# 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!

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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 $$e^{-\lambda}e^{\lambda e^{-\tau}}\quad\text{or equivalently}\quad e^{-\lambda(1-e^{-\tau})}.$$
Thank you. Actually the same argument works for Poisson parameter $c$, exponential parameter $d$. Maybe I'll change the writeup to that. Everybody mean by Poisson($\lambda$) that $\lambda$ is the mean. But there are two usages for Exp($c$)$. The$c$can be the mean, or it can be the reciprocal of the mean. Which is it in your course? – André Nicolas Jan 29 '13 at 16:33 In my course, if X~Exp(c), then E(X), or the expected value / mean equals 1/c, because the density is c*e^(-c). so I guess that c is the reciprocal of the mean. Thank you again for your answer. – Ionut Laceanu Jan 29 '13 at 21:53 You are welcome. In my revised answer, with the$\lambda$and$\tau$, we have$\lambda=a$,$\tau=1/a$, just like you observed in your first comment. The$a$and$1/a\$ do not combine in any interesting way in the formula we get. –  André Nicolas Jan 29 '13 at 22:23