Question towards Poisson random variable

Question

Show that if $X$ is a Poisson random variable with mean 1, then we have $$E(X^n)=E((X+1)^{n-1}).$$ Use this result to compute $Var(3X+1)$ and $E(X^4)$.

I don't know how to show $E(X^n)=E((X+1)^{n-1})$, but I am able to compute $\rm{Var}(3X+1)$ and $E(X^4)$. I really tried my best to find resources but I really cannot compute with this. I would appreciate if someone could help. If there is further notes/videos that can help me to understand the concept of this kind of questions, I would definitely try my best to read through them before I ask the question.

Answer 1

You can use the moment generating function $M_X(t)=E[e^{tX}]=e^{\lambda(e^t-1)}$, of the Poisson distribution, then use the fact that $$E[X^n]= \frac{d^n M_X}{dt^n}(0)$$ Notice $M_{X+1}(t)=E[e^{t(X+1)}]=e^t M_X(t)$. So we have $$\frac{d^{n-1} M_{X+1}}{dt^{n-1}}=\sum_{k=0}^{n-1} {{n-1} \choose k} e^t \frac{d^k M_X}{dt^k }$$ Also note that $$\frac{d^n M_X}{dt^n}=\lambda e^t\Big(\sum_{k=0}^{n-1} {n-1\choose k} \frac{d^k M_X}{dt^k}\Big)$$ Evaluating at zero and using the fact that $M_X(0)=1$ you get the general result: $$E[X^n]=\frac{d^n M_X}{dt^n}(0)=\lambda \sum_{k=0}^{n-1} {n-1\choose k} \frac{d^k M_X}{dt^k}=\lambda\frac{d^{n-1} M_{X+1}}{dt^{n-1}}(0)=\lambda E[(X+1)^{n-1}]$$ Now use that $\lambda=1$ and you get $$E[X^n]=E[(X+1)^{n-1}]$$

Answer 2

$$\mathbb{E}(x^n)=\sum\limits_{i=0}^\infty\frac{( i^n \cdot e^{-\lambda} \cdot \lambda ^i)}{i!} $$ = $\displaystyle\sum\limits_{i=1}^\infty\frac{( i^{(n-1)}\cdot e^{-\lambda} \cdot \lambda ^{(i-1)})}{(i-1)!}$ = $\displaystyle\sum\limits_{i=0}^\infty\frac{ (e^- \lambda\cdot \lambda ^i \cdot(i+1)^{(n-1)})}{i!}$ = $\mathbb{E}\left[(x+1)^{(n-1)}\right]$