# polar transformation of uniform distribution

if $x$ and $y$ are two independent random variables with uniform distribution in (0,1) , then what is the distribution of random variables $r$ and $\theta$ with definitions: $$r=\sqrt({x^2}+{y^2})$$ $$\theta = \arctan \frac yx$$ And what I have done is finding regions of $r$ and $\theta$ . I think: region of $r$ : circle C=((0,0),$\sqrt 2$) region of $\theta$ : (0 , $\frac \pi2$) but I could not proceed anymore. thanks in advance.

• – StubbornAtom Jun 6 at 16:17

For the probability density function of $r$, you need to compute the integral $$p(r)=\int_0^1\int_0^1 dx dy\delta\left(r-\sqrt{x^2+y^2}\right)\ .$$ Calling $f(x)=r-\sqrt{x^2+y^2}$, you first need to identify the roots $f(x^*)=0$ that lie within $(0,1)$. There is only one, $x^*=\sqrt{r^2-y^2}$. Therefore your integral is equivalent to $$p(r)=\int_0^1\int_0^1 dx dy\frac{\delta(x-\sqrt{r^2-y^2})}{|f'(x^*)|}=r\int_0^1\int_0^1 dx dy\frac{\delta(x-\sqrt{r^2-y^2})}{\sqrt{r^2-y^2}}\ .$$ Now you can use the Dirac delta to kill the $x$ integral. Note that, in doing so, you need to impose that $0<\sqrt{r^2-y^2}<1$ and $0<y<1$, resulting in: if $0<r<1$, $0<y<r$, and if $1<r<\sqrt{2}$, then $\sqrt{r^2-1}<y<1$. Hence $$p(r)=r\left[\mathbb{1}_{0<r<1}\int_{0}^r dy\frac{1}{\sqrt{r^2-y^2}}+\mathbb{1}_{1<r<\sqrt{2}}\int_{\sqrt{r^2-1}}^1 dy\frac{1}{\sqrt{r^2-y^2}}\right]$$ $$=r\left[\frac{\pi}{2}\mathbb{1}_{0<r<1}+\mathbb{1}_{1<r<\sqrt{2}}(\mathrm{arccsc}(r)-\mathrm{arctan}\left(\sqrt{r^2-1}\right))\right]\ ,$$ which is correctly normalized, $\int_0^\sqrt{2}dr\ p(r)=1$.
Similarly for $P(\theta)$ $$P(\theta)=\int_0^1\int_0^1 dx dy\delta\left(\theta-\mathrm{arctan}(y/x)\right)\ .$$ Define your function $g(y)=\theta-\arctan(y/x)$, whose only zero for $\theta\in [0,\pi/2]$ is at $y^*=x\tan\theta$. Therefore $$P(\theta)=\int_0^1\int_0^1 dx dy\frac{\delta(y-x\tan\theta)}{|g'(y*)|}=(1+\tan^2\theta)\int_0^1\int_0^1 dx dy\ |x|\delta(y-x\tan\theta)\ .$$ Now, we can kill the $y$-integral by imposing that $0<|x|\tan\theta<1$ and $0<x<1$, resulting in $$P(\theta)=(1+\tan^2\theta)\left[\mathbb{1}_{0<\tan\theta<1}\int_0^1 dx\ x+\mathbb{1}_{\tan\theta>1}\int_0^{\cot\theta} dx\ x\right]=$$ $$=(1+\tan^2\theta)\left[\frac{1}{2}\mathbb{1}_{0<\theta<\pi/4}+\mathbb{1}_{\pi/4<\theta<\pi/2}\frac{1}{2}\cot^2\theta\right]\ ,$$ which is correctly normalized, $\int_0^{\pi/2}d\theta P(\theta)=1$.