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