program because we don’t have to worry about matching each pseudovalue to the correct observation.

Let’s reproduce the previous example.

. sysuse auto, clear
(1978 Automobile Data)

. jackknife, keep: probit foreign  mpg weight
(running probit on estimation sample)

Jackknife replications (74)
----+--- 1 ---+--- 2 ---+--- 3 ---+--- 4 ---+--- 5
..................................................    50
........................

Probit regression                               Number of obs      =        74
Replications       =        74
F(   2,     73)    =     10.36
Prob > F           =    0.0001
Log likelihood = -26.844189                     Pseudo R2          =    0.4039

------------------------------------------------------------------------------
|              Jackknife
foreign |      Coef.   Std. Err.      t    P>|t|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
mpg |  -.1039503   .0831194    -1.25   0.215     -.269607    .0617063
weight |  -.0023355   .0006619    -3.53   0.001    -.0036547   -.0010164
_cons |   8.275464   3.506085     2.36   0.021     1.287847    15.26308
------------------------------------------------------------------------------

. *see how pseudovalues are stored
. describe, fullnames

> dta
obs:            74                          1978 Automobile Data
vars:            15                          13 Apr 2013 17:45
size:         4,070                          (_dta has notes)
--------------------------------------------------------------------------------
storage   display    value
variable name   type    format     label      variable label
--------------------------------------------------------------------------------
make            str18   %-18s                 Make and Model
price           int     %8.0gc                Price
mpg             int     %8.0g                 Mileage (mpg)
rep78           int     %8.0g                 Repair Record 1978
trunk           int     %8.0g                 Trunk space (cu. ft.)
weight          int     %8.0gc                Weight (lbs.)
length          int     %8.0g                 Length (in.)
turn            int     %8.0g                 Turn Circle (ft.)
displacement    int     %8.0g                 Displacement (cu. in.)
gear_ratio      float   %6.2f                 Gear Ratio
foreign         byte    %8.0g      origin     Car type
foreign_b_mpg   float   %9.0g                 pseudovalues: [foreign]_b[mpg]
foreign_b_weight
float   %9.0g                 pseudovalues: [foreign]_b[weight]
foreign_b_cons  float   %9.0g                 pseudovalues: [foreign]_b[_cons]
--------------------------------------------------------------------------------
Sorted by:  foreign
Note:  dataset has changed since last saved

. *verify that all the replications were successful
. assert e(N_misreps)==0

. *compute the dfbeta for each covariate
. local N = e(N)

. foreach var in  mpg weight {
2. gen dfbeta_var' = (foreign_b_var' - _b[var'])/(N'-1)
3. }

. gen dfbeta_cons' = (foreign_b_cons - _b[_cons])/(N'-1)

. *plot deff values for variable weight
. gen obs = _n

. label var obs "observation number"

. label var dfbeta_mpg "dfbeta for mpg"

. scatter dfbeta_mpg obs, mlabel(obs) title("dfbeta values for variable mpg")


### Dfbeta for grouped data

If you have panel data or a situation where each individual is represented by a group of observations (for example, conditional logit or survival models), you might be interested in influential groups. In this case, you would look at the changes on the parameters when each group is suppressed. Let’s see an example with xtlogit.

. webuse towerlondon, clear
. xtset family

. jackknife, cluster(family) idcluster(newclus) keep: xtlogit dtlm difficulty
. assert e(N_misreps)==0


The group-level pseudovalues will be saved on the first observations corresponding to each group, and there will be missing values on the rest. To compute the dfbeta value for the coefficient for difficulty, we type

. local N = e(N_clust)
. gen dfbeta_difficulty = (dtlm_b_difficulty - _b[difficulty])/(N'-1)


We can then plot those values:

. scatter dfbeta_difficulty newclus, mlabel(family) ///
title("dfbeta values for variable difficulty") xtitle("family")


Option idcluster() for jackknife generates a new variable that assigns consecutive integers to the clusters; using this variable produces a plot where families are equally spaced on the horizontal axis.

As before, we can see that some groups are more influential than others. It would require some research to find out whether this is a problem.

### Likelihood displacement

If we want a global measure of influence (that is, not tied to a particular parameter), we can compute the likelihood displacement values. We consider the likelihood displacement value as defined by Cook (1986):

$LD_i = 2[L(\hat\theta) - L(\hat\theta_{(i)})]$

where $$L$$ is the log-likelihood function (evaluated on the full dataset), $$\hat\theta$$ is the set of parameter estimates obtained from the full dataset, and $$\hat\theta_{(i)}$$ is the set of the parameter estimates obtained when leaving out the $$i$$th observation. Notice that what changes is the parameter vector. The log-likelihood function is always evaluated on the whole sample; provided that $$\hat\theta$$ is the set of parameters that maximizes the log likelihood, the log-likelihood displacement is always positive. Cook suggested, as a confidence region for this value, the interval $$[0, \chi^2_p(\alpha))$$, where $$\chi^2_p(\alpha)$$ is the ($$1-\alpha$$) quantile from a chi-squared distribution with $$p$$ degrees of freedom, and $$p$$ is the number of parameters in $$\theta$$.

To perform our assessment based on the likelihood displacement, we will need to do the following:

1. Create an $$N\times p$$ matrix B, where the $$i$$th row contains the vector of parameter estimates obtained by leaving the $$i$$th observation out.
2. Create a new variable L1 such that its $$i$$th observation contains the log likelihood evaluated at the parameter estimates in the $$i$$th row of matrix B.
3. Use variable L1 to obtain the LD matrix, containing the likelihood displacement values.
4. Construct a plot for the values in LD, and add the $$\chi^2_p(\alpha)$$ as a reference.

Let's do it with our probit model.

#### Step 1.

We first create the macro cmdline containing the command line for the model we want to use. We fit the model and save the original log likelihood in macro ll0.

With a loop, the leave-one-out parameters are saved in consecutive rows of matrix B. It is useful to have those values in a matrix, because we will then extract each row to evaluate the log likelihood at those values.

**********Step 1
sysuse auto, clear
set more off
local cmdline probit foreign weight mpg
cmdline'
keep if e(sample)
local ll0 = e(ll)
mat b0 = e(b)
mat b = b0

local N = _N

forvalues i = 1(1)N'{
cmdline' if _n !=i'
mat b1 = e(b)
mat b = b \ b1
}

mat B = b[2...,1...]
mat list B


#### Step 2.

In each iteration of a loop, a row from B is stored as matrix b. To evaluate the log likelihood at these values, the trick is to use them as initial values and invoke the command with 0 iterations. This can be done for any command that is based on ml.

**********Step 2

gen L1 = .

forvalues i = 1(1)N'{
mat b = B[i',1...]
cmdline', from(b) iter(0)
local ll = e(ll)
replace L1 = ll' in i'
}


#### Step 3.

Using variable L1 and the macro with the original log likelihood, we compute Cook's likehood displacement.

**********Step 3

gen LD = 2*(ll0' - L1)


#### Step 4.

Create the plot, using as a reference the 90% quantile for the $$\chi^2$$ distribution. $$p$$ is the number of columns in matrix b0 (or equivalently, the number of columns in matrix B).

**********Step 4

local k = colsof(b0)
gen upper_bound = invchi2tail(k', .1)
gen n = _n

twoway scatter LD n, mlabel(n) || line upper_bound n, ///
title("Likelihood displacement")


We can see that observation 71 is the most influential, and its likelihood displacement value is within the range we would normally expect.

### Reference

Cook, D. 1986. Assessment of local influence. Journal of the Royal Statistical Society, Series B 48: 133–169.

Categories: Statistics Tags:

## Fitting ordered probit models with endogenous covariates with Stata’s gsem command

The new command gsem allows us to fit a wide variety of models; among the many possibilities, we can account for endogeneity on different models. As an example, I will fit an ordinal model with endogenous covariates.

### Parameterizations for an ordinal probit model

The ordinal probit model is used to model ordinal dependent variables. In the usual parameterization, we assume that there is an underlying linear regression, which relates an unobserved continuous variable $$y^*$$ to the covariates $$x$$.

$y^*_{i} = x_{i}\gamma + u_i$

The observed dependent variable $$y$$ relates to $$y^*$$ through a series of cut-points $$-\infty =\kappa_0<\kappa_1<\dots< \kappa_m=+\infty$$ , as follows:

$y_{i} = j {\mbox{ if }} \kappa_{j-1} < y^*_{i} \leq \kappa_j$

Provided that the variance of $$u_i$$ can’t be identified from the observed data, it is assumed to be equal to one. However, we can consider a re-scaled parameterization for the same model; a straightforward way of seeing this, is by noting that, for any positive number $$M$$:

$\kappa_{j-1} < y^*_{i} \leq \kappa_j \iff M\kappa_{j-1} < M y^*_{i} \leq M\kappa_j$

that is,

$\kappa_{j-1} < x_i\gamma + u_i \leq \kappa_j \iff M\kappa_{j-1}< x_i(M\gamma) + Mu_i \leq M\kappa_j$

In other words, if the model is identified, it can be represented by multiplying the unobserved variable $$y$$ by a positive number, and this will mean that the standard error of the residual component, the coefficients, and the cut-points will be multiplied by this number.

