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Homework Help: X and Y identically distributed implies E(X) = E(Y) and var(X) = var(Y)

  1. Oct 13, 2009 #1
    The problem statement, all variables and given/known data
    Let [itex](S, \Sigma, P)[/itex] be a probability space. Let X and Y be two random variables on S that satisfy [itex]P \circ X^{-1} = P \circ Y^{-1}[/itex] (i.e. they are identically distributed) and such that E(X), E(Y), var(X) and var(Y) exist and are finite. Prove that E(X) = E(Y) and var(X) = var(Y).

    The attempt at a solution
    I can only think of proving this the long and tedious way: first when X and Y are simple, then for X and Y bounded, then for X and Y nonnegative, and then finally for X and Y integrable. Is there an easier way?
  2. jcsd
  3. Oct 15, 2009 #2
    Can you easily show X and Y have the same cdf? Then do you have a definition for E[X^n] in terms of the cdf of X?
  4. Oct 15, 2009 #3
    They do have the same CDF but I don't have a definition of E[Xn] in terms of the CDF of X. All I know is that

    E[X^n] = \int_S X^n \, dP

    whenever the integral exists.
  5. Oct 15, 2009 #4
    Do you have a theorem saying something like

    [tex]\int f\, d(P\circ X^{-1})=\int (f\circ X)\, dP[/tex] ?

    If so then you can use f(x)=x, f(x)=x^2, or even f(x)=x^n.

    If you don't have such a theorem, you could prove it. I can't see how to avoid some of the "long" method. Maybe it is shorter than you thought. I don't think you have to approximate X and Y, just the two functions f(x)=x and f(x)=x^2. If you just want to do those two cases, use (or imagine using) a specific sequence of simple (step, in fact) functions.

    Maybe you can get away with observing that for [tex]f= \chi_E[/tex] we have [tex]\chi_E\circ X=\chi_{X^{-1}(E)}[/tex], and then saying a magic phrase like "the result follows by linear combinations and limits." Probably not. I think it might be worth writing it out, just to verify my claim that X and Y don't need to be approximated.
  6. Oct 15, 2009 #5
    I don't know of such a theorem. I'll try to prove that though (seems interesting).
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