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Suppose $X$, $Y$ are random variables with joint density: $$f_{X,Y}(x,y)=\frac{e^{-y/2}}{2\pi\sqrt{x(y-x)}}$$ where $0< x< y$.

(a) Find the distribution of $Y$. Hint: for integration use the substitution $x=ys$

(b) Compute $E(X|Y=1)$

I'm not sure how to use the joint density to find the ditribution of $Y$. And I have no idea how to compute that expectation.

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up vote 2 down vote accepted

An easy approach is to interpret probabilistically the change of variable suggested as a hint, that is, to compute the distribution of $(Z,Y)$, where $X=YZ$. One should find $$ f_{Z,Y}(z,y)=\frac1{\pi\sqrt{z(1-z)}}\mathbf 1_{0\lt z\lt1}\cdot\frac12\mathrm e^{-y/2}\mathbf 1_{y\geqslant0}. $$ Thus, $Z$ and $Y$ are independent with densities $$ f_{Z}(z)=\frac1{\pi\sqrt{z(1-z)}}\mathbf 1_{0\lt z\lt1},\qquad f_Y(y)=\frac12\mathrm e^{-y/2}\mathbf 1_{y\geqslant0}. $$ In passing, this answers (a). Note that $Y$ is exponential with parameter $\frac12$ and that $Z$ follows the arcsine distribution.

To solve (b), note that $E[X\mid Y=1]=E[YZ\mid Y=1]=1\cdot E[Z]$. By the symmetry of $f_Z$ by the transformation $z\to1-z$, one knows that $E[Z]=E[1-Z]$ hence $E[Z]=\frac12$ (but one can also compute directly $E[Z]$ using the explicit form of $f_Z$). Thus, $E[X\mid Y=1]=\frac12$.

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That final density function for $Z$ happens to be the one of the arcsine distribution. – TMM May 1 '13 at 20:09
@TMM Yes. I added this mention to the answer. Thanks. – Did May 1 '13 at 21:07

To find the marginal density from a joint density function, you integrate out the unnecessary variable.

So for example, to find $f_Y(y)$, you integrate $\int_0^y{f(x,y)dx}.$

From here, you know $f_{X|Y}(x|y)=\frac{f_{X,Y}(x,y)}{f_Y(y)}$

Now what is $E[X|Y=y]$? Isn't it just $\int{x\cdot f(x|y=1)dx}$

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I'm getting a ridiculously complicated answer when I integrate to find fY(y) – Steve May 1 '13 at 18:29
@Steve How is that so? – Did May 1 '13 at 21:05

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