Here's my problem:

Assume that positive random variables $X$ and $Y$ are identically distributed with $E[X] = E[Y] = μ < ∞$, and not necessarily independent. Compute

$$Cov\left(X+Y, \frac{X}{X+Y}\right)$$

Here's what I've done:

$$Cov\left(X+Y, \frac{X}{X+Y}\right)= E\left[(X+Y)\left(\frac{X}{X+Y}\right)\right] - E[X+Y] \cdot E\left[\frac{X}{X+Y}\right] = E[X] - (E[X]+E[Y])\cdot E\left[\frac{X} {X+Y}\right]= μ - 2μE\left[\frac{X} {X+Y}\right]$$

but I have no clue what to do with the $E\left[\frac{X}{X+Y}\right]$ term. I can't seem to get a linear term out of $\frac{X} {X+Y}$ (probably impossible), or manipulate the equations to cancel out $E\left[\frac{X} {X+Y}\right]$. Intuitively, it seems that the answer should be in terms of μ, which makes me think that $E\left[\frac{X} {X+Y}\right]$ is a constant, such as 1/2. I cannot seem to figure this out. I think I am supposed to do something with conditional expectation (e.g. $E[A] = E[E[A|B]]$).

Can I please get some insights or hints? Thanks.

  • $\begingroup$ Hopefully $P(X=0)=0$. $\endgroup$ – Alex R. Dec 18 '15 at 3:14

They tell you that $X$ and $Y$ are identically distributed. Then $$1 = E\left[\frac{X+Y}{X+Y}\right] = E\left[\frac{X}{X+Y}\right]+E\left[\frac{Y}{X+Y}\right] = 2E\left[\frac{X}{X+Y}\right].$$ Thus, $$E\left[\frac{X}{X+Y}\right] = \frac{1}{2}.$$

  • 1
    $\begingroup$ That's beautiful $\endgroup$ – sirblobfish Dec 18 '15 at 2:51

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