Moment Generating Function of $X^2$ from MGF of $X$ I have a long and unpleasant MGF: $M_X(t)=E[e^{tX}]$ and I need the MGF of the square: $M_{X^2}(t)=E[e^{tX^2}]$
I can't turn the MGF back into a PDF because mathematica breaks (It is the sum of various exponentials to a variable power that I need to keep). Is there any other way to transform the MGF?
Thanks In Advance
 A: $M_X(t) = \sum_{n=0}^\infty t^n E[X^n]/n!$ while $M_{X^2}(t^2) = \sum_{n=0}^\infty t^{2n} E[X^{2n}]/n!$.  The coefficient of $t^{2n}$ in $M_{X^2}(t^2)$ is the coefficient of $t^{2n}$ in $M_X(t)$ multiplied by $(2n)!/n!$.  Those factorials mess things up a lot.  However, it is possible that we can handle this using Laplace transform and inverse Laplace transform, at least if $X$ is bounded.  Let's suppose $|X| < B$ a.s.
The Laplace transform of $e^{tX}$ is $1/(s-X)$ for $\text{Re}(s) > X$.  This (if it converges) the Laplace transform $\mathscr L M_X(s)$of $M_X(t) = \mathbb E[e^{tX}]$ is $\mathbb E[1/(s-X)]$, for $\text{Re}(s) > B$.  Similarly, the Laplace
transform $\mathscr L M_{-X}(s)$ of $M_{-X}(t) = M_X(-t)$ is $\mathbb E[1/(s+X)]$ for $\text{Re}(s) > B$.
Now $$ \mathbb E \left[\frac{1}{s^2 - X^2}\right] = \frac{1}{2s} \mathbb E \left( \frac{1}{s+X} + \frac{1}{s-X} \right) = \frac{\mathscr L M_X(s) + \mathscr L M_{-X}(s)}{2s}$$
so the inverse Laplace transform of 
$$\dfrac{\mathscr L M_X(\sqrt{s}) + \mathscr L M_{-X}(\sqrt{s})}{2\sqrt{s}} = \mathbb E\left[ \frac{1}{s - X^2}\right]$$
should be $M_{X^2}(t)$.
