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Let $X \sim N(\mu, \sigma^2)$, and $Y \sim Bernoulli(p)$, with $Cov(X, Y) \neq 0$.

Is there an algorithm for generating samples from $X$ and $Y$ given we specify the covariance?

user154510
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1 Answers1

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Without losing generality we assume $\mu=0$ and $\sigma=1$.

\begin{eqnarray} Y|X &\sim & Ber(\Phi(a-X)) \\ X &\sim & N(0,1) \end{eqnarray} where $\Phi$ is cumulative distribution function of normal.

By this we have $E\left(\Phi\left( \frac{a-bX}{c}\right)\right)=\Phi\left( \frac{a}{c\sqrt{1+\frac{b^2}{c^2}}}\right)$ and you can calculate $E(XY)$ and the covariance.

Now generate $x_i$ from $N(0,1)$ and then generate $y_i$ from $Ber(\Phi(a-x_i))$. You can choose $a$ according the value of covariance.

R code, use correlation instead covariance 
n <- 10000
a <- seq(-3, 3, len = 10) ## -2<a<2
cor_1 <- c()
for (i in 1:length(a)) {
  x <- rnorm(n, 0, 1)
  y <- rbinom(n, size = 1, prob = pnorm(a[i] - x))
  cor_1[i] <- round(cor(x, y), 2)
}
> data.frame(a, cor_1)
            a cor_1
1  -3.0000000 -0.23
2  -2.3333333 -0.32
3  -1.6666667 -0.44
4  -1.0000000 -0.52
5  -0.3333333 -0.56
6   0.3333333 -0.57
7   1.0000000 -0.52
8   1.6666667 -0.42
9   2.3333333 -0.32
10  3.0000000 -0.23

R code, use correlation instead covariance 
n <- 10000
a <- seq(-3, 3, len = 10) ## -2<a<2
cor_1 <- c()
for (i in 1:length(a)) {
  x <- rnorm(n, 0, 1)
  y <- rbinom(n, size = 1, prob = pnorm(a[i] + x)) ## here is the change
  cor_1[i] <- round(cor(x, y), 2)
}
> data.frame(a, cor_1)
            a cor_1
1  -3.0000000 0.23
2  -2.3333333 0.32
3  -1.6666667 0.44
4  -1.0000000 0.52
5  -0.3333333 0.56
6   0.3333333 0.57
7   1.0000000 0.52
8   1.6666667 0.42
9   2.3333333 0.32
10  3.0000000 0.23
Masoud
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