Finding the value $k$ for which $p(k)$ (poisson distribution) is at a maximum

Question

Let $ \mu >$ 0 be given and $p(k)=\frac{\mu^k}{k!}e^{-\mu}$ for $k = 0, 1, . . .$ Find the value of $k$ for which $p(k)$ is at a maximum.

My teacher gave two hints:

• there are values of $\mu$ for which there are multiple values of $k$ for which $p(k)$ is at a maximum.

• look at $\frac{p(k+1)}{p(k)}$

I don't know how i could use these hints for this problem.

Answer 1

$$\frac{p(k+1)}{p(k)}=\frac{{\mu}^{k+1}}{(k+1)!}\cdot\frac{k!}{{\mu}^k}\cdot\frac{e^{-\mu}}{e^{-\mu}}=\frac{\mu}{k+1}$$ Therefore, when

1. $k+1 < \mu$, $p(k+1)>p(k)$,
2. $k+1 = \mu$, $p(k+1)=p(k)$,
3. $k+1 > \mu$, $p(k+1)<p(k)$.

If $\mu$ is an positive integer, then there are two maximums, $p(\mu-1)$ and $p(\mu)$. If $\mu <= 1$, $p(0)$ is the maximum. If $\mu > 1$ and $\mu \notin \mathbb{N}$, then the maximum would be $p(\lfloor\mu\rfloor)$.