Why does $P(X=E(X)) = P(X=E(X)-1)$ in the Poisson distribution?

Question

When a discrete random variable follows the Poisson distribution and if $E(X)$ is a natural number, $E(X)$ has the greatest probability, but why does $E(X)-1$ have the same probability?

Answer 1

Fleshing out the first part of Brian Moehring's comment: If $X$ is Poisson with mean $E[X]=\lambda$ where $\lambda$ is an integer, then $P(X=\lambda) = e^{-\lambda} \frac{\lambda^\lambda}{\lambda!}$ while $P(X=\lambda-1) = e^{-\lambda} \frac{\lambda^{\lambda - 1}}{(\lambda-1)!}$.

Answer 2

A comment has made clear the problem is not understanding factorials well. Note that$$(k+1)!=\underbrace{1\cdot2\cdot\dots\cdot k}_{k!}\cdot(k+1)=k!(k+1),$$so$$\frac{P(X=k+1)}{P(X=k)}=\frac{\frac{\lambda^{k+1}}{(k+1)!}e^{-\lambda}}{\frac{\lambda^k}{k!}e^{-\lambda}}=\lambda\frac{k!}{(k+1)!}=\frac{\lambda}{k+1}.$$At first, the ratio of consecutive probabilities is high, but it quickly diminishes. (That makes sense: the probabilities haev to shrink eventually for infinitely many of them to sum to $1$, but we expect very small $k$ to be improbable if $\lambda$ is very large, so probabilities have to rise at first.) In particular:

• $P(X=k+1)>P(X=k)$ if $k<\lambda-1$;
• $P(X=k+1)=P(X=k)$ if $k=\lambda-1$; and
• $P(X=k+1)<P(X=k)$ if $k>\lambda-1$.