Archive for the ‘Mathematics’ Category

Using Stata’s random-number generators, part 4, details

For those interested in how pseudo random number generators work, I just wrote something on Statalist which you can see in the Statalist archives by clicking the link even if you do not subscribe:

To remind you, I’ve been writing about how to use random-number generators in parts 1, 2, and 3, and I still have one more posting I want to write on the subject. What I just wrote on Statalist, however, is about how random-number generators work, and I think you will find it interesting.

To find out more about Statalist, see


How to successfully ask a question on Statalist

Using Stata’s random-number generators, part 3, drawing with replacement

The topic for today is drawing random samples with replacement. If you haven’t read part 1 and part 2 of this series on random numbers, do so. In the series we’ve
discussed that

  1. Stata’s runiform() function produces random numbers over the range [0,1). To produce such random numbers, type
    . generate double u = runiform()


  2. To produce continuous random numbers over [a,b), type
    . generate double u = (b-a)*runiform() + a


  3. To produce integer random numbers over [a,b], type
    . generate ui = floor((b-a+1)*runiform() + a)

    If b > 16,777,216, type

    . generate long ui = floor((b-a+1)*runiform() + a)


  4. To place observations in random order — to shuffle observations — type
    . set seed #
    . generate double u = runiform()
    . sort u


  5. To draw without replacement a random sample of n observations from a dataset of N observations, type
    . set seed #
    . sort variables_that_put_dta_in_unique_order
    . generate double u = runiform()
    . sort u
    . keep in 1/n

    If N>1,000, generate two random variables u1 and u2 in place of u, and substitute sort u1 u2 for sort u.


  6. To draw without replacement a P-percent random sample, type
    . set seed #
    . keep if runiform() <= P/100

I’ve glossed over details, but the above is the gist of it.

Today I’m going to tell you

  1. To draw a random sample of size n with replacement from a dataset of size N, type
    . set seed #
    . drop _all
    . set obs n
    . generate long obsno = floor(N*runiform()+1)
    . sort obsno
    . save obsnos_to_draw
    . use your_dataset, clear
    . generate long obsno = _n
    . merge 1:m obsno using obsnos_to_draw, keep(match) nogen


  2. You need to set the random-number seed only if you care about reproducibility. I’ll also mention that if N ≤ 16,777,216, it is not necessary to specify that new variable obsno be stored as long; the default float will be sufficient.

    The above solution works whether n<N, n=N, or n>N.


Drawing samples with replacement

The solution to sampling with replacement n observations from a dataset of size N is

  1. Draw n observation numbers 1, …, N with replacement. For instance, if N=4 and n=3, we might draw observation numbers 1, 3, and 3.

  2. Select those observations from the dataset of interest. For instance, select observations 1, 3, and 3.

As previously discussed in part 1, to generate random integers drawn with replacement over the range [a, b], use the formula

generate varname = floor((ba+1)*runiform() + a)

In this case, we want a=1 and b=N, and the formula reduces to,

generate varname = floor(N*runiform() + 1)

So the first half of our solution could read

. drop _all
. set obs n
. generate obsno = floor(N*runiform() + 1)

Now we are merely left with the problem of selecting those observations from our dataset, which we can do using merge by typing

. sort obsno
. save obsnos_to_draw
. use dataset_of_interest, clear
. generate obsno = _n
. merge 1:m obsno using obsnos_to_draw, keep(match) nogen

Let’s do an example. In part 2 of this series, I had a dataset with observations corresponding to playing cards:

. use cards

. list in 1/5

     | rank   suit |
  1. |  Ace   Club |
  2. |    2   Club |
  3. |    3   Club |
  4. |    4   Club |
  5. |    5   Club |

There are 52 observations in the dataset; I’m showing you just the first five. Let’s draw 10 cards from the deck, but with replacement.

The first step is to draw the observation numbers. We have N=52 cards in the deck, and we want to draw n=10, so we generate 10 random integers from the integers [1, 52]:

. drop _all

. set obs 10                            // we want n=10
obs was 0, now 10

. gen obsno = floor(52*runiform()+1)    // we draw from N=52

. list obsno                            // let's see what we have

     | obsno |
  1. |    42 |
  2. |    52 |
  3. |    16 |
  4. |     9 |
  5. |    40 |
  6. |    11 |
  7. |    34 |
  8. |    20 |
  9. |    49 |
 10. |    42 |

If you look carefully at the list, you will see that observation number 42 repeats. It will be easier to see the duplicate if we sort the list,

. sort obsno
. list

     | obsno |
  1. |     9 |
  2. |    11 |
  3. |    16 |
  4. |    20 |
  5. |    34 |
  6. |    40 |
  7. |    42 |     <- Obs. 42 repeats
  8. |    42 |     <- See?
  9. |    49 |
 10. |    52 |

An observation didn’t have to repeat, but it’s not surprising that one did because in drawing n=10 from N=52, we would expect one or more repeated cards about 60% of the time.

Anyway, we now know which cards we want, namely cards 9, 11, 16, 20, 34, 40, 42, 42 (again), 49, and 52.

The final step is to select those observations from cards.dta. The way to do that is to perform a one-to-many merge of cards.dta with the list above and keep the matches. Before we can do that, however, we must (1) save the list of observation numbers as a dataset, (2) load cards.dta, and (3) add a variable called obsno to it. Then we will be able to perform the merge. So let’s get that out of the way,

. save obsnos_to_draw                // 1. save the list above
file obsnos_to_draw.dta saved

. use cards                          // 2. load cards.dta

. gen obsno = _n                     // 3.  Add variable obsno to it

Now we can perform the merge:

. merge 1:m obsno using obsnos_to_draw, keep(matched) nogen

    Result                           # of obs.
    not matched                             0
    matched                                10

I’ll list the result, but let me first briefly explain the command

merge 1:m obsno using obsnos_to_draw, keep(matched) nogen

merge …, we are performing the merge command,

1:m …, the merge is one-to-many,

using obsnos_to_draw …, we merge data in memory with obsnos_todraw.dta,

, keep(matched) …, we keep observations that appear in both datasets,

nogen, do not add variable _merge to the resulting dataset; _merge reports the source of the resulting observations; we said keep(matched) so we know each came from both sources.

And here is the result:

. list

     |  rank      suit   obsno |
  1. |     8      Club       9 |
  2. |  Jack      Club      11 |
  3. |   Ace     Spade      16 |
  4. |     2   Diamond      20 |
  5. |     6     Spade      34 |
  6. |     8     Spade      40 |
  7. |     9     Heart      42 |   <- Obs. 42 is here ...
  8. | Queen     Spade      49 |
  9. |  King     Spade      52 |
 10. |     9     Heart      42 |   <- and here

We drew 10 cards — those are the observation numbers on the left. Variable obsno in our dataset records the original observation (card) number and really, we no longer need the variable. Anyway, obsno==42 appears twice, in real observations 7 and 10, and thus we drew the 9 of Hearts twice.


What could go wrong?

Not much can go wrong, it turns out. At this point, our generic solution is

. drop _all
. set obs n
. generate obsno = floor(n*runiform()+1)
. sort obsno
. save obsnos_to_draw

. use our_dataset
. gen obsno = _n
. merge 1:m obsno using obsnos_to_draw, keep(matched) nogen

If you study this code, there are two lines that might cause problems,

. generate obsno = floor(N*runiform()+1)


. generate obsno = _n

When you are looking for problems and see a generate or replace, think about rounding.

Let’s look at the right-hand side first. Both calculations produce integers over the range [1, N]. generate performs all calculations in double and the largest integer that can be stored without rounding is 9,007,199,254,740,992 (see previous blog post on precision). Stata allows datasets up to 2,147,483,646, so we can be sure that N is less than the maximum precise-integer double. There are no rounding issues on the right-hand side.

Next let’s look at the left-hand side. Variable obsno is being stored as a float because we did not instruct otherwise. The largest integer value that can be stored without rounding as a float (also covered in previous blog post on precision) is 16,777,216, and that is less than Stata’s 2,147,483,646 maximum observations. When N exceeds 16,777,216, the solution is to store obsno as a long. We could remember to use long on the rare occasion when dealing with such large datasets, but I’m going to change the generic solution to use longs in all cases, even when it’s unnecessary.

What else could go wrong? Well, we tried an example with n<N and that seemed to work. We should now try examples with n=N and n>N to verify there’s no hidden bug or assumption in our code. I’ve tried examples of both and the code works fine.


We’re done for today

That’s it. Drawing samples with replacement turns out to be easy, and that shouldn’t surprise us because we have a random-number generator that draws with replacement.

We could complicate the discussion and consider solutions that would run a bit more efficiently when n=N, which is of special interest in statistics because it is a key ingredient in bootstrapping, but we will not. The above solution works fine in the n=N case, and I always advise researchers to favor simple-even-if-slower solutions because they will probably save you time. Writing complicated code takes longer than writing simple code, and testing complicated code takes even longer. I know because that’s what we do at StataCorp.

Using Stata’s random-number generators, part 2, drawing without replacement

Last time I told you that Stata’s runiform() function generates rectangularly (uniformly) distributed random numbers over [0, 1), from 0 to nearly 1, and to be precise, over [0, 0.999999999767169356]. And I gave you two formulas,

  1. To generate continuous random numbers between a and b, use

    generate double u = (ba)*runiform() + a

    The random numbers will not actually be between a and b: they will be between a and nearly b, but the top will be so close to b, namely 0.999999999767169356*b, that it will not matter.

  2. To generate integer random numbers between a and b, use

    generate ui = floor((ba+1)*runiform() + a)

I also mentioned that runiform() can solve a variety of problems, including

  • shuffling data (putting observations in random order),
  • drawing random samples without replacement (there’s a minor detail we’ll have to discuss because runiform() itself produces values drawn with replacement),
  • drawing random samples with replacement (which is easier to do than most people realize),
  • drawing stratified random samples (with or without replacement),
  • manufacturing fictional data (something teachers, textbook authors, manual writers, and blog writers often need to do).

Today we will cover shuffling and drawing random samples without replacement — the first two topics on the list — and we will leave drawing random samples with replacement for next time. I’m going to tell you

  1. To place observations in random order — to shuffle the observations — type
    . generate double u = runiform()
    . sort u
  2. To draw without replacement a random sample of n observations from a dataset of N observations, type
    . set seed #
    . generate double u = runiform()
    . sort u
    . keep in 1/n

    I will tell you that there are good statistical reasons for setting the random-number seed even if you don’t care about reproducibility.

