Find $E[X]$ from the function given

Question

From the given probability function of $X$, compute $E[X]$. I know the answer is 8, but not sure how to show the series converges to 8 in this case.

Answer 1

$${\cal E}[x] = {\sum\limits_{x=7}^\infty x\ 2^{6-x} \over \sum\limits_{x=7}^\infty 2^{6-x}} = {8 \over 1} = 8.$$

To perform the sum in the numerator:

Let $y = x-6$ to get:

$$\sum\limits_{y=1}^\infty (y+6) 2^{-y} = 6 \sum\limits_{y=1}^\infty 2^{-y} + \sum\limits_{y=1}^\infty y 2^{-y} = 6 + 2 = 8.$$

Answer 2

Letting $Y = X-6$ we have:

$$p_Y(y) = 2^{-y} \quad \quad \quad \text{for all } y \in \mathbb{N}.$$

This random variable has a geometric distribution $Y \sim \text{Geom}(\tfrac{1}{2})$ so we have $\mathbb{E}(Y) = 2$, which then gives the corresponding expected value $\mathbb{E}(X) = 6 + 2 = 8$. If you want to formally derive the expected value of this distribution you can do so like this:

$$\begin{equation} \begin{aligned} \mathbb{E}(Y) &= \sum_{y} y \cdot p_Y(y) \\[6pt] &= \sum_{y=1}^\infty y \cdot 2^{-y} \\[6pt] &= \sum_{y=1}^\infty \sum_{z=1}^y 2^{-y} \\[6pt] &= \sum_{z=1}^\infty \sum_{y=z}^\infty 2^{-y} \\[6pt] &= \sum_{z=1}^\infty 2^{-z} \sum_{y=0}^\infty 2^{-y} \\[6pt] &= \frac{1}{2} \Bigg( \sum_{z=0}^\infty 2^{-z} \Bigg) \Bigg( \sum_{y=0}^\infty 2^{-y} \Bigg) \\[6pt] &= \frac{1}{2} \times 2 \times 2 = 2. \\[6pt] \end{aligned} \end{equation}$$

(This working uses repeated application of the sum of an infinite geometric series.)