Suppose $N$ is a discrete random variable defined by Poisson distribution of parameter $\lambda$.
Suppose $g$ is a real function. We have proved that when it exists, the expectation
$$ \mathbb{E}(N\cdot g(N))=\lambda\cdot \mathbb{E}(g(N+1)) $$
But I wonder if its converse makes sense :
Let $T$ be a random variable into $\mathbb{N}$. If there exists $\lambda$, for all real function $g$ that makes following expectations exist, we have: $$ \mathbb{E}(T\cdot g(T))=\lambda\cdot \mathbb{E}(g(T+1)) $$ can we say that $T$ is a Poisson distribution?
Thanks a lot~
You can take a set of functions $g_k(x)=\mathbf 1_{\{x=k\}}$ and conclude $$ \mathsf E(T\cdot g_k(T)) = \mathsf E(T\cdot \mathbf 1_{\{T=k\}})=\color{red}{ kp_k} = \lambda\cdot\mathbb E(g(T+1))=\lambda\cdot\mathbb E( \mathbf 1_{\{T+1=k\}}) = \color{red}{\lambda\cdot p_{k-1}}, $$ where $p_k=\mathbb P(T=k).$
Let $p_0=a\geq 0$. Then $p_1=\lambda a$. From $2p_2=\lambda p_1$ get $p_2=\frac{\lambda^2}{2}a$. And so on. Finally, $$ p_k=\dfrac{\lambda^k}{k!}a, \quad\sum_{k=0}^\infty p_k=\sum_{k=0}^\infty \dfrac{\lambda^k}{k!}a = ae^{\lambda}=1, \quad a=e^{-\lambda}.$$