# Find the PDF of $(y\cos(\theta), y\sin(\theta))$

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.

• I think your question would be more interpretable if you used $r$ instead of $y$. Apr 16, 2020 at 19:51
• You cannot multiply the PDFs of $y\cos(\theta)$ and $y\sin(\theta)$ as they are not independent. For example, if $y\cos(\theta) > 0.8$ then $\mathbb P(y\sin(\theta) > 0.8)=0$ while if $0< y\cos(\theta) < 0.1$ then $\mathbb P(y\sin(\theta) > 0.8)>0$ Apr 17, 2020 at 11:12

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 $$D$$.

• how do you conclude the pdf by calculating the expected value? Apr 17, 2020 at 4:23
• $f$ is arbitrary, take for example $f=1_B,$ you will get $$P_X(B)=\frac{1}{\pi}\int_{B}1_D$$ Apr 17, 2020 at 11:18
• another thing how did you get first equality. I know that $E[X]=\int_{\Omega} X(w)dP(w)$ but I don't see how that's the same as your first equality Apr 17, 2020 at 11:31
• $E[f(ycos(\theta),y\sin(\theta))]=\int_{R^2}f(xcos(w),xsin(w))dP_{(y,\theta)}(x,w)$ Apr 17, 2020 at 11:36
• en.wikipedia.org/wiki/Pushforward_measure Apr 17, 2020 at 11:39