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Let $U$ and $V$ be independent variables. Let $V$ be uniformly distributed on interval $[-\pi, \pi]$ and let $U$ be exponentially distributed with pdf $f(u) = \lambda e^{-\lambda u}$.

Let $(X,Y) = (U*\cos(V), U*\sin(V)$. What is the probability density function of $(x,y)$? On which area is $(x,y)$ defined, so that we can derive cumulative distribution function (CDF)?

I am interested in that question because as far as I know, $$f(x,y) = f(u(x,y),v(x,y))* |J(x,y)|$$, where $J(x,y)$ is Jacobian matrix.

The problem is that I cannot express $U(x,y)$.

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Is something wrong with this exercise? – Zenga Aug 15 '14 at 11:47
Hi user160522, I don't know exactly what earned you two down votes without a single comment on it, but I guess it is mainly because you didn't say in the question why you are interested in that problem (is it homework for example? If yes add the homework tag) and what you have tried. Also, it is called cumulative distribution function. Please update your question. I, in the meantime, will give you an answer. – fabee Aug 15 '14 at 12:32
I would be thankful for your help! – Zenga Aug 15 '14 at 12:50
@fabee: While I am personally against anyone using other people to answer their homework without explicitly making clear in the question itself that it is homework, the homework tag is not to be used anymore according to – user21820 Aug 15 '14 at 13:01
Ah, oops. Didn't know that. Thanks user21820. – fabee Aug 15 '14 at 13:07

The usual method yields that the PDF $f_{X,Y}$ of $(X,Y)$ is defined on $\mathbb R^2$ by $$f_{X,Y}(x,y)=\frac{\lambda\,\mathrm e^{-\lambda\sqrt{x^2+y^2}}}{2\pi\,\sqrt{x^2+y^2}}.$$ The CDF can be computed by integration and has no nice form (nor is it particularly interesting in the case of some 2D random variable).

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I remember that I read the usual trick before and I liked it. However, I totally forgot about it. But it is always nice to read it again ... +1. – fabee Aug 15 '14 at 13:16
@fabee Method $\ne$ trick. – Did Aug 15 '14 at 13:17
Unfortunately, I am only allowed to edit comments for 5 minutes. So it will have to stay a trick. I am sorry ... – fabee Aug 15 '14 at 13:40

The first thing to realize is that drawing $V$ uniformly from $[-\pi,\pi)$ and transforming it with $\cos(V)$ and $\sin(V)$ results in a uniform distribution on the sphere. Let $\mathbf S = (\cos(V),\sin(V))^\top$ denote this random variable. Then $\mathbf X = \mathbf S\cdot U$ has a spherically symmetric distribution (by definition a spherically symmetric random variable is the product of a uniformly distributed random variable on the sphere and a positive radial random variable). The general form of spherically symmetric distributions in $n$ dimensions is given by $$p(\mathbf X)=\frac{\varrho(\|X\|_2)\Gamma(\frac{n}{2})}{\|X\|_2^{n-1}2\pi^\frac{n}{2}},$$ where $\varrho$ is the density of $U$.

In your case, this yields $$p(\mathbf X)=\frac{\lambda e^{-\lambda \|X\|_2}\|X\|_2}{2\pi}.$$ $p(\mathbf X)$ is defined on all of $\mathbb R^2$, although the integration for the c.d.f $$P(a,b) \int_{-\infty}^a\int_{-\infty}^b p(x_1,x_2)\mathrm d x_1\mathrm d x_2$$ might be a bit nasty.

You can find a little more general derivation of this and the Jacobian in

Gupta, A. K., & Song, D. (1997). Lp-norm spherical distribution. Journal of Statistical Planning and Inference, 60(2), 241–260.

as well as

Song, D., & Gupta, A. K. (1997). Lp-Norm Uniform Distribution. Proceedings of the American Mathematical Society, 125(2), 595–601.

If found some of the proofs a little inaccessible. This is why I assembled a few notes. You can download them here.

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