Deriving the variance of the Poisson distribution - making an error somewhere?

Question

Having issues deriving the $E[X^2]$ term.

1. Can someone explain lines 5 to 6 here for me? I don't understand how they get $\lambda^2$ in the first term. Seemingly then $n= \lambda$? Which doesn't make sense to me. I'm not sure if they made an error.

Deriving E[X^2]

Lines I don't get:

2. I've tried deriving $E[X^2]$ myself but am making an error somewhere as I lose the $\lambda$ term, maybe in the change of the sum?:

$$\begin{align}E[X^2] &= \sum_{n \in X} n^2P(X=n)\\ &=\sum_{n=0}^\infty n^2 \frac{\lambda^n}{n!}e^{-\lambda} \\ &= \sum_{n=0}^\infty n^2 \frac{\lambda^n}{n!} \bigg(\frac{-\lambda^n}{n!}\bigg) \\ &= \sum_{n=1}^\infty n^2 \frac{\lambda^n}{n(n-1)!} \bigg(\frac{-\lambda^n}{n(n-1)!}\bigg) \\ &= \sum_{n=1}^\infty \frac{\lambda^n}{(n-1)!} \bigg(\frac{-\lambda^n}{(n-1)!}\bigg) \\ &= \sum_{n=0}^\infty \frac{\lambda^{n+1}}{n!} \bigg(\frac{-\lambda^{n+1}}{n!}\bigg) \\ &= \lambda^2 \sum_{n=0}^\infty \frac{ \lambda^{n}}{n!} \bigg(\frac{-\lambda^{n}}{n!}\bigg)\\ &= \lambda^2 e^\lambda e^{-\lambda}\\ & = \lambda^2 \\ \end{align}$$

Using:

$$ e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$$

Answer 1

1. The $\lambda^2$ come from the following \begin{align} \sum_{n=0}^\infty n\frac{\lambda^{n+1}}{n!}e^{-\lambda}&=\sum_{n=1}^\infty n\frac{\lambda^{n+1}}{n!}e^{-\lambda}&\mbox{ the term for }n=0\mbox{ vanishes}\\ &=\sum_{n=1}^\infty \frac{\lambda^{n+1}}{(n-1)!}e^{-\lambda}&\\ &=\sum_{\ell=0}^\infty \frac{\lambda^{\ell+2}}{\ell!}e^{-\lambda}&\mbox{doing the substitution }\ell=n-1 \end{align} then we conclude using $\sum_{\ell=0}^\infty \frac{\lambda^{\ell+2}}{\ell!}=\lambda^2\sum_{\ell=0}^\infty \frac{\lambda^{\ell}}{\ell!}=\lambda^2 e^{\lambda}$.

2. In your attempt, you seem to use (at the third equal sign) that if $b=\sum_{n\geqslant 0}b_n$, then $\sum_{n\geqslant 0}a_nb=\sum_{n\geqslant 0}a_nb_n$ which is not true in general.