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Mar 24, 2016 at 18:24 history edited Bob CC BY-SA 3.0
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Mar 24, 2016 at 18:22 comment added Bob Yeah, I was afraid there might not be a solution. It's related to this one, stats.stackexchange.com/questions/61080/… But I don't see how to generalize to my setting...
Mar 24, 2016 at 18:21 comment added leonbloy Ok, it's clear now, it does not look easy.
Mar 24, 2016 at 18:20 vote accept Bob
Mar 24, 2016 at 18:17 comment added Bob @leonbloy, It's a ratio of two random variables, $Z_1=\Phi(X+Y)$ and $Z_2=\Phi(X)$ where $\Phi$ is the standard normal cdf function, mapping $\mathbb{R} \to [0,1]$. I want to compute $E(Z_1 / Z_2)$
Mar 24, 2016 at 18:06 comment added leonbloy 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 $E(1 +Y/X)=1 + E(Y/X)$
Mar 24, 2016 at 17:34 history edited Bob CC BY-SA 3.0
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Mar 24, 2016 at 17:34 comment added Bob 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)
Mar 24, 2016 at 16:12 answer added Clarinetist timeline score: 1
Mar 24, 2016 at 15:54 comment added Clarinetist @Henry Ah, that's true. Hmm. I might recommend doing a simulation.
Mar 24, 2016 at 15:53 comment added Henry @Clarinetist: note that $\Phi(X+Y)$ and $\Phi(X)$ are not independent
Mar 24, 2016 at 15:37 comment added Clarinetist 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.
Mar 24, 2016 at 15:10 review First posts
Mar 24, 2016 at 15:17
Mar 24, 2016 at 15:10 history asked Bob CC BY-SA 3.0