    If you do care about reproducibility, I will mention (1) that you need to use sort to put the original data in a known, reproducible order, before you generate the random variate u, and I will explain (2) a subtle issue that leads us to use different code for N≤1,000 and N>1,000. The code for for N≤1,000 is

    . set seed #
    . sort variables_that_put_data_in_unique_order
    . generate double u = runiform()
    . sort u
    . keep in 1/n

    and the code for N>1,000 is

    . set seed #
    . sort variables_that_put_data_in_unique_order
    . generate double u1 = runiform()
    . generate double u2 = runiform()
    . sort u1 u2
    . keep in 1/n

    You can use the N>1,000 code for the N≤1,000 case.

  3. To draw without replacement a P-percent random sample, type
    . set seed #
    . keep if runiform() <= P/100

    There’s no issue in this case when N is large.

As I mentioned, we’ll discuss drawing random samples with replacement next time. Today, the topic is random samples without replacement. Let’s start.

Shuffling data

I have a deck of 52 cards, in order, the first four of which are

. list in 1/4

     | rank   suit |
  1. |  Ace   Club |
  2. |    1   Club |
  3. |    2   Club |
  4. |    3   Club |

Well, actually I just have a Stata dataset with observations corresponding to playing cards. To shuffle the deck — to place the observations in random order — type

. generate double u = runiform()

. sort u

Having done that, here’s your hand,

. list in 1/5

     |  rank      suit          u |
  1. | Queen      Club   .0445188 |
  2. |     5   Diamond   .0580662 |
  3. |     7      Club   .0610638 |
  4. |  King     Heart   .0907986 |
  5. |     6     Spade   .0981878 |

and here’s mine:

. list in 6/10

     | rank      suit          u |
  6. |    8   Diamond   .1024369 |
  7. |    5      Club   .1086679 |
  8. |    8     Spade   .1091783 |
  9. |    2     Spade   .1180158 |
 10. |  Ace      Club   .1369841 |

All I did was generate random numbers — one per observation (card) — and then place the observations in ascending order of the random values. Doing that is equivalent to shuffling the deck. I used runiform() random numbers, meaning rectangularly distributed random numbers over [0, 1), but since I’m only exploiting the random-numbers’ ordinal properties, I could have used random numbers from any continuous distribution.

This simple, elegant, and obvious solution to shuffling data will play an important part of the solution to drawing observations without replacement. I have already more than hinted at the solution when I showed you your hand and mine.

Drawing n observations without replacement

Drawing without replacement is exactly the same problem as dealing cards. The solution to the physical card problem is to shuffle the cards and then draw the top cards. The solution to randomly selecting n from N observations is to put the N observations in random order and keep the first n of them.

. use cards, clear

. generate double u = runiform()

. sort u

. keep in 1/5
(47 observations deleted)

. list

     | rank      suit          u |
  1. |  Ace   Diamond   .0064866 |
  2. |    6     Heart   .0087578 |
  3. | King     Spade    .014819 |
  4. |    3     Spade   .0955155 |
  5. | King   Diamond   .1007262 |

. drop u


You might later want to reproduce the analysis, meaning you do not want to draw another random sample, but you want to draw the same random sample. Perhaps you informally distributed some preliminary results and, of course, then discovered a mistake. You want to redistribute updated results and show that your mistake didn’t change results by much, and to drive home the point, you want to use the same samples as you used previously.

Part of the solution is to set the random-number seed. You might type

. set seed 49983882

. use cards, clear

. generate double u = runiform()

. sort u

. keep in 1/5

See help set seed in Stata. As a quick review, when you set the random-number seed, you set Stata’s random-number generator into a fixed, reproducible state, which is to say, the sequence of random numbers that runiform() produces is a function of the seed. Set the seed today to the same value as yesterday, and runiform() will produce the same sequence of random numbers today as it did yesterday. Thus, after setting the seed, if you repeat today exactly what you did yesterday, you will obtain the same results.

So imagine that you set the random number seed today to the value you set it to yesterday and you repeat the above commands. Even so, you might not get the same results! You will not get the same results if the observations in cards.dta are in a different order yesterday and today. Setting the seed merely ensures that if yesterday the smallest value of u was in observation 23, it will again be in observation 23 today (and it will be the same value). If yesterday, however, observation 23 was the 6 of Clubs, and today it’s the 7 of Hearts, then today you will select the 7 of Hearts in place of the 6 of Clubs.

So make sure the data are in the same order. One way to do that is put the dataset in a known order before generating the random values on which you will sort. For instance,

. set seed 49983882

. use cards, clear

. sort rank suit

. generate double u = runiform()

. sort u

. keep in 1/5

An even better solution would add the line

. by rank suit: assert _N==1

just before the generate. That line would check whether sorting on variables rank and suit uniquely orders the observations.

With cards.dta, you can argue that the assert is unnecessary, but not because you know each rank-suit combination occurs once. You have only my assurances about that. I recommend you never trust anyone’s assurances about data. In this case, however, you can argue that the assert is unnecessary because we sorted on all the variables in the dataset and thus uniqueness is not required. Pretend there are two Ace of Clubs in the deck. Would it matter that the first card was Ace of Clubs followed by Ace of Clubs as opposed to being the other way around? Of course it would not; the two states are indistinguishable.

So let’s assume there is another variable in the dataset, say whether there was a grease spot on the back of the card. Yesterday, after sorting, the ordering might have been,

     | rank   suit   grease          u |
  1. |  Ace   Club      yes   .6012949 |
  2. |  Ace   Club       no   .1859054 |

and today,

     | rank   suit   grease          u |
  1. |  Ace   Club       no   .6012949 |
  2. |  Ace   Club      yes   .1859054 |

If yesterday you selected the Ace of Clubs without grease, today you would select the Ace of Clubs with grease.

My recommendation is (1) sort on whatever variables put the data into a unique order, and then verify that, or (2) sort on all the variables in the dataset and then don’t worry whether the order is unique.

Ensuring a random ordering

Included in our reproducible solution but omitted from our base solution was setting the random-number seed,

. set seed 49983882

Setting the seed is important even if you don’t care about reproducibility. Each time you launch Stata, Stata sets the same random-number seed, namely 123456789, and that means that runiform() generates the same sequence of random numbers, and that means that if you generated all your random samples right after launching Stata, you would always select the same observations, at least holding N constant.

So set the seed, but don’t set it too often. You set the seed once per problem. If I wanted to draw 10,000 random samples from the same data, I could code:

  use dataset, clear
  set seed 1702213
  sort variables_that_put_data_in_unique_order
  forvalues i=1(1)10000 {
          generate double u = runiform()
          sort u
          keep in 1/n
          drop u
          save sample`i', replace
          restore, preserve  

In the example I save each sample in a file. In real life, I seldom (never) save the samples; I perform whatever analysis on the samples I need and save the results, which I usually append into a single dataset. I don’t need to save the individual samples because I can recreate them.

And the result still might not be reproducible …

runiform() draws random-numbers with replacement. It is thus possible that two or more observations could have the same random values associated with them. Well yes, you’re thinking, I see that it’s possible, but surely it’s so unlikely that it just doesn’t happen. But it does happen:

. clear all

. set obs 100000
obs was 0, now 100000

. generate double u = runiform() 

. by u, sort: assert _N==1
1 contradiction in 99999 by-groups
assertion is false      

In the 100,000-observation dataset I just created, I got a duplicate! By the way, I didn’t have to look hard for such an example, I got it the first time I tried.

I have three things I want to tell you:

  1. Duplicates happen more often than you might guess.

  2. Do not panic about the duplicates. Because of how Stata is written, duplicates do not lower the quality of the sample selected. I’ll explain.

  3. Duplicates do interfere with reproducibility, however, and there is an easy way around that problem.

Let’s start with the chances of observing duplicates. I mentioned in passing last time that runiform() is a 32-bit random-number generator. That means runiform() can return any of 232 values. Their values are, in order,

          0  =  0
      1/232  =  2.32830643654e-10
      2/232  =  4.65661287308e-10 
      3/232  =  6.98491930962e-10
 (232-2)/232  =  0.9999999995343387
 (232-1)/232  =  0.9999999997671694

So what are the chances that in N draws with replacement from an urn containing these 232 values, that all values are distinct? The probability p that all values are distinct is

      232 * (232-1) * ... *(232-N)
p  =  ----------------------------

Here are some values for various values of N. p is the probability that all values are unique, and 1-p is the probability of observing one or more repeated values.

      N         p            1-p
     50   0.999999715    0.000000285
    500   0.999970955    0.000029045
  1,000   0.999883707    0.000116293
  5,000   0.997094436    0.002905564
 10,000   0.988427154    0.011572846
 50,000   0.747490440    0.252509560
100,000   0.312187988    0.687812012
200,000   0.009498117    0.990501883
300,000   0.000028161    0.999971839
400,000   0.000000008    0.999999992
500,000   0.000000000    1.000000000

In shuffling cards we generated N=52 random values. The probability of a repeated values is infinitesimal. In datasets of N=10,000, I expect to see repeated values 1% of the time. In datasets of N=50,000, I expect to see repeated values 25% of the time. By N=100,000, I expect to see repeated values more often than not. By N=500,000, I expect to see repeated value in virtually all sequences.

Even so, I promised you that this problem does not affect the randomness of the ordering. It does not because of how Stata’s sort command is written. Remember the basic solution,

. use dataset, clear

. generate double u = runiform()

. sort u

. keep in 1/n

Did you know sort has its own, private random-number generator built into it? It does, and sort uses its random-number generator to determine the order of tied observations. In the manuals we at StataCorp are fond of writing, “the ties will be ordered randomly” and a few sophisticated users probably took that to mean, “the ties will be ordered in a way that we at StataCorp do not know and even though they might be ordered in a way that will cause a bias in the subsequent analysis, because we don’t know, we’ll ignore the possibility.” But we meant it when wrote that the ties will be ordered randomly; we know that because we put a random number generator into sort to ensure the result. And that is why I can now write that repeated values of the runiform() function cause a reproducibility issue, but not a statistical issue.

The solution to the reproducibility issue is to draw two random numbers and use the random-number pair to order the observations:

. use dataset, clear

. sort varnames

. set seed #

. generate double u1 = runiform()

. generate double u2 = runiform()

. sort u1 u2

. keep in 1/n

You might wonder if we would ever need three random numbers. It is very unlikely. p, the probability of no problem, equals 1 to at least 5 digits for N=500,000. Of course, the chances of duplication are always nonzero. If you are concerned about this problem, you could add an assert to the code to verify that the two random numbers together do uniquely identify the observations:

. use dataset, clear

. sort varnames

. set seed #

. generate double u1 = runiform()

. generate double u2 = runiform()

. sort u1 u2

. by u1 u2: assert _N==1            // added line

. keep in 1/n

I do not believe that doing that is necessary.

Is using doubles necessary?

