MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose you sample uniformly from the unit vectors in R^n. What are the distributions of the order statistics of the magnitudes of the components of the sampled vectors? That is, for 1 <= i <= n and x in [0,1], what is the probability that the i'th largest component of the vector (in absolute value) is less than or equal to x?

share|cite|improve this question
It doesn't answer your question, but would it help to model the uniform distribution on the sphere by a random vector $(X_1/R, X_2/R, ..., X_n/R)$ where $X_1, \dots, X_N$ are i.i.d. standard Gaussians and R is defined to be $(X_1^2+\dots+X_n^2)^{1/2}$ ? Or have you tried this already? – Yemon Choi Nov 20 '09 at 8:19

There has been some work in the physics community on extreme statistics (i.e. distribution of largest and smallest components) of random vectors. See, link text for example. The largest component is approximately distributed like a Gumbel random variable, while the smallest component is approximately distributed like an exponential random variable.

share|cite|improve this answer

The distribution should be obtainable by integrating over the section of the simplex segment of the surface of the hypersphere bounded by the points (1,0,0,0,...), (1,1,0,0,0...)/sqrt(2), (1,1,1,0,0...)/sqrt(3) etc. along the ith axis.

All the distributions (n,m) have support contained within the unit interval, are piecewise smooth and share the same set of non-smooth points at the reciprocals of the square roots of the natural numbers.

share|cite|improve this answer

Your problem is closely related to order statistics for normal random variables, so you may find this paper useful: Percentage Points and Modes of Order Statistics from the Normal Distribution by Shanti S. Gupta

share|cite|improve this answer

Your Answer


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.