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Expectation of (aX-bY) ?

  1. Dec 17, 2013 #1

    I 'm trying to express the following in integral form:

    E[a/X-b/Y], where E[.] stands for the expectation operator.

    Let a,b be some nonnegative constants and X,Y are independent nonnegative Gamma distributed random variables.

    Any help would be useful.

    Thanks in advance
  2. jcsd
  3. Dec 17, 2013 #2


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    Do you want aX - bY or a/X - b/Y? The first is easy.
    For the second you need to first get the distribution functions for the reciprocals. Then use the addition formula.
  4. Dec 17, 2013 #3
    Thanks for the reply.

    i 'm interested in evaluating the second on (i.e., a/X - b/Y). Can you be more specific ? the addition formula with what integration bounds exactly?
    what if i would use the expectation of X and Y seperately (i.e, E[c/X]-E[b/Y])?
  5. Dec 17, 2013 #4


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    Splitting the expectation by linearity is the obvious first step. The hard question is given a distribution X, find E(1/X). This is non-trivial in general, but luckily for you for a gamma distribution this has already been done

  6. Dec 17, 2013 #5
    This a well-known result for resiprocal PDFs. In fact, this is a generic rule for every distribution function (i.e. taking the inverse variable within the initial PDF and multiplying it with the derivative of the inverse variable gives the corresponding reciprocal PDF).

    I was hoping for a solution by applying the original Gamma PDFs directly..
  7. Dec 18, 2013 #6


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    Let X be a random variable with density function f(x), then E(1/X) = ∫(f(x)/x)dx.
    If there is no density and the distribution function is F(x), you need a Stieljes integral, E(1/X) = ∫(1/x)dF(x)
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