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

  1. Oct 5, 2013 #1
    1. The problem statement, all variables and given/known data
    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?


    2. Relevant equations



    3. 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.
     
  2. jcsd
  3. Oct 5, 2013 #2

    fzero

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    This seems wierd 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.
     
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