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This question already has an answer here:

Why is E(XY)=E(XE(Y|X))?

Is this using the properties of conditional expectation and is there a general formula that can be applied when you have E(...)=E(..(E(Y|X))?

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marked as duplicate by Xi'an, Michael Chernick, Peter Flom May 27 at 14:04

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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By the expression $\mathbb{E}(XY),$ we mean the expectation of XY under their joint distribution. I.e., if these both are continuous, we have that

$$ \mathbb{E}(XY) = \int\int xyf(x, y)dxdy, $$

where $f(x, y)$ is the joint pdf of X and Y. For this reason, we sometimes write $\mathbb{E}_{(X, Y)}(\cdot),$ or $\mathbb{E}_f(\cdot)$ in order to make it explicit which distribution the expectation is to be taken over.

We can factorise distributions in a certain way, specifically we can write the pdf of a $d$-dimensional continuous random vector as $$ f(x_1, \dots, x_d) = g_1(x_1)g_2(x_2\vert x_1)g_3(x_3\vert x_2, x_1)\cdots g_d(x_d\vert x_1, \dots, x_{d-1}). $$ Applying this to $(X, Y),$ we can write $f(x, y) = g(x)h(y\vert x).$ We can then see that \begin{align*} \mathbb{E}_{(X,Y)}(XY) &= \int\int xyf(x, y)dxdy\\ &= \int\int xyg(x)h(y\vert x)dxdy\\ &= \int x\int yh(y\vert x) dy f(x)dx\\ &= \mathbb{E}_X\left(X\int y h(y\vert x) dy\right)\\ &= \mathbb{E}_X\left(\mathbb{E}_{(Y\vert X)}\left(Y\right)\right). \end{align*} This is also true for distributions that are not continuous, and can be generalised to a higher dimension analoguosly to the factorisation property of distributions. To show this in a very general context, you need some measure-theoretic arguments. The general formula that you request is often referred to as the law of iterated expectations, the tower rule, the smoothing theorem, or the law of total expectation.

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Another way to show it is directly using Law of iterated expectations:

$$\begin{align}E[XY] &= E[E[XY|X]]=E[XE[Y|X]]\end{align}$$

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  • $\begingroup$ Thank you. Could we similarly have E(XY)=E[E(XY|Y)]=E[YE(X|Y)] $\endgroup$ – Maths May 27 at 11:29
  • $\begingroup$ yes, exactly, it holds both ways. $\endgroup$ – gunes May 27 at 11:30

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