What is the distribution of the euclidean distance between two random points in 2D

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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(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 uncorrelated Gaussian RVs 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 RV. This has been proven only when the RVs (a and b) are CIRCULAR bivariate RVS (proof here -> Chapter 13, subchapter 13.8.2, page 680, of Mathematical Techniques for Engineers and Scientists - Larry C. Andrews, Ronald L. Phillips - 2003). 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?
 

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  • #2
mathman
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I suspect that the distribution could be transformed to the Rician by first transforming from the x and y means to the center and then scaling using the variances, so that an ellipse becomes a circle.

Note: I have not tried to work it out, but my intuition tells me it should work.
 
  • #3
Stephen Tashi
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X=Xa-Xb and Y=Ya-Yb are themselves RVs with means (xb-xa) and (yb-ya) and variance 2s^2
X and Y are two independent normally distributed random variables with the same variance. If you transform coordinates to make (xb-xa,yb-ya) the origin, then you would have a Rayleigh distribution.
 
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Thank you for your replies, but I am not looking to obtain a Rician distribution. I would like to know the distribution of z, in the given conditions, without making any transformations or other assumptions.
 
  • #5
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Hi there! I have managed to find out that indeed a transformation is needed into polar coordinates. Rotating the mean distance z with a θ angle and using the appropriate notation will results in proving that z is Ricianly distributed. Thanks guys for your help!
 

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