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I've been working on a probability-related problem and I've come across the following integrals of the form $\int_{0}^1 1-x \ \mathrm{d}x$, $\int_0^1 \int_0^{1-x} 1-x-y \ \mathrm{d}y \ \mathrm{d}x$, $\int_0^1 \int_0^{1-x} \int_0^{1-x-y} 1-x-y-z \ \mathrm{d}z \ \mathrm{d}y \ \mathrm{d}x$, $\int_0^1 \int_0^{1-x} \int_0^{1-x-y} \int_0^{1-x-y-z} 1-x-y-z-w \ \mathrm{d}w \ \mathrm{d}z \ \mathrm{d}y \ \mathrm{d}x$, etc. I have evaluated such integrals and found them to be $\frac{1}{2}$, $\frac{1}{6}$, $\frac{1}{24}$, $\frac{1}{120}$ respectively. From this I conclude that the general pattern is that of $\frac{1}{n!}$. I would, however, like to prove such a fact.

I have concluded also that the integrals yield the exact same results if thought of as $\int_{0}^1 1 \ \mathrm{d}x$, $\int_0^1 \int_0^{1-x} 1 \ \mathrm{d}y \ \mathrm{d}x$, $\int_0^1 \int_0^{1-x} \int_0^{1-x-y} 1 \ \mathrm{d}z \ \mathrm{d}y \ \mathrm{d}x$, $\int_0^1 \int_0^{1-x} \int_0^{1-x-y} \int_0^{1-x-y-z} 1 \ \mathrm{d}w \ \mathrm{d}z \ \mathrm{d}y \ \mathrm{d}x$, etc. The general form I am looking for is:

$\frac{{\int \dots \int}_{\prod_{k = 0}^{n-1}[0, 1 - \sum_{i = 1}^kx_i)}1 \ \mathrm{d}x_1 \ \dots \ \mathrm{d}x_n}{{\int \dots \int}_{[0, 1]^n}1 \ \mathrm{d}x_1 \ \dots \ \mathrm{d}x_n}$

And, as the denumerator here is always $=1$, this is just

${\int \dots \int}_{\prod_{k = 0}^{n-1}[0, 1 - \sum_{i = 1}^kx_i)}1 \ \mathrm{d}x_1 \ \dots \ \mathrm{d}x_n$

I have, however, no idea how one may go about proving that such an expression is $= \frac{1}{n!}$. The first idea that comes to mind would be proving so by recursion/induction on the $I_n$ integrals, but I can't really find a way to neatly express $I_{n+1}$ in terms of the preceding $I_k$... Another way might be to make use of Fubini? But I don't think that makes sense seeing how each nested term is dependent on the evaluation of the previous preceding ones, and so "taking them out" and multiplying them would really make no sense.

Intuitively, I can see the geometric interpretation behind what we are doing (going from a line, to a square, to a cube, etc., up to $n$ dimensions) and might have a general idea of why the result (interpeted as length, area, volume, etc.) is so, but, again, no way of actually proving it.

I thank in advance any help or clarifying regarding this matter!

Sho
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1 Answers1

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I think you are essentially computing the volume of the portion of the unit cube in $\mathbb R^n$ defined by an equation $0 < x_1 < x_2 < \ldots < x_n < 1$. The whole cube is divided into $n!$ such areas, that are mapped into each other by the $n!$ permutations of the set of $n$ coordinates. Every such area has the same volume, and the total volume of the cube is 1, so the volume of each single area is $1/n!$.

Misha
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