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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.

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    $\begingroup$ I think your question would be more interpretable if you used $r$ instead of $y$. $\endgroup$ Apr 16, 2020 at 19:51
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    $\begingroup$ 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 $\mathbb P(y\sin(\theta) > 0.8)>0$ $\endgroup$
    – Henry
    Apr 17, 2020 at 11:12

1 Answer 1

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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 $D$.

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  • $\begingroup$ how do you conclude the pdf by calculating the expected value? $\endgroup$
    – Jhon Doe
    Apr 17, 2020 at 4:23
  • $\begingroup$ $f$ is arbitrary, take for example $f=1_B,$ you will get $$P_X(B)=\frac{1}{\pi}\int_{B}1_D$$ $\endgroup$
    – mathex
    Apr 17, 2020 at 11:18
  • $\begingroup$ 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 $\endgroup$
    – Jhon Doe
    Apr 17, 2020 at 11:31
  • $\begingroup$ $E[f(ycos(\theta),y\sin(\theta))]=\int_{R^2}f(xcos(w),xsin(w))dP_{(y,\theta)}(x,w)$ $\endgroup$
    – mathex
    Apr 17, 2020 at 11:36
  • $\begingroup$ en.wikipedia.org/wiki/Pushforward_measure $\endgroup$
    – mathex
    Apr 17, 2020 at 11:39

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