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?

  • 1
    $\begingroup$ I don't see the variable $ r$ in $X = a + b \cos (2\pi\, U)$ and $Y = c + d \sin (2\pi \, U)$. $\endgroup$ Jan 11, 2017 at 23:32
  • 1
    $\begingroup$ 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 $R^2 = sin(2\pi U) ^ 2 + cos(2\pi U) ^2$ $\endgroup$
    – bdeonovic
    Jan 11, 2017 at 23:40
  • 1
    $\begingroup$ 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 $\exp(-x^2/2)$. $\endgroup$
    – whuber
    Jan 11, 2017 at 23:48
  • 1
    $\begingroup$ 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. $\endgroup$
    – Glen_b
    Jan 12, 2017 at 1:08
  • $\begingroup$ @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). $\endgroup$ Jan 12, 2017 at 5:22


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.