## Tuesday, July 6, 2010

### Example 8.1: Digits of Pi

Do the digits of Pi appear in a random order? If so, the trillions of digits of Pi calculated can serve as a useful random number generator. This post was inspired by this entry on Matt Asher's blog.

Generating pseudo-random numbers is a key piece of much of modern statistical practice, whether for Markov chain Monte Carlo applications or simpler simulations of what raw data would look like under given circumstances.

Generating sufficiently random pseudo-random numbers is not trivial and many methods exist for doing so. Unfortunately, there's no way to prove a series of pseudo-random numbers is indistinguishable from a truly random series of numbers. Instead, there are simply what amount to ad hoc tests, one of which might be failed by insufficiently random series of numbers.

The first 10 million digits of Pi can be downloaded from here. Here, we explore couple of simple ad hoc checks of randomness. We'll try something a little trickier in the next entry.

SAS

We start by reading in the data. The file contains one logical line with a length of 10,000,000. We tell SAS to read a digit-long variable and hold the place in the line. For later use, we'll also create a variable with the order of each digit, using the _n_ implied variable (section 1.4.15).

data test;infile "c:\ken\pi-10million.txt" lrecl=10000000;input digit 1. @@;order = _n_;run;

We can do a simple one-way chi-square test for equal probability of each digit in proc freq.

proc freq data = test;tables digit/ chisq;run;      Chi-Square     2.7838      DF                  9      Pr > ChiSq     0.9723      Sample Size = 10000000

We didn't display the counts for each digit, but none was more than 0.11% away from the expected 1,000,000 occurrences.

Another simple check would be to assess autoregression. We can do this in proc autoreg. The dw=2 option calculates the Durbin-Watson statistic for adjacent and alternating residuals. We limit the observations to 1,000,000 digits for compatibility with R.

proc autoreg data=test (obs = 1000000);  model digit = / dw=2 dwprob;run;      Durbin-Watson Statistics      Order            DW    Pr < DW    Pr > DW        1          2.0028     0.9175     0.0825        2          1.9996     0.4130     0.5870

We might want to replicate this set of tests for series of 4 digits instead. To do this, we just tell the data step to read the line line in 4-digit chunks.
data test4;infile "c:\ken\pi-10million.txt" lrecl=10000000;input digit4 4. @@;order = _n_;run;proc freq data = test4;tables digit4/ chisq;run;      Chi-Square  9882.9520      DF               9999      Pr > ChiSq     0.7936      Sample Size = 2500000proc autoreg data=test4 (obs = 1000000);  model digit = / dw=3 dwprob;run;      Durbin-Watson Statistics      Order            DW    Pr < DW    Pr > DW        1          2.0014     0.7527     0.2473        2          1.9976     0.1181     0.8819        3          2.0007     0.6397     0.3603

So far, we see no evidence of a lack of randomness.

R

In R, we use the readLines() function to create a 10,000,000-digit scalar object. In the following line we split the digits using the strsplit() function (as in section 6.4.1). This results in a list object, to which the as.numeric() function (which forces the digit characters to be read as numeric, section 1.4.2) cannot be applied. The unlist() function converts the list into a vector first, so that strsplit() will work. Then the chisq.test() function performs the one-way chi-squared test.

mypi = readLines("c:/ken/pi-10million.txt", warn=FALSE)piby1 = as.numeric(unlist(strsplit(mypi,"")))chisq.test(table(piby1), p=rep(0.1, 10))

This generates the following output:
 Chi-squared test for given probabilitiesdata:  table(piby1) X-squared = 2.7838, df = 9, p-value = 0.9723

Alternatively, it's trivial to write a function to automatically test for equal probabilities of all categories.

onewaychi = function(datavector){   datatable = table(datavector)   expect = rep(length(datavector)/length(datatable),length(datatable))   chi2 = sum(((datatable - expect)^2)/expect)   p = 1- pchisq(chi2,length(datatable)-1)   return(p)}> onewaychi(piby1)[1] 0.972252

The Durbin-Watson test can be generated by the dwtest function, from the lmtest package. Using all 10,000,000 digits causes an error, so we use only the first 1,000,000.

> library(lmtest)> dwtest(lm(piby1[1:1000000] ~ 1))        Durbin-Watson testdata:  lm(piby1[1:1e+06] ~ 1) DW = 2.0028, p-value = 0.9176alternative hypothesis: true autocorrelation is greater than 0

To examine the digits in groups of 4, we read the digit vector as a matrix with 4 columns, then multiply each digit and add the columns together. Alternatively, we could use the paste() function (section 1.4.5) to glue the digits together as a character string, the use the as.numeric() to convert back to numbers.

> pimat = matrix(piby1, ncol = 4,byrow=TRUE)> head(pimat)     [,1] [,2] [,3] [,4][1,]    1    4    1    5[2,]    9    2    6    5[3,]    3    5    8    9[4,]    7    9    3    2[5,]    3    8    4    6[6,]    2    6    4    3> piby4 = pimat[,1] * 1000 + pimat[,2] * 100 + +   pimat[,3] * 10 + pimat[,4]> head(piby4)[1] 1415 9265 3589 7932 3846 2643# alternate approach# piby4_v2 = as.numeric(paste(pimat[,1], pimat[,2], #                          pimat[,3], pimat[,4], sep=""))> onewaychi(piby4)[1] 0.7936358> dwtest(lm(piby4[1:1000000] ~ 1))        Durbin-Watson testdata:  lm(piby4[1:1e+06] ~ 1) DW = 2.0014, p-value = 0.753alternative hypothesis: true autocorrelation is greater than 0