R2sample

library(R2sample)

This package brings together a number of routines for the two sample problem for univariate data. There are two data sets x and y and we want to test whether they were generated by the same probability distribution.

The highlights of this package are:

Example 1

We generate two data sets of size 100 and 120 respectively from standard normal distributions and run the test:

set.seed(123)
x1 = rnorm(100)
y1 = rnorm(120)
twosample_test(x1, y1, B=500)
#> $statistics
#>       KS   Kuiper      CvM       AD       LR       ZA       ZK       ZC 
#>   0.1383   0.1666   0.3363   1.6834   0.6655  -3.2530   2.4087  -3.1940 
#>   Wassp1 ES large ES small EP large EP small 
#>   0.2224  16.0900   5.1700  34.8500   8.8000 
#> 
#> $p.values
#>       KS   Kuiper      CvM       AD       LR       ZA       ZK       ZC 
#>   0.2020   0.3560   0.0920   0.1280   0.1040   0.3540   0.4000   0.3420 
#>   Wassp1 ES large ES small EP large EP small 
#>   0.1540   0.8846   0.5228   0.2482   0.4559

In this case the the null hypothesis is true, both data sets were generated by a standard normal distribution. And indeed, all 13 tests have p values much larger than (say) 0.05 and therefore correctly fail to reject the null hypothesis.

Alternatively one might wish to carry out a select number of tests, but then find a correct p value for the combined test. This can be done as follows:

twosample_test_adjusted_pvalue(x1, y1)
#> p values of individual tests:
#> ES small :  0.5228
#> ZA :  0.6384
#> ZK :  0.5766
#> Wassp1 :  0.8618
#> Kuiper :  0.6124
#> adjusted p value of combined tests: 0.969

By default for continuous data this runs four tests (chi square with a small number of bins, ZA, ZK and Wassp1). It the finds the smallest p value (here 0.02) and adjusts it for the multiple testing problem, resulting in an overall p value of 0.05.

New Tests

Say we wish to use two new tests for continuous data. One is based on the difference in standardized means and the other is based on the difference in standard deviations:

\[ \begin{aligned} &TS_1 = sd(x) - sd(y) \\ &TS_2 = \bar{x}/sd(x) - \bar{y}/sd(y) \\ \end{aligned} \]

myTS1 = function(x, y) {
   out = c(0, 0)
   out[1] = abs(sd(x) - sd(y))
   out[2] = abs(mean(x)/sd(x) - mean(y)/sd(y))
   names(out) = c("std", "std t test")
   out
}

Then we can simply run

twosample_test(x1, y1, TS=myTS1, B=500)
#> $statistics
#>        std std t test 
#>    0.06396    0.18060 
#> 
#> $p.values
#>        std std t test 
#>      0.512      0.192

Arguments and output of new test routine for continuous data

The arguments have to be x and y for the two data sets and (optionally) a list called TSextra for any additional input needed to find test statistic.

Note that the output vector of the routine has to be a named vector. If the routine is written in Rcpp parallel programming is not available.

Example 2

Here we generate two data sets of size 1000 and 1200 respectively from a binomial distribution with 5 trials and success probabilities of 0.5 and 0.55, respectively.

x2 = table(c(0:5,rbinom(1000, 5, 0.5)))-1
y2 = table(c(0:5,rbinom(1200, 5, 0.55)))-1
rbind(x2, y2)
#>     0   1   2   3   4  5
#> x2 28 164 297 345 132 34
#> y2 15 120 335 423 251 56
twosample_test(x2, y2, vals=0:5, B=500)$p.values
#>     KS Kuiper    CvM     AD     LR     ZA Wassp1  large  small 
#>      0      0      0      0      0      0      0      0      0

In this case the the null hypothesis is false, all nine tests for discrete data have p values much smaller than (say) 0.05 and therefore correctly reject the null hypothesis.

Notice, it is the presence of the vals argument that tells the twosample_test command that the data is discrete. The vectors x and y are the counts. Note that the lengths of the three vectors have to be the same and no value of vals is allowed that has both x and y equal to 0.

