7月 222019
Illustration of the 68-95-99.7 rule

Is 4 an extreme value for the standard normal distribution? In high school, students learn the famous 68-95-99.7 rule, which is a way to remember that 99.7 percent of random observation from a normal distribution are within three standard deviations from the mean. For the standard normal distribution, the probability that a random value is bigger than 3 is 0.0013. The probability that a random value is bigger than 4 is even smaller: about 0.00003 or 3 x 10-5.

So, if you draw randomly from a standard normal distribution, it must be very rare to see an extreme value greater than 4, right? Well, yes and no. Although it is improbable that any ONE observation is more extreme than 4, if you draw MANY independent observations, the probability that the sample contains an extreme value increases with the sample size. If p is the probability that one observation is less than 4, then pn is the probability that n independent observations are all less than 4. Thus 1 – pn is the probability that some value is greater than 4. You can use the CDF function in SAS to compute these probabilities in a sample of size n:

/* What is a "large value" of a normal sample? The answer depends on the size of the sample. */
/* Use CDF to find probability that a random value from N(0,1) exceeds 4 */
proc iml;
P_NotGT4 = cdf("Normal", 4);   /* P(x < 4) */
/* Probability of an extreme obs in a sample that contains n independent observations */
n = {1, 100, 1000, 10000};     /* sample sizes */
P_NotGT4 = P_NotGT4**n;        /* P(all values are < 4) */
P_GT4 = 1 - P_NotGT4;          /* P(any value is > 4)   */
print n P_NotGT4 P_GT4;
Probability of a value greater than 4 in a sample that contains n independent observations from the standard normal distribution

The third column of the table shows that the probability of seeing an observation whose value is greater than 4 increases with the size of the sample. In a sample of size 10,000, the probability is 0.27, which implies that about one out of every four samples of that size will contain a maximum value that exceeds 4.

The distribution of the maximum in a Gaussian sample: A simulation approach

You can use a simulation to approximate the distribution of the maximum value of a normal sample of size n. For definiteness, choose n = 1,000 and sample from a standard normal distribution N(0,1). The following SAS/IML program simulates 5,000 samples of size n and computes the maximum value of each sample. You can then graph the distribution of the maximum values to understand how the maximum value varies in random samples of size n.

/* What is the distribution of the maximum value in a 
   sample of size n drawn from the standard normal distribution? */
call randseed(12345);
n = 1000;                  /* sample size */
numSim = 5000;             /* number of simulations */
x = j(n, numSim);          /* each column will be a sample */
call randgen(x, "Normal"); /* generate numSim samples */
max = x[<>, ];             /* find max of each sample (column) */
Title "Distribution of Maximum Value of a Normal Sample";
title2 "n = 1000";
call histogram(max);
/* compute some descriptive statistics */
mean = mean(max`);
call qntl(Q, max`, {0.25 0.5 0.75});
print (mean // Q)[rowname={"Mean" "P25" "Median" "P75"}];
Simulate 5000 samples of size n=1000. Plot the maximum value of each sample.

Based on this simulation, the expected maximum value of a sample of size n = 1,000 is about 3.2. The table indicates that about 25% of the samples have a maximum value that is less than 3. About half have a maximum value less than 3.2. About 75% of the samples have a maximum value that is less than 3.4. The graph shows the distribution of maxima in these samples. The maximum value of a sample ranged from 2.3 to 5.2.

The distribution of a maximum (or minimum) value in a sample is studied in an area of statistics that is known as extreme value theory. It turns out that you can derive the sampling distribution of the maximum of a sample by using the Gumbel distribution, which is also known as the "extreme value distribution of Type 1." You can use the Gumbel distribution to describe the distribution of the maximum of a sample of size n. The Gumbel distribution actually describes the maximum for many distributions, but for simplicity I will only refer to the normal distribution.

The distribution of the maximum in a Gaussian sample: The Gumbel distribution

This section does two things. First, it uses PROC UNIVARIATE to fit the parameters of a Gumbel distribution to the maximum values from the simulated samples. The Gumbel(3.07, 0.29) distribution is the distribution that maximizes the likelihood function for the simulated data. Second, it uses a theoretical formula to show that a Gumbel(3.09, 0.29) distribution is the distribution that models the maximum of a normally distributed sample of size n = 1,000. Thus, the results from the simulation and the theory are similar.

You can write the 5,000 maximum values from the simulation to a SAS data set and use PROC UNIVARIATE to estimate the MLE parameters for the Gumbel distribution, as follows:

create MaxVals var "max"; append; close;
/* Fit a Gumbel distribution, which models the distribution of maximum values */
proc univariate data=MaxVals;
   histogram max / gumbel;
   ods select Histogram ParameterEstimates FitQuantiles;

The output from PROC UNIVARIATE shows that the Gumbel(3.07, 0.29) distribution is a good fit to the distribution of the simulated maxima. But where do those parameter values come from? How are the parameters of the Gumbel distribution related to the sample size of the standard normal distribution?

It was difficult to find an online reference that shows how to derive the Gumbel parameters for a normal sample of size n. I finally found a formula in the book Extreme Value Distributions (Kotz and Nadarajah, 2000, p. 9). For any cumulative distribution F that satisfies certain conditions, you can use the quantile function of the distribution to estimate the Gumbel parameters. The result is that the location parameter (μ) is equal to μ = F-1(1-1/n) and the scale parameter (σ) is equal to σ = F-1(1-1/(ne)) - μ, where e is the base of the natural logarithm. The following SAS/IML program uses the (1 – 1/n)th quantile of the normal distribution to derive the Gumbel parameters for a normal sample of size n:

proc iml;
n = 1000;
/* Compute parameters of Gumbel distribution when the sample is normal of size n.
   SAS calls the parameters (mu, sigma). Wikipedia calls them (mu, beta). Other
   references use (alpha, beta). */
mu_n    = quantile("Normal", 1-1/n);                /* location parameter */
sigma_n = quantile("Normal", 1-1/n*exp(-1)) - mu_n; /* scale parameter */
print n mu_n sigma_n;
/* what is the mean (expected value) and median of this Gumbel distribution? */
gamma = constant("Euler");             /* Euler–Mascheroni constant = 0.5772157 */
mean = mu_n + gamma*sigma_n;           /* expected value of maximum */
median = mu_n - sigma_n * log(log(2)); /* median of maximum value distribution */
print n mean median;

Notice that these parameters are very close to the MLE estimates from the simulated normal samples. The results tell you that the expected maximum in a standard normal sample is 3.26 and about 50% of samples will have a maximum value of 3.19 or less.

