I am working on a homework problem which is related to Poisson processes but it all boils down to finding the probability that a sum of $2$ exponential random variables is greater than a third exponential r.v., all three of them having the same mean and being independent.
I wanted to bash it out, but it seems I am making a mistake somewhere and would be grateful if you could help me.
Let $X,Y,Z$ be $Exp(\lambda)$ independent random variables.
$P(X+Y>Z) = \int _{0}^{\infty}\int _{0}^{\infty}\int _{z-y}^{\infty}\lambda e^{-\lambda x} \lambda e^{-\lambda y} \lambda e^{-\lambda z} dxdydz$ if I am not mistaken.
The problem is that this integral does not converge. Where is my mistake?
Thanks
Edit In the context of the problem we probably have to condition on $Y < Z$, say
As was pointed out by Did, it is easier to use \begin{align} \int_0^\infty\int_0^\infty\int_0^{x+y}\lambda e^{-\lambda z}\lambda e^{-\lambda x}\lambda e^{-\lambda y}\,dz\,dy\,dx &=\int_0^\infty\int_0^\infty\lambda e^{-\lambda x}\lambda e^{-\lambda y} (1-e^{-\lambda(x+y)})\,dx\,dy\\ &=(1-1/4)\int_0^\infty\int_0^\infty\lambda e^{-\lambda x}\lambda e^{-\lambda y}\,dx\,dy\\&=3/4. \end{align}
Your approach does't work because you are integrating from $z-y$, but $z-y$ can be negative and you don't take this into account.