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.
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 |
Continuous version for the smallest one
