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Let $Y$ follow the distribution described by the PDFs $f_Y(y)=2y$ on $(0,1)$. Conditionally on $Y=y$, $X$ follows a uniform distribution on $(0,y)$. Compute $E(X)$ and $E(X/Y)$.

I have calculated the expectation of $X$, which I make as $E(X)=2$, using the formula for iterated conditional expectation. However, I don't know how to proceed from here, since I calculate the combined pdf to also equal $2$, and the pdf of $X$ to equal $2$. Therefore I think I have made a mistake somewhere, but I don't know where, or in reality, how to even tackle that second part of the question. If someone can confirm or refute my answers so far, and show me how to calculate $E(X/Y)$, that would be amazing.

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    $\begingroup$ $$E(XY^{-1})=E(E(XY^{-1}\mid Y))=E(E(X\mid Y)Y^{-1})=E\left(\tfrac12Y\,Y^{-1}\right)=\ldots$$ $\endgroup$
    – Did
    Nov 25, 2015 at 14:06
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    $\begingroup$ $$E(X)=E(E(X\mid Y))=E\left(\tfrac12Y\right)=\int_0^1\tfrac12y\cdot2y\,dy=\ldots$$ $\endgroup$
    – Did
    Nov 25, 2015 at 14:08

1 Answer 1

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Since I already typed it for the other question:

The marginal density of $Y$ is $f_Y(y) = 2y\mathsf 1_{(0,1)}(y)$ and the conditional density of $X\mid Y=y$ is $f_{X\mid Y=y}(x) = \frac1y\mathsf 1_{(0,y)}(x)$. Therefore the joint density is given by the product $$f_{X,Y}(x,y) = f_Y(y)f_{X\mid Y=y}(x) = 2\cdot\mathsf 1_{(0,1)}(y) \mathsf 1_{(0,y)}(x). $$ To find the marginal density of $X$, we integrate over the values of $y$ for which $f_{X,Y}>0$: $$f_X(x) = \int_x^1 2\ \mathsf dy = 2(1-x)\mathsf1_{(0,1)}(x). $$ Therefore the moment-generating function of $X$ is $$M_X(t) = \mathbb E\left[e^{tX}\right] = \int_0^1 e^{tx}2(1-x)\ \mathsf dx = \frac2{t^2}(e^t-1-t). $$

Further, $$\mathbb E[X] = \int_0^1 2x(1-x)\ \mathsf dx = \frac13 $$ and $$\mathbb E\left[\frac XY\right] = \int_0^1\int_x^1 2\left(\frac xy\right)\ \mathsf dy\ \mathsf dx = \frac12. $$

Of course, we could avoid all this computation by using properties of conditional expectation as @Did suggested.

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