Consider the simplex determined by the origin, and $n$ unit basis vectors. The volume of this simplex is $\frac{1}{n!}$, but I am intuitively struggling to see why. I have seen proofs for this and am convinced, but I can't help but think there must be a slicker or more intuitive argument for why this is so than what I have already seen. Any help would be appreciated!
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10If $(X_1, ..., X_n)$ is a random vector with each component independent and uniformly distributed in $[0,1]$, then $P[X_1 < X_2 < ... , X_n]=P[X_2 < X_1 < ... < X_n]$, and so on, so all $n!$ permutations are equally likely. So each probability is $1/n!$. For example, for $n=3$ we get $$P[X_1<X_2<X_3]=P[X_1<X_3<X_2]=P[X_2<X_1<X_3]=P[X_2<X_3<X_1]=P[X_3<X_1<X_2]=P[X_3<X_2<X_1] $$ Notice also that $P[X_i=X_j]=0$ in this scenario, whenever $i \neq j$. – Michael Mar 29 '16 at 00:52
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2I've been wondering recently if there is a way that the volume of a simplex is related to the fact that $\frac{1}{n!}$ comes up in Taylor series. Specifically, the related simplex with sides $x$ would have volume $\frac{x^n}{n!}$. – Thomas Andrews Mar 29 '16 at 01:09
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In $\Bbb R^3$, I've found the volume two different ways and keep getting $\frac1{12} \neq \frac{1}{3!} = \frac16$. Curious. – pjs36 Mar 29 '16 at 15:59
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How are you finding it, @pjs36? Are you using an integral? – Thomas Andrews Mar 29 '16 at 16:23
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2The proof is fairly simple. The simplex is the set of $(x_1,x_2,\dots,x_n)$ with $x_i\geq 0$ and $\sum x_i\leq 1$. Then, by induction, you get that the area volume $V_n$ is $\int_{0}^1 V_{n-1}(1-x_n)^{n-1},dx_n$. That is, each value for $x_n$, the other values are in a similar $n-1$-simplex, scaled by a factor of $(1-x_n)$. So the integral is $V_n=V_{n-1}\frac{1}{n}$. @pjs36 – Thomas Andrews Mar 29 '16 at 16:28
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@ThomasAndrews Just silly mistakes that I didn't see in the moment (no integrals, just elementary geometry). I was playing around hoping to find a symmetry argument. – pjs36 Mar 30 '16 at 14:23
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1https://math.stackexchange.com/questions/769545/volume-of-t-n-x-i-ge0x-1-cdotsx-n-le1/ – StubbornAtom Nov 26 '19 at 18:03
3 Answers
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It's easier to see an alternate simplex has volume $1/n!$: The set of all points $(x_1,x_2,\dots,x_n)$ with $0\leq x_1\leq x_2\leq\cdots\leq x_n\leq 1$. That's because the volume measures the probability that a random sequence of $n$ real numbers is in sorted order, and (except with the probability zero case where some pair of values are equal) there are $n!$ ways to permute a set of $n$ values, and only one of them is sorted.
That this is the same volume as the original simplex is a little harder to see - these two simplices are not congruent, so their equal volume requires a little linear algebra.
There is a linear transformation between the two sending $0\leq x_1\leq x_2\leq \cdots\leq x_n\leq 1$ to $(x_1,x_2-x_1,\dots,x_n-x_{n-1})$. The determinant of this linear transformation is $1$, so it preserves hyper-volumes.
Another approach is to ask how many ways can $n$ natural numbers $a_1,a_2,\dots,a_n$ be chosen so that $a_1+a_2+\cdots+a_n \leq m$. Standard combinatorics says this is $\binom{m+n}{n}$. And we see that $\frac{1}{m^n}\binom{n+m}{n}$ is an approximation of the hypervolume you want as $m$ gets large, by dividing the space into "hypercubes" of side length $\frac{1}{m}$, and this value approaches the volume as $m\to\infty$.
joriki
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Thomas Andrews
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Denote the volume of this simplex by $\sigma_n$. Foliating the simplex with "horizontal" hyperplanes $x_n=z$ $(0\leq z\leq 1)$ and applying Fubini's theorem we obtain $$\sigma_n=\int_0^1(1-z)^{n-1}\sigma_{n-1}\>dz={1\over n}\sigma_{n-1}\ .$$
Christian Blatter
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The unit cube is partitioned by n! such simplicies, all of the same volume.
Indeed, for each permutation $p$ of $S_n$ the set of points $(x_1,\dots,x_n)$ such that $0\leq x_{p(1)}\leq\cdots\leq x_{p(n)}\leq 1$ is a simplex, and it has the same volume as your simplex — which is the one that corresponds to the identity permutation.
Now every point if the cube whose coordinates are pairwise different belongs to exactly one of these cubes, and those that do have two coordinates that coincide determine a subset of volume zero.
Mariano Suárez-Álvarez
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