What is the probability that the first n-1 terms of iid Unif(0,1) random draws are in decreasing order, but the first n terms are not?

Damon Vazquez 2022-10-09 Answered
Probability of first increase in ordering of iid random variables
What is the probability that the first n 1 terms of iid Unif(0,1) random draws are in decreasing order, but the first n terms are not?
I know that due to exchangability, P ( X 1 < X 2 < < X n 1 ) = 1 n 1 ! . Why can't a similar exchangability argument hold to show that P ( X 1 > X 2 > > X n 1 < X n ) = 1 n ! ? My reasoning is that the distribution of ( X 1 , X 2 , , X n ) is symmetric, so no matter which permutation of the random variables exist in the random vector, they will all have the same probability.
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Answers (2)

Cullen Kelly
Answered 2022-10-10 Author has 7 answers
Step 1
In the case of P ( X 1 < X 2 < < X n 1 ) the permuted events all add up to the whole space. But in P ( X 1 > X 2 > > X n 1 < X n ) the permuattions don't exhaust the sample space.
Step 2
For example, ( X 1 > X 2 < X 3 ) ( X 2 > X 3 < X 1 ) ( X 1 > X 3 < X 2 ) ( X 2 > X 1 < X 3 ) ( X 3 > X 2 < X 1 ) ( X 3 > X 1 < X 2 ) does not include the event ( X 1 > X 2 > X 3 ) or ( X 1 < X 2 < X 3 ).
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overrated3245w
Answered 2022-10-11 Author has 3 answers
Step 1
Since P ( X i = X j ) = 0 for every i j, we infer that
P ( X 1 > X 2 > > X n 1 < X n ) =
P ( X 1 > > X n 1 ) P ( X 1 > > X n 1 > X n ) = 1 ( n 1 ) ! 1 n ! = n 1 n ! .
Step 2
Another way to reach the same conclusion is to observe that the event X 1 > X 2 > > X n 1 < X n represents n 1 permutations of the n variables, determined by what is the first k [ 1 , n 1 ] such that X n > X k .
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