You can use these same formulas to find the expected and median values of the maximum in samples of other sizes:

/* If the sample size is n, what is expected maximum */
n = {1E4, 2E4, 1E5, 2E5, 1E6, 2E6};
mu_n    = quantile("Normal", 1-1/n);                /* location parameter */
sigma_n = quantile("Normal", 1-1/n*exp(-1)) - mu_n; /* scale parameter */
mean = mu_n + gamma*sigma_n;           /* expected value of maximum */
median = mu_n - sigma_n * log(log(2)); /* meadian of maximum value distribution */
print n mean median;

The table shows that you would expect to see a maximum value of 4 in a sample of size 20,000. If there are two million observations in a sample, you would expect to see a maximum value of 5!

You can graph this data over a range of sample sizes. The following graph shows the expected value of the maximum value in a sample of size n (drawn from a standard normal distribution) for large values of n.

You can create similar images for quantiles. The p_th quantile for the Gumbel distribution is q = mu_n - sigma_n log(-log(p)).

So, is 4 an unlikely value for the standard normal distribution? Yes, but for sufficiently large samples that value is likely to be observed. You can use the Gumbel distribution to model the distribution of the maximum in a normal sample of size n to determine how likely it is that the sample contains an extreme value. The larger the sample, the more likely it is to observe an extreme value. Although I do not discuss the general case, the Gumbel distribution can also model the maximum value for samples drawn from some non-normal distributions.

The post Extreme values: What is an extreme value for normally distributed data? appeared first on The DO Loop.

5月 202019

Recently I wrote about how to compute the Kolmogorov D statistic, which is used to determine whether a sample has a particular distribution. One of the beautiful facts about modern computational statistics is that if you can compute a statistic, you can use simulation to estimate the sampling distribution of that statistic. That means that instead of looking up critical values of the D statistic in a table, you can estimate the critical value by using empirical quantiles from the simulation.

This is a wonderfully liberating result! No longer are we statisticians constrained by the entries in a table in the appendix of a textbook. In fact, you could claim that modern computation has essentially killed the standard statistical table.

Obtain critical values by using simulation

Before we compute anything, let's recall a little statistical theory. If you get a headache thinking about null hypotheses and sampling distributions, you might want to skip the next two paragraphs!

When you run a hypothesis test, you compare a statistic (computed from data) to a hypothetical distribution (called the null distribution). If the observed statistic is way out in a tail of the null distribution, you reject the hypothesis that the statistic came from that distribution. In other words, the data does not seem to have the characteristic that you are testing for. Statistical tables use "critical values" to designate when a statistic is in the extreme tail. A critical value is a quantile of the null distribution; if the observed statistic is greater than the critical value, then the statistic is in the tail. (Technically, I've described a one-tailed test.)

One of the uses for simulation is to approximate the sampling distribution of a statistic when the true distribution is not known or is known only asymptotically. You can generate a large number of samples from the null hypothesis and compute the statistic on each sample. The union of the statistics approximates the true sampling distribution (under the null hypothesis) so you can use the quantiles to estimate the critical values of the null distribution.

Critical values of the Kolmogorov D distribution

You can use simulation to estimate the critical value for the Kolmogorov-Smirnov statistical test for normality. For the data in my previous article, the null hypothesis is that the sample data follow a N(59, 5) distribution. The alternative hypothesis is that they do not. The previous article computed a test statistic of D = 0.131 for the data (N = 30). If the null hypothesis is true, is that an unusual value to observe? Let's simulate 40,000 samples of size N = 30 from N(59,5) and compute the D statistic for each. Rather than use PROC UNIVARIATE, which computes dozens of statistics for each sample, you can use the SAS/IML computation from the previous article, which is very fast. The following simulation runs in a fraction of a second.

/* parameters of reference distribution: F = cdf("Normal", x, &mu, &sigma) */
%let mu    = 59;
%let sigma =  5;
%let N     = 30;
%let NumSamples = 40000;
proc iml;
call randseed(73);
N = &N;
i = T( 1:N );                           /* ranks */
u = i/N;                                /* ECDF height at right-hand endpoints */
um1 = (i-1)/N;                          /* ECDF height at left-hand endpoints  */
y = j(N, &NumSamples, .);               /* columns of Y are samples of size N */
call randgen(y, "Normal", &mu, &sigma); /* fill with random N(mu, sigma)      */
D = j(&NumSamples, 1, .);               /* allocate vector for results        */
do k = 1 to ncol(y);                    /* for each sample:                   */
   x = y[,k];                           /*    get sample x ~ N(mu, sigma)     */
   call sort(x);                        /*    sort sample                     */
   F = cdf("Normal", x, &mu, &sigma);   /*    CDF of reference distribution   */
   D[k] = max( F - um1, u - F );        /*    D = max( D_minus, D_plus )      */
title "Monte Carlo Estimate of Sampling Distribution of Kolmogorov's D Statistic";
title2 "N = 30; N_MC = &NumSamples";
call histogram(D) other=
     "refline 0.131 / axis=x label='Sample D' labelloc=inside lineattrs=(color=red);";

The test statistic is right smack dab in the middle of the null distribution, so there is no reason to doubt that the sample is distributed as N(59, 5).

How big would the test statistic need to be to be considered extreme? To test the hypothesis at the α significance level, you can compute the 1 – α quantile of the null distribution. The following statements compute the critical value for α = 0.05 and N = 30:

/* estimate critical value as the 1 - alpha quantile */
alpha = 0.05;
call qntl(Dcrit_MC, D, 1-alpha);
print Dcrit_MC;

The estimated critical value for a sample of size 30 is 0.242. This compares favorably with the exact critical value from a statistical table, which gives Dcrit = 0.2417 for N = 30.

You can also use the null distribution to compute a p value for an observed statistic. The p value is estimated as the proportion of statistics in the simulation that exceed the observed value. For example, if you observe data that has a D statistic of 0.28, the estimated p value is obtained by the following statements:

Dobs = 0.28;                        /* hypothetical observed statistic */
pValue = sum(D >= Dobs) / nrow(D);  /* proportion of distribution values that exceed D0 */
print Dobs pValue;

This same technique works for any sample size, N, although most tables critical values only for all N ≤ 30. For N > 35, you can use the following asymptotic formulas, developed by Smirnov (1948), which depend only on α:

The Kolmogorov D statistic does not depend on the reference distribution

It is reasonable to assume that the results of this article apply only to a normal reference distribution. However, Kolmogorov proved that the sampling distribution of the D statistic is actually independent of the reference distribution. In other words, the distribution (and critical values) are the same regardless of the continuous reference distribution: beta, exponential, gamma, lognormal, normal, and so forth. That is a surprising result, which explains why there is only one statistical table for the critical values of the Kolmogorov D statistic, as opposed to having different tables for different reference distributions.

In summary, you can use simulation to estimate the critical values for the Kolmogorov D statistic. In a vectorized language such as SAS/IML, the entire simulation requires only about a dozen statements and runs extremely fast.

The post Critical values of the Kolmogorov-Smirnov test appeared first on The DO Loop.

