My End Goal
To compute the following Expected value: $$ \DeclareMathOperator{\E}{\mathbb{E}} \E \;\left(\frac1{\sum\limits_{i=1}^{B} 2^{-X_i}}\right)$$
where:
- $B$ is a constant (e.g., $B=4096$)
- $X_i \sim Geom(\tfrac{1}{2})$ are independent Geometric random variables with $p=\tfrac{1}{2}$
- $P(X_i =k)=\tfrac{1}{2}^k$, for $k\in \{1, 2, ...\}$
My Plan so far:
We know that:
$$\E \left(\frac{1}{Y}\right)=\int_0^\infty M_Y(-t) \; dt$$
where $M_Y(t) = \E \left(e^{tY}\right)$ is the Moment Generating Function of random variable $Y$.
Therefore, I need to compute the $M_Y(t)$, where $Y=\sum\limits_{i=1}^{B} Y_i=\sum\limits_{i=1}^{B} 2^{-X_i}$.
Due to the i.i.d property of $X_i$s, their moment generating functions are the same, and thus, we need one of them multiplied together $B$ times:
$$M_Y(t) = M_{Y_1}(t)^B$$
So, we just need to compute the term, $M_{Y_1}(t)$, with $Y_1 = 2^{-X_1}$.
For simplification, from now on I will denote $X_1$ as $X$, and $Y_1$ as $Y$
Now, the question is:
How to compute the MGF of $2^{-X}$, where $X \sim Geom(\frac{1}{2})$?
My attempt so far: if I expand the formula for MGF, I get the following:
\begin{align*} M(t) & = \E \left(e^{t \cdot 2^{-X}} \right) \\ & = \sum\limits_{k=1}^{\infty} 2^{-k} \cdot e^{t \cdot 2^{-k}} \\ & = \sum\limits_{k=1}^{\infty} \frac{e^{\frac{t}{2^k}}}{2^k} \\ & = \tfrac{1}{2}e^{\frac{t}{2}} + \tfrac{1}{4}e^{\frac{t}{4}} + \tfrac{1}{8}e^{\frac{t}{8}} + \tfrac{1}{16}e^{\frac{t}{16}} + ... \end{align*}
But, I am not sure how to simplify the above equations !!!!
I thought of the following strategy so far, which is to compute the moments themselves, and then have the MGF represented through these moments.
In other words,
\begin{align*} M_Y(t) & = \E \left(e^{t Y} \right) \\ & = 1 + t \E \left(Y\right) + \frac{t^2}{2!} \E \left(Y^2\right) + \frac{t^3}{3!} \E \left(Y^3\right) + ... \end{align*}
And the fact that these $\E \left(Y^m\right)$ are easily calculable: $$ \E \left(Y^m\right) = \E \left((2^{-X})^m\right) = \sum\limits_{k=1}^{\infty} (2^{-k})^m \cdot 2^{-k} = \sum\limits_{k=1}^{\infty} 2^{-k(m+1)} = \frac{1}{2^{m+1}-1} $$
Plugging it in, \begin{align*} M_Y(t) & = 1 + \left(\frac{1}{3}\right)t + \left(\frac{1}{7}\right)\frac{t^2}{2!} + ... + \left(\frac{1}{2^{n+1}-1}\right)\frac{t^n}{n!} + ... \end{align*}
But, where do I go from there?