# a part of expected value of Poisson distribution $E(X^2)=λ^2+λ$ proof?

a part of expected value of Poisson distribution :

$E(X^2)=λ^2+λ$

What is the proof? (except using the Moment-generating function )

We know:

$Var[X]=E[X^2]-E[X]^2$

So

$E[X^2]=Var[X]+E[X]^2$

For Poisson distribution, $Var[X]=\lambda$ and $E[X]=\lambda$. Therefore:

$E[X^2]=\lambda+\lambda^2$

• Here I used the first and the second moments but not the moment generating functions. Commented Dec 6, 2013 at 22:13

You could, of course, just write this out as a sum: $$\mathbb{E}[X^2]=\sum_{k=0}^{\infty}k^2\,P(X=k)=\sum_{k=0}^{\infty}k^2\frac{e^{-\lambda}\lambda^k}{k!}=\sum_{k=1}^{\infty}k\frac{e^{-\lambda}\lambda^k}{(k-1)!}.$$ Write $k=(k-1)+1$, to get \begin{align*} \mathbb{E}[X^2]&=\sum_{k=1}^{\infty}\frac{(k-1)e^{-\lambda}\lambda^k}{(k-1)!}+\sum_{k=1}^{\infty}\frac{e^{-\lambda}\lambda^k}{(k-1)!}\\ &=\sum_{k=2}^{\infty}\frac{e^{-\lambda}\lambda^k}{(k-2)!}+\sum_{k=1}^{\infty}\frac{e^{-\lambda}\lambda^k}{(k-1)!}. \end{align*} Can you see how to finish it from here?

• result of your answer is $\lambda ^ (2k) + \lambda^k$ not $\lambda ^ 2 + \lambda$ Commented Dec 6, 2013 at 21:32
• @user106557 No... double-check your computations! Commented Dec 7, 2013 at 1:01

Another solution if we know $E[X]= \lambda$ then we can use its derivative in $\lambda$ :

$$\frac{d}{d\lambda} E[X] = \frac{d}{d\lambda} \sum_{k = 0}^{\infty} \frac{e^{-\lambda}\lambda^k}{k!}k$$

$$\frac{d}{d\lambda}\lambda = \sum_{k=0}^{\infty}(-\frac{e^{-\lambda}\lambda^k}{k!}k + \frac{e^{-\lambda}\lambda^{k-1}}{k!}k^2)$$

$$1 = -\sum_{k=0}^{\infty}\frac{e^{-\lambda}\lambda^k}{k!}k + \frac{1}{\lambda}\sum_{k=0}^{\infty} \frac{e^{-\lambda}\lambda^{k}}{k!}k^2$$

$$1 = -E[X] + \frac{1}{\lambda}E[X^2]$$

$$E[X^2]= \lambda (1 + E[X]) = \lambda^2 + \lambda$$

Just bringing this solution on the table because it seems to be a pretty simple way to do it for me.

Let

$X \sim \frac{{{\lambda ^x}{e^\lambda }}}{{x!}}\\\\E({X^2}) = \sum\limits_{x = 0}^\infty {{x^2}\frac{{{\lambda ^x}{e^{ - \lambda }}}}{{x!}}} = {e^{ - \lambda }}\sum\limits_{x = 1}^\infty {{x^2}\frac{{{\lambda ^x}}}{{x!}}} = {e^{ - \lambda }}\sum\limits_{x = 1}^\infty {x\frac{{{\lambda ^{x - 1}}\lambda }}{{(x - 1)!}}} = \lambda {e^{ - \lambda }}\sum\limits_{x = 1}^\infty {x\frac{{{\lambda ^{x - 1}}}}{{(x - 1)!}}}$

Let $\alpha = {\lambda ^x},{\rm{ }}D_\lambda ^1(\alpha ) = x{\lambda ^{x - 1}}$. Thus:

\begin{array}{l}E({X^2}) = \lambda {e^{ - \lambda }}\sum\limits_{x = 1}^\infty {\frac{{D_\lambda ^1(\alpha )}}{{(x - 1)!}}} = \lambda {e^{ - \lambda }}D_\lambda ^1\left( {\sum\limits_{x = 1}^\infty {\frac{\alpha }{{(x - 1)!}}} } \right) = \lambda {e^{ - \lambda }}D_\lambda ^1\left( {\sum\limits_{x = 1}^\infty {\frac{{{\lambda ^x}}}{{(x - 1)!}}} } \right)\ = \lambda {e^{ - \lambda }}D_\lambda ^1\left( {\lambda \sum\limits_{x = 1}^\infty {\frac{{{\lambda ^{x - 1}}}}{{(x - 1)!}}} } \right) = \lambda {e^{ - \lambda }}D_\lambda ^1\left( {\lambda {e^\lambda }} \right) = \lambda {e^{ - \lambda }}\left( {{e^\lambda } + \lambda {e^\lambda }} \right) = \lambda + {\lambda ^2}\end{array}