0
$\begingroup$

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.

$\endgroup$
8
  • $\begingroup$ 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. $\endgroup$ Commented Mar 24, 2016 at 15:37
  • $\begingroup$ @Clarinetist: note that $\Phi(X+Y)$ and $\Phi(X)$ are not independent $\endgroup$ Commented Mar 24, 2016 at 15:53
  • $\begingroup$ @Henry Ah, that's true. Hmm. I might recommend doing a simulation. $\endgroup$ Commented Mar 24, 2016 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) $\endgroup$ Commented Mar 24, 2016 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 $E(1 +Y/X)=1 + E(Y/X)$ $\endgroup$ Commented Mar 24, 2016 at 18:06

1 Answer 1

Reset to default
1
$\begingroup$

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){ set.seed(seed) 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)) } return(means) } #For example: ratio_sim(nsims = 5000, nnorms = 5000, mu_X = 1, mu_Y = 2, sd_X = 2, sd_Y = 4)
$\endgroup$
7
  • $\begingroup$ 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) $\endgroup$ Commented Mar 24, 2016 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?) $\endgroup$ Commented Mar 24, 2016 at 17:54
  • $\begingroup$ @Bob Oh yeah, you're looking for $E$, not the distribution. Give me some time and I'll have it $\endgroup$ Commented Mar 24, 2016 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... $\endgroup$ Commented Mar 24, 2016 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. $\endgroup$ Commented Mar 24, 2016 at 18:04

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.