On 03/29/2010 07:17 PM, Barry Rowlingson wrote:
...
I think the problem as posed doesn't produce a unique ellipse. You
could start with a circle of radius 0 centered on mean(x),mean(y) and
then increase the radius until it has 95% of the points in it. As long
as your points are in continuous space and with no coincident points
then you could do a simple bisection search on the radius.
Similarly you could start with an ellipse of any eccentricity
centered at the same point with fixed angle and do the same. And the
ellipse doesn't even need to be centered at the mean point - it could
be waaay over to the left and eventually as it gets bigger it will
gobble up 95% of the points.
Obviously with bivariate normally-distributed points we tend to show
the ellipse that is numerically derived from the mean and correlation
of the two normals, but that's not the only ellipse that takes 95% of
the points.
So ummm I'm not sure what you should do. What is the question you are
trying to answer?
So, why not begin with a problem that is uniquely soluble and achieve
the viewpoint of its solution?
1) If we assume that the distribution has a barycenter, then that can be
calculated.
2) Calculate the distances of all points from the barycenter, flagging
the 5% most distant.
3) Divide the area covered by the points into an arbitrary number of
equal sectors, say 10.
4) Within each sector, find the most distant "inner" point and the least
distant "outer" point, placing a dot in the middle of the sector at 5%
of their difference beyond the radius of the most distant "inner" point.
5) Join the dots.
6) Look at it.
Jim
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