Probability of first increase in ordering of iid random variablesWhat 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 &lt; X 2 &lt; ⋯ &lt; X n − 1 ) = 1 n − 1 ! . Why can't a similar exchangability argument hold to show that P ( X 1 &gt; X 2 &gt; ⋯ &gt; X n − 1 &lt; 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.

Question

Probability of first increase in ordering of iid random variablesWhat 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        &amp;lt;      X          2        &amp;lt;  ⋯  &amp;lt;      X          n      −      1        )  =      1          n      −      1      !      . Why can&#039;t a similar exchangability argument hold to show that       P    (      X          1        &amp;gt;      X          2        &amp;gt;  ⋯  &amp;gt;      X          n      −      1        &amp;lt;      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.

Cullen Kelly · Accepted Answer

Step 1In the case of       P    (      X          1        &amp;lt;      X          2        &amp;lt;  ⋯  &amp;lt;      X          n      −      1        ) the permuted events all add up to the whole space. But in       P    (      X          1        &amp;gt;      X          2        &amp;gt;  ⋯  &amp;gt;      X          n      −      1        &amp;lt;      X          n        ) the permuattions don&#039;t exhaust the sample space.Step 2For example,   (      X    1    &amp;gt;      X    2    &amp;lt;      X    3    )  ∪  (      X    2    &amp;gt;      X    3    &amp;lt;      X    1    )  ∪  (      X    1    &amp;gt;      X    3    &amp;lt;      X    2    )  ∪  (      X    2    &amp;gt;      X    1    &amp;lt;      X    3    )  ∪  (      X    3    &amp;gt;      X    2    &amp;lt;      X    1    )  ∪  (      X    3    &amp;gt;      X    1    &amp;lt;      X    2    ) does not include the event   (      X    1    &amp;gt;      X    2    &amp;gt;      X    3    ) or   (      X    1    &amp;lt;      X    2    &amp;lt;      X    3    ).

overrated3245w · Accepted Answer

Step 1Since   P  (      X    i    =      X    j    )  =  0 for every   i  ≠  j, we infer that      P    (      X          1        &amp;gt;      X          2        &amp;gt;  ⋯  &amp;gt;      X          n      −      1        &amp;lt;      X          n        )  =      P    (      X          1        &amp;gt;  ⋯  &amp;gt;      X          n      −      1        )  −      P    (      X          1        &amp;gt;  ⋯  &amp;gt;      X          n      −      1        &amp;gt;      X          n        )  =      1          (      n      −      1      )      !        −      1          n      !        =            n      −      1              n      !          .Step 2Another way to reach the same conclusion is to observe that the event       X          1        &amp;gt;      X          2        &amp;gt;  ⋯  &amp;gt;      X          n      −      1        &amp;lt;      X          n       represents   n  −  1 permutations of the n variables, determined by what is the first   k  ∈  [  1  ,  n  −  1  ] such that       X    n    &amp;gt;      X    k  .

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?

Answered question

Answer & Explanation

New Questions in High school probability