# Probability distribution question

Suppose that X is Poisson distributed with mean $\lambda >0$ and Y is geometrically distributed with parameter $p\in(0, 1)$.

Assume that X and Y are mutually independent. How do I show that $P(Y > X) = e^{-\lambda p}$?

• For some basic information about writing math at this site see e.g. here, here, here and here. – Kasper Apr 26 '13 at 13:37

The meaning of "geometrically distributed with parameter $p$" is not uniform among practitioners. Here what is intended is that $Y$ takes on the values $1, 2, 3,\dots$, and the probability that $Y=k$ is $(1-p)^{k-1}p$. So $Y$ is the total number of independent trials until the first success, where the probability of success on any trial is $p$.

In particular, the probability that $Y\gt k$ is the probability of $k$ failures in a row, which is $(1-p)^k$.

We can have $Y\gt X$ is various ways:

Case $0$: We could have $X=0$ and $Y\gt 0$. The probability that $X=0$ is $e^{-\lambda}$, and the probability that $Y\gt 0$ is $1$. So the probability that $X=0$ and $Y\gt 0$ is $e^{-\lambda}$.

Case $1$: We could have $X=1$ and $Y\gt 1$. The probability that $X=1$ is $e^{-\lambda}\frac{\lambda}{1!}$, and the probability that $Y\gt 1$ is $1-p$. So by independence the probability that $X=1$ and $Y\gt 1$ is $e^{-\lambda}\frac{\lambda}{1!}(1-p)$.

Case $2$: We could have $X=2$ and $Y\gt 2$. The probability that $X=2$ is $e^{-\lambda}\frac{\lambda^2}{2!}$, and the probability that $Y\gt 2$ is $(1-p)^2$, so the probability that $X=2$ and $Y\gt 2$ is $e^{-\lambda}\frac{\lambda^2}{2!}(1-p)^2$.

Continue. In general we have:

Case $k$: We could have $X=k$ and $Y\gt k$. The probability that $X=k$ is $e^{-\lambda}\frac{\lambda^k}{k!}$, and the probability that $Y\gt k$ is $(1-p)^k$, so the probability that $X=k$ and $Y\gt k$ is $e^{-\lambda}\frac{\lambda^k}{k!}(1-p)^k$.

Add up, $k=0$ to $\infty$. We get that $$\Pr(Y\gt X)=e^{-\lambda}\sum_{k=0}^\infty \frac{\lambda^k(1-p)^k}{k!}.$$
We recognize the infinite sum as the power series expansion of $e^w$, where $w=\lambda(1-p)$. So the infinite sum is $e^{\lambda(1-p)}$. Multiply by $e^{-\lambda}$, and we get the desired $e^{-\lambda p}$.