# Distribution of Euclidean Distance btwn 2 Non-Centered Points in 2D

• Amiutza
In summary, the conversation discusses the distribution of z as the euclidean distance between two points that are not centered in the origin. It is noted that when the points are circular, the Rician distribution applies, but it is unknown if there is a known parametric distribution for non-circular points. Some suggest transforming the coordinates, while others argue that no assumptions or transformations should be made. Eventually, it is concluded that a transformation into polar coordinates is needed, resulting in a Rician distribution for z.
Amiutza
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?

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.

Amiutza said:
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.

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.

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!

## 1. What is the distribution of Euclidean distance between two non-centered points in 2D?

The distribution of Euclidean distance between two non-centered points in 2D follows a normal distribution, with the mean distance being the distance between the two points and the standard deviation being the average distance from each point to the origin.

## 2. How is the Euclidean distance calculated between two points in 2D?

The Euclidean distance between two points in 2D is calculated by finding the square root of the sum of the squared differences between the coordinates of the two points. This can be represented by the formula: d = √((x2 - x1)^2 + (y2 - y1)^2).

## 3. What is the significance of the distribution of Euclidean distance in 2D?

The distribution of Euclidean distance in 2D has many applications in statistics, data analysis, and machine learning. It is used to measure the similarity or dissimilarity between data points, and can also be used to identify patterns in data.

## 4. How does the distribution of Euclidean distance change with different data sets?

The distribution of Euclidean distance can vary depending on the data set. It can be affected by the number of data points, the range of values, and the distribution of the data. However, it will always follow a normal distribution in 2D.

## 5. Can the distribution of Euclidean distance be applied in higher dimensions?

Yes, the concept of Euclidean distance can be applied in any number of dimensions. The formula for calculating it remains the same, but the distribution may differ depending on the number of dimensions and the data set.

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