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        riddles >> putnam exam (pure math) >> Expected maximum value in [0,1]
        
(Message started by: mistaken_id on Mar 17^th, 2009, 10:16pm)

Title: Expected maximum value in [0,1]
Post by mistaken_id on Mar 17^th, 2009, 10:16pm

Suppose N people choose some value (real number) from the interval (0,1). What is the expected value of the maximum?

Also what is the expected value of the second maximum?

Title: Re: Expected maximum value in [0,1]
Post by ronnodas on Mar 17^th, 2009, 11:29pm

Assuming each person chooses their number independently with uniform probability, the expected maximum is N/(N+1) and the expected second maximum is (N-1)/(N+1). In fact, the expected ith minimum is i/(N+1).

Title: Re: Expected maximum value in [0,1]
Post by mistaken_id on Mar 18^th, 2009, 8:16am

Can you explain how??

Title: Re: Expected maximum value in [0,1]
Post by ronnodas on Mar 18^th, 2009, 8:37am

The maximum having value x (0<x<1) means that one person chose a number in [x, x+dx] and the other N-1 chose numbers in (0, x). So, the probability density function of the maximum f(x)= x^(N-1)/(int 0...1 x^(N-1)dx)

So the expected value is int 0...1 xf(x)dx which comes out to be N/(N+1)

Intuitively, you would expect the numbers to be uniformly distributed in the interval and the N+1 gaps between them to be equal which also conforms to the result.

Title: Re: Expected maximum value in [0,1]
Post by mistaken_id on Mar 19^th, 2009, 10:18am

on 03/18/09 at 08:37:10, ronnodas wrote:

So the expected value is int 0...1 xf(x)dx which comes out to be N/(N+1)

The integral of 0...1 xf(x)dx is 1/(N+1) rite?? How did you get N/N+1

Title: Re: Expected maximum value in [0,1]
Post by Obob on Mar 19^th, 2009, 10:36am

Check the definition of f(x) again.