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Apologies if this is a basic question!

I'm trying to understand a solution to a problem I was solving. The author suggests a trick to calculate expected value by multiplying the expected value series by 0.5 (line 2) and taking the difference (line 3):

$E(X) = 0.5^1 + 2 \cdot0.5^2 + 3\cdot 0.5^3...\\$

$0.5E(X) = 0.5^2 + 2 \cdot0.5^3 + 3\cdot 0.5^4...\\$

$0.5E(X) = 0.5^1 + 0.5^2 + 0.5^3...$

My question: how did he calculate the difference on line 3?

Thanks for your help.

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  • $\begingroup$ This method is called "错位相减法" in Chinese, i.e. "Dislocation Subtraction", which is "almost everywhere" used to calculate the values of seires of the form $\sum P(n)q^n$ where $P$ is a polynoimal. See baike.baidu.com/item/… $\endgroup$ – Feng Shao Aug 7 at 1:12
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We have $$E = 0.5^1 + 2\cdot0.5^2 + 3\cdot0.5^3 + 4\cdot0.5^4+\cdots$$ $$0.5E = 0.5^2 + 2\cdot0.5^3 + 3\cdot0.5^4+\cdots$$ Combining terms with equal powers of $0.5$,$$E - 0.5E = 0.5^1 + 0.5^2(2-1) + 0.5^3(3-2) + 0.5^4 (4-3) \cdots$$ $$\implies 0.5E = 0.5^1 + 0.5^2+0.5^3\cdots$$

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Subtract the second line from the first line

$$0.5E(X)=E(x)-0.5E(X)=$$

$$ 0.5^1 + 2 \cdot0.5^2 + 3\cdot 0.5^3...\\$$

$$- 0.5^2 - 2 \cdot0.5^3 - 3\cdot 0.5^4...=$$

$$ 0.5^1 + 0.5^2 + 0.5^3...$$

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Here's a more "formal" way to write what you have, so it doesn't seem like so much of a trick. Since $$E(X) = \sum_{k \geq 1} \frac{k}{2^k},$$ we have \begin{align*} \frac{E(x)}{2} &= \sum_{k \geq 1} \frac{k}{2^{k + 1}} \\ &= \sum_{k \geq 2} \frac{k - 1}{2^k} \\ &= \sum_{k \geq 2} \frac{k}{2^k} - \sum_{k \geq 2} \frac{1}{2^k} \\ &= E(x) - \frac{1}{2} - \frac{1}{2}. \end{align*} From this we can see that \begin{equation*} E(x) = 2E(x) - 2, \end{equation*} or $E(x) = 2$.

This is something like a particularly nice example of the perturbation method to evaluate sums, which you can learn more about here or in various other sources online.

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    $\begingroup$ I think the reason for writing the OP in the way it was written is: to be understood by someone who does not know the $\Sigma$ notation. $\endgroup$ – GEdgar Aug 7 at 0:32
  • $\begingroup$ @GEdgar Almost certainly. I only hoped to "peel back the curtain" a little bit, though perhaps I should direct OP to the relevant Wikipedia article to learn more about what I've said if they're interested. $\endgroup$ – rwbogl Aug 7 at 0:34
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    $\begingroup$ Something’s not right here. In the OP, E(x) is 2, not 1. $\endgroup$ – Chris Johnson Aug 8 at 12:46
  • $\begingroup$ @ChrisJohnson You were absolutely right; I had left off the $k$ factor from OP's sum. Fortunately this only makes it a slightly more complicated example. $\endgroup$ – rwbogl Aug 8 at 16:43

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