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Consider the example below (from Ross, Introduction to Probability Models, p. 104), with $X$ and $Y$ continuous random variables.

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I found out that the following holds for this example:

$$E[X] = E[X\ |\ Y= E[Y]] $$

$$E[X] = 3,\ \ E[Y] = 1,\ \ E[X\ |\ Y = y] = 2 + y$$

Is this a coincidence (I am inclined to think so) or does the first identity hold somewhat more generally?

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1 Answer 1

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I think you can try $$ f(x,y)=\frac{2}{3}(x+2y) \quad 0\leq x,y \leq 1 $$ and see that in general it's not true. If you consider $X,Y$ discrete, $E[Y]$ may even be non part of the support of $Y$.

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