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Assuming two uncorrelated random variable (RVs) with Gaussian distributions $x\sim N(m_1,s)$ and $y\sim N(m_2,s)$, so with non-zero mean and same variance, what is the distribution of $z=\sqrt{(x^2 + y^2)}$? Is there a known parametric distribution for z?

I have already researched this problem, but I am not sure whether z is a Rician distributed RV. It has been proven that z is Ricianly distributed only when x OR y have a zero mean, because they are considered to be circular bivariate RVS in these demonstrations. I would like to know if the Ricean distribution holds when BOTH uncorrelated Gaussian RVs x and y have non-zero means.

All ideas are welcome! Thank you!

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    The question omits information on the joint distribution, saying only that they are Gaussian and uncorrelated. I can show you pairs of uncorrelated Gaussians that are nowhere near being bivariate Gaussian. If they're bivariate Gaussian and uncorrelated, then they're actually independent.2013-09-13

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Since $Z$ is simply the distance from the origin to a bivariate Gaussian random variable, you can in effect get one of the means to zero by rotation without affecting this distance or the independence of X' and Y', so the answer is yes: one mean goes to zero while the other goes to $\sqrt{\mu_X^2 + \mu_Y^2}$.

Wikipedia sets out this result when talking about a Rice distribution or Rician distribution:

$R \sim \mathrm{Rice}\left(\nu,\sigma\right)$ has a Rice distribution if $R = \sqrt{X^2 + Y^2}$ where $X \sim N\left(\nu\cos\theta,\sigma^2\right)$ and $Y \sim N\left(\nu \sin\theta,\sigma^2\right)$ are statistically independent normal random variables and $\theta$ is any real number.

Clearly you can set $\nu\cos\theta$ and $\nu\sin\theta$ to any real values by choosing suitable $\nu$ and $\theta$.

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    The q$u$estion failed to specify that it was bivariate Gaussian. It said only that it's two uncorrelated Gaussians. I can show you pairs of uncorrelated Gaussians that are nowhere near being bivariate Gaussian.2013-09-13