what is P (X < Y) when X~Poiss(a) and Y~Exp(1/a) with X, Y two independent random variables assume $X,\ Y$ are two independent random variables 
$X \sim \mathrm{Poisson}(a)$
and 
$Y \sim \mathrm{Exponential}(\frac1{a})$
find out $P(X < Y)$
Thank you!
 A: We solve a marginally more general problem. Let $X$ be Poisson with parameter $\lambda$. Let $Y$ be exponential with parameter $\tau$. By this we mean that $Y$ has density function $\tau e^{-\tau y}$. Given that $X$ and $Y$ are independent, we find the probability that $X\lt Y$. 
Given that $X=k$, the probability that $Y\gt X$ is $e^{-k \tau}$. This is because in general, $\Pr(Y\gt y)=e^{-\tau y}$. But the probability that $X=k$ is $e^{-\lambda}\frac{\lambda^k}{k!}$. 
Thus the probability that $X=k$ and $Y\gt X$ is equal to $e^{-\lambda}\frac{\lambda^k}{k!}e^{-k\tau}$. For the probability that $X\lt Y$,  add up  over all $k$. Our required probability is therefore equal to
$$\sum_{k=0}^\infty e^{-\lambda}\frac{\lambda^k}{k!}e^{-k\tau}.$$
This is a correct answer, but it can be greatly simplified. Note that 
$$\lambda^ke^{-k\tau}=(\lambda e^{-\tau})^k.$$
So our answer is 
$$e^{-\lambda}\sum_{k=0}^\infty \frac{b^k}{k!},$$
where $b=\lambda e^{-\tau}$. But we recognize the sum as just $e^b$. So the required probability can be written as 
$$e^{-\lambda}e^{\lambda e^{-\tau}}\quad\text{or equivalently}\quad 
e^{-\lambda(1-e^{-\tau})}.$$