In the generation of random numbers in all of the above, note that I am storing them as doubles. For the reproducibility issue, that is important. As I mentioned in part 1, the 32-bit random numbers that runiform() produces will be rounded if forced into 23-bit floats.

Above I gave you a table of probabilities p that, in creating

. generate double u = runiform()

the values of u would be distinct. Here is what would happen if you instead stored u as a float:

                               u stored as
          -------- double ----------     ----------float ----------
      N         p            1-p               p           1-p
     50   0.999999715    0.000000285     0.999853979    0.000146021
    500   0.999970955    0.000029045     0.985238383    0.014761617
  1,000   0.999883707    0.000116293     0.942190868    0.057809132
  5,000   0.997094436    0.002905564     0.225346930    0.774653070
 10,000   0.988427154    0.011572846     0.002574145    0.997425855
 50,000   0.747490440    0.252509560     0.000000000    1.000000000
100,000   0.312187988    0.687812012     0.000000000    1.000000000
200,000   0.009498117    0.990501883     0.000000000    1.000000000
300,000   0.000028161    0.999971839     0.000000000    1.000000000
400,000   0.000000008    0.999999992     0.000000000    1.000000000
500,000   0.000000000    1.000000000     0.000000000    1.000000000

Drawing without replacement P-percent random samples

We have discussed drawing without replacement n observations from N observations. The number of observations selected has been fixed. Say instead we wanted to draw a 10% random sample, meaning that we independently allow each observation to have a 10% chance of appearing in our sample. In that case, the final number of observations is expected to be 0.1*N, but it may (and probably will) vary from that. The basic solution for drawing a 10% random sample is

. keep if runiform() <= 0.10

and the basic solution for drawing a P% random sample is

. keep if runiform() <= P/100

It is unlikely to matter whether you code <= or < in the comparison. As you now know, runiform() produces values drawn from 232 possible values, and thus the chance of equality is 2-32 or roughly 0.000000000232830644. If you want a P% sample, however, theory says you should code <=.

If you care about reproducibility, you should expand the basic solution to read,

. set seed #

. use data, clear 

. sort variables_that_put_data_in_unique_order

. keep if runiform() <= P/100

Below I draw a 10% sample from the card.dta:

. set seed 838   

. use cards, clear

. sort rank suit

. keep if runiform() <= 10/100
(46 observations deleted)

. list

     |  rank      suit |
  1. |     2   Diamond |
  2. |     2     Heart |
  3. |     3      Club |
  4. |     5     Heart |
  5. |  Jack   Diamond |
  6. | Queen     Spade |

We’re not done, but we’re done for today

In part 3 of this series I will discuss drawing random samples with replacement.

Using Stata’s random-number generators, part 1

I want to start a series on using Stata’s random-number function. Stata in fact has ten random-number functions:

  1. runiform() generates rectangularly (uniformly) distributed random number over [0,1).
  2. rbeta(a, b) generates beta-distribution beta(a, b) random numbers.
  3. rbinomial(n, p) generates binomial(n, p) random numbers, where n is the number of trials and p the probability of a success.
  4. rchi2(df) generates χ2 with df degrees of freedom random numbers.
  5. rgamma(a, b) generates Γ(a, b) random numbers, where a is the shape parameter and b, the scale parameter.
  6. rhypergeometric(N, K, n) generates hypergeometric random numbers, where N is the population size, K is the number of in the population having the attribute of interest, and n is the sample size.
  7. rnbinomial(n, p) generates negative binomial — the number of failures before the nth success — random numbers, where p is the probability of a success. (n can also be noninteger.)
  8. rnormal(μ, σ) generates Gaussian normal random numbers.
  9. rpoisson(m) generates Poisson(m) random numbers.
  10. rt(df) generates Student’s t(df) random numbers.

You already know that these random-number generators do not really produce random numbers; they produce pseudo-random numbers. This series is not about that, so we’ll be relaxed about calling them random-number generators.

You should already know that you can set the random-number seed before using the generators. That is not required but it is recommended. You set the seed not to obtain better random numbers, but to obtain reproducible random numbers. In fact, setting the seed too often can actually reduce the quality of the random numbers! If you don’t know that, then read help set seed in Stata. I should probably pull out the part about setting the seed too often, expand it, and turn it into a blog entry. Anyway, this series is not about that either.

This series is about the use of random-number generators to solve problems, just as most users usually use them. The series will provide practical advice. I’ll stay away from describing how they work internally, although long-time readers know that I won’t keep the promise. At least I’ll try to make sure that any technical details are things you really need to know. As a result, I probably won’t even get to write once that if this is the kind of thing that interests you, StataCorp would be delighted to have you join our development staff.


runiform(), generating uniformly distributed random numbers

Mostly I’m going to write about runiform() because runiform() can solve such a variety of problems. runiform() can be used to solve,

  • shuffling data (putting observations in random order),
  • drawing random samples without replacement (there’s a minor detail we’ll have to discuss because runiform() itself produces values drawn with replacement),
  • drawing random samples with replacement (which is easier to do than most people realize),
  • drawing stratified random samples (with or without replacement),
  • manufacturing fictional data (something teachers, textbook authors, manual writers, and blog writers often need to do).

runiform() generates uniformly, a.k.a. rectangularly distributed, random numbers over the interval, I quote from the manual, “0 to nearly 1”.

Nearly 1? “Why not all the way to 1?” you should be asking. “And what exactly do you mean by nearly 1?”

The answer is that the generator is more useful if it omits 1 from the interval, and so we shaved just a little off. runiform() produces random numbers over [0, 0.999999999767169356].

Here are two useful formulas you should commit to memory.

  1. If you want to generate continuous random numbers between a and b, use

    generate double u = (ba)*runiform() + a

    The random numbers will not actually be between a and b, they will be between a and nearly b, but the top will be so close to b, namely 0.999999999767169356*b, that it will not matter.

    Remember to store continuous random values as doubles.

  2. If you want to generate integer random numbers between a and b, use

    generate ui = floor((ba+1)*runiform() + a)

    In particular, do not even consider using the formula for continuous values but rounded to integers, which is to say, round(u) = round((ba)*runiform() + a). If you use that formula, and if ba>1, then a and b will be under represented by 50% each in the samples you generate!

    I stored ui as a default float, so I am assuming that -16,777,216 ≤ a < b ≤ 16,777,216. If you have integers outside of that range, however, store as a long or double.

I’m going to spend the rest of this blog entry explaining the above.

First, I want to show you how I got the two formulas and why you must use the second formula for generating integer uniform deviates.

Then I want explain why we shaved a little from the top of runiform(), namely (1) while it wouldn’t matter for formula 1, it made formula 2 a little easier, (2) the code would run more quickly, (3) we could more easily prove that we had implemented the random-number generator correctly, and (4) anyone digging deeper into our random numbers would not be misled into thinking they had more than 32 bits of resolution. That last point will be important in a future blog entry.


Continuous uniforms over [a, b)

runiform() produces random numbers over [0, 1). It therefore obviously follows that (ba)*runiform()+a produces number over [a, b). Substitute 0 for runiform() and the lower limit is obtained. Substitute 1 for runiform() and the upper limit is obtained.

I can tell you that in fact, runiform() produces random numbers over [0, (232-1)/232].

Thus (ba)*runiform()+a produces random numbers over [a, ((232-1)/232)*b].

(232-1)/232) approximately equals 0.999999999767169356 and exactly equals 1.fffffffeX-01 if you will allow me to use %21x format, which Stata understands and which you can understand if you see my previous blog posting on precision.

Thus, if you are concerned about results being in the interval [a, b) rather than [a, b], you can use the formula

generate double u = ((ba)*runiform() + a) / 1.fffffffeX-01

There are seven f’s followed by e in the hexadecimal constant. Alternatively, you could type

generate double u = ((ba)*runiform() + a) * ((2^32-1)/2^32)

but multiplying by 1.fffffffeX-01 is less typing so I’d type that. Actually I wouldn’t type either one; the small difference between values lying in [a, b) or [a, b] is unimportant.


Integer uniforms over [a, b]

Whether we produce real, continuous random numbers over [a, b) or [a, b] may be unimportant, but if we want to draw random integers, the distinction is important.

runiform() produces continuous results over [0, 1).

(ba)*runiform()+a produces continuous results over [a, b).

To produce integer results, we might round continuous results over segments of the number line:

           a    a+.5  a+1  a+1.5  a+2  a+2.5       b-1.5  b-1  b-.5    b
real line  +-----+-----+-----+-----+-----+-----------+-----+-----+-----+
int  line  |<-a->|<---a+1--->|<---a+2--->|           |<---b-1--->|<-b->|

In the diagram above, think of the numbers being produced by the continuous formula u=(ba)*runiform()+a as being arrayed along the real line. Then imagine rounding those values, say by using Stata’s round(u) function. If you rounded in that way, then

  • Values of u between a and a+0.5 will be rounded to a.
  • Values of u between a+0.5 and a+1.5 will be rounded to a+1.
  • Values of u between a+1.5 and a+2.5 will be rounded to a+2.
  • Values of u between b-1.5 and b-0.5 will be rounded to b-1.
  • Values of u between b-0.5 and b-1 will be rounded to b.

Note that the width of the first and last intervals is half that of the other intervals. Given that u follows the rectangular distribution, we thus expect half as many values rounded to a and to b as to a+1 or a+2 or … or b-1.

And indeed, that is exactly what we would see:

. set obs 100000
obs was 0, now 100000

. gen double u = (5-1)*runiform() + 1

. gen i = round(u)

. summarize u i 

    Variable |       Obs        Mean    Std. Dev.       Min        Max
           u |    100000    3.005933    1.156486   1.000012   4.999983
           i |    100000     3.00489    1.225757          1          5

. tabulate i

          i |      Freq.     Percent        Cum.
          1 |     12,525       12.53       12.53
          2 |     24,785       24.79       37.31
          3 |     24,886       24.89       62.20
          4 |     25,284       25.28       87.48
          5 |     12,520       12.52      100.00
      Total |    100,000      100.00

To avoid the problem we need to make the widths of all the intervals equal, and that is what the formula floor((ba+1)*runiform() + a) does.

           a          a+1         a+2                     b-1          b          b+1
real line  +-----+-----+-----+-----+-----------------------+-----+-----+-----+-----+
int  line  |<--- a --->|<-- a+1 -->|                       |<-- b-1 -->|<--- b --->)

Our intervals are of equal width and thus we expect to see roughly the same number of observations in each:

. gen better = floor((5-1+1)*runiform() + 1)

. tabulate better

     better |      Freq.     Percent        Cum.
          1 |     19,808       19.81       19.81
          2 |     20,025       20.02       39.83
          3 |     19,963       19.96       59.80
          4 |     20,051       20.05       79.85
          5 |     20,153       20.15      100.00
      Total |    100,000      100.00

So now you know why we shaved a little off the top when we implemented runiform(); it made the formula

floor((ba+1)*runiform() + a):

easier. Our integer [a, b] formula did not have to concern itself that runiform() would sometimes — rarely — return 1. If runiform() did return the occasional 1, the simple formula above would produce the (correspondingly occasional) b+1.