New Test

Again we might want to run a different test, say based again on the difference in standardized means. This time we will implement the new test using Rcpp:

(For reasons to do with submission to CRAN this routine is already part of R2sample)

#include <Rcpp.h>
using namespace Rcpp;
// [[Rcpp::export]]
NumericVector myTS2(IntegerVector x, 
                         IntegerVector y,
                         NumericVector vals) {
  
  Rcpp::CharacterVector methods=CharacterVector::create("std t test");    
  int const nummethods=methods.size();
  int k=x.size(), n=sum(x), i;
  double meanx=0.0, meany=0.0, sdx=0.0, sdy=0.0;
  NumericVector TS(nummethods);
  TS.names() =  methods;
  for(i=0;i<k;++i) {
    meanx = meanx + x[i]*vals[i]/n;
    meany = meany + y[i]*vals[i]/n;
  }
  for(i=0;i<k;++i) {
    sdx = sdx + x[i]*(vals[i] - meanx)*(vals[i] - meanx);
    sdy = sdy + y[i]*(vals[i] - meany)*(vals[i] - meany);
  }  
  sdx = sqrt(sdx/(n-1));
  sdy = sqrt(sdy/(n-1));
  TS(0) = std::abs(meanx/sdx - meany/sdy);
  return TS;
}

Now

twosample_test(x2, y2, vals=0:5, TS=R2sample::myTS2, B=500)
#> $statistics
#> std t test 
#>     0.2714 
#> 
#> $p.values
#> std t test 
#>          0

Arguments and output of new test routine for discrete data

The arguments have to be x and y for the two data sets (the counts) and vals for the set of values where \(P(X=vals)>0\). Also (optionally) a list called TSextra for any additional input needed to find test statistic.

Note that the output vector of the routine has to be a named vector.

If the routine is written in Rcpp parallel programming is not available.

Permutation Method

For all the tests except the chi square variants the p values are found via the permutation method. The idea of a permutation test is simple. Say we have data sets \(x_1,..,x_n\) and \(y_1,..,y_m\). They are combined into one large data set, permuted and split again in sets of sizes n and m. Under the null hypothesis these new data sets come from the same distribution as the actual data. Therefore calculating the tests statistics for them and repeating many times one can build up the distributions of the test statistics and find p values from them.

In the discrete case the situation is somewhat more complicated. Say we have data sets with values \(v_1,..,v_k\) and counts \(x_1,..,x_n\), \(y_1,..,y_k\). One can then mimic the description above by expanding these to yield a large data set with \(x_1+y_1\) \(v_1\)’s, \(x_2+y_2\) \(v_2\)’s and so on. Then this vector is permuted and split as described above. This method is used with the argument samplingmethod=“independence”. The drawback of this sampling method is that its calculation speed increases with the sample size.

Alternatively R2sample provides the following option. One can show that the distribution of the permuted data sets has the following distribution: let \(n=\sum x_i\) and \(m=\sum y_i\), then

\[P(X=a|x,y)=\frac{\prod_{j=1}^k{{x_j+y_j}\choose{a_j}}}{{n+m}\choose{n}}\] for any \(a\) such that \(0\le a_i \le x_i; i=1,..,k\) and \(\sum a=n\). Sampling from this distribution can be done as follows: set \(p=\sum x_i/\sum (x_i+y_i)\), the overall percentage of the data in the x data set. Next generate a new count \(x^{*}_i\) for the \(i^{th}\) bin with \(X^*_i\sim Binom(x_i+y_i,p)\), \(i=1,..,k\). If \(\sum x_i \ne \sum x^{*}_i\), randomly select a bin according to the probabilities \((x_1+y_1,..,x_k+y_k)\) and add or subtract 1 from \(x^{*}_i\) as needed. Repeat this until \(\sum x_i = \sum x^{*}_i\).

The default is Binomial sampling with a simulation size of 5000 for testing and 1000 for power estimation.

Parametric Bootstrap

The user can also provide a routine rnull, which generates new data under the null hypothesis. This is needed in the following situation: we wish to perform a goodness-of-fit test \(H_0:X\sim F(p)\), where \(F\) is some distribution function that depends on unknown parameters \(p\). These parameters can be estimated from the data by \(\hat{p}\). Unfortunately \(F\) is complicated and can not be evaluated directly. We can, however, sample from \(F(\hat{p})\). So we generate a second data set \(y\) from \(F(\hat{p})\). In this setup the permutation method fails, and the p values have to be found via the parametric bootstrap.

