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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?

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

2 Answers 2

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 $E[XY\mid\mathcal F]=YE[X\mid\mathcal F]$.
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I'm not familiar with $\sigma$-algebra. –  hello888 Jan 23 '13 at 10:37
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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.

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