Question

Probability: Expected values for functions of random variables.
help me understand the last line of the proof.
Let \(Y\) be a random variable that is a function of another random variable, \(X\), such that \(Y=g(x)\)
\[\begin{align}\text{definition:}&&E[X]&&=&&\sum_xxp_X(x)\\\text{lemma:}&&p_Y(y)&&=&&\sum_{\{x|g(x)=y\}}p_X(x)\\\text{theorem:}&&E[Y]&&=&&\sum_xg(x)p_X(x)\\\text{proof}:&&E[Y]&&=&&\sum_yyp_Y(y)\\&&&&=&&\sum_yy\sum_{\{x|g(x)=y\}}p_X(x)\\&&&&=&&\sum_y\sum_{\{x|g(x)=y\}}yp_X(x)\\&&&&=&&\sum_y\sum_{\{x|g(x)=y\}}g(x)p_X(x)\\&&&&=&&\sum_xg(x)p_X(x)\end{align}\]

Answer 1

I understand it all except the last line of dropping the outer sum and changing the subscript to sum over all \(x\). The best reasoning I came up with was that since \(\{x|g(x)=y\}\) is a subset of all \(x\) that corresponds to a given \(y\), summing that over each \(y\) amounts to summing all \(x\)'s individually. Is this logic right? Does anyone have a better/different line of reasoning to persuade me?

Answer 2

Never seen this in my whole life. A moderator asking a question? O_O

Answer 3

lol I wasn't a mod when I started using the site. You gotta keep pushing yourself, so questions and doubts should arise, or you aren't working hard enough

Answer 4

I agree with your reasoning and think its correct.

since y is function of x, multiple "x values" can have same "y value". So summing up all x's is equivalent to summing up x|y over all possible y values

Answer 5

thanks, good to know I'm making sense to someone besides myself. Any other justifications you think up though, please don't hesitate to provide :)