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Variable $X$ is a random variable with know $E(X)$ and $Var(X)$. Variable $Y$ is also a random variable with know $E(Y)$ and $Var(Y)$ and $X$ and $Y$ are independent.

What is the variance and expected value of $\dfrac{X}{X+Y}$?

The expected value looks simple, but the variance not... I found this post, and from there the only problem is to get $cov(X, X+Y)$, which I have no idea how to calculate...

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  • $\begingroup$ Any assumptions on the distribution of $X$ and $Y$ $\endgroup$ – IcannotFixThis Aug 16 '16 at 10:34
  • $\begingroup$ They are both geometric distributions. This is related with my question here. However, I'm also interested in the general answer, without assumptions, for curiosity sake :) $\endgroup$ – Diogo Santos Aug 16 '16 at 10:40
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I doubt there is a general formula for any distribution :

n <- 100000

x <- rpois(n, 1)
y <- rchisq(n, df = 2)/2

mean(x/(x+y))
sd(x/(x+y))

x2 <- rexp(n,1)
y2 <- rexp(n,1)

mean(x2/(x2+y2))
sd(x2/(x2+y2))

Though the expected values and variances are all the same:

> mean(x/(x+y))
[1] 0.4179734
> sd(x/(x+y))
[1] 0.3608732

And:

> mean(x2/(x2+y2))
[1] 0.4996955
> sd(x2/(x2+y2))
[1] 0.2889206
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  • $\begingroup$ In the post I linked, there seams to be a more general answer... Given your answer in here I guess I can calculate $cov(X, X+Y) = var(X)$ and use that. $\endgroup$ – Diogo Santos Aug 16 '16 at 11:00

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