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 Expected of each (Posted on 2014-05-15)
Randomly select n numbers from the continuous interval (0,1) and order them from least to greatest.

Find the expected values, in terms of n, of the smallest, second smallest, ..., second greatest, greatest.

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 Continuous version for the smallest one Comment 3 of 3 |
Calculus solution shows:  1/(n+1) for the smallest one.

Select n uniformly distributed, independent random variables between 0 and 1: {a1,a2, ... , an}
F:  Cumulative distribution function
f:  Probability density function
f is the derivative of F
p(event) is the probablitity of some event
E(x) is the Expected value of x

For any such random variable, say ai
F(x) = p(ai<x) = x
F(x) = p(ai>x) = 1-x
let m(i,n) = the i-th smallest of the n random numbers
m(1,n) is the minimum and m(n,n) is the maximum for example.

First consider F(x) for the smallest one:  m(1,n).  I am sure this could be generalized to all i but for now I'm just doing the minimum value:
F(x) = 1 - p{(a1 > x)  AND (a2 > x) AND ... AND ... (an>x)}
F(x) = 1 - (1-x)^(n)
Take the derivative to get little f.
f(x) = - n*(1-x)^(n-1)

E(x) = {integral of  x * f(x) dx} / {integral of  f(x) dx}  each integral is from 0 to 1
Let z = 1-x ; so dz = - dx ;  limits on integral become 1 to 0 instead of 0 to 1
Both numerator and denominator have - signs and an n, so they cancel out.

Numerator:  integral{(1-z)*z^(n-1) dz} = integral{z^(n-1) dz} - integral{z^(n) dz}
Denominator:  integral{z^(n-1) dz}

Numerator: z^n/n - z^(n+1)/(n+1)
Denominator: z^n/n
Evaluate at zero then subtract evaluation at 1:
0 - (1/n - 1/(n+1)) / (-1/n) = 1 - n/(n+1) = 1/(n+1)

So the expected value of the smallest of n such random variables is:
;;;;;;     1/(n+1)     ;;;;;

I'll stop here and see if anyone wants to generalize to all i.
I think big F would have terms like  x^i and (1-x)^(n-i+1) but I'm not totally sure about it

 Posted by Larry on 2014-05-18 13:58:05

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