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I'm learning about computing the expected value for a geometric distribution. This is the proof the book provides.

$$E(X) = \sum_{j=1}^\infty j \times \Pr[X=j] = \sum_{j=1}^\infty \sum_{i=1}^j \Pr[X=j] = \sum_{i=1}^\infty \sum_{j=i}^\infty \Pr[X=j] =\sum_{i=1}^\infty \Pr[X \geq i]$$

How do I interpret the double summation? Is it a "for each j, sum i to j" ?

I understand that

$$\begin{align}E(X) &= (0p_0) + (1p_1) + (2p_2)+ (3p_3) + \dots \\ &= p_1 + (p_2 + p_2) + (p_3 + p_3 + p_3) + \dots\\ &= (p_1 + p_2 + p_3 + \dots) + (p_2 + p_3+ \dots)\\ &= \Pr[X\geq 1] + \Pr[X \geq 2] + \dots \end{align}$$

but I'm confused about the above "closed form" solution.

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1 Answer 1

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Yes, it is exactly that. They use the "trick" that, for every fixed $j\geq 0$, $$ j\cdot \mathbb{P}\{ X = j\} = \left(\sum_{i=1}^j 1\right)\cdot \mathbb{P}\{ X = j\} = \sum_{i=1}^j \mathbb{P}\{ X = j\} $$ and since, by definition (as $X$ takes values in $\{0,1,2,\dots\}$), $$\mathbb{E}[X] = \sum_{j=0}^\infty j\cdot \mathbb{P}\{ X = j\} = \sum_{j=1}^\infty j\cdot \mathbb{P}\{ X = j\}$$ you get what they wrote.

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