Skewness of a random variable that is Poisson distributed Let $X$ be a discrete random variable with $\sum_{x\in  \Omega(X)}|x|^3P[X=x]<\infty$ and $Std(X)>0$.
Then the skewness of $X$ is defined by $$\eta(X)=E\left [\left (\frac{X-E[X]}{Std(X)}\right )^3\right ]$$
For $a,b\in n\mathbb{R}$ with $a\neq 0$ we have that $$\eta(aX+b)=\begin{cases}\eta(X), & a>0\\ -\eta(X), & -a<0\end{cases}$$
We have that $$\eta(X)=\frac{E[X^3]-3E[X]E[X^2]+2(E[X])^3}{(Std(X))^3} \ \ \ \ \ (\star)$$
Calculate the skewness of a random variable that is Poisson distributed with the parameter $\lambda> 0$.
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From $(\star)$ we have that $$\eta(X)=\frac{E[X^3]-3E[X]E[X^2]+2(E[X])^3}{(Std(X))^3}$$
By the Expectation of Poisson Distribution we have that $E(X)=\lambda$.
By the Variance of Poisson Distribution: $Var(X)=\lambda \Rightarrow E[X^2]-(E[X])^2=\lambda\Rightarrow E[X^2]-\lambda^2=\lambda\Rightarrow E[X^2]=\lambda+\lambda^2$.
Then $Std(X)=\sqrt{Var(X)}=\sqrt{\lambda}$.
So far we have $$\eta(X)=\frac{E[X^3]-3E[X]E[X^2]+2(E[X])^3}{(Std(X))^3}=\frac{E[X^3]-3\cdot \lambda \cdot \left (\lambda+\lambda^2\right )+2\lambda^3}{\sqrt{\lambda}^3}=\frac{E[X^3]-3\lambda^2-3\lambda^3+2\lambda^3}{\sqrt{\lambda}^3}=\frac{E[X^3]-3\lambda^2-\lambda^3}{\sqrt{\lambda}^3}$$ How can we calculate $E[X^3]$ ?
 A: Use properties of the exponential generating function:
$$\begin{align}
M_X(t) &= \operatorname{E}[e^{tX}] \\
&= \sum_{x=0}^\infty e^{tx} e^{-\lambda} \frac{\lambda^x}{x!} \\
&= \sum_{x=0}^\infty e^{-\lambda} \frac{(\lambda e^t)^x}{x!} \\
&= e^{\lambda (e^t - 1)} \sum_{x=0}^\infty e^{-\lambda e^t} \frac{(\lambda e^t)^x}{x!} \\
&= e^{\lambda (e^t - 1)}.
\end{align}$$
Now since $$\operatorname{E}[X^k] = \left[\frac{d^k M_X}{dt^k}\right]_{t=0}$$
we obtain the first three moments through differentiation:
$$M_X'(t) = M_X(t) \frac{d}{dt}[\lambda (e^t - 1)] = \lambda e^t M_X(t),$$ using the fact that $\frac{d}{dt}[e^{f(t)}] = f'(t) e^{f(t)}$.  Then by the product rule,
$$M_X''(t) = \lambda \left(\frac{d}{dt}[e^t] M_X(t) + e^t M_X'(t)\right) = \lambda \left(e^t + \lambda e^{2t}\right) M_X(t),$$ where we have substituted the result for the first derivative.  Next,
$$M_X'''(t) = \lambda \left(\frac{d}{dt}[e^t + \lambda e^{2t}] M_X(t) + (e^t + \lambda e^{2t}) M_X'(t)\right) \\ = \lambda (e^t + 2\lambda e^{2t} +  \lambda (e^{2t} + \lambda e^{3t})) M_X(t) \\
= \lambda(e^t + 3\lambda e^{2t} + \lambda^2 e^{3t}) M_X(t).$$
Now evaluating each of these at $t = 0$ yields the desired moments:
$$\operatorname{E}[X] = M_X'(0) = \lambda \\
\operatorname{E}[X^2] = M_X''(0) = \lambda(1+\lambda) \\
\operatorname{E}[X^3] = M_X'''(0) = \lambda(1 + 3\lambda + \lambda^2).
$$
The rest is straightforward.
A: \begin{align}
\operatorname E\big(X(X-1)(X-2)\big) = {} & \sum_{x=0}^\infty x(x-1)(x-2) \cdot \frac{\lambda^x e^{-\lambda}}{x!} \\[10pt]
= {} & \sum_{x=3}^\infty x(x-1)(x-2) \cdot \frac{\lambda^x e^{-\lambda}}{x!} \\ & \text{since the terms with $x=1,2,3$ vanish} \\[10pt]
= {} & \sum_{y=0}^\infty (y+3)(y+2)(y+1)\cdot \frac{\lambda^{y+3}e^{-\lambda}}{(y+3)!} \\
& \text{where $y=x-3,$ so $x=y+3$} \\[10pt]
= {} & \lambda^3 \sum_{y=0}^\infty \frac{\lambda^y e^{-\lambda}}{y!} \\[10pt]
= {} & \lambda^3.
\end{align}
If you know $\lambda^3 = \operatorname E(X(X-1)(X-2)) = \operatorname E(X^3) - 3\operatorname E(X^2) + {}2\operatorname E(X)$ and you know $\operatorname E(X^2)$ and $\operatorname E(X),$ then you can find $\operatorname E(X^3).$
It will follow that $\operatorname E\big((X-\lambda)^3\big) = \lambda.$
