Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
How come noone has answered here? – JohnK Nov 25 '13 at 21:38

The short answer is no you did not misunderstand, in fact your proof and reasoning is correct.

That being said, I would use a bit more explicit set notation to make the proof even more clear (to me at least), although it is not altogether necessary. I prefer this notation because it is more clear as to what exactly you are summing over:

We suppose $X$ has a discrete distribution on countable set $S$, and let $Q \subseteq \mathbb{R}$ denote the range of $g$. Then $Q$ is countable thus $Y$ has a discrete distribution. It follows that $$E(Y) = \sum_{y \in Q} y \cdot P(Y=y) = \sum_{y \in Q} y \sum_{x \in g^{-1}(y)} f(x) = \sum_{y \in Q} \sum_{x \in g^{-1}(y)}g(x)f(x) = \sum_{x \in S} g(x)f(x)$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.