How to derive the Projected normal distribution

Question

Suppose we have a bivariate normal variable $\mathbf{x}= (x_1, x_2)$ with mean $\mu_1$ and $\mu_2$ and variances $\sigma_1^2$ and $\sigma_2^2$ and correlation $\rho$. I need to obtain the pdf of the transformation $\Vert\mathbf{x}\Vert^{-1} \mathbf{x}$: the coordinates of a variable in the unit circle.

In directional statistic this is a well known distribution called the Projected normal, or offset normal. I have already the pdf but i want to derive it by myself. I'm able to do a variable transformation when there is a 1 to 1 relation but this is not the case.

Can someone put me in the right way?

Just for completion: A pdf where it is explained the projected normal distribution.

Answer 1

There is a trick for calculating the pdf of 2-d projected normal, which can be found on pp. 52 of Mardia (1972).

After change of variable $x_1 = r\, \cos\theta$ and $x_2 = r\,\sin\theta$, we have $f(\theta)= \int_{0}^{\infty}\frac{1}{\sqrt{2\pi}\sigma_1\sigma_2\sqrt{1-\rho^2} }r\,\mbox{exp}\{-\frac{\sigma_2^2(r\cos\theta-\mu_1)^2-2\rho(r\cos\theta-\mu_1)(r\sin\theta-\mu_2)+\sigma_1^2(r\sin\theta-\mu_2)^2}{2\sigma_1^2\sigma_2^2(1-\rho^2)}\}\mbox{d}r$,

Use the following result, $d^2\int_{0}^{\infty}r\,\mbox{exp}\{-\frac{1}{2}d^2(r^2-2br)\}\mbox{d}r = 1+ (2\pi)^{\frac{1}{2}}bd\,\mbox{e}^{\frac{1}{2}b^2d^2}\Phi(bd),$

where $\Phi(x)$ is the cdf of $\mbox{N}(0,1)$.

With $d^2 = \frac{\sigma_2^2\cos^2\theta-\rho\sigma_1\sigma_2\sin2\theta+\sigma_1^2\sin^2\theta}{(1-\rho^2)\sigma_1^2\sigma_2^2}$ and $b = \frac{\mu_1\sigma_2(\sigma_2\cos\theta-\rho\sigma_1\sin\theta)+\mu_2\sigma_1(\sigma_1\sin\theta-\rho\sigma_2\cos\theta)}{\sigma_2^2\cos^2\theta-\rho\sigma_1\sigma_2\sin2\theta+\sigma_1^2\sin^2\theta}$, you are going to obtain the pdf listed in the above paper.