On Sep 19, 8:41 pm, Peter <peterp...@hotmail.com> wrote: > I do not know this particular application, but consider that, to keep frequencies equal, high-density events will be binned in smaller intervals. So probability is inversely related to the length of the interval (as in |a2-a1|) that your observation p falls in.
Define pi = P(p in bin i), li = length of interval i, ni = instances in bin i, k = constant of proportionality Then
pi = k * ni / li sum(pi) = sum (k * ni / li) = 1 1 = k * sum (ni / li) k = 1 / sum (ni / li)
therefore pi = ni / (li * sum (ni / li))
I suppose this makes sense from a mathematical derivation point of view, I just find it hard to see the intuitive view of it.