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?


  • 2
    $\begingroup$ 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. $\endgroup$
    – whuber
    Commented May 19, 2019 at 19:37

1 Answer 1


Simplifying the results from the reference given in the question, the joint distribution of $r$ and $\phi$ is given by \begin{equation} 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)} \end{equation} 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*}


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