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

Prove $E[XY]=E[YE[X|Y]]$.

I tried proving it using the definition of covariance, but I ended up going in a circle. Any hints on how to go about the proof?

share|cite|improve this question
I don't think you can use covariance to deal with this, can you? – Tunococ Jan 23 '13 at 10:33

Just use the following:

  • For an integrable random variable $Z$, $E[Z]=E[E[Z\mid\mathcal F]]$ for any $\sigma$-algebra $\mathcal F$.
  • If $Y$ is $\mathcal F$-measurable, then $E[XY\mid\mathcal F]=YE[X\mid\mathcal F]$.
share|cite|improve this answer
I'm not familiar with $\sigma$-algebra. – hello888 Jan 23 '13 at 10:37
How was conditional expectation introduced? – Davide Giraudo Jan 23 '13 at 10:38
$E[X|Y] = \int^\infty_{-\infty}xf_{X|Y}(x|y)dx$ – hello888 Jan 23 '13 at 10:44

Basically $E[XY]=E[E[XY|Y]]=E[YE[X|Y]]$. The first step is the iterated rule of conditional expectation. For the second, use the fact that given Y, Y is like a constant.

However if you are looking for the usage of rigorous definition of conditional expectation, the solution by Davide Giraudo is the one to go for.

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.