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If X and Y are independent, are X^k and Y?

  1. Sep 25, 2011 #1
    The definition of independence of random variables from a measure-theoretic standpoint is so confusing (independence of generated sigma-algebras, etc.) that I cannot answer this seemingly simple question...So if [itex]X,Y[/itex] are independent random variables, does that mean [itex]X,X^2,X^3,X^4,\ldots[/itex] and [itex]Y,Y^2,Y^3,\ldots[/itex] are each pairwise independent?
     
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  3. Sep 25, 2011 #2
    Okay...just looked this up in a book. In fact, if [itex]X,Y[/itex] are independent, then we have [itex]E[f(X)g(Y)] = E[f(X)]E[g(Y)][/itex] for "any" functions f and g. But the proof is from a non-measure-theoretic probability book. Is there anyone who can explain why this holds from a measure-theoretic standpoint?
     
  4. Sep 25, 2011 #3

    mathman

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    Expectations can be expressed as integrals involving the probability density functions. Since X and Y are independent, their joint density is simply the product of their individual densities, so the expectations involving functions of the random variables end up as the product of the individual expectations.
     
  5. Sep 26, 2011 #4

    Stephen Tashi

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    I don't know what that result has to do with your original question. The conclusion deals with the expectations of f(x) and g(y) not with their independence. Random variables can be uncorrelated and still be dependent.

    I'm not an expert on measure theory but I did take the course years ago. I think answering your original post ( which concerns functions of a random variable) in detail is complicated. For example, not all functions are "measureable". The one's you listed are. How do we prove they are? Are you wanting an explanation that begins at elementary points like that? Or do you simply want a theorem from measure theory that answers your question as a special case?
     
  6. Sep 26, 2011 #5
    It's probably easier to use P[X<=x,Y<=y] = P[X<=x]P[Y<=y] (which can be obtained from the measure-theoretic definition of independence by considering generators of the Borel sigma algebras for R and R^2).
     
  7. Dec 24, 2011 #6
    Let X and Y be independent random variable and f and g be borel-measurable functions on R, there is a measure-theoretic way of proof that f(X) and g(Y) are independent r.v. in Shreve's book "Stochastic Calculus for Finance II", see Theorem 2.2.5
     
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