What are the PDFs of polar coordinates given independent Cartesian coordinates? Suppose that
$$
X \sim N(0,1) \\
Y \sim N(0,1)
$$
are i.i.d random variables that represent Cartesian coordinates. What are the marginal distributions of the polar coordinates $r$ and $p$?
What I have so far is
$$
f(r,p) = \frac{r}{2\pi} \cdot e^{-\frac{r^2}{2}}
$$
but I am interested in the marginal PDFs $f(r)$ and $f(p)$.
 A: You can get to the result quickly by partitioning your equation into a product of a function of only $r$ and a function of only $p$.
$$f(r,p) = \frac{r}{2\pi} \cdot e^{-\frac{r^2}{2}} = f_R(r) \cdot f_P(p) \quad \text{for $r \in \mathbb{R}_{\geq 0}$ and $p \in [0,2\pi)$}$$
with
$$\begin{array}{}
f_R(r) &=& r \cdot e^{-\frac{r^2}{2}}& \quad \text{for $r \in \mathbb{R}_{\geq 0}$}\\
f_P(p)& =& \frac{1}{2\pi}& \quad  \text{for  $p \in [0,2\pi)$}
\end{array}$$
These are your marginal distributions.
A: It is simple to show that the Jacobian determinant of the mapping $(x,y) \mapsto (r, p)$ is $r$.  Consequently, you can write the joint density of the polar coordinates as:
$$\begin{align}
f(r, p) 
&= r \times f(x = r \sin p, y = r \cos p) \\[6pt]
&= r \times \frac{1}{2\pi} \cdot \exp \Big( -\frac{r^2 (\sin^2 \theta + \cos^2 \theta)}{2} \Big) \cdot \mathbb{I}(r \geqslant 0, 0 \leqslant p \leqslant 2 \pi) \\[6pt]
&= \frac{r}{2\pi} \cdot \exp \Big( -\frac{r^2}{2}\Big) \cdot \mathbb{I}(r \geqslant 0, 0 \leqslant p \leqslant 2 \pi) \\[6pt]
&= r \cdot \exp \Big( -\frac{r^2}{2}\Big) \cdot \mathbb{I}(r \geqslant 0) \times \frac{1}{2\pi} \cdot \mathbb{I}(0 \leqslant p \leqslant 2 \pi) \\[12pt]
&= \text{Rayleigh}(r|1) \cdot \text{U}(p|0,2\pi). \\[6pt]
\end{align}$$
This demonstrates that $r$ has a Rayleigh distribution with unit scale and $p$ has a uniform distribution over the range of polar angles (and they are independent).
