Determine the PDF from the MGF [closed]

If the moment generating function is given as;

$ \psi_X(s) = e^{s^2}$

How can i determine the PDF of $X$?

Solution 1:

It is quite tedious but the Inverse Laplace Transform of $\psi_X(-s)$ will give you the density. A more straightforward method might be to try to identify the given MGF with known MGFs.

In this example, one might suspect that this is the MGF of the normal distribution. The general Normal MGF takes the form:

$$ m_X(s)=e^{\mu t+\frac{1}{2} \sigma^2s^2} $$

Thus, setting $\mu=0$ and $\sigma^2=2$ gives us:

$$ m_X(s)=e^{\frac{1}{2} (2)~s^2}=e^{s^2} $$

This is then the MGF of X where X~$N(~\mu=0~,~\sigma^2=2~)$

Hope this helps

Solution 2:

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.