How Stata calculates continuous random numbers

I’ve said that we shaved a little off the top, but the fact was that it was easier for us to do the shaving than not.

runiform() is based on the KISS random number generator. KISS produces 32-bit integers, meaning integers the range [0, 232-1], or [0, 4,294,967,295]. You might wonder how we converted that range to being continuous over [0, 1).

Start by thinking of the number KISS produces in its binary form:


The corresponding integer is b31*231 + b31*230 + … + b0*20. All we did was insert a binary point out front:

. b31b30b29b28b27b26b25b24b23b22b21b20b19b18b17b16b15b14b13b12b11b10b9b8b7b6b5b4b3b2b1b0

making the real value b31*2-1 + b30*2-2 + … + b0*2-32. Doing that is equivalent to dividing by 2-32, except insertion of the binary point is faster. Nonetheless, if we had wanted runiform() to produce numbers over [0, 1], we could have divided by 232-1.

Anyway, if the KISS random number generator produced 3190625931, which in binary is


we converted that to


which equals 0.74287549 in base 10.

The largest number the KISS random number generator can produce is, of course,


and 0.11111111111111111111111111111111 equals 0.999999999767169356 in base 10. Thus, the runiform() implementation of KISS generates random numbers in the range [0, 0.999999999767169356].

I could have presented all of this mathematically in base 10: KISS produces integers in the range [0, 232-1], and in runiform() we divide by 232 to thus produce continuous numbers over the range [0, (232-1)/232]. I could have said that, but it loses the flavor and intuition of my longer explanation, and it would gloss over the fact that we just inserted the binary point. If I asked you, a base-10 user, to divide 232 by 10, you wouldn’t actually divide in the same way that they would divide by, say 9. Dividing by 9 is work. Dividing by 10 merely requires shifting the decimal point. 232 divided by 10 is obviously 23.2. You may not have realized that modern digital computers, when programmed by “advanced” programmers, follow similar procedures.

Oh gosh, I do get to say it! If this sort of thing interests you, consider a career at StataCorp. We’d love to have you.


Is it important that runiform() values be stored as doubles?

Sometimes it is important. It’s obviously not important when you are generating random integers using floor((ba+1)*runiform() + a) and -16,777,216 ≤ a < b ≤ 16,777,216. Integers in that range fit into a float without rounding.

When creating continuous values, remember that runiform() produces 32 bits. floats store 23 bits and doubles store 52, so if you store the result of runiform() as a float, it will be rounded. Sometimes the rounding matters, and sometimes it does not. Next time, we will discuss drawing random samples without replacement. In that case, the rounding matters. In most other cases, including drawing random samples with replacement — something else for later — the rounding
does not matter. Rather than thinking hard about the issue, I store all my non-integer
random values as doubles.


Tune in for the next episode

Yes, please do tune in for the next episode of everything you need to know about using random-number generators. As I already mentioned, we’ll discuss drawing random samples without replacement. In the third installment, I’m pretty sure we’ll discuss random samples with replacement. After that, I’m a little unsure about the ordering, but I want to discuss oversampling of some groups relative to others and, separately, discuss the manufacturing of fictional data.

Am I forgetting something?

The Penultimate Guide to Precision

There have recently been occasional questions on precision and storage types on Statalist despite all that I have written on the subject, much of it posted in this blog. I take that as evidence that I have yet to produce a useful, readable piece that addresses all the questions researchers have.

So I want to try again. This time I’ll try to write the ultimate piece on the subject, making it as short and snappy as possible, and addressing every popular question of which I am aware — including some I haven’t addressed before — and doing all that without making you wade with me into all the messy details, which I know I have a tendency to do.

I am hopeful that from now on, every question that appears on Statalist that even remotely touches on the subject will be answered with a link back to this page. If I succeed, I will place this in the Stata manuals and get it indexed online in Stata so that users can find it the instant they have questions.

What follows is intended to provide everything scientific researchers need to know to judge the effect of storage precision on their work, to know what can go wrong, and to prevent that. I don’t want to raise expectations too much, however, so I will entitle it …


  1. Contents

     1. Numeric types
    2. Floating-point types
    3. Integer types
    4. Integer precision
    5. Floating-point precision
    6. Advice concerning 0.1, 0.2, …
    7. Advice concerning exact data, such as currency data
    8. Advice for programmers
    9. How to interpret %21x format (if you care)
    10. Also see

  2. Numeric types

    1.1 Stata provides five numeric types for storing variables, three of them integer types and two of them floating point.

    1.2 The floating-point types are float and double.

    1.3 The integer types are byte, int, and long.

    1.4 Stata uses these five types for the storage of data.

    1.5 Stata makes all calculations in double precision (and sometimes quad precision) regardless of the type used to store the data.

  3. Floating-point types

    2.1 Stata provides two IEEE 754-2008 floating-point types: float and double.

    2.2 float variables are stored in 4 bytes.

    2.3 double variables are stored in 8 bytes.

    2.4 The ranges of float and double variables are

         type             minimum                maximum
         float     -3.40282346639e+ 38      1.70141173319e+ 38
         double    -1.79769313486e+308      8.98846567431e+307
         In addition, float and double can record missing values 
         ., .a, .b, ..., .z.

    The above values are approximations. For those familiar with %21x floating-point hexadecimal format, the exact values are

         type                   minimum                maximum
         float   -1.fffffe0000000X+07f     +1.fffffe0000000X+07e 
         double  -1.fffffffffffffX+3ff     +1.fffffffffffffX+3fe

    Said differently, and less precisely, float values are in the open interval (-2128, 2127), and double values are in the open interval (-21024, 21023). This is less precise because the intervals shown in the tables are closed intervals.

  4. Integer types

    3.1 Stata provides three integer storage formats: byte, int, and long. They are 1 byte, 2 bytes, and 4 bytes, respectively.

    3.2 Integers may also be stored in Stata’s IEEE 754-2008 floating-point storage formats float and double.

    3.3 Integer values may be stored precisely over the ranges

         type                   minimum                 maximum
         byte                      -127                     100
         int                    -32,767                  32,740
         long            -2,147,483,647           2,147,483,620
         float              -16,777,216              16,777,216
         double  -9,007,199,254,740,992   9,007,199,254,740,992
         In addition, all storage types can record missing values
         ., .a, .b, ..., .z.

    The overall ranges of float and double were shown in (2.4) and are wider than the ranges for them shown here. The ranges shown here are the subsets of the overall ranges over which no rounding of integer values occurs.

  5. Integer precision

    4.1 (Automatic promotion.) For the integer storage types — for byte, int, and long — numbers outside the ranges listed in (3.3) would be stored as missing (.) except that storage types are promoted automatically. As necessary, Stata promotes bytes to ints, ints to longs, and longs to doubles. Even if a variable is a byte, the effective range is still [-9,007,199,254,740,992, 9,007,199,254,740,992] in the sense that you could change a value of a byte variable to a large value and that value would be stored correctly; the variable that was a byte would, as if by magic, change its type to int, long, or double if that were necessary.

    4.2 (Data input.) Automatic promotion (4.1) applies after the data are input/read/imported/copied into Stata. When first reading, importing, copying, or creating data, it is your responsibility to choose appropriate storage types. Be aware that Stata’s default storage type is float, so if you have large integers, it is usually necessary to specify explicitly the types you wish to use.

    If you are unsure of the type to specify for your integer variables, specify double. After reading the data, you can use compress to demote storage types. compress never results in a loss of precision.

    4.3 Note that you can use the floating-point types float and double to store integer data.

    4.3.1 Integers outside the range [-2,147,483,647, 2,147,483,620] must be stored as doubles if they are to be precisely recorded.

    4.3.2 Integers can be stored as float, but avoid doing that unless you are certain they will be inside the range [-16,777,216, 16,777,216] not just when you initially read, import, or copy them into Stata, but subsequently as you make transformations.

    4.3.3 If you read your integer data as floats, and assuming they are within the allowed range, we recommend that you change them to an integer type. You can do that simply by typing compress. We make that recommendation so that your integer variables will benefit from the automatic promotion described in (4.1).

    4.4 Let us show what can go wrong if you do not follow our advice in (4.3). For the floating-point types — for float and double — integer values outside the ranges listed in (3.3) are rounded.

    Consider a float variable, and remember that the integer range for floats is [-16,777,216, 16,777,216]. If you tried to store a value outside the range in the variable — say, 16,777,221 — and if you checked afterward, you would discover that actually stored was 16,777,220! Here are some other examples of rounding:

         desired value                            stored (rounded)
         to store            true value             float value 
         maximum             16,777,216              16,777,216 
         maximum+1           16,777,217              16,777,216
         maximum+2           16,777,218              16,777,218
         maximum+3           16,777,219              16,777,220
         maximum+4           16,777,220              16,777,220
         maximum+5           16,777,221              16,777,220
         maximum+6           16,777,222              16,777,222
         maximum+7           16,777,223              16,777,224
         maximum+8           16,777,224              16,777,224
         maximum+9           16,777,225              16,777,224
         maximum+10          16,777,226              16,777,226

    When you store large integers in float variables, values will be rounded and no mention will be made of that fact.

    And that is why we say that if you have integer data that must be recorded precisely and if the values might be large — outside the range ±16,777,216 — do not use float. Use long or use double; or just use the compress command and let automatic promotion handle the problem for you.

    4.5 Unlike byte, int, and long, float and double variables are not promoted to preserve integer precision.

    Float values are not promoted because, well, they are not. Actually, there is a deep reason, but it has to do with the use of float variables for their real purpose, which is to store non-integer values.

    Double values are not promoted because there is nothing to promote them to. Double is Stata’s most precise storage type. The largest integer value Stata can store precisely is 9,007,199,254,740,992 and the smallest is -9,007,199,254,740,992.

    Integer values outside the range for doubles round in the same way that float values round, except at absolutely larger values.

  6. Floating-point precision

    5.1 The smallest, nonzero value that can be stored in float and double is

         type      value          value in %21x         value in base 10
         float     ±2^-127    ±1.0000000000000X-07f   ±5.877471754111e-039
         double    ±2^-1022   ±1.0000000000000X-3fe   ±2.225073858507e-308

    We include the value shown in the third column, the value in %21x, for those who know how to read it. It is described in (9), but it is unimportant. We are merely emphasizing that these are the smallest values for properly normalized numbers.

