This is the explanation from Wikipedia: enter image description here

Is there a more rigorous proof or explanation of how reducing the integration region to these sub-regions introduces a $\frac{1}{n!}$ factor? I am confused about why the number of these sub-regions is $n!$ in the first place. They way the sub-regions are defined it would seem as if there are $n$ of them.

The $\frac{1}{n!}$ factor and the fact that $K$ has to be symmetric suggests that the proof involves looking at different permutations of the time boundaries $t_1, t_2,$ etc. However, the proof given here (and other proofs I've found) never show this step explicitly.


1 Answer 1


There are $n$ time variables. Each way of ordering the times is a permutation of the $n$ variables. There are $n!$ permutations of $n$ time variables ($n$ ways to choose the highest time, $n-1$ ways to choose the second highest time, etc.).

  • $\begingroup$ I see. Now I understand it intuitively, but how do we rigorously prove that the n! number of such integrals S_n is equal to I_n? $\endgroup$
    – pll04
    Commented Feb 27 at 16:14
  • $\begingroup$ @pll04 Have you done it for $n=2$? $\endgroup$ Commented Feb 27 at 16:31
  • $\begingroup$ @TobiasFünke Yes, by setting K(t_1,t_2) = 1. But I couldn't work out a proof for general n, nor for integral with a general symmetric K. $\endgroup$
    – pll04
    Commented Feb 27 at 16:45
  • 1
    $\begingroup$ @pll04 Since you've done it for $n=2$, repeat for $n=3$. That'll give you a feel for using induction to do $m=n+1$, and then you're set. $\endgroup$
    – rob
    Commented Feb 27 at 17:10
  • $\begingroup$ @rob I see, thanks. But how can we extend this to general cases where K isn't 1? $\endgroup$
    – pll04
    Commented Feb 27 at 17:19

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