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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

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

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 $x_1?x_2cdots?x_n$, where the question-marks correspond to $<$ or $>$:

$$|{pi: pi_1?cdots?pi_n}|=n!intcdotsint_{x_1?x_2cdots?x_n; 1>x_i>0} dx_1,dx_2,dx_3.$$

Question: is there a sensible meaning for the integral:

$$n!int_{x_1?x_2cdots?x_n; 1>x_i>0} ,x_1,dx_1,dx_2,dx_3cdots dx_n.$$

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?

This isn't a full answer, but may be a helpful thought.

Consider the order $pi_1>pi_2>dots>pi_n$, so certainly $mathbb{E} pi_1=n$. Also, your integral now becomes $$nint_{0< x_1< 1}biggr(x_1(n-1)!int_{x_2>cdots> x_n: 0<x_i<x_1}dx_2cdots dx_nbiggl) dx_1.$$ Now, based on your observation, $$(n-1)!int_{x_2>cdots>x_n: 0<x_i<x_1}dx_2cdots dx_n=x_1^{n-1}(n-1)!int_{x_2>cdots>x_n: 0<x_i<1}dx_2cdots dx_n=x_1^{n-1},$$ via the substitution $x_imapsto x_i/x_1$.
Thus, the original integral is $nint_0^1 x_1^n dx_1={nover n+1}$.

Similarly, consider the order $pi_1<cdots<pi_n$, so certainly $mathbb{E} pi_1=1$. Now the integral becomes $$nint_{0<x_1<1}biggl(x_1(n-1)!int_{x_2<cdots<x_n: x_1<x_i<1}dx_2cdots dx_nbiggr)dx_1.$$ Again, based on your observation, $$(n-1)!int_{x_2<cdots<x_n: x_1<x_i<1}dx_2cdots dx_n=(1-x_1)^{n-1}(n-1)!int_{x_2<cdots<x_n: 0<x_i<1}dx_2cdots dx_n=(1-x_1)^{n-1},$$ via the substitution $x_imapsto {x_i-x_1over 1-x_1}$. Hence, the original integral is $nint_0^1 x_1(1-x_1)^{n-1}dx_1={1over n+1}$.

Now, I know that these are both very simple examples, but they suggest that the following may be true: $$mathbb{E}_{pisim{piin S_n:pi_1?cdots ?pi_n}} pi_1=(n+1)!int_{x_1?cdots ?x_n: 0<x_i<1}x_1dx_1cdots dx_n.$$

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, $$4!int_{x_1>x_2<x_3: 0<x_i<1}x_1dx_1dx_2dx_3=4!int_{x_2=0}^1biggl(int_{x_2}^1 x_1dx_1int_{x_2}^1dx_3biggr)dx_2={4!over 2}int_0^1(1-x_2)(1-x_2^2)dx_2=5.$$
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 $$mathbb{E}_{pisim{piin S_n:pi_1?cdots ?pi_n}} pi_1={(n+1)!over|{piin S_n:pi_1?cdots ?pi_n}|} int_{x_1?cdots ?x_n: 0<x_i<1}x_1dx_1cdots dx_n.$$

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 $A={xinmathbb{R}^n: x_1?cdots?x_n, 0<x_i<1}$. If we uniformly at random select a point in $A$, then $mathbb{E}_{xsim A} x_1={1over |A|}int_A x_1 dx_1cdots dx_n$. Now, consider a random permutation $pisim{piin S^n:pi_1?cdots ?pi_n}$ and consider the points $x={1over n+1}(pi_1,dots,pi_n)$. 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 ${iover n+1}$ for some $iin[n]$ and $sum_i x_i={nover 2}$. If, however, $x$ is ``close enough'' to being uniform on $A$, then we could get the desired inequality since $${1over|A|}int_A x_1dx_1cdots dx_n={n!over|{piin S_n:pi_1?cdots ?pi_n}|} int_{x_1?cdots ?x_n: 0<x_i<1}x_1dx_1cdots dx_n,$$ 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).