Example 3

Say we have a data set \(x\) and wish to test whether it comes from a normal distribution with unknown mean and standard deviation. Let’s assume we do not know how to evalute the normal cdf, so instead we sample from \(N(\bar{x}, sd(x))\):

rnull=function(dta) {
   list(x=dta$x,
        y=rnorm(length(dta$y), mean(dta$x), sd(dta$x)))
}
twosample_test(x1, y1, rnull=rnull, B=500)
#> $statistics
#>       KS   Kuiper      CvM       AD       LR       ZA       ZK       ZC 
#>   0.1383   0.1666   0.3363   1.6834   0.6655  -3.2530   2.4087  -3.1940 
#>   Wassp1 ES large ES small EP large EP small 
#>   0.2224  16.0900   5.1700  34.8500   8.8000 
#> 
#> $p.values
#>       KS   Kuiper      CvM       AD       LR       ZA       ZK       ZC 
#>   0.0540   0.1700   0.0180   0.0160   0.0240   0.0820   0.1320   0.0780 
#>   Wassp1 ES large ES small EP large EP small 
#>   0.0160   0.8846   0.5228   0.2482   0.4559

The routine rnull has to have as its argument a list with the data sets and return an object of the same type.

The Testing Methods

In the continuous case we have a sample x of size of n and a sample y of size m. In the discussion below both are assumed to be ordered. We denote by z the combined sample. In the discrete case we have a random variable with values v, and x and y are the counts. We denote by k the number of observed values.

We denote the edf’s of the two data sets by \(\widehat{F}\) and \(\widehat{G}\), respectively. Moreover we denote the empirical distribution function of the combined data set by \(\widehat{H}\).

In the case of continuous data, it is binned into nbins[1] equal probability bins, with the default nbins[1]=100. In the case of discrete data the number of classes comes from vals, the values observed.

Then a standard chi square test is run and the p value is found using the usual chi square approximation.

Many previous studies have shown the a chi square test with a large number of classes often has very poor power. So this test in the case of continuous data uses nbins[2] equal probability bins, with the default nbins[2]=10. In the case of discrete data if the number of classes exceeds nbins[2]=10 the classes are combined until there are nbins[2]=10 classes.

In either case the p values are found using the usual chi-square approximation. The degrees of freedom are the number of bins. For a literature review of chi square tests see (W. Rolke 2021).

For all other tests the p values are found using a permutation test. They are

This test is based on the quantity \(\max\{|\widehat{F}(x)-\widehat{G}(x)|:x\in \mathbf{R}\}\). In the continuous case (without ties) the function \(\widehat{F}(x)-\widehat{G}(x)\) either takes a jump of size \(1/n\) up at a point in the x data set, a jump of size \(1/m\) down if x is a point in the y data set, or is flat. Therefore the test statistic is

\[\max\{\sum_{i=1}^{j} \vert \frac1n I\{z_i \in x\}-\frac1m I\{z_i \in y\} \vert:j=1,..,n+m\}\]

In the discrete case the jumps have sizes \(x_i/n\) and \(y_j/m\), respectively and the test statistic is

\[\max\{\sum_{i=1}^{j} \vert \frac{x_i}n -\frac{y_j}m \vert:j=1,..,k\}\]

This test was first proposed in (Kolmogorov 1933), (Smirnov 1939) and is one of the most widely used tests today.

There is a close connection between the edf’s and the ranks of the x and y in the combined data set. Using this one can also calculate the KS statistic, and many statistics to follow, based on these ranks. This is used in many software routines, for example the ks.test routine in R. However, when applied to discrete data these formulas can fail badly because of the many ties, and so we will not use them in our routine.

This test is closely related to Kolmogorov-Smirnov, but it uses the sum of the largest positive and negative differences between the edf’s as a test statistic:

\[T_x=\sum_{i=1}^{j} \vert \frac1n I\{z_i \in x\}-\frac1m I\{z_i \in x\} \vert:j=1,..,n\] \[T_y=\sum_{i=1}^{j} \vert \frac1n I\{z_i \in y\}-\frac1m I\{z_i \in y\} \vert:j=1,..,m\] and the the test statistic is \(T_x-T_y\). The discrete case follows directly. This test was first proposed in (Kuiper 1960).

This test is based on the integrated squared difference between the edf’s:

\[\frac{nm}{(n+m)^2}\sum_{i=1}^{n+m} \left( \hat{F}(z_i)-\hat{G}(z_i) \right)^2 \] the extension to the two-sample problem of the Cramer-vonMises criterion is discussed in (T. W. Anderson 1962).

