# Permutation order statistics integral

Let $U_i$ be $[0,1]$ i.i.d. uniform random variables, for $i=1,\ldots,n$. As an example, let $n=3$. Now pick an ordering, say $x_1>x_2<x_3$. and consider the order statistics integral

$$3!\int\cdots\int_{x_1>x_2<x_3;\ \ 1>x_i>0} dx_1\,dx_2\,dx_3=2.$$

We get that this integral equals the number of permutations $\pi=(\pi_1,\pi_2,\pi_3)$ in $S_3$ with $\pi_1>\pi_2<\pi_3$. The only ones are $(3,1,2)$ and $(2,1,3)$ for a total of 2 as expected. In general, we have for a given fixed ordering $x_1?x_2\cdots?x_n$, where the question-marks correspond to $<$ or $>$:

$$|\{\pi: \pi_1?\cdots?\pi_n\}|=n!\int\cdots\int_{x_1?x_2\cdots?x_n;\ \ 1>x_i>0} dx_1\,dx_2\,dx_3.$$

Question: is there a sensible meaning for the integral:

$$n!\int_{x_1?x_2\cdots?x_n;\ \ 1>x_i>0} \,x_1\,dx_1\,dx_2\,dx_3\cdots dx_n.$$

I want to conclude that it's (related to) the expected value of the first element $\pi_1$ of a uniformly random permutation drawn from the set $\{\pi: \pi_1?\cdots?\pi_n\}$. Unfortunately, this does not seem to be the case. Is there a way to remedy this?

Consider the order $$\pi_1>\pi_2>\dots>\pi_n$$, so certainly $$\mathbb{E}\ \pi_1=n$$. Also, your integral now becomes $$n\int_{0< x_1< 1}\biggr(x_1(n-1)!\int_{x_2>\cdots> x_n:\ 0 Now, based on your observation, $$(n-1)!\int_{x_2>\cdots>x_n:\ 0\cdots>x_n:\ 0 via the substitution $$x_i\mapsto x_i/x_1$$. Thus, the original integral is $$n\int_0^1 x_1^n dx_1={n\over n+1}$$.
Similarly, consider the order $$\pi_1<\cdots<\pi_n$$, so certainly $$\mathbb{E}\ \pi_1=1$$. Now the integral becomes $$n\int_{0 Again, based on your observation, $$(n-1)!\int_{x_2<\cdots via the substitution $$x_i\mapsto {x_i-x_1\over 1-x_1}$$. Hence, the original integral is $$n\int_0^1 x_1(1-x_1)^{n-1}dx_1={1\over n+1}$$.
Now, I know that these are both very simple examples, but they suggest that the following may be true: $$\mathbb{E}_{\pi\sim\{\pi\in S_n:\pi_1?\cdots ?\pi_n\}}\ \pi_1=(n+1)!\int_{x_1?\cdots ?x_n:\ 0
Edit: Unfortunately, this isn't true in the example that you gave of $$\pi_1>\pi_2<\pi_3$$ where we have $$\mathbb{E}\ \pi_1=2.5$$. In this case, $$4!\int_{x_1>x_2 However, there are precisely 2 permutations which have $$\pi_1>\pi_2<\pi_3$$. Thus, a better conjecture (which still agrees with the first two examples I gave) is $$\mathbb{E}_{\pi\sim\{\pi\in S_n:\pi_1?\cdots ?\pi_n\}}\ \pi_1={(n+1)!\over|\{\pi\in S_n:\pi_1?\cdots ?\pi_n\}|} \int_{x_1?\cdots ?x_n:\ 0
Edit #2: Here is some extra partial evidence that the conjecture may be true. For an order $$\pi_1?\cdots?\pi_n$$, consider the set $$A=\{x\in\mathbb{R}^n: x_1?\cdots?x_n,\ 0. If we uniformly at random select a point in $$A$$, then $$\mathbb{E}_{x\sim A}\ x_1={1\over |A|}\int_A x_1 dx_1\cdots dx_n$$. Now, consider a random permutation $$\pi\sim\{\pi\in S^n:\pi_1?\cdots ?\pi_n\}$$ and consider the points $$x={1\over n+1}(\pi_1,\dots,\pi_n)$$. This is "kind of" a uniformly random point of $$A$$ (the reason I scale by $$n+1$$ is that if we scale only by $$n$$, then one of the coordinates of $$x$$ will always be $$1$$). Of course, it's not actually uniform since all of the coordinates of $$x$$ are of the form $${i\over n+1}$$ for some $$i\in[n]$$ and $$\sum_i x_i={n\over 2}$$. If, however, $$x$$ is close enough'' to being uniform on $$A$$, then we could get the desired inequality since $${1\over|A|}\int_A x_1dx_1\cdots dx_n={n!\over|\{\pi\in S_n:\pi_1?\cdots ?\pi_n\}|} \int_{x_1?\cdots ?x_n:\ 0 and then we'd get the extra $$n+1$$ from the scaling. Unfortunately, I don't see how to make this argument precise (especially since the scaling factor of $$n+1$$ seems rather arbitrary).