5
$\begingroup$

Suppose I want to generate $X_1, \ldots, X_n$ Bernoulli random variables such that:

$$ Corr(X_i, X_j) = \rho $$

for all $i \neq j$.

I am wondering what method I might be able to use (I will be trying to do this in R). I am thinking of using Couplas, but could anyone guide me how to implement this? Thanks.

$\endgroup$
8
$\begingroup$

Because this correlation matrix is so symmetric, we might try to solve the problem with a symmetric distribution.

One of the simplest that gives sufficient flexibility in varying the correlation is the following. Given $d\ge 2$ variables, define a distribution on the set of $d$-dimensional binary vectors $X$ by assigning probability $q$ to $X=(1,1,\ldots, 1)$, probability $q$ to $X=(0,0,\ldots, 0)$, and distributing the remaining probability $1-2q$ equally among the $d$ vectors having exactly one $1$; thus, each of those gets probability $(1-2q)/d$. Note that this family of distributions depends on just one parameter $0\le q\le 1/2$.

It's easy to simulate from one of these distributions: output a vector of zeros with probability $q$, output a vector of ones with probability $q$, and otherwise select uniformly at random from the columns of the $d\times d$ identity matrix.

All the components of $X$ are identically distributed Bernoulli variables. They all have common parameter

$$p = \Pr(X_1 = 1) = q + \frac{1-2q}{d}.$$

Compute the covariance of $X_i$ and $X_j$ by observing they can both equal $1$ only when all the components are $1$, whence

$$\Pr(X_i=1=X_j) = \Pr(X=(1,1,\ldots,1)) = q.$$

This determines the mutual correlation as

$$\rho = \frac{d^2q - ((d-2)q + 1)^2}{(1 + (d-2)q)(d-1 - (d-2)q)}.$$

Given $d \ge 2$ and $-1/(d-1)\le \rho \le 1$ (which is the range of all possible correlations of any $d$-variate random variable), there is a unique solution $q(\rho)$ between $0$ and $1/2$.

Simulations bear this out. Beginning with a set of $21$ equally-spaced values of $\rho$, the corresponding values of $q$ were computed (for the case $d=8$) and used to generate $10,000$ independent values of $X$. The $\binom{8}{2}=28$ correlation coefficients were computed and plotted on the vertical axis. The agreement is good.

Figure

I carried out a range of such simulations for values of $d$ between $2$ and $99$, with comparably good results.

A generalization of this approach (namely, allowing for two, or three, or ... values of the $X_i$ simultaneously to equal $1$) would give greater flexibility in varying $E[X_i]$, which in this solution is determined by $\rho$. That combines the ideas related here with the ones in the fully general $d=2$ solution described at https://stats.stackexchange.com/a/285008/919.


The following R code features a function p to compute $q$ from $\rho$ and $d$ and exhibits a fairly efficient simulation mechanism within its main loop.

#
# Determine p(All zeros) = p(All ones) from rho and d.
#
p <- function(rho, d) {
  if (rho==1) return(1/2)
  if (rho <= -1/(d-1)) return(0)
  if (d==2) return((1+rho)/4)
  b <- d-2
  (4 + 2*b + b^2*(1-rho) - (b+2)*sqrt(4 + b^2 * (1-rho)^2)) / (2 * b^2 * (1-rho))
}
#
# Simulate a range of correlations `rho`.
#
d <- 8       # The number of variables.
n.sim <- 1e4 # The number of draws of X in the simulation.

rho.limits <- c(-1/(d-1), 1)
rho <- seq(rho.limits[1], rho.limits[2], length.out=21)
rho.hat <- sapply(rho, function(rho) {
  #
  # Compute the probabilities from rho.
  #
  qd <- q0 <- p(rho, d)
  q1 <- (1 - q0 - qd)
  #
  # First randomly select three kinds of events: all zero, one 1, all ones.
  #
  u <- sample.int(3, n.sim, prob=c(q0,q1,qd), replace=TRUE)
  #
  # Conditionally, when there is to be one 1, uniformly select which 
  # component will equal 1.
  #
  k <- diag(d)[, sample.int(d, n.sim, replace=TRUE)]
  #
  # When there are to be all zeros or all ones, make it so.
  #
  k[, u==1] <- 0
  k[, u==3] <- 1
  #
  # The simulated values of X are the columns of `k`. Return all d*(d-1)/2 correlations.
  #
  cor(t(k))[lower.tri(diag(d))]
})
#
# Display the simulation results.
#
plot(rho, rho, type="n",
     xlab="Intended Correlation",
     ylab="Sample Correlation", 
     xlim=rho.limits, ylim=rho.limits,
     main=paste(d, "Variables,", n.sim, "Iterations"))

abline(0, 1, col="Red", lwd=2)

invisible(apply(rho.hat, 1, function(y) 
  points(rho, y, pch=21, col="#00000010", bg="#00000004")))
$\endgroup$
  • $\begingroup$ Thanks, how did you exactly get the term $\rho = \frac{d^2q - ((d-2)q + 1)^2}{(1 + (d-2)q)(d-1 - (d-2)q)}$ above? Was it by just using the definition of covariance and doing expectations over binary variables? $\endgroup$ – user321627 Jan 24 '18 at 1:56
  • $\begingroup$ Yes, it comes directly from the definition. $\endgroup$ – whuber Jan 24 '18 at 14:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.