Find the PDF of $(y\cos(\theta), y\sin(\theta))$ if $\theta\sim \operatorname{Uniform}[0,2\pi]$ and $y$ has the distribution given by $P(y\in [a,b])=\int_a^b2tdt$ (i.e. the PDF of $y$ is $2t$ where $t\in[0,1]$). Also $y$ and $\theta$ are independent.
My initial idea was to find the product PDFs of $y\cos(\theta)$ and $y\sin(\theta)$ and multiply them together but I am finding it hard to come up with the product PDFs.
For instance the PDF of $\cos(\theta)=l$ is $\frac{1}{\pi\sqrt{1-l^2}}$. Hence, the product PDF of $y\cos(\theta)$ should be $\int_{-1}^1\frac{1}{\pi\sqrt{1-l^2}}\frac{2z}{x}\frac{1}{|x|}$. But this diverges.
Any hints or ideas. I think there should be a faster way to do this.
Best Answer
Let $X:=(y\cos(\theta),y\sin(\theta)).$ Take a bounded function $f:\mathbb{R}^2 \rightarrow \mathbb{R}.$
\begin{align} E[f(y\cos(\theta),y\sin(\theta)]&=\frac{1}{2\pi}\int_{0}^12x\left(\int_{0}^{2\pi}f(x\cos(w),x\sin(w))\,dw\right)dx \\&=\frac{1}{\pi}\int_{0}^1x\left(\int_0^\pi (f(x\cos(u),x\sin(u))+f(-x\cos(u),-x\sin(u)))\,du\right)dx \ \ \ \ (1) \\&=\frac{1}{\pi}\int_0^1x\left(\int_{-x}^x\frac{1}{\sqrt{x^2-w^2}}\left(f(w,\sqrt{x^2-w^2})+f(w,-\sqrt{x^2-w^2})\right)dw\right)dx \ \ \ \ (2) \\&=\frac{1}{\pi}\int_{-1}^1\left(\int_{|u|}^1\frac{x}{\sqrt{x^2-u^2}}\left(f(u,\sqrt{x^2-u^2})+f(u,-\sqrt{x^2-u^2})\right)dx\right)du \ \ \ \ (3) \\&=\frac{1}{\pi}\int_{-1}^1\left(\int_{-\sqrt{1-u^2}}^\sqrt{1-u^2}f(u,w)\,dw\right)du.\ \ \ \ (4) \end{align} $(1)$ Change of variable $u=w-\pi.$
$(2)$ Change of variable $u=\arccos(\frac{w}{x}).$
$(3)$ Fubini.
$(4)$ Change of variable $x=\sqrt{u^2+w^2}.$
In each case we have a $C^1-$ diffeomorphism.
$X$ has a density $$f_{X}(u,v)=\frac{1}{\pi}1_D(u,v)\,,$$ where $$D:=\left\{(u,v) \in [-1;1] \times \mathbb{R};u^2+v^2 \leq1 \right\}$$
This is uniform distribution on $D$.