5月 062019

Here's a simulation tip: When you simulate a fixed-effect generalized linear regression model, don't add a random normal error to the linear predictor. Only the response variable should be random. This tip applies to models that apply a link function to a linear predictor, including logistic regression, Poisson regression, and negative binomial regression.

Recall that a generalized linear model has three components:

  1. A linear predictor η = X β, which is a linear combination of the regressors.
  2. An invertible link function, g, which transforms the linear predictor to the expected value of the response variable. If μ = E(Y), then g(μ) = η or μ = g-1(η).
  3. A random component, which specifies the conditional distribution of the response variable, Y, given the values of the independent variables.

Notice that only the response variable is randomly generated. In a previous article about simulating data from a logistic regression model, I showed that the following SAS DATA step statements can be used to simulate data for a logistic regression model. The statements model a binary response variable, Y, which depends linearly on two explanatory variables X1 and X2:

   /* CORRECT way to simulate data from a logistic model with parameters (-2.7, -0.03, 0.07) */
   eta = -2.7 - 0.03*x1 + 0.07*x2;   /* linear predictor */
   mu = logistic(eta);               /* transform by inverse logit */
   y = rand("Bernoulli", mu);        /* simulate binary response with probability mu */

Notice that the randomness occurs only during the last step when you generate the response variable. Sometimes I see a simulation program in which the programmer adds a random term to the linear predictor, as follows:

   /* WRONG way to simulate logistic data. This is a latent-variable model. */
   eta = -2.7 - 0.03*x1 + 0.07*x2 + RAND("NORMAL", 0, 0.8);   /* WRONG: Leads to a misspecified model! */

Perhaps the programmer copied these statements from a simulation of a linear regression model, but it is not correct for a fixed-effect generalized linear model. When you simulate data from a generalized linear model, use the first set of statements, not the second.

Why is the second statement wrong? Because it has too much randomness. The model that generates the data includes a latent (unobserved) variable. The model you are trying to simulate is specified in a SAS regression procedure as MODEL Y = X1 X2, but the model for the latent-variable simulation (the second one) should be MODEL Y = X1 X2 X3, where X3 is the unobserved normally distributed variable.

What happens if you add a random term to the linear predictor

I haven't figured out all the mathematical ramifications of (incorrectly) adding a random term to the linear predictor prior to applying the logistic transform, but I ran a simulation that shows that the latent-variable model leads to biased parameter estimates when you fit the simulated data.

You can download the SAS program that generates data from two models: from the correct model (the first simulation steps) and from the latent-variable model (the second simulation). I generated 100 samples (each containing 5057 observations), then used PROC LOGISTIC to generate the resulting 100 sets of parameter estimates by using the statement MODEL Y = X1 X2. The results are shown in the following scatter plot matrix.

The blue markers are the parameter estimates from the correctly specified simulation. The reference lines in the upper right cells indicate the true values of the parameters in the simulation: (β0, β1, β2) = (-2.7, -0.03, 0.07). You can see that the true parameter values are in the center of the cloud of blue markers, which indicates that the parameter estimates are unbiased.

In contrast, the red markers show that the parameter estimates for the misspecified latent-variable model are biased. The simulated data does not come from the model that is being fit. This simulation used 0.8 for the standard deviation of the error term in the linear predictor. If you use a smaller value, the center of the red clouds will be closer to the true parameter values. If you use a larger value, the clouds will move farther apart.

For additional evidence that the data from the second simulation does not fit the model Y = X1 X2, the following graphs show the calibration plots for a data sets from each simulation. The plot on the left shows nearly perfect calibration: This is not surprising because the data were simulated from the same model that is fitted! The plot on the right shows the calibration plot for the latent-variable model. The calibration plot shows substantial deviations from a straight line, which indicates that the model is misspecified for the second set of data.

In summary, be careful when you simulate data for a generalized fixed-effect linear model. The randomness only appears during the last step when you simulate the response variable, conditional on the linear predictor. You should not add a random term to the linear predictor.

I'll leave you with a thought that is trivial but important: You can use the framework of the generalized linear model to simulate a linear regression model. For a linear model, the link function is the identity function and the response distribution is normal. That means that a linear model can be simulated by using the following:

   /* Alternative way to simulate a linear model with parameters (-2.7, -0.03, 0.07) */
   eta = -2.7 - 0.03*x1 + 0.07*x2;   /* linear predictor */
   mu = eta;                         /* identity link function */
   y = rand("Normal", mu, 0.7);      /* simulate Y as normal response with RMSE = 0.7 */

Thus simulating a linear model fits into the framework of simulating a generalized linear model, as it should!

Download the SAS program that generates the images in this article.

The post How to simulate data from a generalized linear model appeared first on The DO Loop.

4月 292019

Did you know that SAS provides built-in support for working with probability distributions that are finite mixtures of normal distributions? This article shows examples of using the "NormalMix" distribution in SAS and describes a trick that enables you to easily work with distributions that have many components.

As with all probability distributions, there are four essential functions that you need to know: The PDF, CDF, QUANTILE, and RAND functions.

What is a normal mixture distribution?

A finite mixture distribution is a weighted sum of component distributions. When all of the components are normal, the distribution is called a mixture of normals. If the i_th component has parameters (μi, σi), then you can write the probability density function (PDF) of the normal mixture as
f(x) = Σi wi φ(x; μi, σi)
where φ is the normal PDF and the positive constants wi are the mixing weights. The mixing weights must sum to 1.

The adjacent graph shows the density function for a three-component mixture of normal distributions. The means of the components are -6, 3, and 8, respectively, and are indicated by vertical reference lines. The mixing weights are 0.1, 0.3, and 0.6. The SAS program to create the graph is in the next section.

The "NormalMix" distribution in SAS

The PDF and CDF functions in Base SAS support the "NormalMix" distribution. The syntax is a little unusual because the function needs to support an arbitrary number of components. If there are k components, the PDF and CDF functions require 3k + 3 parameters:

  1. The first parameter is the name of the distribution: "NormalMix". The second parameter is the value, x, at which to evaluate the density function.
  2. The third parameter is the number of component distributions, k > 1.
  3. The next k parameters specify the mixing weights, w1, w2, ..., wk.
  4. The next k parameters specify the component means, μ1, μ2, ..., μk.
  5. The next k parameters specify the component standard deviations, σ1, σ2, ..., σk.

If you are using a model that has many components, it is tedious to explicitly list every parameter in every function call. Fortunately, there is a simple trick that prevents you from having to list the parameters. You can put the parameters into arrays and use the OF operator (sometimes called the OF keyword) to reference the parameter values in the array. This is shown in the next section.

