Moment of order $2$ of $X\sim Poisson(\lambda)$

Question

I'm stuck calculating $E[X^2]$, where $X$ is a random variable with Poisson distribution.

Here's what I have so far:

$$E[X^2]=\sum_{k=1}^{\infty}k^2\cdot\frac{\lambda^{k}}{(e^{\lambda}-1)\cdot k!}=\sum_{k=1}^{\infty}k\cdot\frac{\lambda^{k}}{(e^{\lambda}-1)\cdot (k-1)!}=\frac{1}{(e^{\lambda}-1)}\sum_{k=1}^{\infty}k\cdot\frac{\lambda^{k}}{ (k-1)!}=\frac{\lambda}{(e^{\lambda}-1)}\sum_{k=1}^{\infty}k\cdot\frac{\lambda^{k-1}}{ (k-1)!}$$

This $k$ is bothering me, I can't sum this series. Wolfram said that

$$\sum_{k=1}^{\infty}k\cdot\frac{\lambda^{k-1}}{ (k-1)!}=(\lambda+1)e^{\lambda},$$

but I don't know what Wolfram did to arrive there.

I'd appreciate any help, thanks for your time.

Answer 1

Another way without derivatives is to express $k$ as $(k-1)+1$ and split sum into two sums. In the first sum initial summand is zero when $k=1$, therefore we can begin summation from $k=2$: $$\require{cancel}\sum_{k=1}^{\infty}k\cdot\frac{\lambda^{k-1}}{ (k-1)!}=\sum_{k=1}^{\infty}(k-1+1)\cdot\frac{\lambda^{k-1}}{ (k-1)!}= \sum_{k=\color{red}{\mathbf{2}}}^{\infty}\color{yellow}{\cancel{\,\color{black}{(k-1)}}}\cdot\frac{\lambda^{k-1}}{\color{yellow}{\cancel{\color{black}{(k-1)!}}}}+\sum_{k=1}^{\infty}\frac{\lambda^{k-1}}{ (k-1)!}=$$ Take out one extra $\lambda$ and place it in front of the first sum. Then use exponent expansions: $$=\lambda\sum_{k=2}^{\infty}\frac{\lambda^{k-2}}{ (k-2)!}+\sum_{k=1}^{\infty}\frac{\lambda^{k-1}}{ (k-1)!}=\lambda e^{-\lambda}+e^{-\lambda}=(\lambda+1)e^{\lambda}$$

And $$E[X^2]=\sum_{k=1}^{\infty}k^2\cdot\frac{\lambda^{k-1}}{k!}e^{-\lambda}$$

The term $e^\lambda-1$ could not appear in denominator. Should it mean $(e^\lambda)^{-1}$?

Answer 2

If it had the terms $\lambda^k/k!$ it would be the expansion of an exponential. Use its' derivative.

Answer 3

First you can make an index shift: $j=k-1\Rightarrow k=j+1$

$$\sum_{j=0}^{\infty} (j+1)\cdot \frac{\lambda ^j}{j!}=\sum_{j=0}^{\infty} j\cdot \frac{\lambda ^j}{j!}+\sum_{j=0}^{\infty} \frac{\lambda ^j}{j!}$$

Using the series expansion of the exponential function the latter summand is

$\sum_{j=0}^{\infty} \frac{\lambda ^j}{j!}=\large{e^{\lambda}}$.

It remains to calculate $\sum_{j=0}^{\infty} j\cdot \frac{\lambda ^j}{j!}$. This is equal to $$\sum_{\color{blue}{j=1}}^{\infty} j\cdot \frac{\lambda ^j}{j!},$$ since $0\cdot \frac{\lambda ^0}{0!}=0$

$$\lambda \sum_{j=1}^{\infty} j\cdot \frac{\lambda ^{j-1}}{j!}$$

Cancelling out the factor $j$

$$\lambda \sum_{j=1}^{\infty} \frac{\lambda ^{j-1}}{(j-1)!}$$

Next index shift $m=j-1\Rightarrow j=m+1$

$$\lambda \sum_{m=0}^{\infty} \frac{\lambda ^{m}}{m!}=\lambda\cdot e^{\lambda}$$

In total we get $$\sum_{k=1}^{\infty}k\cdot\frac{\lambda^{k-1}}{ (k-1)!}=e^{\lambda}+\lambda\cdot e^{\lambda}=e^{\lambda}\cdot \left(1+\lambda \right)$$