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The proof seems a little too easy. I am wondering if I misunderstood something.
Let $Y = g(X)$. Prove the $\mathbb E(Y) = \sum_x g(x)f_x(x)$,
provided that the sum converges absolutely.

By definition, $\mathbb E(Y) = \sum_y yf_y(y)$. Since $g^{-1}(y) = \{x_1, x_2 ...\}$,
$f_y(y) = \mathbb P(Y=y) = \sum_i \mathbb P(X=x_i)$, where $g(x_i) = y$.
$\mathbb E(Y) = \sum_y y \sum_i \mathbb P(x_i) = \sum_y \sum_i g(x_i) \mathbb P(x_i)$
$= \sum_i g(x_i) \mathbb P(x_i) = \sum_x g(x)f_x(x)$

My understanding is that you need to capture every single $x_i \in g^{-1}(y)$.
Repeat the process for every $y$ and then add up the terms.

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How come noone has answered here? – JohnK Nov 25 '13 at 21:38

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