The PDF and CDF of a normal mixture

The following example demonstrates how to compute the PDF and CDF for a three-component mixture-of-normals distribution. The DATA step shows two tricks:

  • The parameters (weights, means, and standard deviations) are stored in arrays. In the calls to the PDF and CDF functions, syntax such as OF w[*} enables you to specify the parameter values with minimal typing.
  • A normal density is extremely tiny outside of the interval [μ - 5*σ, μ + 5*σ]. You can use this fact to compute the effective domain for the PDF and CDF functions.
/* PDF and CDF of the normal mixture distribution. This example specifies three components. */
data NormalMix;
array w[3]     _temporary_ ( 0.1, 0.3, 0.6);  /* mixing weights */ 
array mu[3]    _temporary_ (-6,   3,   8);    /* mean for each component */
array sigma[3] _temporary_ (0.5,  0.6, 2.5);  /* standard deviation for each component */
/* For each component, the range [mu-5*sigma, mu+5*sigma] is the effective support. */
minX = 1e308; maxX = -1e308;                  /* initialize to extreme values */
do i = 1 to dim(mu);                          /* find largest interval where density > 1E-6 */
   minX = min(minX, mu[i] - 5*sigma[i]);
   maxX = max(maxX, mu[i] + 5*sigma[i]);
/* Visualize the functions on the effective support. Use arrays and the OF operator to
   specify the parameters. An alternative syntax is to list the arguments, as follows:
   cdf = CDF('normalmix', x, 3, 
              0.1, 0.3, 0.6,    
             -6,   3,   8,    
              0.5, 0.6, 2.5);       */
dx = (maxX - minX)/200;
do x = minX to maxX by dx;
  pdf = pdf('normalmix', x, dim(mu), of w[*], of mu[*], of sigma[*]); 
  cdf = cdf('normalmix', x, dim(mu), of w[*], of mu[*], of sigma[*]); 
keep x pdf cdf;

As shown in the program, the OF operator greatly simplifies and clarifies the function calls. The alternative syntax, which is shown in the comments, is unwieldy.

The following statements create graphs of the PDF and CDF functions. The PDF function is shown at the top of this article. The CDF function, along with a few reference lines, is shown below.

title "PDF function for Normal Mixture Distribution";
title2 "Vertical Lines at Component Means";
proc sgplot data=NormalMix;
   refline -6 3 8 / axis=x;
   series x=x y=pdf;
title "CDF function for Normal Mixture Distribution";
proc sgplot data=NormalMix;
   xaxis grid; yaxis min=0 grid; 
   refline 0.1 0.5 0.7 0.9;
   series x=x y=cdf;

The quantiles of a normal mixture

The quantile function for a continuous distribution is the inverse of the CDF distribution. The graph of the CDF function for a mixture of normals can have flat regions when the component means are far apart relative to their standard deviations. Technically, these regions are not completely flat because the normal distribution has infinite support, but computationally they can be very flat. Because finding a quantile is equivalent to finding the root of a shifted CDF, you might encounter computational problems if you try to compute the quantile that corresponds to an extremely flat region, such as the 0.1 quantile in the previous graph.

The following DATA step computes the 0.1, 0.5, 0.7, and 0.9 quantiles for the normal mixture distribution. Notice that you can use arrays and the OF operator for the QUANTILE function:

data Q;
array w[3]     _temporary_ ( 0.1, 0.3, 0.6);  /* mixing weights */ 
array mu[3]    _temporary_ (-6,   3,   8);    /* mean for each component */
array sigma[3] _temporary_ (0.5,  0.6, 2.5);  /* standard deviation for each component */
array p[4] (0.1, 0.5, 0.7, 0.9);  /* find quantiles for these probabilities */
do i = 1 to dim(p);
   prob = p[i];
   qntl = quantile('normalmix', prob, dim(mu), of w[*], of mu[*], of sigma[*]); 
keep qntl prob;
proc print; run;

The table tells you that 10% of the density of the normal mixture is less than x=-3.824. That is essentially the result of the first component, which has weight 0.1 and therefore is responsible for 10% of the total density. Half of the density is less than x=5.58. Fully 70% of the density lies to the left of x=8, which is the mean of the third component. That result makes sense when you look at the mixing weights.

Random values from a normal mixture

The RAND function does not explicitly support the "NormalMix" distribution. However, as I have shown in a previous article, you can simulate from an arbitrary mixture of distributions by using the "Table" distribution in conjunction with the component distributions. For the three-component mixture distribution, the following DATA step simulates a random sample:

/* random sample from a mixture distribution */
%let N = 1000;
data RandMix(drop=i);
call streaminit(12345);
array w[3]     _temporary_ ( 0.1, 0.3, 0.6);  /* mixing weights */ 
array mu[3]    _temporary_ (-6,   3,   8);    /* mean for each component */
array sigma[3] _temporary_ (0.5,  0.6, 2.5);  /* standard deviation for each component */
do obsNum = 1 to &N;
   i = rand("Table", of w[*]);                /* choose the component by using the mixing weights */
   x = rand("Normal", mu[i], sigma[i]);       /* sample from that component */
title "Random Sample for Normal Mixture Distribution";
proc sgplot data=RandMix;
   histogram x;
   refline -6 3 8 / axis=x;   /* means of component distributions */

The histogram of a random sample looks similar to the graph of the PDF function, as it should.

In summary, SAS provides built-in support for working with the density (PDF), cumulative probability (CDF), and quantiles (QUANTILE) of a normal mixture distribution. You can use arrays and the OF operator to call these Base SAS functions without having to list every parameter. Although the RAND function does not natively support the "NormalMix" distribution, you can use the "Table" distribution to select a component according to the mixing weights and use the RAND("Normal") function to simulate from the selected normal component.

The post The normal mixture distribution in SAS appeared first on The DO Loop.

3月 272019

In simulation studies, sometimes you need to simulate outliers. For example, in a simulation study of regression techniques, you might want to generate outliers in the explanatory variables to see how the technique handles high-leverage points. This article shows how to generate outliers in multivariate normal data that are a specified distance from the center of the distribution. In particular, you can use this technique to generate "regular outliers" or "extreme outliers."

When you simulate data, you know the data-generating distribution. In general, an outlier is an observation that has a small probability of being randomly generated. Consequently, outliers are usually generated by using a different distribution (called the contaminating distribution) or by using knowledge of the data-generating distribution to construct improbable data values.

The geometry of multivariate normal data

A canonical example of adding an improbable value is to add an outlier to normal data by creating a data point that is k standard deviations from the mean. (For example, k = 5, 6, or 10.) For multivaraite data, an outlier does not always have extreme values in all coordinates. However, the idea of an outliers as "far from the center" does generalize to higher dimensions. The following technique generates outliers for multivariate normal data:

  1. Generate uncorrelated, standardized, normal data.
  2. Generate outliers by adding points whose distance to the origin is k units. Because you are using standardized coordinates, one unit equals one standard deviation.
  3. Use a linear transformation (called the Cholesky transformation) to produce multivariate normal data with a desired correlation structure.

