Need help with moment generating function of geometric distribution

Question

The cheat sheet I have tells me the moment generating function for Geometric Distribution is:

$$M(t) = \frac{p}{1-(1-p)e^t} $$

But most resources and me personally working it out I get:

$$M(t) = \frac{pe^t}{1-(1-p)e^t} $$

I believe the second is the correct MGF for Geometric Distribution, but where did my cheat sheet get its function? Why is it missing one $e^t$ on top? I want to make sure I have correct materials for exams, if it is just an error, then I can find another resource for my reference material. Thanks!

Answer 1

There are two common definitions of the geometric distribution, as described on the Wikipedia article:

The geometric distribution is either of two discrete probability distributions:

1. The probability distribution of the number $X$ of Bernoulli trials needed to get one success, supported on the set $\{1, 2, 3, \dots\}$.
2. The probability distribution of the number $Y = X − 1$ of failures before the first success, supported on the set $\{ 0, 1, 2, 3, \dots\}$.

You did your calculation for $X$, and your cheat sheet gives the answer for $Y$. Since $X = Y+1$, $\mathbb E[e^{tX}]$ will be off from $\mathbb E[e^{tY}]$ by a factor of $e^t$.