I am trying to compute the expected value of the ratio of two normal CDFs. Specifically, I like to compute the expected value of $\Phi(X+Y)/\Phi(X)$ where $X$ and $Y$ are independent normally distributed variables with means $\mu_X$ and $\mu_Y$ and variances $\sigma_X^2$ and $\sigma_Y^2$, and where $\Phi(.)$ is the standard normal CDF function.

I am not sure whether there is a closed form solution. In case there isn't any approximation to the derivative of the expected value in $\mu_X$ would be very helpful for my problem, too.

I would appreciate any help or suggestions where to look further.

  • $\begingroup$ If by $\Phi(X+Y)$ and $\Phi(X)$ you mean the CDF of $X+Y$ and the CDF of $X$, aren't continuous CDFs uniformly distributed in $[0, 1]$? If my memory is right, your question is equivalent to finding the expected value of a ratio of uniform $[0, 1]$ distributions. $\endgroup$ – Clarinetist Mar 24 '16 at 15:37
  • $\begingroup$ @Clarinetist: note that $\Phi(X+Y)$ and $\Phi(X)$ are not independent $\endgroup$ – Henry Mar 24 '16 at 15:53
  • $\begingroup$ @Henry Ah, that's true. Hmm. I might recommend doing a simulation. $\endgroup$ – Clarinetist Mar 24 '16 at 15:54
  • $\begingroup$ Thanks for the thoughts and sorry my notation wasn't super clear. With $\Phi(.)$ I meant the standard normal CDF, whereas $X$ and $Y$ can have any mean or variance. (I'll edit the original question to make it more clear) $\endgroup$ – Bob Mar 24 '16 at 17:34
  • $\begingroup$ I don't understand your notation. Your are doing a ratio of variables or of distributions? To write "the expected value of $\Phi(X+Y)/\Phi(X)$" makes no sense to me. If you mean the expected value of $Z=(X+Y)/X$, then it reduces to $E(1 +Y/X)=1 + E(Y/X)$ $\endgroup$ – leonbloy Mar 24 '16 at 18:06

This is too much for a comment, but here's a simulation, for $X \sim \mathcal{N}(1, 4)$ and $Y \sim \mathcal{N}(2, 16)$.

ratio_sim <- function(nsims, nnorms, mu_X, mu_Y, sd_X, sd_Y, seed=30){
means <- vector()
for (i in 1:nsims){
X <- rnorm(nnorms, mean = mu_X, sd = sd_X)
Y <- rnorm(nnorms, mean = mu_Y, sd = sd_Y)
sum <- X+Y
cdf_sum <- pnorm(sum)
cdf_X <- pnorm(X)
ratio <- cdf_sum/cdf_X
means <- c(means, mean(ratio))

#For example:
ratio_sim(nsims = 5000, nnorms = 5000, mu_X = 1, mu_Y = 2, sd_X = 2, sd_Y = 4)
  • $\begingroup$ Thanks again for your help. Unfortunately, I am not used to working with R, so I am not sure if I interpret every bit of the code correctly (in particular the graph...). Here is what I noted. In my original question, I wasn't very clear what I meant with $\Phi(.)$. It is supposed to be the normal cdf. So I think in the code I should replace ecdf(.) by pnorm(.). Then the expectation would be mean(cdf_sum/cdf_X) $\endgroup$ – Bob Mar 24 '16 at 17:50
  • $\begingroup$ @Bob Okay, so are you looking for a three-variable plot? (One axis based on $X$, one based on $Y$, and one on the ratio?) $\endgroup$ – Clarinetist Mar 24 '16 at 17:54
  • $\begingroup$ @Bob Oh yeah, you're looking for $E$, not the distribution. Give me some time and I'll have it $\endgroup$ – Clarinetist Mar 24 '16 at 17:55
  • $\begingroup$ Thanks, don't worry so, I can figure out the code myself. What I was mainly after is to see whether there may exist a closed form solution... $\endgroup$ – Bob Mar 24 '16 at 17:58
  • $\begingroup$ @Bob Yeah, since you're not familiar with R, I've created a function that you might find useful. Modifying it shouldn't take too much work. $\endgroup$ – Clarinetist Mar 24 '16 at 18:04

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