In short, you use the Cholesky transformation to transform standardized uncorrelated data into scaled correlated data. The remarkable thing is that you can specify the covariance and then directly compute the Cholesky transformation that will result in that covariance. This is a special property of the multivariate normal distribution. For a discussion about how to perform similar computations for other multivariate distributions, see Chapter 9 in Simulating Data with SAS (Wicklin 2013).

Let's illustrate this method by using a two-dimensional example. The following SAS/IML program generates 200 standardized uncorrelated data points. In the standardized coordinate system, the Euclidean distance and the Mahalanobis distance are the same. It is therefore easy to generate an outlier: you can generate any point whose distance to the origin is larger than some cutoff value. The following program generates outliers that are 5 units from the origin:

ods graphics / width=400px height=400px attrpriority=none;
proc iml;
/* generate standardized uncorrelated data */
call randseed(123);
N = 200;
x = randfun(N//2, "Normal"); /* 2-dimensional data. X ~ MVN(0, I(2)) */
/* covariance matrix is product of diagonal (scaling) and correlation matrices.
R = {1   0.9,             /* correlation matrix */
     0.9 1  };
D = {4 9};                /* standard deviations */
Sigma = corr2cov(R, D);   /* Covariance matrix Sigma = D*R*D */
print Sigma;
/* U is the Cholesky transformation that scales and correlates the data */
U = root(Sigma);
/* add a few unusual points (outliers) */
pi = constant('pi');
t = T(0:11) * pi / 6;            /* evenly spaced points in [0, 2p) */
outliers = 5#cos(t) || 5#sin(t); /* evenly spaced on circle r=5 */
v = x // outliers;               /* concatenate MVN data and outliers */
w = v*U;                         /* transform from stdized to data coords */
labl = j(N,1," ") // T('0':'11');
Type = j(N,1,"Normal") // j(12,1,"Outliers");
title "Markers on Circle r=5";
title2 "Evenly spaced angles t[i] = i*pi/6";
call scatter(w[,1], w[,2]) grid={x y} datalabel=labl group=Type;

The program simulates standardized uncorrelated data (X ~ MVN(0, I(2))) and then appends 12 points on the circle of radius 5. The Cholesky transformation correlates the data. The correlated data are shown in the scatter plot. The circle of outliers has been transformed into an ellipse of outliers. The original outliers are 5 Euclidean units from the origin; the transformed outliers are 5 units of Mahalanobis distance from the origin, as shown by the following computation:

center = {0 0};
MD = mahalanobis(w, center, Sigma);     /* compute MD based on MVN(0, Sigma) */
obsNum = 1:nrow(w);
title "Mahalanobis Distance of MVN Data and Outliers";
call scatter(obsNum, MD) grid={x y} group=Type;

Adding outliers at different Mahalanobis distances

The same technique enables you to add outliers at any distance to the origin. For example, the following modification of the program adds outliers that are 5, 6, and 10 units from the origin. The t parameter determines the angular position of the outlier on the circle:

MD = {5, 6, 10};               /* MD for outlier */
t = {0, 3, 10} * pi / 6;       /* angular positions */
outliers = MD#cos(t) ||  MD#sin(t);
v = x // outliers;             /* concatenate MVN data and outliers */
w = v*U;                       /* transform from stdized to data coords */
labl = j(N,1," ") // {'p1','p2','p3'};
Type = j(N,1,"Normal") // j(nrow(t),1,"Outlier");
call scatter(w[,1], w[,2]) grid={x y} datalabel=labl group=Type;

If you use the Euclidean distance, the second and third outliers (p2 and p3) are closer to the center of the data than p1. However, if you draw the probability ellipses for these data, you will see that p1 is more probable than p2 and p3. This is why the Mahalanobis distance is used for measuring how extreme an outlier is. The Mahalanobis distances of p1, p2, and p3 are 5, 6, and 10, respectively.

Adding outliers in higher dimensions

In two dimensions, you can use the formula (x(t), y(t)) = r*(cos(t), sin(t)) to specify an observation that is r units from the origin. If t is chosen randomly, you can obtain a random point on the circle of radius r. In higher dimensions, it is not easy to parameterize a sphere, which is the surface of an d-dimensional ball. Nevertheless, you can still generate random points on a d-dimensional sphere of radius r by doing the following:

  1. Generate a point from the d-dimensional multivariate normal distribution: Y = MVN(0, I(d)).
  2. Project the point onto the surface of the d-dimensional sphere of radius r: Z = r*Y / ||Y||.
  3. Use the Cholesky transformation to correlate the variables.

In the SAS/IML language, it would look like this:

/* General technique to generate random outliers at distance r */
d = 2;                              /* use any value of d */
MD = {5, 6, 10};                    /* MD for outlier */
Y = randfun(nrow(MD)//d, "Normal"); /* d-dimensional Y ~ MVN(0, I(d)) */
outliers = MD # Y / sqrt(Y[,##]);   /* surface of spheres of radii MD */
/* then append to data, use Cholesky transform, etc. */

For these particular random outliers, the largest outliers (MD=10) is in the upper right corner. The others are in the lower left corner. If you change the random number seed in the program, the outliers will appear in different locations.

In summary, by understanding the geometry of multivariate normal data and the Cholesky transformation, you can manufacture outliers in specific locations or in random locations. In either case, you can control whether you are adding a "near outlier" or an "extreme outlier" by specifying an arbitrary Mahalanobis distance.

You can download the SAS program that generates the graphs in this article.

The post How to simulate multivariate outliers appeared first on The DO Loop.

1月 282019

This article shows how to use SAS to simulate data that fits a linear regression model that has categorical regressors (also called explanatory or CLASS variables). Simulating data is a useful skill for both researchers and statistical programmers. You can use simulation for answering research questions, but you can also use it generate "fake" (or "synthetic") data for a presentation or blog post when the real data are confidential. I have previously shown how to use SAS to simulate data for a linear model and for a logistic model, but both articles used only continuous regressors in the model.

