Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: version B 2.10.1 (Tek) 9/26/83; site tekchips.UUCP Path: utzoo!watmath!clyde!floyd!vax135!cornell!uw-beaver!tektronix!tekchips!stevev From: stevev@tekchips.UUCP (Steve Vegdahl) Newsgroups: net.math Subject: Re: Yet another probability puzzle Message-ID: <590@tekchips.UUCP> Date: Mon, 27-Feb-84 16:04:25 EST Article-I.D.: tekchips.590 Posted: Mon Feb 27 16:04:25 1984 Date-Received: Wed, 29-Feb-84 07:43:04 EST Organization: Tektronix, Beaverton OR Lines: 64 > What is the expected value of the range of N random points on > a line from 0 to 1? I think that's a concise statement of the > problem. To avoid ambiguity (I'm not a mathematician), I'll > restate it as I conceived it: You have this (finite) 1-dimensional > dart board at which you throw random darts. What is the > expected dispersion of N darts if they must all hit the board, > but any point within is equally probable? Assuming that your definition of "range" and "dispersion" is the distance between the min and max points on the line, I believe that answer is (N-1)/(N+1). Here is my reasoning (this was all conjured up without any stat or caluculus books, so someone please post a note if I goofed up). The density function for the max of N uniformly distributed random varibles on [0,1] is N-1 f (x) = Nx N The density function for the min of N uniformly distributed random varibles on [0,z] is N N-1 g (x,z) = --- (z-x) N N z The density function for the distance between min and max computed by integrating over combinations of points whose difference is y. / 1 | h (y) = | f (x) g (x-y,x) dx N | N N-1 / y (Please excuse the "ascii" integral sign.) This integral basically sums the probablities of the max of the N points being at x, and the min of the remaining N-1 points (which are now limited to being <= x) being at x-y. (Note that random variables for the min and max points are not independent, so it doesn't work to compute their expectations independently and then subtract.) This integration is easy to do because lots of terms cancel out. the result is N-2 h (y) = N(N-1)(1-y)y N The expected value can then be computed by integrating y h (y) dy N over the interval [0,1], giving our result of (N-1)/(N+1). As I said before, I don't have a lot of time to verify this. A sanity check however, indicates that it works for N = 1, where the range should obviously be zero, and approaches 1 as N approaches infinity, again consistent with intuition. Finally, the density function h integrated over [0,1] is 1, and is clearly always non-negative for positive N, hence it is a feasible density function. Would someone like to corroborate or contradict? Steve Vegdahl Tektronix Inc. Beaverton, Oregon