    5.2 The smallest value of epsilon such that 1+epsilon ≠ 1 is

         type      epsilon       epsilon in %21x        epsilon in base 10
         float      ±2^-23     ±1.0000000000000X-017    ±1.19209289551e-07
         double     ±2^-52     ±1.0000000000000X-034    ±2.22044604925e-16

    Epsilon is the distance from 1 to the next number on the floating-point number line. The corresponding unit roundoff error is u = ±epsilon/2. The unit roundoff error is the maximum relative roundoff error that is introduced by the floating-point number storage scheme.

    The smallest value of epsilon such that x+epsilon ≠ x is approximately |x|*epsilon, and the corresponding unit roundoff error is ±|x|*epsilon/2.

    5.3 The precision of the floating-point types is, depending on how you want to measure it,

         Measurement                           float              double
         # of binary digits                       23                  52
         # of base 10 digits (approximate)         7                  16 
         Relative precision                   ±2^-24              ±2^-53
         ... in base 10 (approximate)      ±5.96e-08           ±1.11e-16

    Relative precision is defined as

                           |x - x_as_stored|
                  ± max   ------------------    
                     x            x

    performed using infinite precision arithmetic, x chosen from the subset of reals between the minimum and maximum values that can be stored. It is worth appreciating that relative precision is a worst-case relative error over all possible numbers that can be stored. Relative precision is identical to roundoff error, but perhaps this definition is easier to appreciate.

    5.4 Stata never makes calculations in float precision, even if the data are stored as float.

    Stata makes double-precision calculations regardless of how the numeric data are stored. In some cases, Stata internally uses quad precision, which provides approximately 32 decimal digits of precision. If the result of the calculation is being stored back into a variable in the dataset, then the double (or quad) result is rounded as necessary to be stored.

    5.5 (False precision.) Double precision is 536,870,912 times more accurate than float precision. You may worry that float precision is inadequate to accurately record your data.

    Little in this world is measured to a relative accuracy of ±2-24, the accuracy provided by float precision.

    Ms. Smith, it is reported, made $112,293 this year. Do you believe that is recorded to an accuracy of ±2-24*112,293, or approximately ±0.7 cents?

    David was born on 21jan1952, so on 27mar2012 he was 21,981 days old, or 60.18 years old. Recorded in float precision, the precision is ±60.18*2-24, or roughly ±1.89 minutes.

    Joe reported that he drives 12,234 miles per year. Do you believe that Joe’s report is accurate to ±12,234*2-24, equivalent to ±3.85 feet?

    A sample of 102,400 people reported that they drove, in total, 1,252,761,600 miles last year. Is that accurate to ±74.7 miles (float precision)? If it is, each of them is reporting with an accuracy of roughly ±3.85 feet.

    The distance from the Earth to the moon is often reported as 384,401 kilometers. Recorded as a float, the precision is ±384,401*2-24, or ±23 meters, or ±0.023 kilometers. Because the number was not reported as 384,401.000, one would assume float precision would be accurate to record that result. In fact, float precision is more than sufficiently accurate to record the distance because the distance from the Earth to the moon varies from 356,400 to 406,700 kilometers, some 50,300 kilometers. The distance would have been better reported as 384,401 ±25,150 kilometers. At best, the measurement 384,401 has relative accuracy of ±0.033 (it is accurate to roughly two digits).

    Nonetheless, a few things have been measured with more than float accuracy, and they stand out as crowning accomplishments of mankind. Use double as required.

  7. Advice concerning 0.1, 0.2, …

    6.1 Stata uses base 2, binary. Popular numbers such as 0.1, 0.2, 100.21, and so on, have no exact binary representation in a finite number of binary digits. There are a few exceptions, such as 0.5 and 0.25, but not many.

    6.2 If you create a float variable containing 1.1 and list it, it will list as 1.1 but that is only because Stata’s default display format is %9.0g. If you changed that format to %16.0g, the result would appear as 1.1000000238419.

    This scares some users. If this scares you, go back and read (5.5) False Precision. The relative error is still a modest ±2-24. The number 1.1000000238419 is likely a perfectly acceptable approximation to 1.1 because the 1.1 was never measured to an accuracy of less than ±2-24 anyway.

    6.3 One reason perfectly acceptable approximations to 1.1 such as 1.1000000238419 may bother you is that you cannot select observations containing 1.1 by typing if x==1.1 if x is a float variable. You cannot because the 1.1 on the right is interpreted as double precision 1.1. To select the observations, you have to type if x==float(1.1).

    6.4 If this bothers you, record the data as doubles. It is best to do this at the point when you read the original data or when you make the original calculation. The number will then appear to be 1.1. It will not really be 1.1, but it will have less relative error, namely, ±2-53.

    6.5 If you originally read the data and stored them as floats, it is still sometimes possible to recover the double-precision accuracy just as if you had originally read the data into doubles. You can do this if you know how many decimal digits were recorded after the decimal point and if the values are within a certain range.

    If there was one digit after the decimal point and if the data are in the range [-1,048,576, 1,048,576], which means the values could be -1,048,576, -1,048,575.9, …, -1, 0, 1, …, 1,048,575.9, 1,048,576, then typing

    . gen double y = round(x*10)/10

    will recover the full double-precision result. Stored in y will be the number in double precision just as if you had originally read it that way.

    It is not possible, however, to recover the original result if x is outside the range ±1,048,576 because the float variable contains too little information.

    You can do something similar when there are two, three, or more decimal digits:

         # digits to
         right of 
         decimal pt.   range     command
             1      ±1,048,576   gen double y = round(x*10)/10
             2      ±  131,072   gen double y = round(x*100)/100
             3      ±   16,384   gen double y = round(x*1000)/1000
             4      ±    1,024   gen double y = round(x*10000)/10000
             5      ±      128   gen double y = round(x*100000)/100000
             6      ±       16   gen double y = round(x*1000000)/1000000
             7      ±        1   gen double y = round(x*10000000)/10000000

    Range is the range of x over which command will produce correct results. For instance, range = ±16 in the next-to-the-last line means that the values recorded in x must be -16 ≤ x ≤ 16.

  8. Advice concerning exact data, such as currency data

    7.1 Yes, there are exact data in this world. Such data are usually counts of something or are currency data, which you can think of as counts of pennies ($0.01) or the smallest unit in whatever currency you are using.

    7.2 Just because the data are exact does not mean you need exact answers. It may still be that calculated answers are adequate if the data are recorded to a relative accuracy of ±2-24 (float). For most analyses — even of currency data — this is often adequate. The U.S. deficit in 2011 was $1.5 trillion. Stored as a float, this amount has a (maximum) error of ±2-24*1.5e+12 = ±$89,406.97. It would be difficult to imagine that ±$89,406.97 would affect any government decision maker dealing with the full $1.5 trillion.

    7.3 That said, you sometimes do need to make exact calculations. Banks tracking their accounts need exact amounts. It is not enough to say to account holders that we have your money within a few pennies, dollars, or hundreds of dollars.

    In that case, the currency data should be converted to integers (pennies) and stored as integers, and then processed as described in (4). Assuming the dollar-and-cent amounts were read into doubles, you can convert them into pennies by typing

    . replace x = x*100

    7.4 If you mistakenly read the currency data as a float, you do not have to re-read the data if the dollar amounts are between ±$131,072. You can type

    . gen double x_in_pennies = round(x*100)

    This works only if x is between ±131,072.

  9. Advice for programmers

    8.1 Stata does all calculations in double (and sometimes quad) precision.

    Float precision may be adequate for recording most data, but float precision is inadequate for performing calculations. That is why Stata does all calculations in double precision. Float precision is also inadequate for storing the results of intermediate calculations.

    There is only one situation in which you need to exercise caution — if you create variables in the data containing intermediate results. Be sure to create all such variables as doubles.

    8.2 The same quad-precision routines StataCorp uses are available to you in Mata; see the manual entries [M-5] mean, [M-5] sum, [M-5] runningsum, and [M-5] quadcross. Use them as you judge necessary.

  10. How to interpret %21x format (if you care)

    9.1 Stata has a display format that will display IEEE 754-2008 floating-point numbers in their full binary glory but in a readable way. You probably do not care; if so, skip this section.

    9.2 IEEE 754-2008 floating-point numbers are stored as a pair of numbers (a, b) that are given the interpretation

    z = a * 2b

    where -2 < a < 2. In double precision, a is recorded with 52 binary digits. In float precision, a is recorded with 23 binary digits. For example, the number 2 is recorded in double precision as

    a = +1.0000000000000000000000000000000000000000000000000000
    b = +1

    The value of pi is recorded as

    a = +1.1001001000011111101101010100010001000010110100011000
    b = +1

    9.3 %21x presents a and b in base 16. The double-precision value of 2 is shown in %21x format as


    and the value of pi is shown as


    In the case of pi, the interpretation is

    a = +1.921fb54442d18 (base 16)
    b = +001             (base 16)

    Reading this requires practice. It helps to remember that one-half corresponds to 0.8 (base 16). Thus, we can see that a is slightly larger than 1.5 (base 10) and b = 1 (base 10), so _pi is something over 1.5*21 = 3.

    The number 100,000 in %21x is


    which is to say

    a = +1.86a0000000000 (base 16)
    b = +010             (base 16)

    We see that a is slightly over 1.5 (base 10), and b is 16 (base 10), so 100,000 is something over 1.5*216 = 98,304.

    9.4 %21x faithfully presents how the computer thinks of the number. For instance, we can easily see that the nice number 1.1 (base 10) is, in binary, a number with many digits to the right of the binary point:

    . display %21x 1.1

    We can also see why 1.1 stored as a float is different from 1.1 stored as a double:

    . display %21x float(1.1)

    Float precision assigns fewer digits to the mantissa than does double precision, and 1.1 (base 10) in base 16 is a repeating hexadecimal.

    9.5 %21x can be used as an input format as well as an output format. For instance, Stata understands

    . gen x = 1.86ax+10

    Stored in x will be 100,000 (base 10).

    9.6 StataCorp has seen too many competent scientific programmers who, needing a perturbance for later use in their program, code something like

    epsilon = 1e-8

    It is worth examining that number:

    . display %21x 1e-8

    That is an ugly number that can only lead to the introduction of roundoff error in their program. A far better number would be

    epsilon = 1.0x-1b

    Stata and Mata understand the above statement because %21x may be used as input as well as output. Naturally, 1.0x-1b looks just like what it is,

    . display %21x 1.0x-1b

    and all those pretty zeros will reduce numerical roundoff error.

    In base 10, the pretty 1.0x-1b looks like

    . display %20.0g 1.0x-1b

    and that number may not look pretty to you, but you are not a base-2 digital computer.