This discussion and SAS program are based on Chapter 11 in Simulating Data with SAS (Wicklin, 2013, p. 208-209). The main steps in the simulation are as follows. The comments in the SAS DATA program indicate each step:

  1. Macro variables are used to define the number of continuous and categorical regressors. Another macro variable is used to specify the number of levels for each categorical variable. This program simulates eight continuous regressors (x1-x8) and four categorical regressors (c1-c4). Each categorical regressor in this simulation has three levels.
  2. The continuous regressors are simulated from a normal distribution, but you can use any distribution you want. The categorical levels are 1, 2, and 3, which are generated uniformly at random by using the "Integer" distribution. The discrete "Integer" distribution was introduced in SAS 9.4M5; for older versions of SAS, use the %RandBetween macro as shown in the article "How to generate random integers in SAS." You can also generate the levels non-uniformly by using the "Table" distribution.
  3. The response variable, Y, is defined as a linear combination of some explanatory variables. In this simulation, the response depends linearly on the x1 and x8 continuous variables and on the levels of the C1 and C4 categorical variables. Noise is added to the model by using a normally distributed error term.
/* Program based on Simulating Data with SAS, Chapter 11 (Wicklin, 2013, p. 208-209) */
%let N = 10000;                  /* 1a. number of observations in simulated data */
%let numCont =  8;               /* number of continuous explanatory variables */
%let numClass = 4;               /* number of categorical explanatory variables */
%let numLevels = 3;              /* (optional) number of levels for each categorical variable */
data SimGLM; 
  call streaminit(12345); 
  /* 1b. Use macros to define arrays of variables */
  array x[&numCont]  x1-x&numCont;   /* continuous variables named x1, x2, x3, ... */
  array c[&numClass] c1-c&numClass;  /* CLASS variables named c1, c2, ... */
  /* the following statement initializes an array that contains the number of levels
     for each CLASS variable. You can hard-code different values such as (2, 3, 3, 2, 5) */
  array numLevels[&numClass] _temporary_ (&numClass * &numLevels);  
  do k = 1 to &N;                 /* for each observation ... */
    /* 2. Simulate value for each explanatory variable */ 
    do i = 1 to &numCont;         /* simulate independent continuous variables */
       x[i] = round(rand("Normal"), 0.001);
    do i = 1 to &numClass;        /* simulate values 1, 2, ..., &numLevels with equal prob */
       c[i] = rand("Integer", numLevels[i]);        /* the "Integer" distribution requires SAS 9.4M5 */
    /* 3. Simulate response as a function of certain explanatory variables */
    y = 4 - 3*x[1] - 2*x[&numCont] +                /* define coefficients for continuous effects */
       -3*(c[1]=1) - 4*(c[1]=2) + 5*c[&numClass]    /* define coefficients for categorical effects */
        + rand("Normal", 0, 3);                     /* normal error term */
  drop i k;
proc glm data=SimGLM;
   class c1-c&numClass;
   model y = x1-x&numCont c1-c&numClass / SS3 solution;
   ods select ModelANOVA ParameterEstimates;
Parameter estimates for synthetic (simulated) data that follows a regression model.

The ModelANOVA table from PROC GLM (not shown) displays the Type 3 sums of squares and indicates that the significant terms in the model are x1, x8, c1, and c4.

The parameter estimates from PROC GLM are shown to the right. You can see that each categorical variable has three levels, and you can use PROC FREQ to verify that the levels are uniformly distributed. I have highlighted parameter estimates for the significant effects in the model:

  • The estimates for the significant continuous effects are highlighted in red. The estimates for the coefficients of x1 and x8 are about -3 and -2, respectively, which are the parameter values that are specified in the simulation.
  • The estimates for the levels of the C1 CLASS variable are highlighted in green. They are close to (-3, -4, 0), which are the parameter values that are specified in the simulation.
  • The estimates for the Intercept and the C4 CLASS variable are highlighted in magenta. Notice that they seem to differ from the parameters in the simulation. As discussed previously, the simulation and PROC GLM use different parameterizations of the C4 effect. The simulation assigns Intercept = 4. The contribution of the first level of C4 is 5*1, the contribution for the second level is 5*2, and the contribution for the third level is 5*3. As explained in the previous article, the GLM parameterization reparameterizes the C4 effect as (4 + 15) + (5 - 15)*(C4=1) + (10 - 15)*(C4=2). The estimates are very close to these parameter values.

Although this program simulates a linear regression model, you can modify the program and simulate from a generalized linear model such as the logistic model. You just need to compute the linear predictor, eta (η), and then use the link function and the RAND function to generate the response variable, as shown in a previous article about how to simulate data from a logistic model.

In summary, this article shows how to simulate data for a linear regression model in the SAS DATA step when the model includes both categorical and continuous regressors. The program simulates arbitrarily many continuous and categorical variables. You can define a response variable in terms of the explanatory variables and their interactions.

The post Simulate data for a regression model with categorical and continuous variables appeared first on The DO Loop.

10月 032018

It is sometimes necessary for researchers to simulate data with thousands of variables. It is easy to simulate thousands of uncorrelated variables, but more difficult to simulate thousands of correlated variables. For that, you can generate a correlation matrix that has special properties, such as a Toeplitz matrix or a first-order autoregressive (AR(1)) correlation matrix. I have previously written about how to generate a large Toeplitz matrix in SAS. This article describes three useful results for an AR(1) correlation matrix:

  • How to generate an AR(1) correlation matrix in the SAS/IML language
  • How to use a formula to compute the explicit Cholesky root of an AR(1) correlation matrix.
  • How to efficiently simulate multivariate normal variables with AR(1) correlation.

Generate an AR(1) correlation matrix in SAS

The AR(1) correlation structure is used in statistics to model observations that have correlated errors. (For example, see the documentation of PROC MIXED in SAS.) If Σ is AR(1) correlation matrix, then its elements are constant along diagonals. The (i,j)th element of an AR(1) correlation matrix has the form Σ[i,j] = ρ|i – j|, where ρ is a constant that determines the geometric rate at which correlations decay between successive time intervals. The exponent for each term is the L1 distance between the row number and the column number. As I have shown in a previous article, you can use the DISTANCE function in SAS/IML 14.3 to quickly evaluate functions that depend on the distance between two sets of points. Consequently, the following SAS/IML function computes the correlation matrix for a p x p AR(1) matrix:

proc iml;
/* return p x p matrix whose (i,j)th element is rho^|i - j| */
start AR1Corr(rho, p); 
   return rho##distance(T(1:p), T(1:p), "L1");
/* test on 10 x 10 matrix with rho = 0.8 */
rho = 0.8;  p = 10;
Sigma = AR1Corr(rho, p);
print Sigma[format=Best7.];

A formula for the Cholesky root of an AR(1) correlation matrix

Every covariance matrix has a Cholesky decomposition, which represents the matrix as the crossproduct of a triangular matrix with itself: Σ = RTR, where R is upper triangular. In SAS/IML, you can use the ROOT function to compute the Cholesky root of an arbitrary positive definite matrix. However, the AR(1) correlation matrix has an explicit formula for the Cholesky root in terms of ρ. This explicit formula does not appear to be well known by statisticians, but it is a special case of a general formula developed by V. Madar (Section 5.1, 2016), who presented a poster at a Southeast SAS Users Group (SESUG) meeting a few years ago. An explicit formula means that you can compute the Cholesky root matrix, R, in a direct and efficient manner, as follows:

/* direct computation of Cholesky root for an AR(1) matrix */
start AR1Root(rho, p);
   R = j(p,p,0);                /* allocate p x p matrix */
   R[1,] = rho##(0:p-1);        /* formula for 1st row */
   c = sqrt(1 - rho**2);        /* scaling factor: c^2 + rho^2 = 1 */
   R2 = c * R[1,];              /* formula for 2nd row */
   do j = 2 to p;               /* shift elements in 2nd row for remaining rows */
      R[j, j:p] = R2[,1:p-j+1]; 
   return R;
R = AR1Root(rho, p);   /* compare to R = root(Sigma), which requires forming Sigma */
print R[L="Cholesky Root" format=Best7.];

You can compute an AR(1) covariance matrix from the correlation matrix by multiplying the correlation matrix by a positive scalar, σ2.

