# The Problem

Let $U$ and $R$ be independent random variables where $U\sim \text{Uniform}(0,1)$ and $R$ has pdf $f_R(r)=r e^{-r^2/2}$ on $(0,\infty)$. Define the random variables \begin{align} X &= a + b \cos (2\pi\, U), \\ Y &= c + d \sin (2\pi \, U). \end{align} where $b,d \ne 0$. Find the marginal distributions of $X$ and $Y$.

# Attempt at a Solution

We can express the joint distribution as $$f_{X,Y}=f(g^{-1})\,\left|J\left(g^{-1}\right)\right|,$$ where $g(U,R)=(X,Y)$. I have tried to work this out but it seems to get pretty ugly, and I suspect there's a much easier way to do this.

To avoid extra constants, let's take $U \sim \text{Uniform}(0,2\pi)$ and drop the $2\pi$ in the trig functions. To find $g^{-1}$, define $$H_X = \frac{X-a}{b}\quad \text{and} \quad H_Y = \frac{Y-c}{d}.$$ Then $$R = H_X^2 + H_Y^2$$ and $$U = \arccos \left( \frac{H_X}{H_X^2 + H_Y^2} \right).$$ We next compute the Jacobian. First, $$\frac{dR}{dX} = \frac{2H_X}{b} \quad \text{and} \quad \frac{dR}{dY} = \frac{2H_Y}{d}.$$ After some simplification, $$\frac{dU}{dX} = \frac{1}{b}\frac{1}{H_X} - \frac{2}{b}\frac{H_X}{H_X^2 + H_Y^2}$$ and $$\frac{dU}{dY} = -\frac{2}{d}\frac{H_Y}{H_X^2 + H_Y^2}.$$ The Jacobian simplifies to $$J = \left|\frac{2}{bd}\frac{H_Y}{H_X}\right|.$$ Because $U$ and $R$ are independent, $$f_{U,R}(u,r) = f_U(u)f_R(r) = \frac{1}{2\pi} r e^{-r^2/2} \ I_{(0,2\pi)}(u) \ I_{(0,\infty)}(r)$$ We could then substitute all of these results into the expression for $f_{X,Y}$ and integrate to obtain the marginal distributions, but this seems to be a huge mess. Is there something simple I am missing?

• I don't see the variable $r$ in $X = a + b \cos (2\pi\, U)$ and $Y = c + d \sin (2\pi \, U)$. Jan 11, 2017 at 23:32
• From his equations it looks like $R = sin(2\pi U) ^ 2 + cos(2\pi U) ^2$, but my spidey-senses are telling me it should be $R^2 = sin(2\pi U) ^ 2 + cos(2\pi U) ^2$ Jan 11, 2017 at 23:40
• After ignoring the linear transformations (that is, focusing on $H_X$ and $H_Y$ compared to $R$ and $2\pi U$), convert from polar coordinates to Cartesian coordinates. From the equality $$\exp(-r^2/2)rdrd\theta = \exp(-(x^2+y^2)/2)dxdy,$$ it is immediate that both the marginal distributions have densities proportional to $\exp(-x^2/2)$.
– whuber
Jan 11, 2017 at 23:48
• In addition to the above comments, if you still have trouble make your life easier by focusing on a=c=0 and b=d=1. Then consider transforming the resulting variables to the general case. Jan 12, 2017 at 1:08
• @whuber That is exactly the kind of thing I had hoped for. I'm reworking the question to try to answer it using this method, but it may take a bit as I have an impending exam (prompting this question). Jan 12, 2017 at 5:22