This test is based on

\[nm\sum_{i=1}^{n+m-1} \frac{\left( \hat{F}(z_i)-\hat{G}(z_i) \right)^2}{i(n-i)} \]

It was first proposed in (Theodore W. Anderson, Darling, et al. 1952).

Let \(r_i\) and \(s_i\) be the ranks of x and y in the combined sample, then the test statistic is given by

\[\frac1{nm(n+m)}\left[n\sum_{i=1}^n(r_i-1)^2+m\sum_{i=1}^m(s_i-1)^2\right]\] (Lehmann 1951) and (Rosenblatt 1952)

These tests are based on the paper (Zhang 2006). Note that the calculation of ZC is the one exception from the rule: even in the discrete data case the calculation of the test statistic does require loops of lengths n and m. The calculation time will therefore increase with the sample sizes, and for very large data sets this test might have to be excluded.

A test based on the Wasserstein p1 metric. It compares the quantiles of the x and y in the combined data set. If n=m the test statistic in the continuous case is very simple:

\[ \frac1n\sum_{i=1}^n |x_i-y_i|\] If \(n\ne m\) it is first necessary to find the respective quantiles of the x’s and y’s in the combined data set. In the discrete case the test statistic is

\[\sum_{i=1}^{k-1} \vert \sum_{j=1}^i\frac{x_j}{n}-\sum_{j=1}^i\frac{y_j}{m} \vert(v_{i+1}-v_i)\]

For a discussion of the Wasserstein distance see (Vaserstein 1969).

The Arguments of twosample_test

The routine has the following arguments:

Say for example we want to use only the KS and AD tests:

x=rnorm(10)
y=rnorm(12)
twosample_test(x, y, B=500, doMethods=c("KS","AD"))
#> $statistics
#>       KS       AD 
#> 0.450000 1.490646 
#> 
#> $p.values
#>    KS    AD 
#> 0.150 0.142

The Output of twosample_test

A list with vectors statistics (the test statistics) and p values.

shiny app

The tests can also be run via a shiny app. To do so run

run_shiny()

twosample_power and plot_power

The routine twosample_power allows the estimation of the power of the various tests, and plot_power draws the corresponding power graph with the methods sorted by their mean power.

Note that due to the use of permutation tests the power calculations are fairly slow. Because of the time constraints imposed by CRAN the following routines are not run. A full power study with (say) 25 alternatives and fairly large data sets might take several hours to run.

New tests can be used in the same way as above.

Example 4

Say we wish to find the powers of the tests when one data set comes from a standard normal distribution with sample size 100 and the other from a t distribution with 1-10 degrees of freedom and sample size 200.

plot_power(twosample_power(
  f=function(df) list(x=rnorm(100), y=rt(200, df))
)

The Arguments of twosample_power

The arguments TS, TSextra, nbins, minexpcount , UseLargeSample, samplingmethod, rnull, SuppressMessages, maxProcessor and B are the same as in twosample_test. In addtion we have

  • f A function that generates data sets x and y, either continuous or the counts of discrete variables. In the discrete case care needs to be taken that the resulting vectors have the same length as vals. The output has to be a list with elements x and y. In the case of discrete data the list also has to include a vector vals, the values of the discrete variable.

  • … arguments passed to f. The most common case would be a single vector, but two vectors or none is also possible.

  • alpha=0.05 type I error probability to use in power calculation

The arguments of plot_power are a matrix of powers created by twosample_power and (optional) Smooth=FALSE if no smoothing is desired and a name of the variable on the x axis.

Example 5

We wish to find the powers of the tests when the data set x are 100 observations comes from a binomial n=10, p=0.5 and y 120 observations from from a binomial n=10 and a number of different success probabilities p. Note that in either case not all the possible values 0-10 will actually appear in every data set, so we need to take some care in assuring that x, y and vals match every time:

plot_power(twosample_power(
  function(p) {
    vals=0:10
    x=table(c(vals, rbinom(100, 10, 0.5))) - 1 #Make sure each vector has length 11
    y=table(c(vals, rbinom(120, 10, p))) - 1  #and names 1-10
    vals=vals[x+y>0] # only vals with at least one count
    list(x=x[as.character(vals)],y=y[as.character(vals)], vals=vals)
  },
  p=seq(0.5, 0.6, length=5)
), "p")

Example 6

We want to assess the effect of the sample size when one data set comes from a standard normal and the other from a normal with mean 1 and standard deviation 1:

plot_power(twosample_power(
  f=function(n) list(x=rnorm(n), y=rnorm(n, 1)),
  n=10*1:10), "n")

Adjusted p values for Several Tests

As no single test can be relied upon to consistently have good power, it is reasonable to employ several of them. We would then reject the null hypothesis if any of the tests does so, that is, if the smallest p-value is less than the desired type I error probability \(\alpha\).

