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

For the context: I am trying to calculate $E(X^2)$, where X is a poisson distributed random variable.

All my calculations lead to a dead end. Is there a trick to process the $k^2$? The only thing I see worth doing is pulling out $1/e^\lambda$.

Edit: Considering @Srivatsan's hint I got:

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

And here we are! Thank you very much, @Srivatsan!

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Related post: math.stackexchange.com/questions/44113 (but not quite a duplicate). –  Srivatsan Dec 12 '11 at 0:17

2 Answers 2

up vote 5 down vote accepted

HINT: Write $k^2 = k(k-1) + k$, and split the sum into two. [Update: The OP has added the complete solution to the post.]


If you are interested in a general $d^{th}$ moment of this distribution, then we are faced with an expression of the form $$ \mathrm e^{-\lambda} \sum_{k=0}^{\infty} k^d \frac{\lambda^k}{k!}. $$ To proceed, we first write $x^d$ as a linear combination of the $d$ falling factorial polynomials (also called Pochhammer symbols): $$ \begin{align*} (x)_1 &= x \\ (x)_2 &= x(x-1) \\ (x)_3 &= x(x-1)(x-2) \\ &\vdots \\ (x)_d &= x(x-1)(x-2) \cdots (x-d+1) \end{align*} $$ [I am following the notation used in the wikipedia article.] After this step, the rest of the manipulations is very similar to the $d=2$ case that the OP showed in the question.

The falling factorial polynomials appear in multiple contexts. First, the Pochhammer symbol $(x)_i$ is obviously related to the binomial coefficient $\binom{x}{i}$. Further, as seen above, these polynomials are very useful in manipulating summations. See also this answer of robjohn.

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Thanks for the hint. I'll try this out first. Then I'll come back if I need more help! –  Aufwind Dec 12 '11 at 0:06
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Note: Sometimes people employ the reverse of wiki's notation. In other words, $(x)_n$ is taken to be $\prod\limits_{i=0}^{n-1}(x+i)$, while $x^{(n)}=n!\binom{x}{n}$. Be careful. :) –  J. M. Dec 12 '11 at 3:14
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As a matter of fact, the polynomials these sums generate are called Bell polynomials. The formula is essentially the generalization of the usual Dobinski formula for Bell numbers, since $\mathrm{B}_k(1)=\mathrm{B}_k$. –  J. M. Dec 12 '11 at 3:29
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In fact, the $d$th moment of the Poisson distribution is $\sum_{k=0}^n \lambda^k \left\{ n \atop k\right\}$, where $\left\{ n \atop k\right\}$ is a Stirling number of the second kind. Thus in the special case $\lambda = 1$ we get that the $d$th moment is the $d$th Bell number. See, for example, this answer to "Bell numbers and moments of the Poisson distribution" and the Wikipedia article on the factorial moment. –  Mike Spivey Dec 12 '11 at 3:36
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@all: You guys are awesome. I had only one simple question and I got so much worthful and interesting answers. Thanks! I have never seen this phenomenon more often as I have seen it here on math.stackexchange. Maybe it is a mathematician-related thing... :-) –  Aufwind Dec 12 '11 at 20:40

The moments of the Poisson distribution, as functions of the first moment, are given by the Touchard polynomials, also called "exponential polynomials". Their coefficients are related to the enumeration of set partitions. See Dobinski's formula.

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Interesting; until this answer I hadn't realized that Bell polynomials and Touchard polynomials are precisely the same thing... –  J. M. Dec 14 '11 at 2:10

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