    Perhaps the programmer feels that epsilon really needs to be closer to 1e-8. In %21x, we see that 1e-8 is +1.5798ee2308c3aX-01b, so if we want to get closer, perhaps we use

    epsilon = 1.6x-1b

    9.7 %21x was invented by StataCorp.

  11. Also see

    If you wish to learn more, see

    How to read the %21x format

    How to read the %21x format, part 2

    Precision (yet again), Part I

    Precision (yet again), Part II

Precision (yet again), Part II

In part I, I wrote about precision issues in English. If you enjoyed that, you may want to stop reading now, because I’m about to go into the technical details. Actually, these details are pretty interesting.

For instance, I offered the following formula for calculating error due to float precision:

maximum_error = 2-24 X

I later mentioned that the formula is an approximation, and said that the true formula is,

maximum_error = 2-24 2floor(log2 X)

I didn’t explain how I got either formula.

I need to be more precise today than I was in my previous posting. For instance, I previously used x for two concepts, the true value and the rounded-after-storage value. Today I need to distinguish those concepts.

X is the true value.

x is the value after rounding due to storage.

The issue is the difference between x and X when X is stored in 24-binary-digit float precision.

Base 10

Although I harp on the value of learning to think in binary and hexadecimal, I admit that I, too, find it easier to think in base 10. So let’s start that way.

Say we record numbers to two digits of accuracy, which I will call d=2. Examples of d=2 numbers include

52*10^1 (i.e, 520, but with only two significant digits)

To say that we record numbers to two digits of accuracy is to say that, coming upon the recorded number 1, we know only that the number lies between 0.95 and 1.05; or coming upon 12, that the true number lies between 11.5 and 12.5, and so on. I assume that numbers are rounded efficiently, which is to say, stored values record midpoints of intervals.

Before we get into the math, let me note that most us would be willing to say that numbers recorded this way are accurate to 1 part in 10 or, if d=3, to 1 part in 100. If numbers are accurate to 1 part in 10^(d-1), then couldn’t we must multiply the number by 1/(10^(d-1)) to obtain the width of the interval? Let’s try:

Assume X=520 and d=2. Then 520/(10^(2-1)) = 52. The true interval, however, is (515, 525] and it has width 10. So the simple formula does not work.

The simple formula does not work yet I presented its base-2 equivalent in Part 1 and I even recommended its use! We will get to that. It turns out the smaller the base, the more accurately the simple formula approximates the true formula, but before I can show that, I need the true formula.

Let’s start by thinking about d=1.

  1. The recorded number 0 will contain all numbers between [-0.5, 0.5). The recorded number 1 will contain all numbers between [0.5, 1.5), and so on. For 0, 1, …, 9, the width of the intervals is 1.

  2. The recorded number 10 will contain all numbers between [5, 15). The recorded number 20 will contain all numbers between [15, 25), and so on, For 10, 20, …, 90, the width of the intervals is 10.

The derivation for the width of interval goes like this:

  1. If we recorded the value of X to one decimal digit, the recorded digit will will be b, the recorded value will be x = b*10p, and the power of ten will be p = floor(log10X). More importantly, W1 = 10p will be the width of the interval containing X.

  2. It therefore follows that if we recorded the value of X to two decimal digits, the interval length would be W2 = W1/10. What ever the width with one digit, adding another must reduce width by one-tenth.

  3. If we recorded the value of X to three decimal digits, the interval length would be W3 = W2/10.

  4. Thus, if d is the number of digits to which numbers are recorded, the width of the interval is 10p where p = floor(log10X) – (d-1).

The above formula is exact.

Base 2

Converting the formula

interval_width = 10floor(log10X)-(d-1)

from base 10 to base 2 is easy enough:

interval_width = 2floor(log2X)-(d-1)

In Part 1, I presented this formula for d=24 as

maximum_error = 2floor(log2X)-24 = 2 -24 2floor(log2 X)

In interval_width, it is d-1 and not d that appears in the formula. You might think I made an error and should have put -23 where I put -24 in the maximum_error formula. There is no mistake. In Part 1, the maximum error was defined as a plus-or-minus quantity and is thus half the width of the overall interval. So I divided by 2, and in effect, I did put -23 into the maximum_error formula, at least before I subracted one more from it, making it -24 again.

I started out this posting by considering and dismissing the base-10 approximation formula

interval_width = 10-(d-1) X

which in maximum-error units is

maximum_error = 10-d X

and yet in Part 1, I presented — and even recommended — its base-2, d=24 equivalent,

maximum_error = 2-24 X

It turns out that the approximation formula is not as inaccurate in base 2 and it would be in base 10. The correct formula,

maximum_error = 2floor(log2X)-d

can be written

maximum_error = 2-d 2floor(log2X

so the question becomes about the accuracy of substituting X for 2^floor(log2X). We know by examination that X ≥ 2^floor(log2X), so making the substitution will overstate the error and, in that sense, is a safe thing to do. The question becomes how much the error is overstated.

X can be written 2^(log2X) and thus we need to compare 2^(log2X) with 2^floor(log2X). The floor() function cannot reduce its argument by more than 1, and thus 2^(log2X) cannot differ from 2^floor(log2X) by more than a factor of 2. Under the circumstances, this seems a reasonable approximation.

In the case of base 10, the the floor() function reducing its argument by up to 1 results in a decrease of up to a factor of 10. That, it seems to me, is not a reasonable amount of error.

Categories: Numerical Analysis Tags: ,

Precision (yet again), Part I

I wrote about precision here and here, but they were pretty technical.

“Great,” coworkers inside StataCorp said to me, “but couldn’t you explain these issues in a way that doesn’t get lost in the details of how computers store binary and maybe, just maybe, write about floats and doubles from a user’s perspective instead of programmer’s perspective?”

“Mmmm,” I said clearly.

Later, when I tried, I liked the result. It contains new material, too. What follows is what I now wish I had written first. I’d would have still written the other two postings, but as technical appendices.

In Part 2 (forthcoming), I provide the mathematical derivations underlying what follows. There are a few interesting issues underlying what follows.

Please excuse the manualish style of what follows, but I suspect that what follows will eventually work its way into Stata’s help files or manuals, so I wrote it that way.



. generate x = 1.1

. list
  (Stata displays output showing x is 1.1 in all observations)

. count if x==1.1

Solution 1:

. count if x==float(1.1)

Solution 2:

. generate double x = 1.1

. count if x==1.1

Solution 3:

. set type double

. generate x = 1.1

. count if x==1.1


Stata works in binary. Stata stores data in float precision by default. Stata preforms all calculations in double precision. Sometimes the combination results in surprises until you think more carefully about what happened.


Remarks are presented under the headings

Why count==1.1 produces 0
How count==float(1.1) solves the problem
How storing data as double appears to solve the problem (and does)
Float is plenty accurate to store most data
Why don’t I have the problems using Excel?


Justifications for all statements made appear in the sections below. In summary,

  1. It sometimes appears that Stata is inaccurate. That is not true and, in fact, the appearance of inaccuracy happens in part because Stata is so accurate.

  2. You can cover up this appearance of inaccuracy by storing all your data in double precision. This will double or more the size of your dataset, and so I do not recommend the double-precision solution unless your dataset is small relative to the amount of memory on your computer. In that case, there is nothing wrong with storing all your data in double precision.

    The easiest way to implement the double-precision solution is by typing set type double. After that, Stata will default to to creating all new variables as doubles, at least for the remainder of the session. If all your datasets are small relative to the amount of memory on your computer, you can set type double, permanently.

  3. The double-precision solution is needlessly wasteful of memory. It is difficult to imagine data that are accurate to more than float precision. Regardless of how your data are stored, Stata does all calculations in double precision, and sometimes in quad precision.

  4. The issue of 1.1 not being equal to 1.1 arises only with “nice” decimal numbers. You just have to remember to use Stata’s float() function when dealing with such numbers.

Why count x==1.1 produces 0

Let’s trace through what happens when you type the commands

. generate x = 1.1

. count if x==1.1

Here is how it works:

  1. Some numbers have no exact finite-digit binary representation just as some numbers have no exact finite-digit decimal representation. One-third, 0.3333… (base 10), is an example of a number with no exact finite-digit decimal representation. In base 12, one-third does have an exact finite-digit representation, namely 0.4 (base 12). In base 2 (binary), base 10 numbers such as 0.1, 0.2, 0.3, 0.4, 0.6, … have no exact finite-digit representation.

  2. Computers store numbers with a finite number of binary digits. In float precision, numbers have 24 binary digits. In double precision, they have 53 binary digits.

    The decimal number 1.1 in binary is 1.000110011001… (base 2). The 1001 on the end repeats forever. Thus, 1.1 (base 10) is stored by a computer as


    in float, or as


    in double. There are 24 and 53 digits in the numbers above.

  3. Typing generate x = 1.1 results in 1.1 being interpreted as the longer binary number Stata performs all calculations in double precision. New variable x is created as a float by default. When the more precise number is stored in x, it is rounded to the shorter number.

  4. Thus when you count if x==1.1 the result is 0 because 1.1 is again interpreted as the longer binary number and the longer number is compared to shorter number stored in x, and they are not equal.

How count x==float(1.1) solves the problem

One way to fix the problem is to change count if x==1.1 to read count if x==float(1.1):

. generate x = 1.1

. count if x==float(1.1)

Function float() rounds results to float precision. When you type float(1.1), the 1.1 is converted to binary, double precision, namely,

 1.0001100110011001100110011001100110011001100110011010 (base 2)

and float() then rounds that long binary number to

 1.00011001100110011001101 (base 2)

or more correctly, to

 1.0001100110011001100110000000000000000000000000000000 (base 2)

because the number is still stored in double precision. Regardless, this new value is equal to the value stored in x, and so count reports that 100 observations contain float(1.1).

As an aside, when you typed generate x = 1.1, Stata acted as if you typed generate x = float(1.1). Whenever you type generate x = … and x is a float, Stata acts if if you typed generate x = float(…).

How storing data as double appears to solve the problem (and does)

When you type

. generate double x = 1.1

. count if x==1.1

it should be pretty obvious how the problem was solved. Stata stores

1.0001100110011001100110011001100110011001100110011010 (base 2)

in x, and then compares the stored result to

1.0001100110011001100110011001100110011001100110011010 (base 2)

and of course they are equal.

In the Summary above, I referred to this as a cover up. It is a cover up because 1.1 (base 10) is not what is stored in x. What is stored in x is the binary number just shown, and to be equal to 1.1 (base 10), the binary number needs to suffixed with 1001, and then another 1001, and then another, and so on without end.

Stata tells you that x is equal to 1.1 because Stata converted the 1.1 in count to the same inexact binary representation as Stata previously stored in x, and those two values are equal, but neither is equal to 1.1 (base 10). This leads to an important property of digital computers:

If storage and calculation are done to the same precision, it will appear to the user as if all numbers that the user types are stored without error.

