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   Expected maximum value in [0,1]
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   Author  Topic: Expected maximum value in [0,1]  (Read 2083 times)
mistaken_id
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Expected maximum value in [0,1]  
« on: Mar 17th, 2009, 10:16pm »
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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?
« Last Edit: Mar 17th, 2009, 10:17pm by mistaken_id » IP Logged
Ronno
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Re: Expected maximum value in [0,1]  
« Reply #1 on: Mar 17th, 2009, 11:29pm »
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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).
« Last Edit: Mar 17th, 2009, 11:36pm by Ronno » IP Logged

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mistaken_id
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Re: Expected maximum value in [0,1]  
« Reply #2 on: Mar 18th, 2009, 8:16am »
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Can you explain how??
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Ronno
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Re: Expected maximum value in [0,1]  
« Reply #3 on: Mar 18th, 2009, 8:37am »
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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.
« Last Edit: Mar 18th, 2009, 8:42am by Ronno » IP Logged

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mistaken_id
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Re: Expected maximum value in [0,1]  
« Reply #4 on: Mar 19th, 2009, 10:18am »
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on Mar 18th, 2009, 8:37am, 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
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Obob
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Re: Expected maximum value in [0,1]  
« Reply #5 on: Mar 19th, 2009, 10:36am »
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Check the definition of f(x) again.
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