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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 $U(x,y)$.

Zenga
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  • Hi there, it is not homework, the case it that I got a note from my colleague, where they simply changed u for x. But that is wrong in my opinion. Otherwise I am preparing myself for the course on probability. – Zenga Aug 15 '14 at 14:06
  • Ah I see. Anyway I didn't downvote, but I guess you should make clear the source of your question so that others don't make the wrong assumption. All the best in your preparation! – user21820 Aug 16 '14 at 03:15

2 Answers2

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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).

Community
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Did
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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
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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.

fabee
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