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This is problem 6.9 in Ross.

I have the joint pdf $$f(x,y) = 6/7(x^2+xy/2), 0<x<1, 0<y<2$$ marginal pdf $f(x) = (6/7)x(2x+1)$ conditional pdf $$f(y|x) = f(x,y)/f(x) = (x+y/2)/(2x+1)$$ conditional expectation $$E(y|x) = (3x+16)/(6(2x+1))$$

Then my professor has $$E(xy) = E(xE(y|x)) = \int_0^1 E(y|x) f(x) dx$$

This is the part I don't understand. I thought $E(xE(y|x))$ would be $$\int_0^1 xE(y|x)dx$$ but instead of the $x$, she has $f(x)$.

However, in another example I have $$ E(Y|X) = \frac 12 x. $$

Then my professor has $$ E(XE(Y|X)) = E(\frac 12 x^2) $$ I don't understand why in one case you multiply by $x$ and in the other by $f(x)$

asahi
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1 Answers1

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I would distinguish between capital $X$ and $Y$ and lower-case $x$ and $y$, the former being the random variables, and the latter being the variables used in expressions like $f(x,y)$ and in $\int\cdots\cdots\,dx$, etc.

Then we have $$ E(X E(Y\mid X)) = \int_0^1 x E(Y\mid X=x) \, f(x) \, dx. $$

More generally, for any function $g$, $$ E(g(X)) = \int_0^1 g(x) f(x) \, dx. $$

Throughout, you should carefully note where I've put capital $X$ and $Y$, and where I've put lower-case $x$ and $y$.

begin quote

However, in another example I have $$ E(Y|X) = \frac 12 x. $$

Then my professor has $$ E(XE(Y|X)) = E(\frac 12 x^2) $$ I don't understand why in one case you multiply by $x$ and in the other by $f(x)$

end quote

Again, being careful about capital and lower case, I'd write:

However, in another example I have $$ E(Y\mid X) = \frac 12 X $$ (with a capital $X$, since this is a random variable).

$$ E(XE(Y\mid X)) = E\left(\frac 12 X^2\right) $$

After that, you can write $$ E\left(\frac 1 2 X^2 \right) = \int_0^1 \left(\frac12 x^2 \right) f(x)\, dx. $$

Michael Hardy
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