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a part of expected value of Poisson distribution :

$E(X^2)=λ^2+λ$

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

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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:

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

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  • $\begingroup$ Here I used the first and the second moments but not the moment generating functions. $\endgroup$ – hhsaffar Dec 6 '13 at 22:13
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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?

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  • $\begingroup$ result of your answer is $\lambda ^ (2k) + \lambda^k$ not $\lambda ^ 2 + \lambda$ $\endgroup$ – user106557 Dec 6 '13 at 21:32
  • $\begingroup$ @user106557 No... double-check your computations! $\endgroup$ – Nick Peterson Dec 7 '13 at 1:01
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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.

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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}

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