Let me show you an example; I will first fit a standard ordinal probit model, both with oprobit and with gsem. Then, I will use gsem to fit an ordinal probit model where the residual term for the underlying linear regression has a standard deviation equal to 2. I will do this by introducing a latent variable $$L$$, with variance 1, and coefficient $$\sqrt 3$$. This will be added to the underlying latent residual, with variance 1; then, the ‘new’ residual term will have variance equal to $$1+((\sqrt 3)^2\times Var(L))= 4$$, so the standard deviation will be 2. We will see that as a result, the coefficients, as well as the cut-points, will be multiplied by 2.

. sysuse auto, clear
(1978 Automobile Data)

. oprobit rep mpg disp , nolog

Ordered probit regression                         Number of obs   =         69
LR chi2(2)      =      14.68
Prob > chi2     =     0.0006
Log likelihood = -86.352646                       Pseudo R2       =     0.0783

------------------------------------------------------------------------------
rep78 |      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
mpg |   .0497185   .0355452     1.40   0.162    -.0199487    .1193858
displacement |  -.0029884   .0021498    -1.39   0.165     -.007202    .0012252
-------------+----------------------------------------------------------------
/cut1 |  -1.570496   1.146391                      -3.81738    .6763888
/cut2 |  -.7295982   1.122361                     -2.929386     1.47019
/cut3 |   .6580529   1.107838                     -1.513269    2.829375
/cut4 |    1.60884   1.117905                     -.5822132    3.799892
------------------------------------------------------------------------------

. gsem (rep <- mpg disp, oprobit), nolog

Generalized structural equation model             Number of obs   =         69
Log likelihood = -86.352646

--------------------------------------------------------------------------------
|      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]
---------------+----------------------------------------------------------------
rep78 <-       |
mpg |   .0497185   .0355452     1.40   0.162    -.0199487    .1193858
displacement |  -.0029884   .0021498    -1.39   0.165     -.007202    .0012252
---------------+----------------------------------------------------------------
rep78          |
/cut1 |  -1.570496   1.146391    -1.37   0.171     -3.81738    .6763888
/cut2 |  -.7295982   1.122361    -0.65   0.516    -2.929386     1.47019
/cut3 |   .6580529   1.107838     0.59   0.553    -1.513269    2.829375
/cut4 |    1.60884   1.117905     1.44   0.150    -.5822132    3.799892
--------------------------------------------------------------------------------

. local a = sqrt(3)

. gsem (rep <- mpg disp L@a'), oprobit var(L@1) nolog

Generalized structural equation model             Number of obs   =         69
Log likelihood = -86.353008

( 1)  [rep78]L = 1.732051
( 2)  [var(L)]_cons = 1
--------------------------------------------------------------------------------
|      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]
---------------+----------------------------------------------------------------
rep78 <-       |
mpg |    .099532     .07113     1.40   0.162    -.0398802    .2389442
displacement |  -.0059739   .0043002    -1.39   0.165    -.0144022    .0024544
L |   1.732051  (constrained)
---------------+----------------------------------------------------------------
rep78          |
/cut1 |  -3.138491   2.293613    -1.37   0.171     -7.63389    1.356907
/cut2 |  -1.456712   2.245565    -0.65   0.517    -5.857938    2.944513
/cut3 |   1.318568    2.21653     0.59   0.552     -3.02575    5.662887
/cut4 |   3.220004   2.236599     1.44   0.150     -1.16365    7.603657
---------------+----------------------------------------------------------------
var(L)|          1  (constrained)
--------------------------------------------------------------------------------

### Ordinal probit model with endogenous covariates

This model is defined analogously to the model fitted by -ivprobit- for probit models with endogenous covariates; we assume an underlying model with two equations,

$\begin{eqnarray} y^*_{1i} =& y_{2i} \beta + x_{1i} \gamma + u_i & \\ y_{2i} =& x_{1i} \pi_1 + x_{2i} \pi_2 + v_i & \,\,\,\,\,\, (1) \end{eqnarray}$

where $$u_i \sim N(0, 1)$$, $$v_i\sim N(0,s^2)$$, and $$corr(u_i, v_i) = \rho$$.

We don’t observe $$y^*_{1i}$$; instead, we observe a discrete variable $$y_{1i}$$, such as, for a set of cut-points (to be estimated) $$\kappa_0 = -\infty < \kappa_1 < \kappa_2 \dots < \kappa_m = +\infty$$,

$y_{1i} = j {\mbox{ if }} \kappa_{j-1} < y^*_{1i} \leq \kappa_j$

### The parameterization we will use

I will re-scale the first equation, preserving the correlation. That is, I will consider the following system:

$\begin{eqnarray} z^*_{1i} =& y_{2i}b +x_{1i}c + t_i + \alpha L_i &\\ y_{2i} = &x_{1i}\pi_1 + x_{2i}\pi_2 + w_i + \alpha L_i & \,\,\,\,\,\, (2) \end{eqnarray}$

where $$t_i, w_i, L_i$$ are independent, $$t_i \sim N(0, 1)$$ , $$w_i \sim N(0,\sigma^2)$$, $$L_i \sim N(0, 1)$$

$y_{1i} = j {\mbox{ if }} \lambda_{j-1} < z^*_{1i} \leq \lambda_j$

By introducing a latent variable in both equations, I am modeling a correlation between the error terms. The fist equation is a re-scaled version of the original equation, that is, $$z^*_1 = My^*_1$$,

$y_{2i}b +x_{1i}c + t_i + \alpha_i L_i = M(y_{2i}\beta) +M x_{1i}\gamma + M u_i$

This implies that
$M u_i = t_i + \alpha_i L_i,$
where $$Var(u_i) = 1$$ and $$Var(t_i + \alpha L_i) = 1 + \alpha^2$$, so the scale is $$M = \sqrt{1+\alpha^2}$$.

The second equation remains the same, we just express $$v_i$$ as $$w_i + \alpha L_i$$. Now, after estimating the system (2), we can recover the parameters in (1) as follows:

$\beta = \frac{1}{\sqrt{1+ \alpha^2}} b$
$\gamma = \frac{1}{\sqrt{1+ \alpha^2}} c$
$\kappa_j = \frac{1}{\sqrt{1+ \alpha^2}} \lambda_j$

$V(v_i) = V(w_i + \alpha L_i) =V(w_i) + \alpha^2$.

$\rho = Cov(t_i + \alpha L_i, w_i + \alpha L_i) = \frac{\alpha^2}{(\sqrt{1+\alpha^2}\sqrt{V(w_i)+\alpha^2)}}$

Note: This parameterization assumes that the correlation is positive; for negative values of the correlation, $$L$$ should be included in the second equation with a negative sign (that is, L@(-a) instead of L@a). When trying to perform the estimation with the wrong sign, the model most likely won’t achieve convergence. Otherwise, you will see a coefficient for L that is virtually zero. In Stata 13.1 we have included features that allow you to fit the model without this restriction. However, this time we will use the older parameterization, which will allow you to visualize the different components more easily.

### Simulating data, and performing the estimation

clear
set seed 1357
set obs 10000
forvalues i = 1(1)5 {
gen xi' =2* rnormal() + _n/1000
}

mat C = [1,.5 \ .5, 1]
drawnorm z1 z2, cov(C)

gen y2 = 0
forvalues i = 1(1)5 {
replace y2 = y2 + xi'
}
replace y2 = y2 + z2

gen y1star = y2 + x1 + x2 + z1
gen xb1 = y2 + x1 + x2

gen y1 = 4
replace y1 = 3 if xb1 + z1 <=.8
replace y1 = 2 if xb1 + z1 <=.3
replace y1 = 1 if xb1 + z1 <=-.3
replace y1 = 0 if xb1 + z1 <=-.8

gsem (y1 <- y2 x1 x2 L@a, oprobit) (y2 <- x1 x2 x3 x4 x5 L@a), var(L@1)

local y1 y1
local y2 y2

local xaux  x1 x2 x3 x4 x5
local xmain  y2 x1 x2

local s2 sqrt(1+_b[y1':L]^2)
foreach v in xmain'{
local trans trans' (y1'_v': _b[y1':v']/s2')
}

foreach v in xaux' _cons {
local trans trans' (y2'_v': _b[y2':v'])
}

qui tab y1' if e(sample)
local ncuts = r(r)-1
forvalues i = 1(1) ncuts'{
local trans trans' (cut_i': _b[y1'_cuti':_cons]/s2')
}

local s1 sqrt(  _b[var(e.y2'):_cons]  +_b[y1':L]^2)

local trans trans' (sig_2: s1')
local trans trans' (rho_12: _b[y1':L]^2/(s1'*s2'))
nlcom trans'


### Results

This is the output from gsem:

Generalized structural equation model             Number of obs   =      10000
Log likelihood = -14451.117

