Warren said: Now let us ask, what if we assume exponential not uniform distribution - which has the advantage of being self-similar, and with no high-cutoff necessary, causing the formula we shall get to be valid everywhere - and ask for y so that integral(from A to y) (1-A/x)*exp(-K*x) dx = integral(from y to B) (B/x-1)*exp(-K*x) dx ? then we get a nasty transcendental equation involving "exponential integrals" (higher transcendental fns) to solve numerically.
I reply: Tha's why I wouldn't use an exponential. I'd use B/(x+A). It would do. Warren continues: I had derived this equation in an earlier email to Ossipoff some days ago. I reply: Warren sent me pages and pages of gibberish. I confess that I didn't read it. His writing doesn't inspire much confidence. Mike Ossipoff _________________________________________________________________ Your Hotmail address already works to sign into Windows Live Messenger! Get it now http://clk.atdmt.com/MSN/go/msnnkwme0020000001msn/direct/01/?href=http://get.live.com/messenger/overview ---- election-methods mailing list - see http://electorama.com/em for list info