# E[g(Y)] proof question

This is one of the theorems in my stats text, and I need some help understanding the proof.

1. How can the summand($g_{i}$) be out of its summation sign when multiplying?
I thought you can never take out a summand when $\sum_{i}^{}$ depends on it.
2. I think there are several steps missing in achieving that final result from the one above.
Can someone please explain how 2 summations became 1 with a different index?
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1. The fact that

$$\sum_{i=1}^m g_i \left\{ \sum_{\substack{\text{all y_j such that} \\ g(y_j)=g_i}} p(y_j) \right\} = \sum_{i=1}^m \sum_{\substack{\text{all y_j such that} \\ g(y_j)=g_i}} g_i p(y_j)$$

is just the distributive property of the sum. Here is a more transparent example: \begin{align*} & \sum_i \left\{ a_i \sum_j b_j \right\} \\ & = \sum_i \left\{ \vphantom{\sum_i} a_i \; (b_1 + b_2 + \dotsb)\right\} \\ & = \sum_i \left\{ \vphantom{\sum_i} (a_i b_1 + a_i b_2 + \dotsb) \right\} \\ & = \sum_i \left\{ \sum_j a_ib_j\right\} \end{align*}

2.

$$\sum_{i=1}^m \sum_{\substack{\text{all y_j such that} \\ g(y_j)=g_i}} g_i p(y_j) = \sum_{j=1}^n g(y_j) p(y_j)$$

As the 2nd sum is over all $y_j$ such that $g(y_j) = g_i$, you can replace $g_i$ by $g(y_j)$ in its summand:

$$\sum_{i=1}^m \sum_{\substack{\text{all y_j such that} \\ g(y_j)=g_i}} g_i p(y_j) = \sum_{i=1}^m \sum_{\substack{\text{all y_j such that} \\ g(y_j)=g_i}} g(y_j) p(y_j)$$

Note that $i$ only appears in the index of the 2nd sum. For $i=1$, the 2nd sum is over all $j$'s such that $g(y_j)=g_1$. For $i=2$, the 2nd sum is over all $j$'s (different from before) such that $g(y_j)=g_2$. Etc. Thus, you can replace the double summation by a single summation running over all $j$'s.

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Awesome, the examples on 1 and 2 really hit the spot! Thank you so much for your explanation!! – Belphegor Feb 22 '14 at 5:22
You're welcome :-) – ocram Feb 22 '14 at 5:26