Probability: Proof of E[X/X+Y]=(X+Y)/2, X and Y are i.i.d. r.v.

[Sandbo]

Homework Statement

Given that X and Y are i.i.d. r.v, and E[X]<infinity.
Any idea how to prove that E[X/X+Y]=E[Y/X+Y]=(X+Y)/2?

Homework Equations

The Attempt at a Solution

I know that E[X] or E[Y] are constants, how come it now says E[X/X+Y] = (X+Y)/2 which is a r.v.?
and for X and Y being independent, E[XY]=E[X]E[Y], but now it is E[X/X+Y], which I think it means E[X/(X+Y)], the upper and lower aren't independent, we cannot write E[X]/E[X+Y]...no idea where to begin with.

Any help is appreciated.

 

[fzero]

This seems weird to me, both for the reason you mention, that $E[f(X,Y)]$ can't be a function of the random variables, and also because you can easily show that

$$ E[X/(X+Y)] + E[Y/(X+Y)] = 1$$

for a normalized distribution. So the only way we can have $E[X/(X+Y)] = E[Y/(X+Y)]$ is if they are both 1/2.

Aside from those concerns, one way to approach the reciprocal of a random variable is through a power series. Namely, given a random variable $X$, we can define a new random variable $X'$ by

$$ X = \bar{X} + X',~~~~ E[X']=0,$$

where $\bar{X} = E[X]$. Then, if $\bar{X}\neq 0$, we can use the geometric series to define

$$ \frac{1}{X} = \frac{1}{\bar{X}} \sum_{n\geq 0} (-1)^n \left( \frac{X'}{\bar{X}} \right)^n ,$$

which is convergent if $|X'| \leq |\bar{X}|$. A similar construction would serve to define $1/(X+Y)$ or similar rational functions.

If you weren't already given a seemingly misleading answer to the question, this is the approach I would suggest.