# Thread: Finding the center of a statistical sample

1. ## Finding the center of a statistical sample

Suppose I have a large sample of the function f(x,y), for many x's and y's.
I believe that the function should be radial, i.e: f(r^2=(x-x0)^2+(y-y0)^2) = constant.
It is difficult to estimate the errors in each point in the sample, so I don't particularly trust a least square analysis to find x0,y0.
Is there a different more robust way to estimate x0 and y0 (center of the radial function), using some sort of symmetry methods?

2. Originally Posted by Marril
Suppose I have a large sample of the function f(x,y), for many x's and y's.
I believe that the function should be radial, i.e: f(r^2=(x-x0)^2+(y-y0)^2) = constant.
It is difficult to estimate the errors in each point in the sample, so I don't particularly trust a least square analysis to find x0,y0.
Is there a different more robust way to estimate x0 and y0 (center of the radial function), using some sort of symmetry methods?
Can you provide more background information. This looks like you have a different problem which you have done some work on to reach this point.

It is often advantageous to post the original question which is why we often ask for maore background or the original problem.

CB

3. For all intended purposes, the sample f(x,y) can be regarded as an experimental result, which basically it is. The system is as follows:
Put a detector (for example, a CCD plate) in front of a light source, which has a radial intensity profile (the radial function is not known). From this experiment you get the intensity at each pixel, caused from the light source and other noise generators (not generally white noise).
The CCD like device has an unknown efficiency, and it gets randomly saturated at the high intensities directly in front of the light source. Therefore, the units and errors of the data points are pretty arbitrary.
The "good" points are located not directly in front of the source.
My problem is to find the center of the light source on the image. Then I would be able to perform more analyses on the data.