Skewness of poisson distribution I want to calculate the skewness of a Poisson distributed random variable but I can't find my mistake.
$$\begin{align}
\mathbb{E}(X^3)&= (-1)^3 \frac{d^3}{d\alpha^3} \mathrm{e}^{\lambda(\mathrm{e}^{-\alpha}-1)}\bigg|_{\alpha=0}\\
&= - \frac{d^2}{d\alpha^2} \mathrm{e}^{\lambda(e^{-\alpha}-1)} \cdot (\lambda \mathrm{e^{-\alpha}})\bigg|_{\alpha=0} \\
&= - \frac{d}{d\alpha} \mathrm{e}^{\lambda(e^{-\alpha}-1)} \cdot (\lambda \mathrm{e^{-\alpha}})^2 -  \mathrm{e}^{\lambda(e^{-\alpha}-1)} (\lambda \mathrm{e^{-\alpha}})\bigg|_{\alpha=0}\\
&= - \frac{d}{d\alpha} \mathrm{e}^{\lambda(e^{-\alpha}-1)} \left[(\lambda \mathrm{e^{-\alpha}})^2-(\lambda \mathrm{e^{-\alpha}})\right]\bigg|_{\alpha=0} \\
&= -\bigg[\mathrm{e}^{\lambda(e^{-\alpha}-1)} (\lambda \mathrm{e^{-\alpha}}) \big((\lambda \mathrm{e^{-\alpha}})^2-(\lambda \mathrm{e^{-\alpha}})\big)+\mathrm{e}^{\lambda(e^{-\alpha}-1)} \big(-2\lambda^2\mathrm{e}^{-2\alpha}+\lambda\mathrm{e}^{-\alpha}\big)\bigg]\bigg|_{\alpha=0}\\
&= -\left[\lambda(\lambda^2-\lambda)-2\lambda^2+\lambda\right] \\
&= -\lambda^3+3\lambda^2-\lambda\\
\end{align}$$
Furthermore one knows that:
$$\mathbb{E}(X^2)=\lambda^2+\lambda, \quad \mathbb{E}(X)=\lambda, \quad \sigma=\sqrt{\lambda}$$
So one get:
$$\begin{align} \text{Skew}(X)&=\frac{\mathbb{E}(X^3)-3\mathbb{E}(X)\mathbb{E}(X^2)+2\mathbb{E}^3}{\sigma^3}\\
&= \frac{-\lambda^3+3\lambda^2-\lambda-3\lambda^3-3\lambda^2+2\lambda^2}{\lambda^{\frac{3}{2}}}
\end{align}$$
Thanks for helping!
 A: Alternatively, one may observe that
$$
\begin{align}
&\sum_{n=3}^\infty n(n-1)(n-2) \cdot \frac{\lambda^n}{n!}\cdot e^{-\lambda}=\lambda^3\sum_{n=3}^\infty \frac{\lambda^{n-3}}{(n-3)!}\cdot e^{-\lambda}=\lambda^3\sum_{k=0}^\infty \frac{\lambda^{k}}{k!}\cdot e^{-\lambda}=\lambda^3 \tag1
\\
&\sum_{n=2}^\infty n(n-1)\cdot \frac{\lambda^n}{n!}\cdot e^{-\lambda}=\lambda^2\sum_{n=2}^\infty \frac{\lambda^{n-2}}{(n-2)!}\cdot e^{-\lambda}=\lambda^2\sum_{k=0}^\infty \frac{\lambda^{k}}{k!}\cdot e^{-\lambda}=\lambda^2 \tag2
\\
&\sum_{n=1}^\infty n\cdot \frac{\lambda^n}{n!}\cdot e^{-\lambda}=\lambda\sum_{n=1}^\infty \frac{\lambda^{n-1}}{(n-1)!}\cdot e^{-\lambda}=\lambda\sum_{k=0}^\infty \frac{\lambda^{k}}{k!}\cdot e^{-\lambda}=\lambda \tag3
\end{align}
$$ 
Then by writing $$n^3=n(n-1)(n-2)+3n(n-1)+n$$
and combining it with $(1)$, $(2)$, $(3)$, one gets$$
\begin{align}
\mathbb{E}(X^3)=&\sum_{n=0}^\infty n^3 \cdot \frac{\lambda^n}{n!}\cdot e^{-\lambda}
\\\mathbb{E}(X^3)=&\sum_{n=0}^\infty \left[n(n-1)(n-2)+3n(n-1)+n\right]\cdot \frac{\lambda^n}{n!}\cdot e^{-\lambda}
\\\mathbb{E}(X^3)=&\lambda^3+3\lambda^2+\lambda
\end{align}
$$ giving, with $\mathbb{E}(X^2)=\lambda^2+\lambda, \, \mathbb{E}(X)=\lambda, \, \sigma=\sqrt{\lambda}$,
$$\begin{align} \text{Skew}(X)&=\frac{\mathbb{E}(X^3)-3\mathbb{E}(X)\mathbb{E}(X^2)+2\mathbb{E}^3(X)}{\sigma^3}\\
&= \frac{\lambda^3+3\lambda^2+\lambda-3\lambda(\lambda^2+\lambda)+2\lambda^3}{\lambda^{\frac{3}{2}}}
\\&= \frac{\lambda}{\lambda^{\frac{3}{2}}}
\end{align}$$ that is 

$$
\text{Skew}(X)=\frac{1}{\sqrt{\lambda}}
$$ 

as expected.
A: It is worth noting that $$P_X(t) = \operatorname{E}[t^X] = \sum_{x=0}^\infty t^x e^{-\lambda} \frac{\lambda^x}{x!} = e^{\lambda(t-1)} \sum_{x=0}^\infty e^{-\lambda t} \frac{(\lambda t)^x}{x!} = e^{\lambda(t-1)}$$ is the  probability generating function of $X$.  Then we have $$\operatorname{E}[X(X-1)\cdots(X-k+1)] = P^{(k)}_X(1).$$  By Taylor's theorem, we have $$\sum_{k=0}^\infty \frac{P^{(k)}_X(1)}{k!} (t-1)^k = \sum_{k=0}^\infty \frac{(\lambda(t-1))^k}{k!},$$ thus $$\operatorname{E}[X(X-1)\cdots(X-k+1)] = \lambda^k,$$ from which we trivially obtain $$\operatorname{E}[X] = \lambda, \quad \operatorname{E}[X(X-1)] = \lambda^2, \operatorname{E}[X(X-1)(X-2)] = \lambda^3.$$  The use of the PGF in this case is clearly more tractable than the MGF.  We can backsolve to get $$\operatorname{E}[X^2] = \lambda^2 + \lambda, \quad \operatorname{E}[X^3] = \lambda^3 + 3\lambda^2 + \lambda,$$ hence $$\operatorname{Skew}[X] = \lambda^{-3/2} \left((\lambda^3 + 3\lambda^2 + \lambda) - 3(\lambda^2 + \lambda)\lambda + 3\lambda (\lambda^2) - \lambda^3\right) = \lambda^{-1/2}.$$