That is, it appears to you as if there is no inaccuracy in storing 1.1 in x when x is a double because Stata performs calculations in double. And it is equally true that it would appear to you as if there were no accuracy issues storing 1.1 when x is stored in float precision if Stata, observing that x is float, performed calculations involving x in float. The fact is that there are accuracy issues in both cases.

“Wait,” you are probably thinking. “I understand your argument, but I’ve always heard that float is inaccurate and double is accurate. I understand from your argument that it is only a matter of degree but, in this case, those two degrees are on opposite sides of an important line.”

“No,” I respond.

What you have heard is right with respect to calculation. What you have heard might apply to data storage too, but that is unlikely. It turns out that float provides plenty of precision to store most real measurements.

Float is plenty accurate to store most data

The misconception that float precision is inaccurate comes from the true statement that float precision is not accurate enough when it comes to making calculations with stored values. Whether float precision is accurate enough for storing values depends solely on the accuracy with which the values are measured.

Float precision provides 24 base-2 (binary) digits, and thus values stored in float precision have a maximum relative error error of plus-or-minus 2^(-24) = 5.96e-08, or less than +/-1 part in 15 million.

  1. The U.S. deficit in 2011 is projected to be $1.5 trillion. Stored as a float, the number has a (maximum) error of 2^(-24) * 1.5e+12 = $89,407. That is, if the true number is 1.5 trillion, the number recorded in float precision is guaranteed to be somewhere in the range [(1.5e+12)-89,407, (1.5e+14)+89,407]. The projected U.S. deficit is not known to an accuracy of +/-$89,407.

  2. People in the US work about 40 hours per week, or roughly 0.238 of the hours in the week. 2^(-24) * 0.238 = 1.419e-09 of a week, or 0.1 milliseconds. Time worked in a week is not known to an accuracy of +/-0.1 milliseconds.

  3. A cancer survivor might live 350 days. 2^(-24) * 350 = .00002086, or 1.8 seconds. Time of death is rarely recorded to an accuracy of +/-1.8 seconds. Time of diagnosis is never recorded to such accuracy, nor could it be.

  4. The moon is said to be 384,401 kilometers from the Earth. 2^(-24) * 348,401 = 0.023 kilometers, or 23 meters. At its closest and farthest, the moon is 356,400 and 406,700 kilometers from Earth.

  5. Most fundamental constants of the universe are known to a few parts in a million, which is to say, less than 1 part in 15 million, the accuracy float precision can provide. An exception is the speed of light, measured to be 299,793.458 kilometers per second. Record that as a float and you will be off by 0.01 km/s.

In all the examples except the last, quoted are worst-case scenarios. The actual errors depend on the exact number and is a more tedious calculation (not shown):

  1. For the U.S. deficit, the exact error for 1.5 trillion is -$26,624, which is within the plus or minus $89,407 quoted.

  2. For fraction of the week, at 0.238 the error is -0.04 milliseconds, which is within the +/-0.1 milliseconds quoted.

  3. For cancer survival time, at 350 days the actual error is 0, which is within the +/-1.8 seconds quoted.

  4. For the distance between the Earth and moon, the actual error is 0, which is within within the +/-23 meters quoted.

The actual errors may be interesting, but the maximum errors are more useful. Remember the multiplier 2^(-24). All you have to do is multiply a measurement by 2^(-24) and compare the result with the inherent error in the measurement. If 2^(-24) multiplied by the measurement is less than the inherent error, you can use float precision to store your data. Otherwise, you need to use double.

By the way, the formula

maximum_error = 2^(-24) * x

is an approximation. The true formula is

maximum_error = 2^(-24) * 2^(floor(log2(x)))

It can be readily proven that x ≥ 2^(floor(log2(x))) and thus the approximation formula overstates the maximum error. The approximation formula can overstate the maximum error by as much as a factor of 2. Float precision is adequate for most data. There is one kind of data, however, where float precision may not be adequate, and that is financial data such as sales data, general ledgers, and the like. People working with dollar-and-cent data, or Euro-and-Eurocent data, or Pound Stirling-and-penny data, or any other currency data, usually find it best to use doubles. To avoid rounding issues, it is preferable to store the data as pennies. Float precision binary cannot store 0.01, 0.02, and the like, exactly. Integer values, however, can be stored exactly, at least up to certain 16,777,215.

Floats can store up to 16,777,215 exactly. If stored your data in pennies, that would correspond to $167,772.15.

Doubles can store up to 9,007,199,254,740,991 exactly. If you stored your data in pennies, the would correspond to $90,071,992,547,409.91, or just over $90 trillion.

Why don’t I have these problems using Excel?

You do not have these problems when you use Excel because Excel stores numeric values in double precision. As I explained in How float(1.1) solves the problem above,

If storage and calculation are done to the same precision, it will appear to the user as if all numbers that the user types are stored without error.

You can adopt the Excel solution in Stata by typing

. set type double, permanently

You will double (or more) the amount of memory Stata uses to store your data, but if that is not of concern to you, there are no other disadvantages to adopting this solution. If you adopt this solution and later wish to change your mind, type

. set type float, permanently

That’s all for today

If you enjoyed the above, you may want to see Part II (forthcoming). As I said, There are a few technical issues underlying what is written above that may interest those interested in computer science as it applies to statistical computing.

Categories: Numerical Analysis Tags: ,

Pi is (still) wrong

See this video, by Vi Hart:

This link was passed on to me by my friend Marcello. I’ve been bold enough to make up words such as eigenaxis and eigenpoint, but it takes real courage to suggest redefining π, even when you’re right!

After seeing the video, you can go here and here to learn more about what is being proposed.

Don’t click on comments until you’ve seen the video. Ms. Hart does a better job presenting the proposal than any of us can.

Categories: Mathematics Tags: ,

Understanding matrices intuitively, part 2, eigenvalues and eigenvectors

Last time, I showed you a way to graph and to think about matrices. This time, I want to apply the technique to eigenvalues and eigenvectors. The point is to give you a picture that will guide your intuition, just as it was previously.

Before I go on, several people asked after reading part 1 for the code I used to generate the graphs. Here it is, both for part 1 and part 2:

The eigenvectors and eigenvalues of matrix A are defined to be the nonzero x and λ values that solve

Ax = λx

I wrote a lot about Ax in the last post. Just as previously, x is a point in the original, untransformed space and Ax is its transformed value. λ on the right-hand side is a scalar.

Multiplying a point by a scalar moves the point along a line that passes through the origin and the point:

The figure above illustrates y=λx when λ>1. If λ were less than 1, the point would move toward the origin and if λ were also less than 0, the point would pass right by the origin to land on the other side. For any point x, y=λx will be somewhere on the line passing through the origin and x.

Thus Ax = λx means the transformed value Ax lies on a line passing through the origin and the original x. Points that meet that restriction are eigenvectors (or more correctly, as we will see, eigenpoints, a term I just coined), and the corresponding eigenvalues are the λ‘s that record how far the points move along the line.

Actually, if x is a solution to Ax = λx, then so is every other point on the line through 0 and x. That’s easy to see. Assume x is a solution to Ax = λx and substitute cx for x: Acx = λcx. Thus x is not the eigenvector but is merely a point along the eigenvector.

And with that prelude, we are now in a position to interpret Ax = λx fully. Ax = λx finds the lines such that every point on the line, say, x, transformed by Ax moves to being another point on the same line. These lines are thus the natural axes of the transform defined by A.

The equation Ax = λx and the instructions “solve for nonzero x and λ” are deceptive. A more honest way to present the problem would be to transform the equation to polar coordinates. We would have said to find θ and λ such that any point on the line (r, θ) is transformed to (λr, θ). Nonetheless, Ax = λx is how the problem is commonly written.

However we state the problem, here is the picture and solution for A = (2, 1 \ 1, 2)

I used Mata’s eigensystem() function to obtain the eigenvectors and eigenvalues. In the graph, the black and green lines are the eigenvectors.

The first eigenvector is plotted in black. The “eigenvector” I got back from Mata was (0.707 \ 0.707), but that’s just one point on the eigenvector line, the slope of which is 0.707/0.707 = 1, so I graphed the line y = x. The eigenvalue reported by Mata was 3. Thus every point x along the black line moves to three times its distance from the origin when transformed by Ax. I suppressed the origin in the figure, but you can spot it because it is where the black and green lines intersect.

The second eigenvector is plotted in green. The second “eigenvector” I got back from Mata was (-0.707 \ 0.707), so the slope of the eigenvector line is 0.707/(-0.707) = -1. I plotted the line y = –x. The eigenvalue is 1, so the points along the green line do not move at all when transformed by Ax; y=λx and λ=1.

Here’s another example, this time for the matrix A = (1.1, 2 \ 3, 1):

The first “eigenvector” and eigenvalue Mata reported were… Wait! I’m getting tired of quoting the word eigenvector. I’m quoting it because computer software and the mathematical literature call it the eigenvector even though it is just a point along the eigenvector. Actually, what’s being described is not even a vector. A better word would be eigenaxis. Since this posting is pedagogical, I’m going to refer to the computer-reported eigenvector as an eigenpoint along the eigenaxis. When you return to the real world, remember to use the word eigenvector.

The first eigenpoint and eigenvalue that Mata reported were (0.640 \ 0.768) and λ = 3.45. Thus the slope of the eigenaxis is 0.768/0.640 = 1.2, and points along that line — the green line — move to 3.45 times their distance from the origin.

The second eigenpoint and eigenvalue Mata reported were (-0.625 \ 0.781) and λ = -1.4. Thus the slope is -0.781/0.625 = -1.25, and points along that line move to -1.4 times their distance from the origin, which is to say they flip sides and then move out, too. We saw this flipping in my previous posting. You may remember that I put a small circle and triangle at the bottom left and bottom right of the original grid and then let the symbols be transformed by A along with the rest of space. We saw an example like this one, where the triangle moved from the top-left of the original space to the bottom-right of the transformed space. The space was flipped in one of its dimensions. Eigenvalues save us from having to look at pictures with circles and triangles; when a dimension of the space flips, the corresponding eigenvalue is negative.

We examined near singularity last time. Let’s look again, and this time add the eigenaxes:

The blue blob going from bottom-left to top-right is both the compressed space and the first eigenaxis. The second eigenaxis is shown in green.

Mata reported the first eigenpoint as (0.789 \ 0.614) and the second as (-0.460 \ 0.888). Corresponding eigenvalues were reported as 2.78 and 0.07. I should mention that zero eigenvalues indicate singular matrices and small eigenvalues indicate nearly singular matrices. Actually, eigenvalues also reflect the scale of the matrix. A matrix that compresses the space will have all of its eigenvalues be small, and that is not an indication of near singularity. To detect near singularity, one should look at the ratio of the largest to the smallest eigenvalue, which in this case is 0.07/2.78 = 0.03.

