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, $\mathbb{P}({X}_{1}<{X}_{2}<\cdots <{X}_{n-1})=\frac{1}{n-1!}$. Why can't a similar exchangability argument hold to show that $\mathbb{P}({X}_{1}>{X}_{2}>\cdots >{X}_{n-1}<{X}_{n})=\frac{1}{n!}$? My reasoning is that the distribution of $({X}_{1},{X}_{2},\dots ,{X}_{n})$ is symmetric, so no matter which permutation of the random variables exist in the random vector, they will all have the same probability.

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

Beginner2022-10-10Added 7 answers

Step 1

In the case of $\mathbb{P}({X}_{1}<{X}_{2}<\cdots <{X}_{n-1})$ the permuted events all add up to the whole space. But in $\mathbb{P}({X}_{1}>{X}_{2}>\cdots >{X}_{n-1}<{X}_{n})$ the permuattions don't exhaust the sample space.

Step 2

For example, $({X}_{1}>{X}_{2}<{X}_{3})\cup ({X}_{2}>{X}_{3}<{X}_{1})\cup ({X}_{1}>{X}_{3}<{X}_{2})\cup ({X}_{2}>{X}_{1}<{X}_{3})\cup ({X}_{3}>{X}_{2}<{X}_{1})\cup ({X}_{3}>{X}_{1}<{X}_{2})$ does not include the event $({X}_{1}>{X}_{2}>{X}_{3})$ or $({X}_{1}<{X}_{2}<{X}_{3})$.

In the case of $\mathbb{P}({X}_{1}<{X}_{2}<\cdots <{X}_{n-1})$ the permuted events all add up to the whole space. But in $\mathbb{P}({X}_{1}>{X}_{2}>\cdots >{X}_{n-1}<{X}_{n})$ the permuattions don't exhaust the sample space.

Step 2

For example, $({X}_{1}>{X}_{2}<{X}_{3})\cup ({X}_{2}>{X}_{3}<{X}_{1})\cup ({X}_{1}>{X}_{3}<{X}_{2})\cup ({X}_{2}>{X}_{1}<{X}_{3})\cup ({X}_{3}>{X}_{2}<{X}_{1})\cup ({X}_{3}>{X}_{1}<{X}_{2})$ does not include the event $({X}_{1}>{X}_{2}>{X}_{3})$ or $({X}_{1}<{X}_{2}<{X}_{3})$.

overrated3245w

Beginner2022-10-11Added 3 answers

Step 1

Since $P({X}_{i}={X}_{j})=0$ for every $i\ne j$, we infer that

$$\mathbb{P}({X}_{1}>{X}_{2}>\cdots >{X}_{n-1}<{X}_{n})=$$

$$\mathbb{P}({X}_{1}>\cdots >{X}_{n-1})-\mathbb{P}({X}_{1}>\cdots >{X}_{n-1}>{X}_{n})=\frac{1}{(n-1)!}-\frac{1}{n!}=\frac{n-1}{n!}\phantom{\rule{thinmathspace}{0ex}}.$$

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 $k\in [1,n-1]$ such that ${X}_{n}>{X}_{k}$.

Since $P({X}_{i}={X}_{j})=0$ for every $i\ne j$, we infer that

$$\mathbb{P}({X}_{1}>{X}_{2}>\cdots >{X}_{n-1}<{X}_{n})=$$

$$\mathbb{P}({X}_{1}>\cdots >{X}_{n-1})-\mathbb{P}({X}_{1}>\cdots >{X}_{n-1}>{X}_{n})=\frac{1}{(n-1)!}-\frac{1}{n!}=\frac{n-1}{n!}\phantom{\rule{thinmathspace}{0ex}}.$$

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 $k\in [1,n-1]$ such that ${X}_{n}>{X}_{k}$.

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