Finitely many random variables $p_1,p_2,...,p_n$ are independently and uniformly distributed on interval $[0,1]$. They form an ascending sequence $p_{i_1} \le p_{i_2} \le ... \le p_{i_n}$. For example, if $n=3$, $p_1=0.1, p_2=0.7, p_3 = 0.3$, the sequence is $p_1 \le p_3 \le p_2$.
In the case when two points coincide, we may let the one with lower index go first. For example, if $p_1=p_2=0.6,p_3=0.2$, then the sequence is $p_3\le p_1\le p_2$ since $p_1$ has lower index than $p_2$.
Prove that any sequence $i_1,i_2,\dots,i_n$ is equally likely to occur.
If the distribution is not uniform but still independent and identical, is every sequence still equally likely?
I have trouble proving this intuitive claim. Hope someone can help. Thank you!
PS: I realized this might need measure theory to get rigorously proved. For example, it can start by define probability space $([0,1],\cal F, \Bbb P)$ where $\cal F$ is the set of all Lebesgue measurable sets on $[0,1]$, and probability measure $\Bbb P$ is uniform (equivalent to Lebesgue measure in this case). Each $p_i$ is a $\cal F$-measurable function st. $p_i(\omega) = \omega$ for any $\omega\in[0,1]$.
Then this problem is equivalent to that in the product probability space $([0,1]^n,\cal F, \Bbb P)$, $\Bbb P\{p_{i_1} \le p_{i_2} \le ... \le p_{i_n}\}=\Bbb P\{p_{j_1} \le p_{j_2} \le ... \le p_{j_n}\}$ for any two index sequences $i_1,i_2,...,i_n$ and $j_1,j_2,...,j_n$. In the two dimensional case it is just $\Bbb P\{p_{1} \le p_{2}\}=\Bbb P\{p_{2} \le p_{1}\}$.
Anyone can help complete such proof? Thank you.
If the distribution is discrete, there may be a problem. For example, imagine tossing a fair coin that has a $0$ on one side and a $1$ on the other. Then all possibilities $(0,0)$, $(0,1)$, $(1,0)$, and $(1,1)$ are equally likely.
When we order, the result $(0,1)$ is twice as likely as either $(0,0)$ or $(1,1)$.
Remark: If the underlying distribution is continuous, after reordering all strictly increasing sequences are equally likely.