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From: stevev@tekchips.UUCP
Newsgroups: net.math
Subject: Re: Yet another probability puzzle
Message-ID: <589@tekchips.UUCP>
Date: Mon, 27-Feb-84 15:41:28 EST
Article-I.D.: tekchips.589
Posted: Mon Feb 27 15:41:28 1984
Date-Received: Wed, 29-Feb-84 12:38:07 EST
Organization: Tektronix, Beaverton OR
Lines: 50

> 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 leftmost and rightmost 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 h(y) 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.
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

The expected value of h(y) (which was the original question) is then found
easily by integrating
			y h(y) dy
over the interval [0,1], giving