( 1)  [y1]L - [y2]L = 0
( 2)  [var(L)]_cons = 1
------------------------------------------------------------------------------
|      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
y1 <-        |
y2 |   1.379511   .0775028    17.80   0.000     1.227608    1.531414
x1 |   1.355687   .0851558    15.92   0.000     1.188785    1.522589
x2 |   1.346323   .0833242    16.16   0.000      1.18301    1.509635
L |   .7786594   .0479403    16.24   0.000     .6846982    .8726206
-------------+----------------------------------------------------------------
y2 <-        |
x1 |   .9901353   .0044941   220.32   0.000      .981327    .9989435
x2 |   1.006836   .0044795   224.76   0.000      .998056    1.015615
x3 |   1.004249   .0044657   224.88   0.000     .9954963    1.013002
x4 |   .9976541   .0044783   222.77   0.000     .9888767    1.006431
x5 |   .9987587   .0044736   223.26   0.000     .9899907    1.007527
L |   .7786594   .0479403    16.24   0.000     .6846982    .8726206
_cons |   .0002758   .0192417     0.01   0.989    -.0374372    .0379887
-------------+----------------------------------------------------------------
y1           |
/cut1 |  -1.131155   .1157771    -9.77   0.000    -1.358074   -.9042358
/cut2 |  -.5330973   .1079414    -4.94   0.000    -.7446585    -.321536
/cut3 |   .2722794   .1061315     2.57   0.010     .0642654    .4802933
/cut4 |     .89394   .1123013     7.96   0.000     .6738334    1.114047
-------------+----------------------------------------------------------------
var(L)|          1  (constrained)
-------------+----------------------------------------------------------------
var(e.y2)|   .3823751    .074215                      .2613848    .5593696
------------------------------------------------------------------------------


These are the results we obtain when we transform the values reported by gsem to the original parameterization:

------------------------------------------------------------------------------
|      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
y1_y2 |   1.088455   .0608501    17.89   0.000     .9691909    1.207719
y1_x1 |   1.069657   .0642069    16.66   0.000      .943814    1.195501
y1_x2 |   1.062269   .0619939    17.14   0.000      .940763    1.183774
y2_x1 |   .9901353   .0044941   220.32   0.000      .981327    .9989435
y2_x2 |   1.006836   .0044795   224.76   0.000      .998056    1.015615
y2_x3 |   1.004249   .0044657   224.88   0.000     .9954963    1.013002
y2_x4 |   .9976541   .0044783   222.77   0.000     .9888767    1.006431
y2_x5 |   .9987587   .0044736   223.26   0.000     .9899907    1.007527
y2__cons |   .0002758   .0192417     0.01   0.989    -.0374372    .0379887
cut_1 |   -.892498   .0895971    -9.96   0.000    -1.068105   -.7168909
cut_2 |  -.4206217   .0841852    -5.00   0.000    -.5856218   -.2556217
cut_3 |   .2148325   .0843737     2.55   0.011     .0494632    .3802018
cut_4 |    .705332   .0905974     7.79   0.000     .5277644    .8828997
sig_2 |   .9943267    .007031   141.42   0.000     .9805462    1.008107
rho_12 |   .4811176   .0477552    10.07   0.000     .3875191     .574716
------------------------------------------------------------------------------

The estimates are quite close to the values used for the simulation. If you try to perform the estimation with the wrong sign for the coefficient for L, you will get a number that is virtually zero (if you get convergence at all). In this case, the evaluator is telling us that the best value it can find, provided the restrictions we have imposed, is zero. If you see such results, you may want to try the opposite sign. If both give a zero coefficient, it means that this is the solution, and there is not endogeneity at all. If one of them is not zero, it means that the non-zero value is the solution. As stated before, in Stata 13.1, the model can be fitted without this restriction.

Categories: Statistics Tags:

## Measures of effect size in Stata 13

Today I want to talk about effect sizes such as Cohen’s d, Hedges’s g, Glass’s Δ, η2, and ω2. Effects sizes concern rescaling parameter estimates to make them easier to interpret, especially in terms of practical significance.

Many researchers in psychology and education advocate reporting of effect sizes, professional organizations such as the American Psychological Association (APA) and the American Educational Research Association (AERA) strongly recommend their reporting, and professional journals such as the Journal of Experimental Psychology: Applied and Educational and Psychological Measurement require that they be reported.

Anyway, today I want to show you

1. What effect sizes are.
2. How to calculate effect sizes and their confidence intervals in Stata.
3. How to calculate bootstrap confidence intervals for those effect sizes.
4. How to use Stata’s effect-size calculator.

## 1. What are effect sizes?

The importance of research results is often assessed by statistical significance, usually that the p-value is less than 0.05. P-values and statistical significance, however, don’t tell us anything about practical significance.

What if I told you that I had developed a new weight-loss pill and that the difference between the average weight loss for people who took the pill and the those who took a placebo was statistically significant? Would you buy my new pill? If you were overweight, you might reply, “Of course! I’ll take two bottles and a large order of french fries to go!”. Now let me add that the average difference in weight loss was only one pound over the year. Still interested? My results may be statistically significant but they are not practically significant.

Or what if I told you that the difference in weight loss was not statistically significant — the p-value was “only” 0.06 — but the average difference over the year was 20 pounds? You might very well be interested in that pill.

The size of the effect tells us about the practical significance. P-values do not assess practical significance.

All of which is to say, one should report parameter estimates along with statistical significance.

In my examples above, you knew that 1 pound over the year is small and 20 pounds is large because you are familiar with human weights.

In another context, 1 pound might be large, and in yet another, 20 pounds small.

Formal measures of effects sizes are thus usually presented in unit-free but easy-to-interpret form, such as standardized differences and proportions of variability explained.

### The “d” family

Effect sizes that measure the scaled difference between means belong to the “d” family. The generic formula is

The estimators differ in terms of how sigma is calculated.

Cohen’s d, for instance, uses the pooled sample standard deviation.

Hedges’s g incorporates an adjustment which removes the bias of Cohen’s d.

Glass’s Δ was originally developed in the context of experiments and uses the “control group” standard deviation in the denominator. It has subsequently been generalized to nonexperimental studies. Because there is no control group in observational studies, Kline (2013) recommends reporting Glass’s Δ using the standard deviation for each group. Glass’s Delta_1 uses one group’s standard deviation and Delta_2 uses the other group’s.

Although I have given definitions to Cohen’s d, Hedges’s g, and Glass’s Δ, different authors swap the definitions around! As a result, many authors refer to all of the above as just Delta.

Be careful when using software to know which Delta you are getting. I have used Stata terminology, of course.

Anyway, the use of a standardized scale allows us to assess of practical significance. Delta = 1.5 indicates that the mean of one group is 1.5 standard deviations higher than that of the other. A difference of 1.5 standard deviations is obviously large, and a difference of 0.1 standard deviations is obviously small.

### The “r” family

The r family quantifies the ratio of the variance attributable to an effect to the total variance and is often interpreted as the “proportion of variance explained”. The generic estimator is known as eta-squared,

η2 is equivalent to the R-squared statistic from linear regression.

ω2 is a less biased variation of η2 that is equivalent to the adjusted R-squared.

Both of these measures concern the entire model.

Partial η2 and partial ω2 are like partial R-squareds and concern individual terms in the model. A term might be a variable or a variable and its interaction with another variable.

Both the d and r families allow us to make an apples-to-apples comparison of variables measured on different scales. For example, an intervention could affect both systolic blood pressure and total cholesterol. Comparing the relative effect of the intervention on the two outcomes would be difficult on their original scales.

How does one compare mm/Hg and mg/dL? It is straightforward in terms of Cohen’s d or ω2 because then we are comparing standard deviation changes or proportion of variance explained.

## 2. How to calculate effect sizes and their confidence intervals in Stata

Consider a study where 30 school children are randomly assigned to classrooms that incorporated web-based instruction (treatment) or standard classroom environments (control). At the end of the school year, the children were given tests to measure reading and mathematics skills. The reading test is scored on a 0-15 point scale and, the mathematics test, on a 0-100 point scale.

Let’s download a dataset for our fictitious example from the Stata website by typing:

. use http://www.stata.com/videos13/data/webclass.dta

Contains data from http://www.stata.com/videos13/data/webclass.dta
obs:            30                          Fictitious web-based learning
experiment data
vars:             5                          5 Sep 2013 11:28
size:           330                          (_dta has notes)
-------------------------------------------------------------------------------
storage   display    value
variable name   type    format     label      variable label
-------------------------------------------------------------------------------
id              byte    %9.0g                 ID Number
treated         byte    %9.0g      treated    Treatment Group
agegroup        byte    %9.0g      agegroup   Age Group
math            float   %9.0g                 Math Score
-------------------------------------------------------------------------------

. notes

_dta:
1.  Variable treated records 0=control, 1=treated.
2.  Variable agegroup records 1=7 years old, 2=8 years old, 3=9 years old.


We can compute a t-statistic to test the null hypothesis that the average math scores are the same in the treatment and control groups.

. ttest math, by(treated)

Two-sample t test with equal variances
------------------------------------------------------------------------------
Group |     Obs        Mean    Std. Err.   Std. Dev.   [95% Conf. Interval]
---------+--------------------------------------------------------------------
Control |      15    69.98866    3.232864    12.52083    63.05485    76.92246
Treated |      15    79.54943    1.812756    7.020772    75.66146     83.4374
---------+--------------------------------------------------------------------
combined |      30    74.76904    2.025821    11.09588    70.62577    78.91231
---------+--------------------------------------------------------------------
diff |           -9.560774    3.706412               -17.15301   -1.968533
------------------------------------------------------------------------------
diff = mean(Control) - mean(Treated)                          t =  -2.5795
Ho: diff = 0                                     degrees of freedom =       28

Ha: diff < 0                 Ha: diff != 0                 Ha: diff > 0
Pr(T < t) = 0.0077         Pr(|T| > |t|) = 0.0154          Pr(T > t) = 0.9923


The treated students have a larger mean, yet the difference of -9.56 is reported as negative because -ttest- calculated Control minus Treated. So just remember, negative differences mean Treated > Control in this case.

