# If $\Bbb E[{X \choose r}]=\frac{\lambda^r}{r!}$ for all positive integer $r$, show $\Bbb E[X^r]$ is the $r$th moment of a Poisson random variable

Let $X$ be a random variable with support over positive integers. If $\Bbb E[{X \choose r}]=\frac{\lambda^r}{r!}$ for every integer $r\ge 1$, how to show that $\Bbb E[X^r]$ equals the $r$th moment of Poisson distribution for every $r$?

The condition is equivalent to $\Bbb E[X(X-1)...(X-r+1)]=\lambda^r$ for $r=1,2,...$. We can check indeed $\Bbb E X=\lambda,\Bbb E[X^2-X]=\lambda^2\Rightarrow \Bbb E[X^2]=\lambda^2+\lambda$, but then it looks very hard to prove it by simple induction as the higher moments of Poisson distribution look complicated. Do we need to use properties of Poisson distribution or some polynomial techniques?

• You can write $X^n$ in terms of $\binom Xk$ for $k\leqslant n$. Have you tried to come up with a formula? – Pedro Tamaroff Jan 21 '18 at 23:55
• @PedroTamaroff I think I currently do not know a combination formula to express $X^n$ with $X$ at the top. Could you help give a hint or reference. Thanks! – River Jan 22 '18 at 0:46

So if $Y \sim \text{Possion}(\lambda)$, then from the given conditions, you can show that $X$ and $Y$ equal in factorial moments, i.e.
$$\mathbb{E}\left[\prod_{i=1}^r(X-i+1)\right] = \lambda^r = \mathbb{E}\left[\prod_{i=1}^r(Y-i+1)\right]$$
for every integer $r \geq 1$ as the factorial moment of Poisson is easy to calculate. For $r = 1$ we have
$$\mathbb{E}[X] = \mathbb{E}[Y]$$ and for $r > 1$, expanding it you obtain $$\mathbb{E}[X^r] - \mathbb{E}[Y^r] + \sum_{i=1}^{r-1}c_{r,i}(\mathbb{E}[X^i] - \mathbb{E}[Y^i]) = 0$$ for some non-zero constants $c_{r,i}$. Then the remaining part is induction.