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.


1 Answer 1


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.


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.
# 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")))
  • $\begingroup$ 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
    Commented Jan 24, 2018 at 1:56
  • $\begingroup$ Yes, it comes directly from the definition. $\endgroup$
    – whuber
    Commented Jan 24, 2018 at 14:21

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