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I would like to know the distribution of z as the euclidean distance between 2 points which are not centred in the origin. If I assume 2 points in the 2D plane $A = (X_a,Y_a)$ and $B = (X_b,Y_b)$, where the $X_a \sim \mathcal N(\mu_a,s^2)$, $X_b\sim\mathcal N(\mu_b,s^2)$, $Y_a\sim\mathcal N(\nu_a,s^2)$, $Y_b\sim\mathcal N(\nu_b,s^2)$, then the Euclidean distance between $A$ and $B$, would be $z= \sqrt{(X_a-X_b)^2 + (Y_a-Y_b)^2}$. Now: $X=X_a-X_b$ and $Y=Y_a-Y_b$ are themselves random variables with means $(\mu_b-\mu_a)$ and $(\nu_b-\nu_a)$ 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 uncorrelated Gaussian random variables with non-zero means and the same variance, $2s^2$.

The Rician distribution applies when $z$ is the distance from the origin to a bivariate random variable. This has been proven only when the random variables ($A$ and $B$) are circular bivariate random variables (A proof can be found in L. C. Andres and R. L. Phillips, Mathematical Techniques for Engineers and Scientists, 2003, Ch. 13, Sec. 13.8.2, p. 680). I would like to know the pdf/cdf of $z$ as a distance between two points (none of them being centred in the origin) when they are not circular. Is there a known parametric distribution for $z$? What would this distribution look like if it is a generalized form of the Rician distribution?

kjetil b halvorsen
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Ana Maria Popescu
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    possible duplicate of [What is the distribution of the euclidean distance between two normally distributed random variables](http://stats.stackexchange.com/questions/9220/what-is-the-distribution-of-the-euclidean-distance-between-two-normally-distribu) – Andy W Mar 27 '12 at 13:24
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    The answer is here: http://en.wikipedia.org/wiki/Chi_distribution – whuber Mar 27 '12 at 13:32
  • Please note that Ana posted this as an answer yesterday to the "possible duplicate" listed above. I and perhaps a moderator too suggested she repost it as a separate question instead which she has done here. So, there is a doubt lurking that was not completely addressed by the other question. – cardinal Mar 27 '12 at 13:38
  • @whuber: Can you clarify/expand briefly? The answer to which question is at the link you provide? I only skimmed it, but it doesn't seem to directly address what seems to be the primary question of the OP. – cardinal Mar 27 '12 at 13:47
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    Cardinal: "determining the pdf of z=sqrt(X^2 +Y^2), knowing that X and Y are 2 uncorrelated Gaussian RVs with NON-ZERO MEANS and the same variance, 2s^2" defines the chi distribution, doesn't it (up to scaling by 2s^2)? When the means are unequal it should be a non-central chi distribution. – whuber Mar 27 '12 at 13:49
  • @whuber: The noncentral chi distribution in this case is the Rice distribution, which I think the OP recognizes. The wording of the question is a little unclear but I believe what is *really* being asked is: What happens to the pdf of this quantity if we go from the uncorrelated case to the correlated one? Maybe my reading of it is off-base though. – cardinal Mar 27 '12 at 14:01
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    Thanks for that, Cardinal. Your reading matches my interpretation of the OP's previous question, but this question explicitly refers to "uncorrelated" RVs (at the end of the first paragraph). Ana Maria: please clarify this and help us understand how the present question differs from your previous one on this topic. – whuber Mar 27 '12 at 14:05
  • My question refers to the uncorrelated RVs, just as the previous post did. However, it is my understanding that the non-central chi-squared distribution is appropriate for z^2 not for z=sqrt(...). Maybe I am wrong, but even if I were to consider the noncentral chi distribution, instead of the chi-squared, it still does not make sense to me as the variable z does not have the expression of sqrt((xi/sigmai)^2) as here: http://en.wikipedia.org/wiki/Noncentral_chi_distribution – Ana Maria Popescu Mar 27 '12 at 14:15
  • The difference between this post and the previous thread (where I had my comment posted), is that the variance can be the same for the 2 RVs. – Ana Maria Popescu Mar 27 '12 at 14:18
  • Why the reference to "circular" distributions at the end if your post? I think that is what threw me off as I interpreted that as asking what happens when the coordinates of each vector were correlated. – cardinal Mar 27 '12 at 14:20
  • It has been proven that z is Ricianly distributed only when X and Y are considered to be circular bivariate RVS -> X∼N(νcosθ,σ2) and Y∼N(νsinθ,σ2). In this case it can be considered that one of the points is set in the origin, but what I want to know is what happens when none of the points is set in the origin? Does the distribution remain Rician? Is it a generalized form and if so how would it look like? – Ana Maria Popescu Mar 27 '12 at 14:30
  • I think you may simply be confused by the parameterization of the mean, which is simply in polar form. – cardinal Mar 27 '12 at 14:33
  • In other words, if $X$ has (arbitrary!) mean $\mu_x$ and $Y$ has (arbitrary!) mean $\mu_y$, then there exists a unique $\theta \in [0,2\pi)$ and $\nu = \sqrt{\mu_x^2 + \mu_y^2}$ such that $\mu_x = \nu \cos \theta$ and $\mu_y = \nu \sin \theta$. – cardinal Mar 27 '12 at 14:48
  • That may be so, but could you recommend a reference where this demonstration is made, for both RVs with non-zero means? In the reference that I mention in the post, there is such a demonstration but only for one non-zero mean RV. – Ana Maria Popescu Mar 27 '12 at 14:49
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    Hi Ana. I'm sorry. What part, precisely, remains unclear? I (or perhaps someone else) can then post a short proof to remove all doubt. Cheers. – cardinal Mar 27 '12 at 14:55
  • If you know the distribution of $Z^2$, you can immediately deduce the distribution of $Z=\sqrt{Z^2}$. What does remain unclear? – Xi'an Mar 28 '12 at 16:58
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    Thank you for your answers! They have been useful to me. Took me a while to figure it out but I think I understood. – Ana Maria Popescu Mar 29 '12 at 11:41
  • Good! You might consider expanding on your thoughts in this space. They may help others in the future. I hope to see more questions (and answers) from you in the future. Cheers. – cardinal Mar 29 '12 at 11:51
  • My understanding is, that given the problem above, by rotating the mean distance between A and B by the original angle θ, we will obtain 2 transformed Gaussian random variables (we use the Rotation matrix for the transformation). With these new RVs (one with non-zero mean and one with zero mean), of same variance, it is easy to demonstrate that the distribution of z is Rician (has been done in Mathematical Techniques for Engineers and Scientists). – Ana Maria Popescu Mar 30 '12 at 14:49

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