1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
    [/tex]


    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

    Mute

    User Avatar
    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

    User Avatar
    Science Advisor
    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);
    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]
    where
    [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]:
    http://www.pnas.org/content/107/51/22096.full.pdf+html?with-ds=yes

    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].
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Expected value of g(x)=1/(x+c) under gamma distribution
Loading...