expectation of function of geometric random variable

Question

This is question 3a from this sample test

Let $X$ be a geometric RV with mean 2. Compute $E[e^{−X}]$

If I could convert this to a sum (by knowing the probabilities of each term?), then perhaps I could do this problem, but I only know the mean of X.

Answer 1

We know from definition of the moment generating function of $X$ which is Geometrically distributed, that: $$E[e^{tX}]=\sum_{k=1}^{\infty}e^{tk}\left (\frac 12 \right)^k=\frac {\frac 12e^t} {{1-e^t \left(\frac 12 \right)}}$$

Now $t=-1$ and we simplify...

Answer 2

As a 1-shifted geometric random variable, $X$ has probability mass function for some rate parameter $p$ : $$\mathsf P(X=k)=(1-p)^{k-1} p~\mathbf 1_{k\in\Bbb N^+}$$

The expectation of $X$ being $2$ should tell you what value of $p$ you have.

Now, by definition of expectation (and the Law of the Unconcious Statastitian):

$$\mathsf E(\mathsf e^{-X}) ~{= \sum_{k=1}^\infty e^{-k} \mathsf P(X=k) \\ = \frac p{1-p}\sum_{k=1}^\infty \left(\frac{1-p}e\right)^{k} \\ = \phantom{~ \sum_{k=1}^\infty \left(\frac{1}{2e}\right)^{k}}}$$