Quote:
Originally Posted by twoheadedboy
Could you try explaining in words why the minimum of exponentials is independent of their ordering?
Perhaps I do not understand your question, since what you seem to be describing has nothing to do with exponentials. The minimum of any iid collection of random variables is independent of their ordering, provided their distribution has no atoms (which is the case if they have a density, for example).
More precisely, let X
1,...,X
n be iid with common cumulative distribution function F(x). Let M = min(X
1,...,X
n) and let
σ be the random permutation describing the ordering of the X's. (That is,
σ is a random variable taking values in the symmetric group S
n, satisfying X
σ(1) < ... < X
σ(n).) If F is continuous, then M and
σ are independent.
In words, I think it is pretty easy to be convinced that this should be true. Just consider two random variables X and Y. Initially, by symmetry, we have
P(X < Y) = P(Y < X) = 1/2.
Now suppose we discover that min(X,Y) = 7, for example. Then again by symmetry, we know that the two possibilities
(1) X < Y and X = 7,
(2) Y < X and Y = 7,
are equally likely. In other words,
P(X < Y | min(X,Y) = 7) = P(Y < X | min(X,Y) = 7) = 1/2.
So learning the value of the minimum tells us nothing about the ordering of the variables.