Find the mode of a Poisson random variable

Question

If $X$ is a Poisson random variable with parameter $λ$, what value of $k \in \{0, 1, 2, . . .\}$ maximizes $P(X = k)$?

First, $$f(k) = P(X = k) = e^{-λ}\frac{λ^k}{k!}$$

Thus, attempting to get a ratio:

$$\frac{f(t+1)}{f(t)} = \frac{e^{-λ}\frac{λ^{t+1}}{t+1!}}{e^{-λ}\frac{λ^t}{t!}}$$

Simplifying this gives the result:

$$λ \cdot \frac{1}{t+1}$$

However, I am not too sure where to go from here. I understand I can try and plug in values for t. I note that for values greater than one, the entire expression starts to decrease. Thus, it seems that the ratio is highest when t is equivalent to 1. Aside from that, I cannot tie a connection between this reasoning and what the problem wants.

Answer 1

Following your lead, consider the ratio $$R(t) := \frac{f(t+1)}{f(t)} = \frac{\lambda}{t+1}$$ Note that $R(t)$ is strictly decreasing in $t$.

If $R(t) > 1$, then $f(t+1) > f(t)$ and thus the maximiser $k^* > t$. Similarly, if $R(t)<1$, then $k^* \le t$. In other words, the maximiser $k^*$ is found at the value $t$ where $R(t) - 1$ changes sign from positive to non-negative. (This includes the special case $R(t) = 1$.)