The t-statistic equals -2.58 and its two-sided p-value of 0.0154 indicates that the difference between the math scores in the two groups is statistically significant.

Next, let’s calculate effect sizes from the d family:

. esize twosample math, by(treated) cohensd hedgesg glassdelta

Effect size based on mean comparison

Obs per group:
Control =     15
Treated =     15
---------------------------------------------------------
Effect Size |   Estimate     [95% Conf. Interval]
--------------------+------------------------------------
Cohen's d |  -.9419085    -1.691029   -.1777553
Hedges's g |   -.916413    -1.645256   -.1729438
Glass's Delta 1 |  -.7635896     -1.52044    .0167094
Glass's Delta 2 |  -1.361784    -2.218342   -.4727376
---------------------------------------------------------


Cohen’s d and Hedges’s g both indicate that the average reading scores differ by approximately -0.93 standard deviations with 95% confidence intervals of (-1.69, -0.18) and (-1.65, -0.17) respectively.

Since this is an experiment, we are interested in Glass’s Delta 1 because it is calculated using the control group standard deviation. Average reading scores differ by -0.76 and the confidence interval is (-1.52, 0.02).

The confidence intervals for Cohen’s d and Hedges’s g do not include the null value of zero but the confidence interval for Glass’s Delta 1 does. Thus we cannot completely rule out the possibility that the treatment had no effect on math scores.

Next we could incorporate the age group of the children into our analysis by using a two-way ANOVA to test the null hypothesis that the mean math scores are equal for all groups.

. anova math treated##agegroup

Number of obs =      30     R-squared     =  0.2671
Root MSE      = 10.4418     Adj R-squared =  0.1144

Source |  Partial SS    df       MS           F     Prob > F
-----------------+----------------------------------------------------
Model |  953.697551     5   190.73951       1.75     0.1617
|
treated |  685.562956     1  685.562956       6.29     0.0193
agegroup |  47.7059268     2  23.8529634       0.22     0.8051
treated#agegroup |  220.428668     2  110.214334       1.01     0.3789
|
Residual |  2616.73825    24  109.030761
-----------------+----------------------------------------------------
Total |   3570.4358    29  123.118476


The F-statistic for the entire model is not statistically significant (F=1.75, ndf=5, ddf=24, p=0.1617) but the F-statistic for the main effect of treatment is statistically significant (F=6.29, ndf=1, ddf=24, p=0.0193).

We can compute the η2 and partial η2 estimates for this model using the estat esize command immediately after our anova command (note that estat esize works after the regress command too).

. estat esize

Effect sizes for linear models

---------------------------------------------------------------------
Source |   Eta-Squared     df     [95% Conf. Interval]
----------------------+----------------------------------------------
Model |   .2671096         5            0    .4067062
|
treated |   .2076016         1     .0039512    .4451877
agegroup |   .0179046         2            0    .1458161
treated#agegroup |   .0776932         2            0     .271507
---------------------------------------------------------------------


The overall η2 indicates that our model accounts for approximately 26.7% of the variablity in math scores though the 95% confidence interval includes the null value of zero (0.00%, 40.7%). The partial η2 for treatment is 0.21 (21% of the variability explained) and its 95% confidence interval excludes zero (0.3%, 20%).

We could calculate the alternative r-family member ω2 rather than η2 by typing

. estat esize, omega

Effect sizes for linear models

---------------------------------------------------------------------
Source | Omega-Squared     df     [95% Conf. Interval]
----------------------+----------------------------------------------
Model |   .1144241         5            0    .2831033
|
treated |    .174585         1            0    .4220705
agegroup |          0         2            0    .0746342
treated#agegroup |   .0008343         2            0    .2107992
---------------------------------------------------------------------


The overall ω2 indicates that our model accounts for approximately 11.4% of the variability in math scores and treatment accounts for 17.5%. This perplexing result stems from the way that ω2 and partial ω2 are calculated. See Pierce, Block, & Aguinis (2004) for a thorough explanation.

Except for the η2 for treatment, the confidence intervals include 0 so we cannot rule out the possibility that there is no effect. Whether results are practically significant is generically a matter context and opinion. In some situations, accounting for 5% of the variability in an outcome could be very important and in other situations accounting for 30% may not be.

We could repeat the same analyses for the reading scores using the following commands:

. ttest reading, by(treated)
. esize twosample reading, by(treated) cohensd hedgesg glassdelta
. estat esize
. estat esize, omega


None of the t- or F-statistics for reading scores were statistically significant at the 0.05 level.

Even though the reading and math scores were measured on two different scales, we can directly compare the relative effect of the treatment using effect sizes:

        Effect Size   |     Reading Score          Math Score
------------------------------------------------------------
Cohen's d     |   -0.23 (-0.95 - 0.49)  -0.94 (-1.69 - -0.18)
Hedges's g    |   -0.22 (-0.92 - 0.48)  -0.92 (-1.65 - -0.17)
Glass's Delta |   -0.21 (-0.93 - 0.51)  -0.76 (-1.52 -  0.02)
Eta-squared   |    0.02 ( 0.00 - 0.20)   0.21 ( 0.00 -  0.44)
Omega-squared |    0.00 ( 0.00 - 0.17)   0.17 ( 0.00 -  0.42)


The results show that the average reading scores in the treated and control groups differ by approximately 0.22 standard deviations while the average math scores differ by approximately 0.92 standard deviations. Similarly, treatment status accounted for almost none of the variability in reading scores while it accounted for roughly 17% of the variability in math scores. The intervention clearly had a larger effect on math scores than reading scores. We also know that we cannot completely rule out an effect size of zero (no effect) for both reading and math scores because several confidence intervals included zero. Whether or not the effects are practically significant is a matter of interpretation but the effect sizes provide a standardized metric for evaluation.

## 3. How to calculate bootstrap confidence intervals

Simulation studies have shown that bootstrap confidence intervals for the d family may be preferable to confidence intervals based on the noncentral t distribution when the variable of interest does not have a normal distribution (Kelley 2005; Algina, Keselman, and Penfield 2006). We can calculate bootstrap confidence intervals for Cohen’s d and Hedges’s g using Stata’s bootstrap prefix:

. bootstrap r(d) r(g), reps(500) nowarn:  esize twosample reading, by(treated)
(running esize on estimation sample)

Bootstrap replications (500)
----+--- 1 ---+--- 2 ---+--- 3 ---+--- 4 ---+--- 5
..................................................    50
..................................................   100
..................................................   150
..................................................   200
..................................................   250
..................................................   300
..................................................   350
..................................................   400
..................................................   450
..................................................   500

Bootstrap results                               Number of obs      =        30
Replications       =       500

_bs_1:  r(d)
_bs_2:  r(g)

------------------------------------------------------------------------------
|   Observed   Bootstrap                         Normal-based
|      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
_bs_1 |   -.228966   .3905644    -0.59   0.558    -.9944582    .5365262
_bs_2 |  -.2227684   .3799927    -0.59   0.558    -.9675403    .5220036
------------------------------------------------------------------------------


The bootstrap estimate of the 95% confidence interval for Cohen’s d is -0.99 to 0.54 which is slightly wider than the earlier estimate based on the non-central t distribution (see [R] esize for details). The bootstrap estimate is slightly wider for Hedges’s g as well.

## 4. How to use Stata’s effect-size calculator

You can use Stata’s effect size calculators to estimate them using summary statistics. If we know that the mean, standard deviation and sample size for one group is 70, 12.5 and 15 respectively and 80, 7 and 15 for another group, we can use esizei to estimate effect sizes from the d family:

. esizei 15 70 12.5 15 80 7, cohensd hedgesg glassdelta

Effect size based on mean comparison

Obs per group:
Group 1 =     15
Group 2 =     15
---------------------------------------------------------
Effect Size |   Estimate     [95% Conf. Interval]
--------------------+------------------------------------
Cohen's d |  -.9871279    -1.739873   -.2187839
Hedges's g |  -.9604084    -1.692779   -.2128619
Glass's Delta 1 |        -.8    -1.561417   -.0143276
Glass's Delta 2 |  -1.428571    -2.299112   -.5250285
---------------------------------------------------------


We can estimate effect sizes from the r family using esizei with slightly different syntax. For example, if we know the numerator and denominator degrees of freedom along with the F statistic, we can calculate η2 and ω2 using the following command:

. esizei 1 28 6.65

Effect sizes for linear models

---------------------------------------------------------
Effect Size |   Estimate     [95% Conf. Interval]
--------------------+------------------------------------
Eta-Squared |   .1919192     .0065357    .4167874
Omega-Squared |   .1630592            0    .3959584
---------------------------------------------------------


## Video demonstration

Stata has dialog boxes that can assist you in calculating effect sizes. If you would like a brief introduction using the GUI, you can watch a demonstration on Stata’s YouTube Channel:

Tour of effect sizes in Stata

## Final thoughts and further reading

Most older papers and many current papers do not report effect sizes. Nowadays, the general consensus among behavioral scientists, their professional organizations, and their journals is that effect sizes should always be reported in addition to tests of statistical significance. Stata 13 now makes it easy to compute most popular effects sizes.

Some methodologists believe that effect sizes with confidence intervals should always be reported and that statistical hypothesis tests should be abandoned altogether; see Cumming (2012) and Kline (2013). While this may sound like a radical notion, other fields such as epidemiology have been moving in this direction since the 1990s. Cumming and Kline offer compelling arguments for this paradigm shift as well as excellent introductions to effect sizes.

