Question

Let (X, Y) be uniformly distributed in the circle of radius 1 centered at the origin. Its joint density is thus

$f(x,y)=\frac{1}{\mathrm{\pi }}0\le x^2+y^2\le 1$

Let R = (X2 + Y2)1/2 and = tan−1(Y/X) denote

the polar coordinates of (X, Y). Show that R and are

independent, with R2 being uniform on (0, 1) and being

uniform on (0, 2π).

Short answer

The statement is proved and explained below.

Step-by-step solution

Given Information

We have given the function

$f(x,y)=\frac{1}{\mathrm{\pi }}0\le x^2+y^2\le 1$.

Simplify

Considering the transformation $g:c\to (0,1)\times (0,2\mathrm{\pi }),$where $C$is a unit circle. Transformation $g$is defined by

$g(x,y)=(r^2,\theta )=\left(x^2+y^2,{\tan }^{-1}\left(\frac{y}{x}\right)\right)$

We are interested in the distribution of random vector $(R^2,\theta )=g(X,Y).$Using the theorem about the density of transformation of a random vector, we have that

$f{R}^2,\theta (r^2,\theta )=fXY,\left(g^{-1}\right(r^2,\theta \left)\right)·\left|det\nabla g^{-1}(r^2,\theta )\right|$

We have $fX,Y\left(g^{-1}\right(r^2,\theta \left)\right)=\frac{1}{\pi }\chi g^{-1}(r^2,\theta )\in C=\frac{1}{\mathrm{\pi }}\chi (r^2,\theta )\in (0,1)\times (0,2\mathrm{\pi })$and

$\nabla g(x,y)=\left[-\begin{array}{cc}2x& 2y\\ \frac{y}{x^2+y^2}& \frac{x}{x^2+y^2}\end{array}\right]$

which implies

$\left|det\nabla g(x,y)\right|=2\Rightarrow \left|det\nabla g^{-1}(r^2,\theta )\right|=\frac{1}{2}$

Hence, we have obtained

$f{R}^2,\theta (r^2,\theta )=\frac{1}{2\mathrm{\pi }}·\chi (r^2,\theta )\in (0,1)\times (0,2\mathrm{\pi })$

$fX,Y\left(g^{\_1}\right(r^2,\theta \left)\right)=\frac{1}{\mathrm{\pi }}\chi g^{-1}(r^2,\theta )\in C=\frac{1}{\mathrm{\pi }}\chi (r^2,\theta )\in (0,1)\times (0,2\mathrm{\pi })$

Explanation

We have seen that $R^2$and $\theta$are independent since their joint distribution can be factorized as

$f{R}^2,\theta (r^2,\theta )=\chi R^2\in (0,1)·\frac{1}{2\mathrm{\pi }}\chi \theta \in (0,2\mathrm{\pi })$

and we also have that

$R^2~Unif(0,1),\theta ~Unif(0,2\mathrm{\pi })$