Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

What is the distribution of difference of two Gamma Distributions ?

  1. Sep 19, 2011 #1
    What is the distribution of the difference of two gamma distributions with same scale parameter, and shape parameter of the first one is k(1+e), e -> 0 and second one is k.

    What i exactly want to know is the following.
    X~Gamma(K(1+e),\theta)
    Y~Gamma(K,\theta)
    Prob (X>Y) or P(X-Y)>0.

    While trying to integrate i am stuck at the following intermediate step.

    int (0,inf) (y)^(Ke-1) exp(-y/"\theta") Gamma(K,y/"\thata") dy.

    Please suggest any way out.
     
  2. jcsd
  3. Sep 19, 2011 #2

    Stephen Tashi

    User Avatar
    Science Advisor

    I don't know if I can solve the problem, but let me see if I can state it.

    You want to integrate the function
    [tex] f(x,y) = \frac{ x^{\epsilon}exp(-x/\theta) y^k exp(-y/\theta)}{\theta^{1 + \epsilon} \Gamma(1+\epsilon) \theta^k \Gamma(k) } [/tex]

    over the region in the first quadrant defined by X > Y

    Then you want to take the limit of the answer as [itex] \epsilon [/itex] approaches zero.
     
  4. Sep 19, 2011 #3
    Thanks for your interest.

    In the function you have written, in place of \eps it is k*(1+\eps).
    But this is the original function. I tried to solve it and went couple of rounds ahead, when i got stuck at the point i mentioned in my first post.
     
  5. Sep 19, 2011 #4

    Stephen Tashi

    User Avatar
    Science Advisor

    So the problem is to find:

    [tex] \lim_{\epsilon \rightarrow 0 } \int_0^\infty \int_y^\infty \frac{ x^{k(1+\epsilon)}exp(-x/\theta) y^k exp(-y/\theta)}{\theta^{k(1 + \epsilon)} \Gamma(k(1+\epsilon)) \theta^k \Gamma(k) } dx dy [/tex]
     
  6. Sep 19, 2011 #5
    for the integral dx, the limit is from 0 to y, instead of y to \inf [\tex]
     
  7. Sep 19, 2011 #6

    Stephen Tashi

    User Avatar
    Science Advisor

    Why would the limit for dx be 0 to y ? Do we want to compute P(X > Y) or do we want to compute P(X < Y) ?

    Intuitively, I would expect the answer to this problem to be 1/2.
     
  8. Sep 19, 2011 #7
    Yes, you are right. It will be from y to \inf.

    We are interested to show that it is greater than 1/2 by a small constant factor that is a function of \epsilon and may be K and \theta.
     
  9. Sep 20, 2011 #8

    Stephen Tashi

    User Avatar
    Science Advisor

    I don't understand how the integration that you asked about arises in solving the original problem.

    [tex] \int_0^\infty y^{ke-1} exp( - y/\theta) \Gamma(k, y/\theta) dy [/tex]

    Have I interpreted the integral correctly?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: What is the distribution of difference of two Gamma Distributions ?
  1. Gamma distribution (Replies: 1)

Loading...