Efficient simulation of multivariate normal variables with AR(1) correlation

An efficient way to simulate data from a multivariate normal population with covariance Σ is to use the Cholesky decomposition to induce correlation among a set of uncorrelated normal variates. This is the technique used by the RandNormal function in SAS/IML software. Internally, the RandNormal function calls the ROOT function, which can compute the Cholesky root of an arbitrary positive definite matrix.

When there are thousands of variables, the Cholesky decomposition might take a second or more. If you call the RandNormal function thousands of times during a simulation study, you pay that one-second penalty during each call. For the AR(1) covariance structure, you can use the explicit formula for the Cholesky root to save a considerable amount of time. You also do not need to explicitly form the p x p matrix, Σ, which saves RAM. The following SAS/IML function is an efficient way to simulate N observations from a p-dimensional multivariate normal distribution that has an AR(1) correlation structure with parameter ρ:

/* simulate multivariate normal data from a population with AR(1) correlation */
start RandNormalAR1( N, Mean, rho );
    mMean = rowvec(Mean);
    p = ncol(mMean);
    U = AR1Root(rho, p);      /* use explicit formula instead of ROOT(Sigma) */
    Z = j(N,p);
    call randgen(Z,'NORMAL');
    return (mMean + Z*U);
call randseed(12345);
p = 1000;                  /* big matrix */
mean = j(1, p, 0);         /* mean of MVN distribution */
/* simulate data from MVN distribs with different values of rho */
v = do(0.01, 0.99, 0.01);  /* choose rho from list 0.01, 0.02, ..., 0.99 */
t0 = time();               /* time it! */
do i = 1 to ncol(v);
   rho = v[i];
   X = randnormalAR1(500, mean, rho);  /* simulate 500 obs from MVN with p vars */
t_SimMVN = time() - t0;     /* total time to simulate all data */
print t_SimMVN;

The previous loop generates a sample that contains N=500 observations and p=1000 variables. Each sample is from a multivariate normal distribution that has an AR(1) correlation, but each sample is generated for a different value of ρ, where ρ = 0.01. 0.02, ..., 0.99. On my desktop computer, this simulation of 100 correlated samples takes about 4 seconds. This is about 25% of the time for the same simulation that explicitly forms the AR(1) correlation matrix and calls RandNormal.

In summary, the AR(1) correlation matrix is an easy way to generate a symmetric positive definite matrix. You can use the DISTANCE function in SAS/IML 14.3 to create such a matrix, but for some applications you might only require the Cholesky root of the matrix. The AR(1) correlation matrix has an explicit Cholesky root that you can use to speed up simulation studies such as generating samples from a multivariate normal distribution that has an AR(1) correlation.

The post Fast simulation of multivariate normal data with an AR(1) correlation structure appeared first on The DO Loop.

7月 112018

In a previous article, I showed how to find the intersection (if it exists) between two line segments in the plane. There are some fun problems in probability theory that involve intersections of line segments. One is "What is the probability that two randomly chosen chords of a circle intersect?" This article shows how to create a simulation in SAS to estimate the probability.

An exact answer

For this problem, a "random chord" is defined as the line segment that joins two points chosen at random (with uniform probability) on the circle. The probability that two random chords intersect can be derived by using a simple counting argument. Suppose that you pick four points at random on the circle. Label the points according to their polar angle as p1, p2, p3, and p4. As illustrated by the following graphic, the points are arranged on the circle in one of the following three ways. Consequently, the probability that two random chords intersect is 1/3 because the chords intersect in only one of the three possible arrangements.

Possible arrangements of two random chords in the circle

A simulation in SAS

You can create a simulation to estimate the probability that two random chords intersect. The intersection of two segments can be detected by using either of the two SAS/IML modules in my article about the intersection of line segments. The following simulation generates four angles chosen uniformly at random in the interval (0, 2π). It converts those points to (x,y) coordinates on the unit circle. It then computes whether the chord between the first two points intersects the chord between the third and fourth points. It repeats this process 100,000 times and reports the proportion of times that the chords intersect.

proc iml;
/* Find the intersection between 2D line segments [p1,p2] and [q1,q2].
   This function assumes that the line segments have different slopes (A is nonsingular) */
start IntersectSegsSimple(p1, p2, q1, q2);
   b = colvec(q1 - p1); 
   A = colvec(p2-p1) || colvec(q1-q2); /* nonsingular when segments have different slopes */
   x = solve(A, b);                    /* x = (s,t) */
   if all(0<=x && x<=1) then           /* if x is in [0,1] x [0,1] */
      return (1-x[1])*p1 + x[1]*p2;    /* return intersection */
   else                                /* otherwise, segments do not intersect */
      return ({. .});                  /* return missing values */
/* Generate two random chords on the unit circle.
   Simulate the probability that they intersect  */
N = 1e5;
theta = j(N, 4);
call randseed(123456);
call randgen(theta, "uniform", 0, 2*constant('pi'));
intersect = j(N,1,0);
do i = 1 to N;
   t = theta[i,]`;                 /* 4 random U(0, 2*pi) */
   pts = cos(t) || sin(t);         /* 4 pts on unit circle */
   p1 = pts[1,];    p2 = pts[2,];
   q1 = pts[3,];    q2 = pts[4,];
   intersect[i] = all(IntersectSegsSimple(p1, p2, q1, q2) ^= .);
prob = mean(intersect);
print prob;

This simulation produces an estimate that is close to the exact probability of 1/3.

Connection to Bertrand's Paradox

This problem has an interesting connection to Bertrand's Paradox. Bertrand's paradox shows the necessity of specifying the process that is used to define the random variables in a probability problem. It turns out that there are multiple ways to define "random chords" in a circle, and the different definitions can lead to different answers to probability questions. See the Wikipedia article for an example.

For the definition of "random chords" in this problem, the density of the endpoints is uniform on the circle. After you make that choice, other distributions are determined. For example, the distribution of the lengths of 1,000 random chords is shown below. The lengths are NOT uniformly distributed! The theoretical density of the chord lengths is overlaid on the distribution of the sample. If you change the process by which chords are randomly chosen (for example, you force the lengths to be uniformly distributed), you might also change the answer to the problem, as shown in Bertrand's Paradox.

Distribution of lengths of random chords in the unit circle, generated by choosing uniformly distributed endpoints

The post The probability that two random chords of a circle intersect appeared first on The DO Loop.

