Damon Vazquez

2022-10-09

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, $\mathbb{P}\left({X}_{1}<{X}_{2}<\cdots <{X}_{n-1}\right)=\frac{1}{n-1!}$. Why can't a similar exchangability argument hold to show that $\mathbb{P}\left({X}_{1}>{X}_{2}>\cdots >{X}_{n-1}<{X}_{n}\right)=\frac{1}{n!}$? My reasoning is that the distribution of $\left({X}_{1},{X}_{2},\dots ,{X}_{n}\right)$ 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

Step 1
In the case of $\mathbb{P}\left({X}_{1}<{X}_{2}<\cdots <{X}_{n-1}\right)$ the permuted events all add up to the whole space. But in $\mathbb{P}\left({X}_{1}>{X}_{2}>\cdots >{X}_{n-1}<{X}_{n}\right)$ the permuattions don't exhaust the sample space.
Step 2
For example, $\left({X}_{1}>{X}_{2}<{X}_{3}\right)\cup \left({X}_{2}>{X}_{3}<{X}_{1}\right)\cup \left({X}_{1}>{X}_{3}<{X}_{2}\right)\cup \left({X}_{2}>{X}_{1}<{X}_{3}\right)\cup \left({X}_{3}>{X}_{2}<{X}_{1}\right)\cup \left({X}_{3}>{X}_{1}<{X}_{2}\right)$ does not include the event $\left({X}_{1}>{X}_{2}>{X}_{3}\right)$ or $\left({X}_{1}<{X}_{2}<{X}_{3}\right)$.

overrated3245w

Step 1
Since $P\left({X}_{i}={X}_{j}\right)=0$ for every $i\ne j$, we infer that
$\mathbb{P}\left({X}_{1}>{X}_{2}>\cdots >{X}_{n-1}<{X}_{n}\right)=$
$\mathbb{P}\left({X}_{1}>\cdots >{X}_{n-1}\right)-\mathbb{P}\left({X}_{1}>\cdots >{X}_{n-1}>{X}_{n}\right)=\frac{1}{\left(n-1\right)!}-\frac{1}{n!}=\frac{n-1}{n!}\phantom{\rule{thinmathspace}{0ex}}.$
Step 2
Another way to reach the same conclusion is to observe that the event ${X}_{1}>{X}_{2}>\cdots >{X}_{n-1}<{X}_{n}$ represents $n-1$ permutations of the n variables, determined by what is the first $k\in \left[1,n-1\right]$ such that ${X}_{n}>{X}_{k}$.

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