Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

(Random number with uniform distribution over [0, 1])

For clarification, in the case where N = 2, we can use geometric probability. On the coordinate plane consider points with 0<=x,y<=1. The condition is satisfied on a diagonal band of area 3/4 from the origin to (1,1). Similarly, with N = 3 the volume of the space in which the condition is satisfied is 7/27.

share|improve this question
    
How did you compute 7/27, and why did your argument not generalize to arbitrary N? –  Steven Landsburg Jan 5 '11 at 21:23
    
What is meant by "the range of a set"? Is that just max minus min? –  Gerry Myerson Jan 6 '11 at 4:03

1 Answer 1

The keyword is the order statistics. The distributions of the maximum and minimum values of a sample of $n$ independent uniformly distributed random variables are given respectively by the laws

$$U_{max}\sim \mbox{Beta}(n,1),\qquad U_{min}\sim \mbox{Beta}(1,n).$$

The range $U_{max}-U_{min}$ has a $\mbox{Beta}(n-1,2)$ distribution (see, e.g., Section 2.5 of A First Course in Order Statistics) so $$\mathbb P\{U_{max}-U_{min} < a\}=\frac{1}{B(n-1,2)}\int_{0}^{a}x^{n-2}(1-x)dx=na^{n-1}-(n-1)a^n.$$

share|improve this answer
2  
=$(n^2-n+1)n^{-n}$ which agrees with the examples given (however arrived at) –  Mark Bennet Jan 5 '11 at 22:23
    
(Could I suggest writing $U_\max$ and $U_\min$ instead of $U_{max}$ and $U_{min}$? The TeX code is U_\max and U_\min. Using \max and \min not only prevents italicization, but also causes standard formatting conventions to be followed in some contexts.) –  Michael Hardy Jan 6 '11 at 18:40
    
@Michael Hardy: I agree with your suggestion and I will try to stick to it in the future. I just don't think it is worth bumping up the question to make the edit. –  Andrey Rekalo Jan 6 '11 at 19:34

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.