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?


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 $$n\int_{0< x_1< 1}\biggr(x_1(n-1)!\int_{x_2>\cdots> x_n:\ 0<x_i<x_1}dx_2\cdots dx_n\biggl) dx_1.$$ Now, based on your observation, $$(n-1)!\int_{x_2>\cdots>x_n:\ 0<x_i<x_1}dx_2\cdots dx_n=x_1^{n-1}(n-1)!\int_{x_2>\cdots>x_n:\ 0<x_i<1}dx_2\cdots dx_n=x_1^{n-1},$$ 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 $$n\int_{0<x_1<1}\biggl(x_1(n-1)!\int_{x_2<\cdots<x_n:\ x_1<x_i<1}dx_2\cdots dx_n\biggr)dx_1.$$ Again, based on your observation, $$(n-1)!\int_{x_2<\cdots<x_n:\ x_1<x_i<1}dx_2\cdots dx_n=(1-x_1)^{n-1}(n-1)!\int_{x_2<\cdots<x_n:\ 0<x_i<1}dx_2\cdots dx_n=(1-x_1)^{n-1},$$ 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: $$\mathbb{E}_{\pi\sim\{\pi\in S_n:\pi_1?\cdots ?\pi_n\}}\ \pi_1=(n+1)!\int_{x_1?\cdots ?x_n:\ 0<x_i<1}x_1dx_1\cdots dx_n.$$

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, $$4!\int_{x_1>x_2<x_3:\ 0<x_i<1}x_1dx_1dx_2dx_3=4!\int_{x_2=0}^1\biggl(\int_{x_2}^1 x_1dx_1\int_{x_2}^1dx_3\biggr)dx_2={4!\over 2}\int_0^1(1-x_2)(1-x_2^2)dx_2=5.$$ 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 $$\mathbb{E}_{\pi\sim\{\pi\in S_n:\pi_1?\cdots ?\pi_n\}}\ \pi_1={(n+1)!\over|\{\pi\in S_n:\pi_1?\cdots ?\pi_n\}|} \int_{x_1?\cdots ?x_n:\ 0<x_i<1}x_1dx_1\cdots dx_n.$$

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 $A=\{x\in\mathbb{R}^n: x_1?\cdots?x_n,\ 0<x_i<1\}$. 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 $${1\over|A|}\int_A x_1dx_1\cdots dx_n={n!\over|\{\pi\in S_n:\pi_1?\cdots ?\pi_n\}|} \int_{x_1?\cdots ?x_n:\ 0<x_i<1}x_1dx_1\cdots dx_n,$$ 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).


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