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There is a proof of $$ E(E(Y|x)) = E(Y) $$ $Proof:$

WLOG, suppose X and Y are two continuous random variables.

Let $E(Y|x)=m(x) =\int_{-\infty}^{\infty} yf(y|x)\, dy$

Then $$ E(E(Y|x))=E(m(x))= \int_{-\infty}^{\infty} m(x) f(x) \, dx =\int_{-\infty}^{\infty} \Big(\int_{-\infty}^{\infty} yf(y|x)\, dy\Big) f(x)\, dx $$ $$ =\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} y\frac{f(x,y)}{f(x)}f(x)\, dy \, dx =\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} yf(x,y)\, dy \, dx = E(Y) $$

My question is why $$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} yf(x,y)\, dy \, dx=\int_{-\infty}^{\infty} yf(y)\, dy=E(Y)$$

son520804
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wanchihsin
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1 Answers1

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Change the order of integration and use that the marginal density can be obtained via $$ f(y)=\int_{-\infty}^\infty f(x,y)\,\mathrm dx. $$

The "WLOG" statement is however a bit odd, since there is lost a lot of generality by assuming $(X,Y)$ to be continuous (which by the way is different from assuming that $X$ and $Y$ marginally are continuous random variables).

Stefan Hansen
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  • Ok,I got it. Can I directly omit to say which kind of variables X and Y are ? – wanchihsin Aug 12 '14 at 06:53
  • You can, but that requires a different, more general definition of the conditional expectation ${\rm E}[Y\mid X]$ (one not using densities, of course). Are you familiar with such? – Stefan Hansen Aug 12 '14 at 06:55