What is the distribution of difference of two Gamma Distributions ?

[ashish1789kgp]

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.

 

[Stephen Tashi]

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
f(x,y) = \frac{ x^{\epsilon}exp(-x/\theta) y^k exp(-y/\theta)}{\theta^{1 + \epsilon} \Gamma(1+\epsilon) \theta^k \Gamma(k) }

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

Then you want to take the limit of the answer as \epsilon approaches zero.

 

[ashish1789kgp]

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.

 

[Stephen Tashi]

So the problem is to find:

\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

 

[ashish1789kgp]

for the integral dx, the limit is from 0 to y, instead of y to \inf [\tex]

 

[Stephen Tashi]

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.

 

[ashish1789kgp]

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.

 

[Stephen Tashi]

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

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

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

Have I interpreted the integral correctly?