Despite appearances, computers do not find 0.03 to be small and thus do not think of this matrix as being nearly singular. This matrix gives computers no problem; Mata can calculate the inverse of this without losing even one binary digit. I mention this and show you the picture so that you will have a better appreciation of just how squished the space can become before computers start complaining.

When do well-programmed computers complain? Say you have a matrix A and make the above graph, but you make it really big — 3 miles by 3 miles. Lay your graph out on the ground and hike out to the middle of it. Now get down on your knees and get out your ruler. Measure the spread of the compressed space at its widest part. Is it an inch? That’s not a problem. One inch is roughly 5*10-6 of the original space (that is, 1 inch by 3 miles wide). If that were a problem, users would complain. It is not problematic until we get around 10-8 of the original area. Figure about 0.002 inches.

There’s more I could say about eigenvalues and eigenvectors. I could mention that rotation matrices have no eigenvectors and eigenvalues, or at least no real ones. A rotation matrix rotates the space, and thus there are no transformed points that are along their original line through the origin. I could mention that one can rebuild the original matrix from its eigenvectors and eigenvalues, and from that, one can generalize powers to matrix powers. It turns out that A-1 has the same eigenvectors as A; its eigenvalues are λ-1 of the original’s. Matrix AA also has the same eigenvectors as A; its eigenvalues are λ2. Ergo, Ap can be formed by transforming the eigenvalues, and it turns out that, indeed, A½ really does, when multiplied by itself, produce A.

Understanding matrices intuitively, part 1

I want to show you a way of picturing and thinking about matrices. The topic for today is the square matrix, which we will call A. I’m going to show you a way of graphing square matrices, although we will have to limit ourselves to the 2 x 2 case. That will be, as they say, without loss of generality. The technique I’m about to show you could be used with 3 x 3 matrices if you had a better 3-dimensional monitor, and as will be revealed, it could be used on 3 x 2 and 2 x 3 matrices, too. If you had more imagination, we could use the technique on 4 x 4, 5 x 5, and even higher-dimensional matrices.

But we will limit ourselves to 2 x 2. A might be

delim{[}{matrix{2}{2}{2 1 1.5 2}}{]}

From now on, I’ll write matrices as

A = (2, 1 \ 1.5, 2)

where commas are used to separate elements on the same row and backslashes are used to separate the rows.

To graph A, I want you to think about

y = Ax


y: 2 x 1,

A: 2 x 2, and

x: 2 x 1.

That is, we are going to think about A in terms of its effect in transforming points in space from x to y. For instance, if we had the point

x = (0.75 \ 0.25)


y = (1.75 \ 1.625)

because by the rules of matrix multiplication y[1] = 0.75*2 + 0.25*1 = 1.75 and y[2] = 0.75*1.5 + 0.25*2 = 1.625. The matrix A transforms the point (0.75 \ 0.25) to (1.75 \ 1.625). We could graph that:

To get a better understanding of how A transforms the space, we could graph additional points:

I do not want you to get lost among the individual points which A could transform, however. To focus better on A, we are going to graph y = Ax for all x. To do that, I’m first going to take a grid,

One at a time, I’m going to take every point on the grid, call the point x, and run it through the transform y = Ax. Then I’m going to graph the transformed points:

Finally, I’m going to superimpose the two graphs:

In this way, I can now see exactly what A = (2, 1 \ 1.5, 2) does. It stretches the space, and skews it.

I want you to think about transforms like A as transforms of the space, not of the individual points. I used a grid above, but I could just as well have used a picture of the Eiffel tower and, pixel by pixel, transformed it by using y = Ax. The result would be a distorted version of the original image, just as the the grid above is a distorted version of the original grid. The distorted image might not be helpful in understanding the Eiffel Tower, but it is helpful in understanding the properties of A. So it is with the grids.

Notice that in the above image there are two small triangles and two small circles. I put a triangle and circle at the bottom left and top left of the original grid, and then again at the corresponding points on the transformed grid. They are there to help you orient the transformed grid relative to the original. They wouldn’t be necessary had I transformed a picture of the Eiffel tower.

I’ve suppressed the scale information in the graph, but the axes make it obvious that we are looking at the first quadrant in the graph above. I could just as well have transformed a wider area.

Regardless of the region graphed, you are supposed to imagine two infinite planes. I will graph the region that makes it easiest to see the point I wish to make, but you must remember that whatever I’m showing you applies to the entire space.

We need first to become familiar with pictures like this, so let’s see some examples. Pure stretching looks like this:

Pure compression looks like this:

Pay attention to the color of the grids. The original grid, I’m showing in red; the transformed grid is shown in blue.

A pure rotation (and stretching) looks like this:

Note the location of the triangle; this space was rotated around the origin.

Here’s an interesting matrix that produces a surprising result: A = (1, 2 \ 3, 1).

This matrix flips the space! Notice the little triangles. In the original grid, the triangle is located at the top left. In the transformed space, the corresponding triangle ends up at the bottom right! A = (1, 2 \ 3, 1) appears to be an innocuous matrix — it does not even have a negative number in it — and yet somehow, it twisted the space horribly.

So now you know what 2 x 2 matrices do. They skew,stretch, compress, rotate, and even flip 2-space. In a like manner, 3 x 3 matrices do the same to 3-space; 4 x 4 matrices, to 4-space; and so on.

Well, you are no doubt thinking, this is all very entertaining. Not really useful, but entertaining.

Okay, tell me what it means for a matrix to be singular. Better yet, I’ll tell you. It means this:

A singular matrix A compresses the space so much that the poor space is squished until it is nothing more than a line. It is because the space is so squished after transformation by y = Ax that one cannot take the resulting y and get back the original x. Several different x values get squished into that same value of y. Actually, an infinite number do, and we don’t know which you started with.

A = (2, 3 \ 2, 3) squished the space down to a line. The matrix A = (0, 0 \ 0, 0) would squish the space down to a point, namely (0 0). In higher dimensions, say, k, singular matrices can squish space into k-1, k-2, …, or 0 dimensions. The number of dimensions is called the rank of the matrix.

Singular matrices are an extreme case of nearly singular matrices, which are the bane of my existence here at StataCorp. Here is what it means for a matrix to be nearly singular:

Nearly singular matrices result in spaces that are heavily but not fully compressed. In nearly singular matrices, the mapping from x to y is still one-to-one, but x‘s that are far away from each other can end up having nearly equal y values. Nearly singular matrices cause finite-precision computers difficulty. Calculating y = Ax is easy enough, but to calculate the reverse transform x = A-1y means taking small differences and blowing them back up, which can be a numeric disaster in the making.

So much for the pictures illustrating that matrices transform and distort space; the message is that they do. This way of thinking can provide intuition and even deep insights. Here’s one:

In the above graph of the fully singular matrix, I chose a matrix that not only squished the space but also skewed the space some. I didn’t have to include the skew. Had I chosen matrix A = (1, 0 \ 0, 0), I could have compressed the space down onto the horizontal axis. And with that, we have a picture of nonsquare matrices. I didn’t really need a 2 x 2 matrix to map 2-space onto one of its axes; a 2 x 1 vector would have been sufficient. The implication is that, in a very deep sense, nonsquare matrices are identical to square matrices with zero rows or columns added to make them square. You might remember that; it will serve you well.

Here’s another insight:

In the linear regression formula b = (XX)-1Xy, (XX)-1 is a square matrix, so we can think of it as transforming space. Let’s try to understand it that way.

Begin by imagining a case where it just turns out that (XX)-1 = I. In such a case, (XX)-1 would have off-diagonal elements equal to zero, and diagonal elements all equal to one. The off-diagonal elements being equal to 0 means that the variables in the data are uncorrelated; the diagonal elements all being equal to 1 means that the sum of each squared variable would equal 1. That would be true if the variables each had mean 0 and variance 1/N. Such data may not be common, but I can imagine them.

If I had data like that, my formula for calculating b would be b = (XX)-1Xy = IXy = Xy. When I first realized that, it surprised me because I would have expected the formula to be something like b = X-1y. I expected that because we are finding a solution to y = Xb, and b = X-1y is an obvious solution. In fact, that’s just what we got, because it turns out that X-1y = Xy when (XX)-1 = I. They are equal because (XX)-1 = I means that XX = I, which means that X‘ = X-1. For this math to work out, we need a suitable definition of inverse for nonsquare matrices. But they do exist, and in fact, everything you need to work it out is right there in front of you.

Anyway, when correlations are zero and variables are appropriately normalized, the linear regression calculation formula reduces to b = Xy. That makes sense to me (now) and yet, it is still a very neat formula. It takes something that is N x k — the data — and makes k coefficients out of it. Xy is the heart of the linear regression formula.

Let’s call b = Xy the naive formula because it is justified only under the assumption that (XX)-1 = I, and real XX inverses are not equal to I. (XX)-1 is a square matrix and, as we have seen, that means it can be interpreted as compressing, expanding, and rotating space. (And even flipping space, although it turns out the positive-definite restriction on XX rules out the flip.) In the formula (XX)-1Xy, (XX)-1 is compressing, expanding, and skewing Xy, the naive regression coefficients. Thus (XX)-1 is the corrective lens that translates the naive coefficients into the coefficient we seek. And that means XX is the distortion caused by scale of the data and correlations of variables.

Thus I am entitled to describe linear regression as follows: I have data (y, X) to which I want to fit y = Xb. The naive calculation is b = Xy, which ignores the scale and correlations of the variables. The distortion caused by the scale and correlations of the variables is XX. To correct for the distortion, I map the naive coefficients through (XX)-1.

Intuition, like beauty, is in the eye of the beholder. When I learned that the variance matrix of the estimated coefficients was equal to s2(XX)-1, I immediately thought: s2 — there’s the statistics. That single statistical value is then parceled out through the corrective lens that accounts for scale and correlation. If I had data that didn’t need correcting, then the standard errors of all the coefficients would be the same and would be identical to the variance of the residuals.

If you go through the derivation of s2(XX)-1, there’s a temptation to think that s2 is merely something factored out from the variance matrix, probably to emphasize the connection between the variance of the residuals and standard errors. One easily loses sight of the fact that s2 is the heart of the matter, just as Xy is the heart of (XX)-1Xy. Obviously, one needs to view both s2 and Xy though the same corrective lens.

I have more to say about this way of thinking about matrices. Look for part 2 in the near future. Update: part 2 of this posting, “Understanding matrices intuitively, part 2, eigenvalues and eigenvectors”, may now be found at