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Take two discrete random variables $X$ and $Y$. $X$ is an indicator random variable with $\mathbb{E}(X \mid Y) = 1/Y$. I want to compute

$$ \mathbb{E}((1 + 3X)Y^2). $$

The random variables are not independent.

The method I have seen does the following. First fix $Y$ and compute

$$\mathbb{E}((1 + 3X)Y^2) = Y^2 + 3Y.$$

Then compute

$$ \mathbb{E}((1 + 3X)Y^2) = \mathbb{E}(Y^2) + 3\mathbb{E}(Y). $$

How can this method be justified?

Simd
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1 Answers1

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By the tower property $\mathbb E((1+3X)Y^{2})=\mathbb E (\mathbb E((1+3X)Y^{2})|Y)$. You can pull out $Y^{2}$ from the conditional expectation, so you get $\mathbb E((1+3X)Y^{2})=\mathbb E Y^{2}(\mathbb E((1+3X)|Y)$ and $\mathbb E((1+3X)|Y)=1+\frac 3Y$.

geetha290krm
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  • Thank you. I am really asking about why the method of fixing one variable, taking the mean and then taking the mean wrt the previously fixed variable is justifiable. – Simd Oct 11 '23 at 12:05
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    That is a basic property of conditional expectations called the tower property. You can find a discussion in Wikipedia. @Simd – geetha290krm Oct 11 '23 at 12:07
  • @Simd This link may also be useful: https://math.stackexchange.com/questions/41536/intuitive-explanation-of-the-tower-property-of-conditional-expectation – geetha290krm Oct 11 '23 at 12:09