Find the moment generating function of a discrete random variable given its probability mass function

Question

Say $Y$ is a random variable with support on $(1, 2, 3, \ldots ) $ and has the PMF:

$$\frac{C}{2^x}$$

How would you find the MGF from here and how would you arrange your answer so that it is not in sum notation?

Answer 1

First we consider that PMF must sum up to unity so $(P_y(Y) = \frac{1}{2^y})$ for $y \in \mathbb N$: $$\sum_{y=1}^{\infty}\frac{C}{2^y}=1 \Rightarrow C \frac{\frac{1}{2}}{1-\frac{1}{2}}=1 \Rightarrow C = 1$$

Then by definition of MGF we see that ($i=\sqrt{-1}$): $$\Phi_Y(\omega) = \mathbb{E}\{ e^{i\omega y} \}= \sum_{y=1}^{\infty} \frac{e^{i\omega y}}{2^y} = \frac{\frac{e^{i\omega}}{2}}{1-\frac{e^{i\omega}}{2}}=\frac{e^{i\omega}}{2-e^{i\omega}}$$

Also we have the geometric series formula: for a complex number $z$ with $|z|<1$ we have: $$\sum_{i=a}^{N}z^n = S = z^a+z^{a+1} +...z^N$$ $$\sum_{i=a}^{N}z^{n+1} = zS = z^{a+1}+z^{a+2} +...z^{N+1}$$ $$zS-S = (z-1)S = z^{N+1}-z^a \Rightarrow S = \frac{z^{N+1}-z^a}{z-1}= \frac{z^a-z^{N+1}}{1-z}$$ When $N\rightarrow\infty$ we have $|z|^{N+1}\rightarrow 0$ and the formula becomes $$S=\frac{z^a}{1-z}$$ For above consider the case $a=1$. In first one $z=\frac{1}{2}$ and in the second $z = \frac{e^{i\omega}}{2}$

Also it can be interpreted as the discrete Fourier's transform of the sequence of the PMF which can be obtained from z-transform, also.

Answer 2

By equating the total probability to unity, it can be shown that $C=1$.

Then, by definition, \begin{eqnarray*} M_{X}(t)&=& E(e^{tX})=\sum_{x=1}^{\infty}e^{tX}P(X=x)=\sum_{x=1}^{\infty}\left(\dfrac{e^t}{2}\right)^x\\ &=&\left(\dfrac{e^t}{2} + \left(\dfrac{e^t}{2}\right)^{2} + \left(\dfrac{e^t}{2}\right)^{3} + \cdots \right)\\ &=& \dfrac{e^t}{2} \left( 1 + \left(\dfrac{e^t}{2}\right) + \left(\dfrac{e^t}{2}\right)^{2} + \cdots \right)\\ &=&\dfrac{e^t}{2}\cdot \dfrac{1}{1-e^t/2}\\ M_{X}(t)&=&\dfrac{e^t}{2-e^t} \end{eqnarray*}