6月 152018

My 2018 SAS Global Forum paper was about "how to use the random-number generators (RNGs) in SAS." You can read the paper for details, but I recently recorded a short video that summarizes the main ideas in the paper. In particular, the video gives an overview of the new RNGs in SAS, which include the following:

  • MTHYBRID, MT2002, and MT64: Variants of the Mersenne twister RNG. The MTHYBRID method is the default RNG in SAS, beginning with the SAS 9.4M3.
  • PCG: A 64-bit permuted congruential generator
  • RDRAND: A hardware-based RNG that generates random numbers from thermal noise in the chip
  • TF2 and TF4: Counter-based Threefry RNGs

If your browser does not support embedded video, you can go directly to the video on YouTube.

The following references provide more information about the random number generators in SAS:

The post Video: New random number generators in SAS appeared first on The DO Loop.

6月 062018

The SURVEYSELECT procedure in SAS 9.4M5 supports the OUTRANDOM option, which causes the selected items in a simple random sample to be randomly permuted after they are selected. This article describes several statistical tasks that benefit from this option, including simulating card games, randomly permuting observations in a DATA step, assigning a random ID to patients in a clinical study, and generating bootstrap samples. In each case, the new OUTRANDOM option reduces the number of statements that you need to write. The OUTRANDOM option can also be specified by using OUTORDER=RANDOM.

Sample data with PROC SURVEYSELECT

Often when you draw a random sample (with or without replacement) from a population, the order in which the items were selected is not important. For example, if you have 10 patients in a clinical trial and want to randomly assign five patients to the control group, the control group does not depend on the order in which the patients were selected. Similarly, in simulation studies, many statistics (means, proportions, standard deviations,...) depend only on the sample, not on the order in which the sample was generated. For these reasons, and for efficiency, the SURVEYSELECT procedure in SAS uses a "one-pass" algorithm to select observations in the same order that they appear in the "population" data set.

However, sometimes you might require the output data set from PROC SURVEYSELECT to be in a random order. For example, in a poker simulation, you might want the output of PROC SURVEYSELECT to represent a random shuffling of the 52 cards in a deck. To be specific, the following DATA step generates a deck of 52 cards in order: Aces first, then 2s, and so on up to jacks, queens, and kings. If you use PROC SURVEYSELECT and METHOD=SRS to select 10 cards at random (without replacement), you obtain the following subset:

data CardDeck;
length Face $2 Suit $8;
do Face = 'A','2','3','4','5','6','7','8','9','10','J','Q','K';
   do Suit = 'Clubs', 'Diamonds', 'Hearts', 'Spades';
      CardNumber + 1;
/* Deal 10 cards. Order is determined by input data */
proc surveyselect data=CardDeck out=Deal noprint
     seed=1234 method=SRS /* sample w/o replacement */
     sampsize=10;         /* number of observations in sample */
proc print data=Deal; run;
Random sample without replacement from a card deck. The sample is in the same order as the data.

Notice that the call to PROC SURVEYSELECT did not use the OUTRANDOM option. Consequently, the cards are in the same order as they appear in the input data set. This sample is adequate if you want to simulate dealing hands and estimate probabilities of pairs, straights, flushes, and so on. However, if your simulation requires the cards to be in a random order (for example, you want the first five observations to represent the first player's cards), then clearly this sample is inadequate and needs an additional random permutation of the observations. That is exactly what the OUTRANDOM option provides, as shown by the following call to PROC SURVEYSELECT:

/* Deal 10 cards in random order */
proc surveyselect data=CardDeck out=Deal2 noprint
     seed=1234 method=SRS /* sample w/o replacement */
     sampsize=10          /* number of observations in sample */
     OUTRANDOM;           /* SAS/STAT 14.3: permute order */
proc print data=Deal2; run;
Random sample without replacement from a card deck. The sample is then permuted into a random order.

You can use this sample when the output needs to be in a random order. For example, in a poker simulation, you can now assign the first five cards to the first player and the second five cards to a second player.

Permute the observations in a data set

A second application of the OUTRANDOM option is to permute the rows of a SAS data set. If you sample without replacement and request all observations (SAMPRATE=1), you obtain a copy of the original data in random order. For example, the students in the Sashelp.Class data set are listed in alphabetical order by their name. The following statements use the OUTRANDOM option to rearrange the students in a random order:

/* randomly permute order of observations */
proc surveyselect data=Sashelp.Class out=RandOrder noprint
     seed=123 method=SRS /* sample w/o replacement */
     samprate=1          /* proportion of observations in sample */
     OUTRANDOM;          /* SAS/STAT 14.3: permute order */
proc print data=RandOrder; run;
Use PROC SURVEYSELECT to produce a random ordering of a SAS data set

There are many other ways to permute the rows of a data set, such as adding a uniform random variable to the data and then sorting. The two methods are equivalent, but the code for the SURVEYSELECT procedure is shorter to write.

Assign unique random IDs to patients in a clinical trial

Another application of the OUTRANDOM option is to assign a unique random ID to participants in an experimental trial. For example, suppose that four-digit integers are used for an ID variable. Some clinical trials assign an ID number sequentially to each patient in the study, but I recently learned from a SAS discussion forum that some companies assign random ID values to subjects. One way to assign random IDs is to sample randomly without replacement from the set of all ID values. The following DATA step generates all four-digit IDs, selects 19 of them in random order, and then merges those IDs with the participants in the study:

data AllIDs;
do ID = 1000 to 9999;  /* create set of four-digit ID values */
/* randomly select 19 unique IDs */
proc surveyselect data=AllIDs out=ClassIDs noprint
     seed=12345 method=SRS  /* sample w/o replacement */
     sampsize=19            /* number of observations in sample */
     OUTRANDOM;             /* SAS/STAT 14.3: permute order */
data Class;
   merge ClassIDs Sashelp.Class;   /* merge ID variable and subjects */
proc print data=Class; var ID Name Sex Age; run;

Random order for other sampling methods

The OUTRANDOM option also works for other sampling schemes, such as sampling with replacement (METHOD=URS, commonly used for bootstrap sampling) or stratified sampling. If you use the REPS= option to generate multiple samples, each sample is randomly ordered.

It is worth mentioning that the SAMPLE function in SAS/IML also can to perform a post-selection sort. Suppose that X is any vector that contains N elements. Then the syntax SAMPLE(X, k, "NoReplace") generates a random sample of k elements from the set of N. The documentation states that "the elements ... might appear in the same order as in X." This is likely to happen when k is almost equal to N. If you need the sample in random order, you can use the syntax SAMPLE(X, k, "WOR") which adds a random sort after the sample is selected, just like PROC SURVEYSELECT does when you use the OUTRANDOM option.

The post Sample and obtain the results in random order appeared first on The DO Loop.