This procedure clearly suffers from the problem of simultaneous inference, and the true type I error probability will be much larger than \(\alpha\). It is however possible to adjust the p value so it does achieve the desired \(\alpha\). This can be done as follows:

We generate a number of data sets under the null hypothesis. Generally about 1000 will be sufficient. Then for each simulated data set we apply the tests we wish to include, and record the smallest p value. Here is an example. .

pvals=matrix(0,1000,13)
for(i in 1:1000) 
  pvals[i, ]=R2sample::twosample_test(runif(100), runif(100), B=1000)$p.values

Next we find the smallest p value in each run for two selections of four methods. One is the selection found to be best above, namely Zhang’s ZC and ZK methods, the methods by Kuiper and Wasserstein as well as a chi square test with a small number of bins. As a second selection we use the methods by Kolmogorov-Smirnov, Kuiper, Anderson-Darling, Cramer-vonMises and Lehman-Rosenblatt, which for this null data turn out to be highly correlated.

colnames(pvals)=names(R2sample::twosample_test(runif(100), runif(100), B=1000)$p.values)
p1=apply(pvals[, c("ZK", "ZC", "Wassp1", "Kuiper", "ES small" )], 1, min)
p2=apply(pvals[, c("KS", "Kuiper", "AD", "CvM", "LR")], 1, min)

Next we find the empirical distribution function for the two sets of p values and draw their graphs. We also add the curves for the cases of four identical tests and the case of four independent tests, which of course is the Bonferroni correction. The data for the cdfs is in the inst/extdata directory of the package.

tmp=R2sample::pvaluecdf
Tests=factor(c(rep("Identical Tests", nrow(tmp)),
        rep("Correlated Selection", nrow(tmp)),
        rep("Best Selection", nrow(tmp)),
        rep("Independent Tests", nrow(tmp))),
        levels=c("Identical Tests",  "Correlated Selection", 
                 "Best Selection", "Independent Tests"),
        ordered = TRUE)
dta=data.frame(x=c(tmp[,1],tmp[,1],tmp[,1],tmp[,1]),
          y=c(tmp[,1],tmp[,3],tmp[,2],1-(1-tmp[,1])^4),
          Tests=Tests)
ggplot2::ggplot(data=dta, ggplot2::aes(x=x,y=y,col=Tests))+
  ggplot2::geom_line(linewidth=1.2)+
  ggplot2::labs(x="p value", y="CDF")+
  ggplot2::scale_color_manual(values=c("blue","red", "Orange", "green"))

Case Studies

The package includes the routine run.studies, which provides 20 case studies each for the continuous and the discrete case. This allows the user to easily compare the power of the methods included in the package to a different one of their choice.

Say a user wishes to study the performance of the chi-square test, implemented in chitest. Of course this test is already implemented in R2sample, so this is for illustration only. Say we wish to see its performance when compared to the tests implemented in R2sample for the case where the null hypothesis specifies a uniform distribution but the data actually comes from a linear distribution with slope s:

chitest = function(x, y, TSextra) {
   nbins=TSextra$nbins
   nx=length(x);ny=length(y);n=nx+ny
   xy=c(x,y)
   bins=quantile(xy, (0:nbins)/nbins)
   Ox=hist(x, bins, plot=FALSE)$counts
   Oy=hist(y, bins, plot=FALSE)$counts
   tmp=sqrt(sum(Ox)/sum(Oy))
   chi = sum((Ox/tmp-Oy*tmp)^2/(Ox+Oy))
   pval=1-pchisq(chi, nbins-1)
   out=ifelse(TSextra$statistic,chi,pval)
   names(out)="ChiSquare"
   out
}
TSextra=list(nbins=5,statistic=FALSE)
pwr=R2sample::run.studies(chitest, "uniform.linear", TSextra=TSextra, With.p.value = TRUE)
#> Running case study uniform.linear.cont ...
R2sample::plot_power(pwr, "slope")

Arguments of run.studies:

Arguments of routine to calculate new test:

The output of the routine has to be a named vector with either the test statistic(s) or the p value(s).

