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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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A similar question is being discussed on stats.SE right now –  Dilip Sarwate Mar 26 '12 at 17:26
    
The title does not describe the question. –  Did Feb 28 '13 at 13:58
    
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. –  Michael Hardy Sep 13 '13 at 18:01
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2 Answers 2

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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Thank you for your reply, but I might not have explained my problem correctly. I would like to know the distribution of z as the euclidean distance between 2 points which are not centred in the origin, so when x and y have non-zero means. The Rician distribution applies, as you said, when z is the distance from the origin to a bivariate RV. I need the pdf/cdf of z as a distance between two points (none of them being centred in the origin). –  Ana Maria Popescu Mar 27 '12 at 12:29
    
If I assume 2 points in the 2D plane A(Xa,Ya) and B(Xb,Yb), where the Xa~N(xa,s^2), Xb~N(xb,s^2), Ya~N(ya,s^2), Yb~N(yb,s^2), then the distance between A and B, would be z=sqrt[(Xa-Xb)^2 + (Ya-Yb)^2]. Now: X=Xa-Xb and Y=Ya-Yb are themselves RVs with means (xb-xa) and (yb-ya) and variance 2s^2, so the problem that I have is determining the pdf of z=sqrt(X^2 +Y^2), knowing that X and Y are 2 Gaussian RVs with non-zero means and 2s^2 variance. –  Ana Maria Popescu Mar 27 '12 at 12:41
    
@Ana: Let $\theta = \arctan\left(\frac{y_a-y_b}{x_a-x_b}\right)$ [with a suitable adjustment for signs] be the angle of the mean. Then rotating by $\theta$: $X' = (X_a-X_b)\cos(\theta) + (Y_a-Y_b)\sin(\theta)$ is a Gaussian random variable with mean $\sqrt{(x_a-x_b)^2+(y_a-y_b)^2}$ and variance $s^2$ while $Y' = (Y_a-Y_b)\cos(\theta) - (X_a-X_b)\sin(\theta)$ is a Gaussian random variable with mean $0$ and variance $s^2$, so $R=\sqrt{X'^2+Y'^2}$ is a Rice distribution with parameters $\sqrt{(x_a-x_b)^2+(y_a-y_b)^2}$ and $s$. But $R$ is also $\sqrt{(X_a-X_b)^2+(Y_a-Y_b)^2}$. E&OE –  Henry Mar 27 '12 at 15:16
    
Thank you Henry! You answer is the most useful I got so far! –  Ana Maria Popescu Mar 29 '12 at 11:36
    
The question 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. –  Michael Hardy Sep 13 '13 at 18:00
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R∼Rice(ν,σ) has a Rice distribution, L~Raylaiegh(σ), so the problem that I have is determining the pdf of z = R-L

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