Q.6.52 - A First Course in Probability

Question

Let X and Y denote the coordinates of a point uniformly chosen in the circle of radius $1$ centered at the origin. That is, their joint density is $f(x, y) = \frac{1}{\mathrm{\pi }}{x}^2+y^2\le 1$.

Find the joint density function of the polar coordinates $R = {(X^2 + Y^2)}^{1/2}$ and $\theta ={\tan }^{-1}Y/X$. 

Step-by-Step Solution

Answer

Joint density function of the polar coordinates,

$f_{R,\theta }(r,\theta )=\frac{r}{\mathrm{\pi }}$

Step 1 : Radius :

A straight line drawn from the center of a circle or sphere to the circumference.

Step 2 : Explanation :

The random vector's density function,

$f(x,y)=\frac{1}{\mathrm{\pi }}$ 

For $x^2+y^2\le 1$

Polar coordinates: $R={(X^2+Y^2)}^{1/2}$ and $\theta ={\tan }^{-1}Y/X$.

The density function of a random vector, according to the assertion,

$f_{X.Y}(x,y)=\frac{1}{\mathrm{\pi }}$ for $x^2+y^2\le 1$.

Now, apply the transformation 

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

We can express the density function of a random vector using the theorem about variable transformations $(R,\theta )=g(X,Y)$ as

$f_{R,\theta }(r,\theta )=f_{X,Y}(x,y).{\left|det(\nabla g(x,y\left)\right)\right|}^{-1}$

Then calculate,

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

That yields

$\left|det(\nabla g(x,y\left)\right)\right|=\frac{x^2+y^2}{{(x^2+y^2)}^{{\displaystyle \frac{3}{2}}}}=\frac{1}{\sqrt{x^2+y^2}}=\frac{1}{r}$

Hence,

$f_{R,\theta }(r,\theta )=r.f_{X.Y}(x,y)=\frac{r}{\mathrm{\pi }}$ for $r\in (0,1)$

And  $\theta \in (0,2\mathrm{\pi })$