$\mathbb E[Xf(X)] = \gamma \mathbb E[f(X+1)]$ for every $f: \mathbb Z^{\ge 0} \to \mathbb R \implies X$ has distribution $Po(\gamma)$

Question

If $f:= \mathbb 1_{k}$ for $k \in \mathbb Z^{>0}$, then $$k\mathbb P(X=k) = \gamma \mathbb P(X = k-1).$$

I used Characteristic Root Technique by letting $\mathbb P(X=k) = t^k$, but I do not think this can solve the problem.

How to show that $\mathbb P(X=k) = \frac{\gamma^k}{k!} e^{-\gamma}$?

Answer 1

Let $c=P(X=0)$. From the expression you wrote, you can infer by induction that $$P(X=k)=c\gamma^k/k! $$ for all integers $k \ge 0$. To find the value of $c$, sum this equation over all $k \ge 0$ and recall the Taylor series for $e^\gamma$.