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I'm trying to solve the following question:

Let $(X,Y)$ be a randomly chosen point on the unit disk, and let $R={(X^2+Y^2)}^{\frac12}$.

Find the joint density of the vector $(X,R)$.

I'm not sure whether my solution is correct.

First, I've calculated the Jacobian of the transformation:

$$J=det(\frac{\partial X, R}{\partial X,Y})= \frac{Y}{{(X^2+Y^2)}^{\frac12}}=\frac{Y}{R}$$

Now, the joint density should be

$$f_{X,R}(x,r)=\frac{1}{|J|}f_{X,Y}(X,{(R^2-X^2)}^{\frac12})=\frac{r}{{(r^2-x^2)}^{\frac12}}f_{X,Y}(X,{(R^2-X^2)}^{\frac12})$$

Is there any way to simplify this expression without any further information on the distributions of $X$ and $Y$? Should I suppose that $X$ and $Y$ distribute uniformly?

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Brassican
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  • Sorry but P(R=1)=1, no? – Did Jan 02 '16 at 20:57
  • Yes, $P(R=1)=1$. – Brassican Jan 02 '16 at 20:59
  • Thus the vector (X,R) cannot have a density since (X,R) is on a given line of the plane with full probability. How come you are asked to compute a nonexistent PDF? – Did Jan 02 '16 at 21:03
  • You are right. I mistakenly thought that the point is on the unit circle, but it's within the unit disk. – Brassican Jan 02 '16 at 21:07
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    Now the text of the question makes sense but you ought to simplify the partial result in your question: for example, the factor $y$ (which should be $|y|$) should not appear and should be replaced by $|y|=\sqrt{r^2-x^2}$ and indicator functions of the domain $|x|<r<1$ should appear in $f_{X,R}(x,r)$. – Did Jan 02 '16 at 21:14

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