American Psychological Association (2009). Publication Manual of the American Psychological Association, 6th Ed. Washington, DC: American Psychological Association.

Algina, J., H. J. Keselman, and R. D. Penfield. (2006). Confidence interval coverage for Cohen’s effect size statistic. Educational and Psychological Measurement, 66(6): 945–960.

Cumming, G. (2012). Understanding the New Statistics: Effect Sizes, Confidence Intervals, and Meta-Analysis. New York: Taylor & Francis.

Kelley, K. (2005). The effects of nonnormal distributions on confidence intervals around the standardized mean difference: Bootstrap and parametric confidence intervals. Educational and Psychological Measurement 65: 51–69.

Kirk, R. (1996). Practical significance: A concept whose time has come. Educational and Psychological Measurement, 56, 746-759.

Kline, R. B. (2013). Beyond Significance Testing: Statistics Reform in the Behavioral Sciences. 2nd ed. Washington, DC: American Psychological Association.

Pierce, C.A., Block, R. A., and Aguinis, H. (2004). Cautionary note on reporting eta-squared values from multifactor ANOVA designs. Educational and Psychological Measurement, 64(6) 916-924

Thompson, B. (1996) AERA Editorial Policies regarding Statistical Significance Testing: Three Suggested Reforms. Educational Researcher, 25(2) 26-30

Wilkinson, L., & APA Task Force on Statistical Inference. (1999). Statistical methods in psychology journals: Guidelines and explanations. American Psychologist, 54, 594-604

Categories: Statistics Tags:

## Multilevel linear models in Stata, part 2: Longitudinal data

In my last posting, I introduced you to the concepts of hierarchical or “multilevel” data. In today’s post, I’d like to show you how to use multilevel modeling techniques to analyse longitudinal data with Stata’s xtmixed command.

Last time, we noticed that our data had two features. First, we noticed that the means within each level of the hierarchy were different from each other and we incorporated that into our data analysis by fitting a “variance component” model using Stata’s xtmixed command.

The second feature that we noticed is that repeated measurement of GSP showed an upward trend. We’ll pick up where we left off last time and stick to the concepts again and you can refer to the references at the end to learn more about the details.

## The videos

Stata has a very friendly dialog box that can assist you in building multilevel models. If you would like a brief introduction using the GUI, you can watch a demonstration on Stata’s YouTube Channel:

Introduction to multilevel linear models in Stata, part 2: Longitudinal data

## Longitudinal data

I’m often asked by beginning data analysts – “What’s the difference between longitudinal data and time-series data? Aren’t they the same thing?”.

The confusion is understandable — both types of data involve some measurement of time. But the answer is no, they are not the same thing.

Univariate time series data typically arise from the collection of many data points over time from a single source, such as from a person, country, financial instrument, etc.

Longitudinal data typically arise from collecting a few observations over time from many sources, such as a few blood pressure measurements from many people.

There are some multivariate time series that blur this distinction but a rule of thumb for distinguishing between the two is that time series have more repeated observations than subjects while longitudinal data have more subjects than repeated observations.

Because our GSP data from last time involve 17 measurements from 48 states (more sources than measurements), we will treat them as longitudinal data.

## Random intercept models

As I mentioned last time, repeated observations on a group of individuals can be conceptualized as multilevel data and modeled just as any other multilevel data. We left off last time with a variance component model for GSP (Gross State Product, logged) and noted that our model assumed a constant GSP over time while the data showed a clear upward trend.

If we consider a single observation and think about our model, nothing in the fixed or random part of the models is a function of time.

Let’s begin by adding the variable year to the fixed part of our model.

As we expected, our grand mean has become a linear regression which more accurately reflects the change over time in GSP. What might be unexpected is that each state’s and region’s mean has changed as well and now has the same slope as the regression line. This is because none of the random components of our model are a function of time. Let’s fit this model with the xtmixed command:

. xtmixed gsp year, || region: || state:

------------------------------------------------------------------------------
gsp |      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
year |   .0274903   .0005247    52.39   0.000     .0264618    .0285188
_cons |  -43.71617   1.067718   -40.94   0.000    -45.80886   -41.62348
------------------------------------------------------------------------------

------------------------------------------------------------------------------
Random-effects Parameters  |   Estimate   Std. Err.     [95% Conf. Interval]
-----------------------------+------------------------------------------------
region: Identity             |
sd(_cons) |   .6615238   .2038949      .3615664    1.210327
-----------------------------+------------------------------------------------
state: Identity              |
sd(_cons) |   .7805107   .0885788      .6248525    .9749452
-----------------------------+------------------------------------------------
sd(Residual) |   .0734343   .0018737      .0698522    .0772001
------------------------------------------------------------------------------

The fixed part of our model now displays an estimate of the intercept (_cons = -43.7) and the slope (year = 0.027). Let’s graph the model for Region 7 and see if it fits the data better than the variance component model.

predict GrandMean, xb
label var GrandMean "GrandMean"
predict RegionEffect, reffects level(region)
predict StateEffect, reffects level(state)
gen RegionMean = GrandMean + RegionEffect
gen StateMean = GrandMean + RegionEffect + StateEffect

twoway  (line GrandMean year, lcolor(black) lwidth(thick))      ///
(line RegionMean year, lcolor(blue) lwidth(medthick))   ///
(line StateMean year, lcolor(green) connect(ascending)) ///
(scatter gsp year, mcolor(red) msize(medsmall))         ///
if region ==7,                                          ///
ytitle(log(Gross State Product), margin(medsmall))      ///
legend(cols(4) size(small))                             ///
title("Multilevel Model of GSP for Region 7", size(medsmall))

That looks like a much better fit than our variance-components model from last time. Perhaps I should leave well enough alone, but I can’t help noticing that the slopes of the green lines for each state don’t fit as well as they could. The top green line fits nicely but the second from the top looks like it slopes upward more than is necessary. That’s the best fit we can achieve if the regression lines are forced to be parallel to each other. But what if the lines were not forced to be parallel? What if we could fit a “mini-regression model” for each state within the context of my overall multilevel model. Well, good news — we can!

## Random slope models

By introducing the variable year to the fixed part of the model, we turned our grand mean into a regression line. Next I’d like to incorporate the variable year into the random part of the model. By introducing a fourth random component that is a function of time, I am effectively estimating a separate regression line within each state.

Notice that the size of the new, brown deviation u1ij. is a function of time. If the observation were one year to the left, u1ij. would be smaller and if the observation were one year to the right, u1ij.would be larger.

It is common to “center” the time variable before fitting these kinds of models. Explaining why is for another day. The quick answer is that, at some point during the fitting of the model, Stata will have to compute the equivalent of the inverse of the square of year. For the year 1986 this turns out to be 2.535e-07. That’s a fairly small number and if we multiply it by another small number…well, you get the idea. By centering age (e.g. cyear = year – 1978), we get a more reasonable number for 1986 (0.01). (Hint: If you have problems with your model converging and you have large values for time, try centering them. It won’t always help, but it might).

So let’s center our year variable by subtracting 1978 and fit a model that includes a random slope.

gen cyear = year - 1978
xtmixed gsp cyear, || region: || state: cyear, cov(indep)

I’ve color-coded the output so that we can match each part of the output back to the model and the graph. The fixed part of the model appears in the top table and it looks like any other simple linear regression model. The random part of the model is definitely more complicated. If you get lost, look back at the graphic of the deviations and remind yourself that we have simply partitioned the deviation of each observation into four components. If we did this for every observation, the standard deviations in our output are simply the average of those deviations.

Let’s look at a graph of our new “random slope” model for Region 7 and see how well it fits our data.

predict GrandMean, xb
label var GrandMean "GrandMean"
predict RegionEffect, reffects level(region)
predict StateEffect_year StateEffect_cons, reffects level(state)

gen RegionMean = GrandMean + RegionEffect
gen StateMean_cons = GrandMean + RegionEffect + StateEffect_cons
gen StateMean_year = GrandMean + RegionEffect + StateEffect_cons + ///
(cyear*StateEffect_year)

twoway  (line GrandMean cyear, lcolor(black) lwidth(thick))             ///
(line RegionMean cyear, lcolor(blue) lwidth(medthick))          ///
(line StateMean_cons cyear, lcolor(green) connect(ascending))   ///
(line StateMean_year cyear, lcolor(brown) connect(ascending))   ///
(scatter gsp cyear, mcolor(red) msize(medsmall))                ///
if region ==7,                                                  ///
ytitle(log(Gross State Product), margin(medsmall))              ///
legend(cols(3) size(small))                                     ///
title("Multilevel Model of GSP for Region 7", size(medsmall))

The top brown line fits the data slightly better, but the brown line below it (second from the top) is a much better fit. Mission accomplished!

## Where do we go from here?

I hope I have been able to convince you that multilevel modeling is easy using Stata’s xtmixed command and that this is a tool that you will want to add to your kit. I would love to say something like “And that’s all there is to it. Go forth and build models!”, but I would be remiss if I didn’t point out that I have glossed over many critical topics.

In our GSP example, we would still like to consider the impact of other independent variables. I haven’t mentioned choice of estimation methods (ML or REML in the case of xtmixed). I’ve assessed the fit of our models by looking at graphs, an approach important but incomplete. We haven’t thought about hypothesis testing. Oh — and, all the usual residual diagnostics for linear regression such as checking for outliers, influential observations, heteroskedasticity and normality still apply….times four! But now that you understand the concepts and some of the mechanics, it shouldn’t be difficult to fill in the details. If you’d like to learn more, check out the links below.

