# Find $E[X]$ from the function given

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.

• I could be wrong but say $Y \sim Geometric(1/2)$ and $E[Y] = 2$ and X = 6+Y so $E[X] = 6+2 =8$ – HJ_beginner Jan 15 at 3:58

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

To perform the sum in the numerator:

Let $$y = x-6$$ to get:

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

• Thanks for you answer David. I have a question... why is the denominator necessary... correct me if I'm wrong but that is showing it's a pmf that adds to $1$? Thanks for your help. – HJ_beginner Jan 15 at 4:01
• Yes... it happens to sum to $1$ since you gave it as a true $PDF$, but in some problems the given function will not, so you must normalize. You can ignore it here, if you like. – David G. Stork Jan 15 at 4:03
• Ah I see, thanks! – HJ_beginner Jan 15 at 4:03
• But how do you know that the numerator converges to 8? – Ryan Greyling Jan 15 at 4:07
• @DavidG.Stork How did the second summation term turns to 2? – dembrownies Jan 15 at 22:33

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{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}

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

• Very nice!!!!! +1 – Mikey Spivak Jan 15 at 4:17