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HELP MEEEEEEE

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.

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  • $\begingroup$ 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 $ $\endgroup$ – HJ_beginner Jan 15 at 3:58
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$${\cal E}[x] = {\sum\limits_{x=7}^\infty x\ 2^{6-x} \over \sum\limits_{x=7}^\infty 2^{6-x}} = {8 \over 1} = 8.$$

To perform the sum in the numerator:

Let $y = x-6$ to get:

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

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  • $\begingroup$ 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. $\endgroup$ – HJ_beginner Jan 15 at 4:01
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    $\begingroup$ 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. $\endgroup$ – David G. Stork Jan 15 at 4:03
  • $\begingroup$ Ah I see, thanks! $\endgroup$ – HJ_beginner Jan 15 at 4:03
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    $\begingroup$ But how do you know that the numerator converges to 8? $\endgroup$ – Ryan Greyling Jan 15 at 4:07
  • $\begingroup$ @DavidG.Stork How did the second summation term turns to 2? $\endgroup$ – dembrownies Jan 15 at 22:33
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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.)

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    $\begingroup$ Very nice!!!!! +1 $\endgroup$ – Mikey Spivak Jan 15 at 4:17

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