I hope this was helpful…thanks for stopping by.

Multilevel and Longitudinal Modeling Using Stata, Third Edition
Volume I: Continuous Responses
Volume II: Categorical Responses, Counts, and Survival
by Sophia Rabe-Hesketh and Anders Skrondal

or sign up for our popular public training course Multilevel/Mixed Models Using Stata.

Categories: Statistics Tags:

## Multilevel linear models in Stata, part 1: Components of variance

In the last 15-20 years multilevel modeling has evolved from a specialty area of statistical research into a standard analytical tool used by many applied researchers.

Stata has a lot of multilevel modeling capababilities.

I want to show you how easy it is to fit multilevel models in Stata. Along the way, we’ll unavoidably introduce some of the jargon of multilevel modeling.

I’m going to focus on concepts and ignore many of the details that would be part of a formal data analysis. I’ll give you some suggestions for learning more at the end of the post.

The videos

Stata has a friendly dialog box that can assist you in building multilevel models. If you would like a brief introduction using the GUI, you can watch a demonstration on Stata’s YouTube Channel:

Introduction to multilevel linear models in Stata, part 1: The xtmixed command

Multilevel data

Multilevel data are characterized by a hierarchical structure. A classic example is children nested within classrooms and classrooms nested within schools. The test scores of students within the same classroom may be correlated due to exposure to the same teacher or textbook. Likewise, the average test scores of classes might be correlated within a school due to the similar socioeconomic level of the students.

You may have run across datasets with these kinds of structures in your own work. For our example, I would like to use a dataset that has both longitudinal and classical hierarchical features. You can access this dataset from within Stata by typing the following command:

use http://www.stata-press.com/data/r12/productivity.dta

We are going to build a model of gross state product for 48 states in the USA measured annually from 1970 to 1986. The states have been grouped into nine regions based on their economic similarity. For distributional reasons, we will be modeling the logarithm of annual Gross State Product (GSP) but in the interest of readability, I will simply refer to the dependent variable as GSP.

. describe gsp year state region

storage  display     value
variable name   type   format      label      variable label
-----------------------------------------------------------------------------
gsp             float  %9.0g                  log(gross state product)
year            int    %9.0g                  years 1970-1986
state           byte   %9.0g                  states 1-48
region          byte   %9.0g                  regions 1-9

Let’s look at a graph of these data to see what we’re working with.

twoway (line gsp year, connect(ascending)), ///
by(region, title("log(Gross State Product) by Region", size(medsmall)))

Each line represents the trajectory of a state’s (log) GSP over the years 1970 to 1986. The first thing I notice is that the groups of lines are different in each of the nine regions. Some groups of lines seem higher and some groups seem lower. The second thing that I notice is that the slopes of the lines are not the same. I’d like to incorporate those attributes of the data into my model.

Components of variance

Let’s tackle the vertical differences in the groups of lines first. If we think about the hierarchical structure of these data, I have repeated observations nested within states which are in turn nested within regions. I used color to keep track of the data hierarchy.

We could compute the mean GSP within each state and note that the observations within in each state vary about their state mean.

Likewise, we could compute the mean GSP within each region and note that the state means vary about their regional mean.

We could also compute a grand mean and note that the regional means vary about the grand mean.

Next, let’s introduce some notation to help us keep track of our mutlilevel structure. In the jargon of multilevel modelling, the repeated measurements of GSP are described as “level 1″, the states are referred to as “level 2″ and the regions are “level 3″. I can add a three-part subscript to each observation to keep track of its place in the hierarchy.

Now let’s think about our model. The simplest regression model is the intercept-only model which is equivalent to the sample mean. The sample mean is the “fixed” part of the model and the difference between the observation and the mean is the residual or “random” part of the model. Econometricians often prefer the term “disturbance”. I’m going to use the symbol μ to denote the fixed part of the model. μ could represent something as simple as the sample mean or it could represent a collection of independent variables and their parameters.

Each observation can then be described in terms of its deviation from the fixed part of the model.

If we computed this deviation of each observation, we could estimate the variability of those deviations. Let’s try that for our data using Stata’s xtmixed command to fit the model:

. xtmixed gsp

Mixed-effects ML regression                     Number of obs      =       816

Wald chi2(0)       =         .
Log likelihood = -1174.4175                     Prob > chi2        =         .

------------------------------------------------------------------------------
gsp |      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
_cons |   10.50885   .0357249   294.16   0.000     10.43883    10.57887
------------------------------------------------------------------------------

------------------------------------------------------------------------------
Random-effects Parameters  |   Estimate   Std. Err.     [95% Conf. Interval]
-----------------------------+------------------------------------------------
sd(Residual) |   1.020506   .0252613      .9721766    1.071238
------------------------------------------------------------------------------

The top table in the output shows the fixed part of the model which looks like any other regression output from Stata, and the bottom table displays the random part of the model. Let’s look at a graph of our model along with the raw data and interpret our results.

predict GrandMean, xb
label var GrandMean "GrandMean"
twoway  (line GrandMean year, lcolor(black) lwidth(thick))              ///
(scatter gsp year, mcolor(red) msize(tiny)),                    ///
ytitle(log(Gross State Product), margin(medsmall))              ///
legend(cols(4) size(small))                                     ///
title("GSP for 1970-1986 by Region", size(medsmall))

The thick black line in the center of the graph is the estimate of _cons, which is an estimate of the fixed part of model for GSP. In this simple model, _cons is the sample mean which is equal to 10.51. In “Random-effects Parameters” section of the output, sd(Residual) is the average vertical distance between each observation (the red dots) and fixed part of the model (the black line). In this model, sd(Residual) is the estimate of the sample standard deviation which equals 1.02.

At this point you may be thinking to yourself – “That’s not very interesting – I could have done that with Stata’s summarize command”. And you would be correct.

. summ gsp

Variable |       Obs        Mean    Std. Dev.       Min        Max
-------------+--------------------------------------------------------
gsp |       816    10.50885    1.021132    8.37885   13.04882

But here’s where it does become interesting. Let’s make the random part of the model more complex to account for the hierarchical structure of the data. Consider a single observation, yijk and take another look at its residual.

The observation deviates from its state mean by an amount that we will denote eijk. The observation’s state mean deviates from the the regionals mean uij. and the observation’s regional mean deviates from the fixed part of the model, μ, by an amount that we will denote ui... We have partitioned the observation’s residual into three parts, aka “components”, that describe its magnitude relative to the state, region and grand means. If we calculated this set of residuals for each observation, wecould estimate the variability of those residuals and make distributional assumptions about them.

These kinds of models are often called “variance component” models because they estimate the variability accounted for by each level of the hierarchy. We can estimate a variance component model for GSP using Stata’s xtmixed command:

xtmixed gsp, || region: || state:

------------------------------------------------------------------------------
gsp |      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
_cons |   10.65961   .2503806    42.57   0.000     10.16887    11.15035
------------------------------------------------------------------------------

------------------------------------------------------------------------------
Random-effects Parameters  |   Estimate   Std. Err.     [95% Conf. Interval]
-----------------------------+------------------------------------------------
region: Identity             |
sd(_cons) |   .6615227   .2038944       .361566    1.210325
-----------------------------+------------------------------------------------
state: Identity              |
sd(_cons) |   .7797837   .0886614      .6240114    .9744415
-----------------------------+------------------------------------------------
sd(Residual) |   .1570457   .0040071       .149385    .1650992
------------------------------------------------------------------------------

The fixed part of the model, _cons, is still the sample mean. But now there are three parameters estimates in the bottom table labeled “Random-effects Parameters”. Each quantifies the average deviation at each level of the hierarchy.

Let’s graph the predictions from our model and see how well they fit the data.

predict GrandMean, xb
label var GrandMean "GrandMean"
predict RegionEffect, reffects level(region)
predict StateEffect, reffects level(state)
gen RegionMean = GrandMean + RegionEffect
gen StateMean = GrandMean + RegionEffect + StateEffect

twoway  (line GrandMean year, lcolor(black) lwidth(thick))      ///
(line RegionMean year, lcolor(blue) lwidth(medthick))   ///
(line StateMean year, lcolor(green) connect(ascending)) ///
(scatter gsp year, mcolor(red) msize(tiny)),            ///
ytitle(log(Gross State Product), margin(medsmall))      ///
legend(cols(4) size(small))                             ///
by(region, title("Multilevel Model of GSP by Region", size(medsmall)))

Wow – that’s a nice graph if I do say so myself. It would be impressive for a report or publication, but it’s a little tough to read with all nine regions displayed at once. Let’s take a closer look at Region 7 instead.

twoway  (line GrandMean year, lcolor(black) lwidth(thick))      ///
(line RegionMean year, lcolor(blue) lwidth(medthick))   ///
(line StateMean year, lcolor(green) connect(ascending)) ///
(scatter gsp year, mcolor(red) msize(medsmall))         ///
if region ==7,                                          ///
ytitle(log(Gross State Product), margin(medsmall))      ///
legend(cols(4) size(small))                             ///
title("Multilevel Model of GSP for Region 7", size(medsmall))

The red dots are the observations of GSP for each state within Region 7. The green lines are the estimated mean GSP within each State and the blue line is the estimated mean GSP within Region 7. The thick black line in the center is the overall grand mean for all nine regions. The model appears to fit the data fairly well but I can’t help noticing that the red dots seem to have an upward slant to them. Our model predicts that GSP is constant within each state and region from 1970 to 1986 when clearly the data show an upward trend.

