TITLE: Determine the PDF from the MGF
QUESTION [6 upvotes]: If the moment generating function is given as;
$ \psi_X(s) = e^{s^2}$
How can i determine the PDF of $X$?

REPLY [4 votes]: Let me expand upon the hint I gave in comments above. Recall that the moment-generating function (MGF) of $X$ is defined as $\mathbb{E}[e^{s X}]$ for real $s$. By comparison, the characteristic function (CF) is defined as $\mathbb{E}[e^{ i s X}]$. Since for the problem at hand we are given the MGF as $e^{s^2}$, we may analytically continue $s\to i s$ to obtain the CF as $e^{-s^2}$ i.e. a centered Gaussian.
To go from here to the probability density, observe that $\mathbb{E}[e^{ i s X}]$ represents the inverse Fourier transform of the PDF $f_X(x)$. Consequently we may obtain the PDF by taking the Fourier transform of the CF i.e. $\displaystyle f_X(x)=\frac{1}{2\pi}\int_{-\infty}^\infty \mathbb{E}[e^{ i s X}] e^{-i s x}\,ds$. This can be computed explicitly, but let me 'guess' the answer by recalling that the Fourier transform of a centered Gaussian is also a centered Gaussian. But a centered Gaussian PDF depends only on the variance, which we get from the MGF as $$\sigma^2=\mathbb{E}[X^2]=\dfrac{d^2}{ds^2} \mathbb{E}[e^{s X}]_{s=0}=\dfrac{d^2}{ds^2} \left(1+s^2+\frac{1}{2}s^4\cdots\right)_{s=0}=2.$$ Hence we have enough information to express the PDF, a task I leave for the reader to complete.