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If $Z = \sqrt{X^2+Y^2}$, and $X$ and $Y$ are zero-mean i.i.d. normally-distributed random variables, then $Z$ is Rayleigh distributed.

What is the distribution of $Z$ if $X$ and $Y$ are correlated (but still zero-mean, normally-distributed random variables)?

Can't seem to find info on this scenario, even though it should be relatively common.

Lord Soth
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eyio
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    Hopefully this time: http://math.stackexchange.com/questions/238057/general-non-centered-chi-distribution-having-correlated-random-variables – Lord Soth Apr 22 '13 at 19:54
  • The reference is http://faculty.kfupm.edu.sa/EE/naffouri/publications/isit09-indef-quad-forms.pdf "On the Distribution of Indefinite Quadratic Forms in Gaussian Random Variables" by T. Y. Al-Naffouri and B. Hassibi. Look at Section III-C. There may not be a closed form solution, but you can at least express the CDF as a one-dimensional integral. – Lord Soth Apr 22 '13 at 22:11
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    If you have a hypothesis that $X,Y$ are JOINTLY Gaussian, i.e. so distributed that every linear combination of the two with constant coefficients is univariate Gaussian, then you might get somewhere with this. If you say only that they're individually Gaussian and possibly correlated, then the set of possible distributions that that function of them could have is vast. – Michael Hardy Apr 22 '13 at 22:22

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Assuming $X$ and $Y$ follow a bivariate normal distribution with correlation $\rho\in[0,1)$, zero-means, and variances $\sigma_x^2$ and $\sigma_y^2$, respectively. Then, the pdf of the random variable $Z = \sqrt{X^2+Y^2}$ follows the Hoyt distribution, that is \begin{align*} f_Z(z) &=\frac{z}{\sqrt{1-\rho ^2} \sigma_x \sigma_y} \exp\!\left(-\frac{z^2 \left(\sigma_x^2+\sigma_y^2\right)}{4 \left(1-\rho ^2\right) \sigma_x^2 \sigma_y^2}\right) I_0\!\left(\frac{z^2 \sqrt{4 \rho ^2 \sigma_x^2 \sigma_y^2+\left(\sigma_x^2-\sigma_y^2\right)^2}}{4 \left(1-\rho ^2\right) \sigma_x^2 \sigma_y^2}\right),~ z>0, \end{align*} where $I_0$ is the modified Bessel function of the first kind.

Indeed, if $\sigma_x = \sigma_y = \sigma$ and $\rho = 0$, we have the Rayleigh distribution: $f_Z(z) = \frac{z}{\sigma^2} e^{-z^2/(2\sigma^2)}$.

Additional comment: Note that we can always find $X'$ and $Y'$ with variances $\sigma_x'^2$ and $\sigma_y'^2$, respectively, that are uncorrelated and satisfies $\sqrt{X'^2+Y'^2} = \sqrt{X^2+Y^2} = Z$. Then, the pdf of $Z$ can be written in the simpler form \begin{align*} f_Z(z) &=\frac{z}{\sigma_x' \sigma_y'} \exp\!\left( -\frac{z^2}{4}\left(\frac{1}{\sigma_x'^2} + \frac{1}{\sigma_y'^2}\right) \right) I_0\!\left( \frac{z^2}{4} \left(\frac{1}{\sigma_x'^2} - \frac{1}{\sigma_y'^2}\right)\right),~ z>0. \end{align*}

P.S. Dester
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