# Joint distribution of the magnitude/phase of a circular bivariate normal distribution?

A bivariate normal distribution with no correlation and identical variance in both dimensions can be written as $$P(x,y|\mu_x, \mu_y, \sigma) = \frac{1}{2\pi\sigma^2}\exp{\left(-\frac{1}{2}\left(\frac{x-\mu_x}{\sigma}\right)^2 -\frac{1}{2}\left(\frac{y-\mu_y}{\sigma}\right)^2\right)}$$ I'm interested in corresponding distributions for the magnitude $$r = \sqrt{x^2 + y^2}$$ and phase $$\phi = \mathrm{arctan}(y/x)$$.

I believe the marginal distribution for the magnitude is given by the Rice distribution, but I'm looking for the joint distribution $$P(r, \phi|...)$$ or the marginal phase distribution $$P( \phi|...)$$.

I did find this reference, which gives expressions for the completely general case of a bivariate normal distribution, but I suspect simpler results exist for this special case.

Anyone happen to know what they are?

Thanks!

• Just apply the results in your reference for the bivariate normal distribution. The simplification is that $b-a=2\gamma=0,$ which makes all the $\cos(2\theta)$ terms disappear.
– whuber
May 19 '19 at 19:37

Simplifying the results from the reference given in the question, the joint distribution of $$r$$ and $$\phi$$ is given by $$$$P(r,\phi|\mu_x, \mu_y, \sigma) = \frac{r}{2\pi\sigma^2} \exp{\left(-\frac{1}{2\sigma^2} \left(r^2 -2r(\mu_x \cos{\phi} +\mu_y \sin{\phi}) + \mu_x^2 + \mu_y^2\right) \right)}$$$$ We can re-express the above in terms of the central magnitude $$r_0 = \sqrt{\mu_x^2 + \mu_y^2}$$ and and central phase $$\phi_0 = \mathrm{arctan}(\mu_y/\mu_x)$$ to give \begin{align*} P(r,\phi|r_0, \phi_0, \sigma) &= \frac{r}{2\pi\sigma^2} \exp{\left(-\frac{1}{2\sigma^2} \left(r^2 + r_0^2 -2r r_0(\cos{\phi_0} \cos{\phi} +\sin{\phi_0} \sin{\phi})\right) \right)} \\ &= \frac{r}{2\pi\sigma^2} \exp{\left(-\frac{1}{2\sigma^2} \left(r^2 + r_0^2 -2r r_0 \cos{(\phi-\phi_0)} \right) \right)} \end{align*}