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Homework Help: Expected value of g(x)=1/(x+c) under gamma distribution

  1. Nov 28, 2012 #1
    1. The problem statement, all variables and given/known data

    Let [itex] g(x) = \frac{1}{x+c}[/itex], where [itex]c[/itex] is a positive constant, and [itex]x [/itex] is a random variable distributed according to the Gamma distribution
    [itex]x\sim f(x)=\frac{1}{\Gamma(\alpha) \beta^\alpha} x^{\alpha \,-\, 1} e^{-\frac{x}{\beta}}[/itex].

    I wish to calculate the expected value of [itex]g(x)[/itex] with respect to the probability density function [itex]f(x)[/itex].

    2. Relevant equations

    The expected value can be calculated as
    [tex]E(g(x))=\int_0^∞g(x)f(x)dx = \int_0^∞ \frac{1}{x+c} \frac{1}{\Gamma(\alpha) \beta^\alpha} x^{\alpha \,-\, 1} e^{-\frac{x}{\beta}}dx

    3. The attempt at a solution
    I have problems with calculating the integral. If [itex]g(x)=\frac{1}{x}[/itex], then the integral would not be too difficult. But the constant in the denominator gives me problems. I have attempted with variable substitutions and integration by parts. However, I have not been able to come up with a solution.
    Actually, this is not a homework problem. I posted it here since it is `homework style', so I do not know if it is event possible to calculate the expectation.

    Any help is much appreciated.
  2. jcsd
  3. Nov 28, 2012 #2


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    Homework Helper

    The integral may not have a closed form. Do you know for sure if it does?

    I'm not sure if this will help much, but you could try splitting up the integral into two, one from ##[0,c)## and the other on ##[c,\infty)##, and then expand the 1/(x+c) as a power series and integrate term by term to get a series expression. In the first integral, since x is smaller than c, you would use

    $$\frac{1}{x+c} = \frac{1}{c} \frac{1}{1+x/c} = \frac{1}{c} \sum_{n=0}^\infty \left(\frac{x}{c}\right)^n,$$

    while in the second term x is greater than c, so you would write

    $$\frac{1}{x+c} = \frac{1}{x} \frac{1}{1+c/x} = \frac{1}{x} \sum_{n=0}^\infty \left(\frac{c}{x}\right)^n.$$

    In principle one needs to be careful about switching the integral and the sum since the geometric series is only uniformly convergent on a domain ##|z-c| \leq R## for R < 1, so if you want to be rigorous I guess you could split up the integral into three regions, ##[0,c-\epsilon), [c-\epsilon,c+\epsilon], (c+\epsilon,\infty)## and do the series expansion for the first and the last, then estimate the error from neglecting the ##[c-\epsilon,c+\epsilon]## integral when taking ##\epsilon \rightarrow 0##, but my guess is that it won't cause problems.
  4. Nov 28, 2012 #3

    Ray Vickson

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    Homework Helper

    Using b instead of 1/β, Maple evaluates this integral in terms of the incomplete Gamma function:

    f = b^a*x^(a-1)/GAMMA(a)*exp(-b*x)

    J1:=int(f/(x+c),x=0..infinity) assuming a>0,b>0,c>0;
    J1 = b^a*c^(a-1)*exp(c*b)*GAMMA(1-a,c*b)

    That is,
    [tex] \int_0^{\infty} \frac{b^a}{\Gamma(a)} \frac{x^{a-1}}{x+c} e^{-bx} \, dx
    = b^a c^{a-1} e^{bc} \Gamma(1-a,bc),[/tex]
    [tex] \Gamma(u,z) = \int_z^{\infty} e^{-t} t^{u-1} \, dt \text{ if } z > 0[/tex]
    is the incomplete Gamma function. I believe it has been proven that the incomplete Gamma function is non-elementary if u is not a positive integer.
    Last edited: Nov 28, 2012
  5. Dec 4, 2012 #4
    Thank you for your help:smile:. So Maple was able to provide an answer in terms of a closed form solution (I tried in Maxima but did not succeed).
  6. Aug 19, 2014 #5
    A more general problem has been solved here, for [itex]1/(1+cx^n)[/itex] and [itex]1/(1+cx)^n[/itex], for any [itex]n[/itex]:

    You can find there the explicit formula for the probability density function, the monotonicity properties of the pdf, the first and second moments for [itex]n=1[/itex] and [itex]n=2[/itex].
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