If $X$ is a Poisson distribution with mean $\lambda$ how is $X^2$ distributed? Any explanation would be very appreciated.
If $X$ is a Poisson distribution with mean $\lambda$ how is $X^2$ distributed?
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2You 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. – 2012-03-01
1 Answers
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".
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0Thank you both very much! this is very helpful – 2012-03-01