So we’ve tackled the first feature of our data. We’ve succesfully incorporated the basic hierarchical structure into our model by fitting a variance componentis using Stata’s xtmixed command. But our graph tells us that we aren’t finished yet.

Next time we’ll tackle the second feature of our data — the longitudinal nature of the observations.

Multilevel and Longitudinal Modeling Using Stata, Third Edition
Volume I: Continuous Responses
Volume II: Categorical Responses, Counts, and Survival
by Sophia Rabe-Hesketh and Anders Skrondal

or sign up for our popular public training course “Multilevel/Mixed Models Using Stata“.

There’s a course coming up in Washington, DC on February 7-8, 2013.

Categories: Statistics Tags:

## Using Stata’s SEM features to model the Beck Depression Inventory

I just got back from the 2012 Stata Conference in San Diego where I gave a talk on Psychometric Analysis Using Stata and from the 2012 American Psychological Association Meeting in Orlando. Stata’s structural equation modeling (SEM) builder was popular at both meetings and I wanted to show you how easy it is to use. If you are not familiar with the basics of SEM, please refer to the references at the end of the post. My goal is simply to show you how to use the SEM builder assuming that you already know something about SEM. If you would like to view a video demonstration of the SEM builder, please click the play button below:

The data used here and for the silly examples in my talk were simulated to resemble one of the most commonly used measures of depression: the Beck Depression Inventory (BDI). If you find these data too silly or not relevant to your own research, you could instead imagine it being a set of questions to measure mathematical ability, the ability to use a statistical package, or whatever you wanted.

The Beck Depression Inventory

Originally published by Aaron Beck and colleagues in 1961, the BDI marked an important change in the conceptualization of depression from a psychoanalytic perspective to a cognitive/behavioral perspective. It was also a landmark in the measurement of depression shifting from lengthy, expensive interviews with a psychiatrist to a brief, inexpensive questionnaire that could be scored and quantified. The original inventory consisted of 21 questions each allowing ordinal responses of increasing symptom severity from 0-3. The sum of the responses could then be used to classify a respondent’s depressive symptoms as none, mild, moderate or severe. Many studies have demonstrated that the BDI has good psychometric properties such as high test-retest reliability and the scores correlate well with the assessments of psychiatrists and psychologists. The 21 questions can also be grouped into two subscales. The affective scale includes questions like “I feel sad” and “I feel like a failure” that quantify emotional symptoms of depression. The somatic or physical scale includes questions like “I have lost my appetite” and “I have trouble sleeping” that quantify physical symptoms of depression. Since its original publication, the BDI has undergone two revisions in response to the American Psychiatric Association’s (APA) Diagnostic and Statistical Manuals (DSM) and the BDI-II remains very popular.

The Stata Depression Inventory

Since the BDI is a copyrighted psychometric instrument, I created a fictitious instrument called the “Stata Depression Inventory”. It consists of 20 questions each beginning with the phrase “My statistical software makes me…”. The individual questions are listed in the variable labels below.

. describe qu1-qu20

variable  storage  display    value
name       type   format     label      variable label
------------------------------------------------------------------------------
qu1         byte   %16.0g     response   ...feel sad
qu2         byte   %16.0g     response   ...feel pessimistic about the future
qu3         byte   %16.0g     response   ...feel like a failure
qu4         byte   %16.0g     response   ...feel dissatisfied
qu5         byte   %16.0g     response   ...feel guilty or unworthy
qu6         byte   %16.0g     response   ...feel that I am being punished
qu7         byte   %16.0g     response   ...feel disappointed in myself
qu8         byte   %16.0g     response   ...feel am very critical of myself
qu9         byte   %16.0g     response   ...feel like harming myself
qu10        byte   %16.0g     response   ...feel like crying more than usual
qu11        byte   %16.0g     response   ...become annoyed or irritated easily
qu12        byte   %16.0g     response   ...have lost interest in other people
qu13        byte   %16.0g     qu13_t1    ...have trouble making decisions
qu14        byte   %16.0g     qu14_t1    ...feel unattractive
qu15        byte   %16.0g     qu15_t1    ...feel like not working
qu16        byte   %16.0g     qu16_t1    ...have trouble sleeping
qu17        byte   %16.0g     qu17_t1    ...feel tired or fatigued
qu18        byte   %16.0g     qu18_t1    ...makes my appetite lower than usual
qu19        byte   %16.0g     qu19_t1    ...concerned about my health
qu20        byte   %16.0g     qu20_t1    ...experience decreased libido

The responses consist of a 5-point Likert scale ranging from 1 (Strongly Disagree) to 5 (Strongly Agree). Questions 1-10 form the affective scale of the inventory and questions 11-20 form the physical scale. Data were simulated for 1000 imaginary people and included demographic variables such as age, sex and race. The responses can be summarized succinctly in a matrix of bar graphs:

Classical statistical analysis

The beginning of a classical statistical analysis of these data might consist of summing the responses for questions 1-10 and referring to them as the “Affective Depression Score” and summing questions 11-20 and referring to them as the “Physical Depression Score”.

egen Affective = rowtotal(qu1-qu10)
label var Affective "Affective Depression Score"
egen physical = rowtotal(qu11-qu20)
label var physical "Physical Depression Score"

We could be more sophisticated and use principal components to create the affective and physical depression score:

pca qu1-qu20, components(2)
predict Affective Physical
label var Affective "Affective Depression Score"
label var Physical "Physical Depression Score"

We could then ask questions such as “Are there differences in affective and physical depression scores by sex?” and test these hypotheses using multivariate statistics such as Hotelling’s T-squared statistic. The problem with this analysis strategy is that it treats the depression scores as though they were measured without error and can lead to inaccurate p-values for our test statistics.

Structural equation modeling

Structural equation modeling (SEM) is an ideal way to analyze data where the outcome of interest is a scale or scales derived from a set of measured variables. The affective and physical scores are treated as latent variables in the model resulting in accurate p-values and, best of all….these models are very easy to fit using Stata! We begin by selecting the SEM builder from the Statistics menu:

In the SEM builder, we can select the “Add Measurement Component” icon:

which will open the following dialog box:

In the box labeled “Latent Variable Name” we can type “Affective” (red arrow below) and we can select the variables qu1-qu10 in the “Measured variables” box (blue arrow below).

When we click “OK”, the affective measurement component appears in the builder:

We can repeat this process to create a measurement component for our physical depression scale (images not shown). We can also allow for covariance/correlation between our affective and physical depression scales using the “Add Covariance” icon on the toolbar (red arrow below).

I’ll omit the intermediate steps to build the full model shown below but it’s easy to use the “Add Observed Variable” and “Add Path” icons to create the full model:

Now we’re ready to estimate the parameters for our model. To do this, we click the “Estimate” icon on the toolbar (duh!):

And the flowing dialog box appears:

Let’s ignore the estimation options for now and use the default settings. Click “OK” and the parameter estimates will appear in the diagram:

Some of the parameter estimates are difficult to read in this form but it is easy to rearrange the placement and formatting of the estimates to make them easier to read.

If we look at Stata’s output window and scroll up, you’ll notice that the SEM Builder automatically generated the command for our model:

sem (Affective -> qu1) (Affective -> qu2) (Affective -> qu3)
(Affective -> qu4) (Affective -> qu5) (Affective -> qu6)
(Affective -> qu7) (Affective -> qu8) (Affective -> qu9)
(Affective -> qu10) (Physical -> qu11) (Physical -> qu12)
(Physical -> qu13) (Physical -> qu14) (Physical -> qu15)
(Physical -> qu16) (Physical -> qu17) (Physical -> qu18)
(Physical -> qu19) (Physical -> qu20) (sex -> Affective)
(sex -> Physical), latent(Affective Physical) cov(e.Physical*e.Affective)

We can gather terms and abbreviate some things to make the command much easier to read:

sem (Affective -> qu1-qu10) ///
(Physical -> qu11-qu20) ///
(sex -> Affective Physical) ///
, latent(Affective Physical ) ///
cov( e.Physical*e.Affective)

We could then calculate a Wald statistic to test the null hypothesis that there is no association between sex and our affective and physical depression scales.

test sex

( 1)  [Affective]sex = 0
( 2)  [Physical]sex = 0

chi2(  2) =    2.51
Prob > chi2 =    0.2854

Final thoughts
This is an admittedly oversimplified example – we haven’t considered the fit of the model or considered any alternative models. We have only included one dichotomous independent variable. We might prefer to use a likelihood ratio test or a score test. Those are all very important issues and should not be ignored in a proper data analysis. But my goal was to demonstrate how easy it is to use Stata’s SEM builder to model data such as those arising from the Beck Depression Inventory. Incidentally, if these data were collected using a complex survey design, it would not be difficult to incorporate the sampling structure and sample weights into the analysis. Missing data can be handled easily as well using Full Information Maximum Likelihood (FIML) but those are topics for another day.

If you would like view the slides from my talk, download the data used in this example or view a video demonstration of Stata’s SEM builder using these data, please use the links below. For the dataset, you can also type use followed by the URL for the data to load it directly into Stata.

