Show that $P(X=r)=e^{-2}\cdot\frac{2^r}{r!}$, where $r=0,1,2,3,\cdots$ is a probability distribution function.

Question

Hint: you may use the Maclaurin series for $e^x$

Answer 1

For the given function to be a valid probability mass function, then it must satisfy two criteria: 1) it is nonnegative and 2) the total probability sums to 1. It is not hard to see that, when $r$ is nonnegative, $2^r$, $r!$, and $e^{-2}$ are all positive, so the entire expression is positive, satisfying the first condition. To show that the series sums to $1$, recall that $e^x = \sum_{r\geq 0}\frac{x^r}{r!}$. Plugging in $2$ gives $e^{2} = \sum_{r\geq 0}\frac{2^r}{r!}$. Thus $\sum_{r\geq 0}e^{-2}\frac{2^r}{r!} = e^{-2}\sum_{r\geq 0}\frac{2^r}{r!} = e^{-2}e^{2} = 1$, which satisfies the second condition.