Determine pdf from mgf

Question

If I am given a moment generating function $M(t)$, such that $$M(t) = \frac14 + \frac34\cdot\frac1{1-t}$$ where $t>1$, how can I know its probability density function?

Answer 1

In general it is not that easy to identify the PDF from an MGF; in general the method is to consider instead the Characteristic function (i.e the MGF evaluated at $it$), and then take the inverse Fourier transform.

Note however that in your example there is a significantly easier way. If we define $X \sim \text{Exp}(1)$ and $Y \sim \text{Bin}(3/4)$ independently of each other, then consider

$$Z = YX$$

Then this will have the moment generating function you provide; from this you can easily identify the PDF.

To see that this has the relevant MGF note

\begin{align*} \mathbf E[e^{tZ}] & = \mathbf E[e^{tZ} \, | \, Y = 0]\mathbf P[Y = 0] + \mathbf E[e^{tZ} \, | \, Y = 1] \mathbf P[Y = 1] \\ & = \frac14 \mathbf E[e^{t0}] + \frac34 \mathbf E[ e^{tX}] \\ & = \frac14 + \frac34 \frac{1}{1-t}, \end{align*} where we rely on the fact that $1/(1-t)$ is the MGF of a standard exponential distribution.

Therefore the pdf of $Z$ is given to be

$$f_Z(z) = \frac14 \delta_0(z) + \frac34 e^{-z},$$ where $\delta_0$ is the Dirac-delta function.