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Let $\Theta$ and $R$ be two independent random variables, where $R$ has density $f_{R}(r)=re^{-\frac{1}{2}r^2}$ for $r>0$ (zero otherwise) and $\Theta$ is uniform on $(-\pi,\pi)$. Let $X=R \cos(\Theta)$ and $Y= R \sin(\Theta)$.

Calculate $E[X|Y=y]$.

I tried getting this by calculating $f(X,Y)$ and using it to get the conditional distribution but I get stuck trying to find the joint distribution. I also tried conditioning on different events using the law of total expectation, but that has not lead to anything fruitful. I also found $Cov(X,Y)=0$ but I don't see how that is helpful.

Wintermute
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2 Answers2

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Notice that $$ \mathsf{E}\left(X \mid Y = y\right) = \mathsf{E}\left(R \cos \Theta \mid R \sin \Theta = y\right) = \mathsf{E}\left( R \cdot \mathsf{E}\left(\cos \Theta \mid \sin \Theta = \frac{y}{R}, R\right) \right) $$ The equation $\sin\Theta = y/R$ has two solutions, $\Theta = \arcsin\left(y/R\right)$ and $\Theta = \pi - \arcsin\left(y/R\right)$ for which $\cos(\Theta) = \pm \sqrt{1-y^2/R^2}$. These two solutions are equally likely, hence $$ \mathsf{E}\left(\cos \Theta \mid \sin \Theta = \frac{y}{R}, R\right) = \frac{1}{2} \sqrt{1-y^2/R^2} - \frac{1}{2} \sqrt{1-y^2/R^2} = 0 $$ The remaining expectation over $R$ is now trivial.

Incidentally, you might find the answer to the following question of mine helpful in establishing that $X$ and $Y$ are independent standard normal random variables, giving $$ \mathsf{E}(X \mid Y = y) = \mathsf{E}(X) = 0 $$
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Sasha
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  • I do not understand the second = sign in the first displayed equation. To begin with, what are the symbols $E_X$ supposed to mean? – Did Jan 10 '14 at 07:58
  • @Did The second equality is the law of total expectation with inner expectation perform with $R$ fixed, and the outer is averaging over $R$. I removed subscripts from expectations. – Sasha Jan 10 '14 at 13:07
  • This seems to rely on the fact that, considering some random variables $X$, $Y$ and $R$, if $E(X|Y)=g(Y)$ and $E(X|Y,R)=h(Y,R)$ then $g(y)=E(h(y,R))$, $y$-almost surely. This fact holds when $Y$ and $R$ are independent but not in general. Does it hold in the present case? – Did Jan 10 '14 at 13:19
  • @Did My intuition tells me that it holds. Geometrically, fixing $\sin(\Theta)$ constraints $\cos(\Theta)$ to have two equally probable and opposite in sign values. I am sure I can make a similar argument based on some parity condition, but the present approached seemed easier to explain. – Sasha Jan 10 '14 at 14:48
  • I would be very annoyed if I had to transform this intuition into a proof, being unsure about which features of the distribution of $(X,Y,R)$ make the argument miraculously valid in the case at hand. – Did Jan 10 '14 at 16:42
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I get stuck trying to find the joint distribution.

It would help if you described how you got stuck, since the result follows from a straightforward application of the change of variable formula.

One considers the transformation $(r,\theta)\mapsto(x,y)=(r\cos\theta,r\sin\theta)$. The Jacobian is $J(r,\theta)=r$, hence $$ f_{X,Y}(x,y)=\frac1{|J(r,\theta)|}f_R(r)f_\Theta(\theta)=\frac1r\cdot r\mathrm e^{-r^2/2}\cdot\frac1{2\pi}=\frac1{2\pi}\mathrm e^{-r^2/2}, $$ that is, $$ f_{X,Y}(x,y)=\frac1{\sqrt{2\pi}}\mathrm e^{-x^2/2}\cdot\frac1{\sqrt{2\pi}}\mathrm e^{-y^2/2}. $$ It follows that $X$ and $Y$ are independent and $E(X)=0$ hence $E(X\mid Y)=E(X)=0$ almost surely.

Did
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