References

Beck AT, Ward CH, Mendelson M, Mock J, Erbaugh J (June 1961). An inventory for measuring depression. Arch. Gen. Psychiatry 4 (6): 561–71.

Beck AT, Ward C, Mendelson M (1961). Beck Depression Inventory (BDI). Arch Gen Psychiatry 4 (6): 561–571

Beck AT, Steer RA, Ball R, Ranieri W (December 1996). Comparison of Beck Depression Inventories -IA and -II in psychiatric outpatients. Journal of Personality Assessment 67 (3): 588–97
Bollen, KA. (1989). Structural Equations With Latent Variables. New York, NY: John Wiley and Sons

Kline, RB (2011). Principles and Practice of Structural Equation Modeling. New York, NY: Guilford Press

Raykov, T & Marcoulides, GA (2006). A First Course in Structural Equation Modeling. Mahwah, NJ: Lawrence Erlbaum

Schumacker, RE & Lomax, RG (2012) A Beginner’s Guide to Structural Equation Modeling, 3rd Ed. New York, NY: Routledge

Categories: Statistics Tags:

## Comparing predictions after arima with manual computations

Some of our users have asked about the way predictions are computed after fitting their models with arima. Those users report that they cannot reproduce the complete set of forecasts manually when the model contains MA terms. They specifically refer that they are not able to get the exact values for the first few predicted periods. The reason for the difference between their manual results and the forecasts obtained with predict after arima is the way the starting values and the recursive predictions are computed. While Stata uses the Kalman filter to compute the forecasts based on the state space representation of the model, users reporting differences compute their forecasts with a different estimator that is based on the recursions derived from the ARIMA representation of the model. Both estimators are consistent but they produce slightly different results for the first few forecasting periods.

When using the postestimation command predict after fitting their MA(1) model with arima, some users claim that they should be able to reproduce the predictions with

where

However, the recursive formula for the Kalman filter prediction is based on the shrunk error (See section 13.3 in Hamilton (1993) for the complete derivation based on the state space representation):

where

: is the estimated variance of the white noise disturbance

: corresponds to the unconditional mean for the error term

Let’s use one of the datasets available from our website to fit a MA(1) model and compute the predictions based on the Kalman filter recursions formulated above:

** Predictions with Kalman Filter recursions (obtained with -predict- **
use http://www.stata-press.com/data/r12/lutkepohl, clear
arima dlinvestment, ma(1)
predict double yhat

** Coefficient estimates and sigma^2 from ereturn list **
scalar beta = _b[_cons]
scalar theta = [ARMA]_b[L1.ma]
scalar sigma2 = e(sigma)^2

** pt and shrinking factor for the first two observations**
generate double pt=sigma2 in 1/2
generate double sh_factor=(sigma2)/(sigma2+theta^2*pt) in 2

** Predicted series and errors for the first two observations **
generate double my_yhat = beta
generate double myehat = sh_factor*(dlinvestment - my_yhat) in 2

** Predictions with the Kalman filter recursions **
quietly {
forvalues i = 3/91 {
replace my_yhat = my_yhat + theta*l.myehat in i'
replace pt= (sigma2*theta^2*L.pt)/(sigma2+theta^2*L.pt) in i'
replace sh_factor=(sigma2)/(sigma2+theta^2*pt)          in i'
replace myehat=sh_factor*(dlinvestment - my_yhat)       in i'
}
}


List the first 10 predictions (yhat from predict and my_yhat from the manual computations):

. list qtr yhat my_yhat pt sh_factor in 1/10

+--------------------------------------------------------+
|    qtr        yhat     my_yhat          pt   sh_factor |
|--------------------------------------------------------|
1. | 1960q1   .01686688   .01686688   .00192542           . |
2. | 1960q2   .01686688   .01686688   .00192542   .97272668 |
3. | 1960q3   .02052151   .02052151   .00005251   .99923589 |
4. | 1960q4   .01478403   .01478403   1.471e-06   .99997858 |
5. | 1961q1   .01312365   .01312365   4.125e-08    .9999994 |
|--------------------------------------------------------|
6. | 1961q2   .00326376   .00326376   1.157e-09   .99999998 |
7. | 1961q3   .02471242   .02471242   3.243e-11           1 |
8. | 1961q4   .01691061   .01691061   9.092e-13           1 |
9. | 1962q1   .01412974   .01412974   2.549e-14           1 |
10. | 1962q2   .00643301   .00643301   7.147e-16           1 |
+--------------------------------------------------------+


Notice that the shrinking factor (sh_factor) tends to 1 as t increases, which implies that after a few initial periods the predictions produced with the Kalman filter recursions become exactly the same as the ones produced by the formula at the top of this entry for the recursions derived from the ARIMA representation of the model.

Reference:

Hamilton, James. 1994. Time Series Analysis. Princeton University Press.

Categories: Statistics Tags:

## Building complicated expressions the easy way

Have you every wanted to make an “easy” calculation–say, after fitting a model–and gotten lost because you just weren’t sure where to find the degrees of freedom of the residual or the standard error of the coefficient? Have you ever been in the midst of constructing an “easy” calculation and was suddenly unsure just what e(df_r) really was? I have a solution.

It’s called Stata’s expression builder. You can get to it from the display dialog (Data->Other Utilities->Hand Calculator)

In the dialog, click the Create button to bring up the builder. Really, it doesn’t look like much:

I want to show you how to use this expression builder; if you’ll stick with me, it’ll be worth your time.

Let’s start over again and assume you are in the midst of an analysis, say,

. sysuse auto, clear
. regress price mpg length

Next invoke the expression builder by pulling down the menu Data->Other Utilities->Hand Calculator. Click Create. It looks like this:

Now click on the tree node icon (+) in front of “Estimation results” and then scroll down to see what’s underneath. You’ll see

Click on Scalars:

The middle box now contains the scalars stored in e(). N happens to be highlighted, but you could click on any of the scalars. If you look below the two boxes, you see the value of the e() scalar selected as well as its value and a short description. e(N) is 74 and is the “number of observations”.

It works the same way for all the other categories in the box on the left: Operators, Functions, Variables, Coefficients, Estimation results, Returned results, System parameters, Matrices, Macros, Scalars, Notes, and Characteristics. You simply click on the tree node icon (+), and the category expands to show what is available.

You have now mastered the expression builder!

Let’s try it out.

Say you want to verify that the p-value of the coefficient on mpg is correctly calculated by regress–which reports 0.052–or more likely, you want to verify that you know how it was calculated. You think the formula is

or, as an expression in Stata,

2*ttail(e(df_r), abs(_b[mpg]/_se[mpg]))

But I’m jumping ahead. You may not remember that _b[mpg] is the coefficient on variable mpg, or that _se[mpg] is its corresponding standard error, or that abs() is Stata’s absolute value function, or that e(df_r) is the residual degrees of freedom from the regression, or that ttail() is Stata’s Student’s t distribution function. We can build the above expression using the builder because all the components can be accessed through the builder. The ttail() and abs() functions are in the Functions category, the e(df_r) scalar is in the Estimation results category, and _b[mpg] and _se[mpg] are in the Coefficients category.

What’s nice about the builder is that not only are the item names listed but also a definition, syntax, and value are displayed when you click on an item. Having all this information in one place makes building a complex expression much easier.

Another example of when the expression builder comes in handy is when computing intraclass correlations after xtmixed. Consider a simple two-level model from Example 1 in [XT] xtmixed, which models weight trajectories of 48 pigs from 9 successive weeks:

. use http://www.stata-press.com/data/r12/pig
. xtmixed weight week || id:, variance

The intraclass correlation is a nonlinear function of variance components. In this example, the (residual) intraclass correlation is the ratio of the between-pig variance, var(_cons), to the total variance, between-pig variance plus residual (within-pig) variance, or var(_cons) + var(residual).

The xtmixed command does not store the estimates of variance components directly. Instead, it stores them as log standard deviations in e(b) such that _b[lns1_1_1:_cons] is the estimated log of between-pig standard deviation, and _b[lnsig_e:_cons] is the estimated log of residual (within-pig) standard deviation. So to compute the intraclass correlation, we must first transform log standard deviations to variances:

exp(2*_b[lns1_1_1:_cons])
exp(2*_b[lnsig_e:_cons])

The final expression for the intraclass correlation is then

exp(2*_b[lns1_1_1:_cons]) / (exp(2*_b[lns1_1_1:_cons])+exp(2*_b[lnsig_e:_cons]))

The problem is that few people remember that _b[lns1_1_1:_cons] is the estimated log of between-pig standard deviation. The few who do certainly do not want to type it. So use the expression builder as we do below:

In this case, we’re using the expression builder accessed from Stata’s nlcom dialog, which reports estimated nonlinear combinations along with their standard errors. Once we press OK here and in the nlcom dialog, we’ll see

. nlcom (exp(2*_b[lns1_1_1:_cons])/(exp(2*_b[lns1_1_1:_cons])+exp(2*_b[lnsig_e:_cons])))

_nl_1:  exp(2*_b[lns1_1_1:_cons])/(exp(2*_b[lns1_1_1:_cons])+exp(2*_b[lnsig_e:_cons]))

------------------------------------------------------------------------------
weight |      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
_nl_1 |   .7717142   .0393959    19.59   0.000     .6944996    .8489288
------------------------------------------------------------------------------

The above could easily be extended to computing different types of intraclass correlations arising in higher-level random-effects models. The use of the expression builder for that becomes even more handy.

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