3
$\begingroup$

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...

Improve this question
$\endgroup$
2
  • $\begingroup$ Any assumptions on the distribution of $X$ and $Y$ $\endgroup$
    – IcannotFixThis
    Commented Aug 16, 2016 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
    Commented Aug 16, 2016 at 10:40

1 Answer 1

Reset to default
0
$\begingroup$

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
Improve this answer
$\endgroup$
1
  • $\begingroup$ 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
    Commented Aug 16, 2016 at 11:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.