The case studies are as follows. In each the first term specifies the model under the null hypothesis and the second the model under the alternative hypothesis.

  1. uniform.linear\(\hspace{3cm}\) U[0,1] vs a linear model on [0,1] with slope s.

  2. uniform.quadratic\(\hspace{2.5cm}\) U[0,1] vs a quadratic model with vertex at 0.5 and some curvature a.

  3. uniform.bump\(\hspace{3.1cm}\) U[0,1] vs U[0,1]+N(0.5,0.05).

  4. uniform.sine\(\hspace{3.3cm}\) U[0,1] vs U[0,1]+Sine wave

  5. beta22.betaaa\(\hspace{3cm}\) Beta(2,2) vs Beta(a,a)

  6. beta22.beta2a\(\hspace{3cm}\) Beta(2,2) vs Beta(2,a)

  7. normal.shift\(\hspace{3.5cm}\) N(0,1) vs N(\(\mu\),1)

  8. normal.stretch\(\hspace{3.1cm}\) N(0,1) vs N(0, \(\sigma\))

  9. normal.t\(\hspace{4.2cm}\) N(0,1) vs t(df)

  10. normal.outlier1\(\hspace{3.1cm}\) N(0,1) vs N(0,1)+U[2,3]

  11. normal.outlier2\(\hspace{3.1cm}\) N(0,1) vs N(0,1)+U[-3,-2]+U[2,3]

  12. exponential.gamma\(\hspace{2.3cm}\) Exp(1) vs Gamma(1,b)

  13. exponential.weibull\(\hspace{2.5cm}\) Exp(1) vs Weibull(1,b)

  14. exponential.bump\(\hspace{2.7cm}\) Exp(1) vs Exp(1)+N(0.5,0.05)

  15. gamma.normal\(\hspace{3.1cm}\) Gamma(\(\mu\)) vs N(\(\bar{x}\), sd(x)), here the mean of the normal distribution are the sample mean and sample standard deviation of the x data set.

  16. normal.normalmixture\(\hspace{2.1cm}\) N(0,1) vs N(\(-\mu\),1)+N(\(\mu\),1)

  17. uniform.uniformmixture\(\hspace{1.9cm}\) U[0,1] vs. \(\alpha\)U[0,1/2]+(1-\(\alpha\))U[1/2,1]

  18. uniform.betamixture\(\hspace{2.5cm}\) U[0,1] vs. \(\alpha\)U[0,1/2]+(1-\(\alpha\))Beta(2,2)

  19. chisquare.noncentral\(\hspace{2.3cm}\) \(\chi^2(5)\) vs. \(\chi^2(5, \tau)\)

  20. uniform.triangular\(\hspace{2.9cm}\) U[0,1] vs. triangular

Generally a user will wish to apply their test to all the case studies. This can be done easily with:

R2sample::run.studies(chitest, TSextra=TSextra, With.p.value=TRUE)

run.studies can also be used to run the studies for the included methods but for different values of param_alt, nsample and/or alpha. For example, say we wish to find the powers of the included continuous methods for the cases uniform.linear and uniform.quadratic as well as samples of size 2000 and a true type I error of 0.1. Moreover for the case uniform/linear we want the slopes to be 0.1 and 0.2 and for the case uniform.quadratic we want param_alt to be 1.3:

R2sample::run.studies(TRUE, # continuous data/model
        study=c("uniform.linear", "uniform.quadratic"),
        param_alt=list(c(0.1, 0.2), 1.3),
        nsample=2000,
        alpha=0.1)

References

Anderson, T. W. 1962. “On the Distribution of the Two_sample Cramer-von Mises Criterion.” The Annals of Mathematical Statistics 33(3): 1148–59.
Anderson, Theodore W, Donald A Darling, et al. 1952. “Asymptotic Theory of Certain Goodness of Fit Criteria Based on Stochastic Processes.” The Annals of Mathematical Statistics 23 (2): 193–212.
Kolmogorov, A. N. 1933. “Sulla Determinazione Empirica Di Una Legge Di Distribuzione.” Giornale Dell’Instituto Italiano Degli Attuari 4: 83–91.
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