Recent content by 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.

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

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.

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...