If $X$ is a Poisson distribution with mean $\lambda$ how is $X^2$ distributed? Any explanation would be very appreciated.

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    $\begingroup$ You should be able to work this out yourself (Hint: $P\{X^2 = k\} = 0$ unless $k$ is the square of a non-negative integer). As far as I know, there is no specific name for this distribution: call it Squared-Poisson if you like. $\endgroup$ – Dilip Sarwate Mar 1 '12 at 16:44

Your title would be better phrased as "If $X$ is a Poisson-distributed random variable with mean $\lambda$, how is $X^2$ distributed?" or "If $X$ has a Poisson distribution with mean $\lambda$, how is $X^2$ distributed?".

I don't think much can be said beyond the fact that $\Pr(X^2 = x^2) = \dfrac{\lambda^x e^{-\lambda}}{x!}$.

One thing of possible interest is that $\operatorname{E}(X^2) = \lambda+\lambda^2.\,$ This is an instance of a pattern: $$ \begin{align} \operatorname{E}(X^3) & = \lambda + 3\lambda^2 + \lambda^3 \\ \operatorname{E}(X^4) & = \lambda + 7\lambda^2 + 6 \lambda^3 + \lambda^4 \\ \operatorname{E}(X^5) & = \lambda + 15\lambda^2 + 25 \lambda^3 + 10\lambda^4 + \lambda^5 \\ & \,\,\, \vdots \end{align} $$ The coefficient of $\lambda^k$ in the expansion of $\operatorname{E}(X^n)$ is the number of (un-ordered) partitions of a set of size $n$ into $k$ parts. I.e., it's $\left\{\begin{array}{c} n \\ k \end{array}\right\} = {}$a Stirling number of the second kind. These are called the "exponential polynomials" or "Touchard polynomials".

  • $\begingroup$ Thank you very much! Could to elaborate how you found the expectation of $X^2$? $\endgroup$ – Freeman Mar 1 '12 at 17:50
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    $\begingroup$ @LHS: I think the easiest way to find $E(X^2)$ is to first find $E(X(X-1))$. $\endgroup$ – André Nicolas Mar 1 '12 at 17:59
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    $\begingroup$ @LHS : I think the comment by André Nicolas is what you need. It's easier to find the "factorial moments" $\mathbb{E}(X(X-1)(X-2)\cdots(X-n+1))$ than the moments $\mathbb{E}(X^n)$. But if you know the first $n$ factorial moments, you can use them to find the first $n$ moments. That doesn't prove that the assertion about partitions above will continue to hold for all $n$. For that, you might start with the article on Stirling numbers. $\endgroup$ – Michael Hardy Mar 1 '12 at 19:07
  • $\begingroup$ Thank you both very much! this is very helpful $\endgroup$ – Freeman Mar 1 '12 at 23:46

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