# Volume of $T_n=\{x_i\ge0:x_1+\cdots+x_n\le1\}$

Let $T_n=\{x_i\ge0:x_1+\cdots+x_n\le1\}$. I know $T_n$ is tetrahedron.

My question: How can I compute the volume of $T_n$ for every $n$?

• Using this change of variables also leads to a direct answer in terms of an integral. Commented Apr 8, 2019 at 15:17

What is the probability that a random sequence of $$n$$ numbers in $$[0,1]$$ is sorted (lowest to highest)? that is, let $$y_1,y_2,\dots,y_n$$ be independent uniform random variables in $$[0,1]$$. Then the probability $$P(y_1\leq y_2\leq \dots \leq y_n)=\frac{1}{n!}$$.

Let $$A=(a_{ij})$$ be the matrix such that $$a_{ij}=1$$ if $$i\geq j$$ and $$a_{ij}=0$$ if $$i. Then $$\det A = 1$$ since $$A$$ is lower-triangular and $$a_{ii}=1$$ for all $$i$$. On the other hand, if you take the two sets:

$$S=\{(x_1,\dots,x_n)^T: 0\leq x_i\leq 1: \sum x_i \leq 1\}$$

and $$T=\{(x_1,\dots,x_n)^T: 0\leq x_1\leq x_2\dots \leq x_n\leq 1\}$$

Then $$T$$ is the image of $$S$$ under the transformation $$A$$, that is, $$AS=T$$. So $$\frac{1}{n!}=\mathrm{vol}(T) = \det A\cdot\mathrm{vol}(S) = \mathrm{vol}(S)$$

• It's 1/n!, but I'm having trouble seeing how that relates to volume Commented Feb 26, 2013 at 22:01
• it is actually the same one, but with different form, and I doubt if it is easier to see this. Commented Feb 26, 2013 at 22:01
• @MathPhys137, just think of Monte Carlo method Commented Feb 26, 2013 at 22:01
• Oh I see. Thanks Thomas! Commented Feb 26, 2013 at 22:03
• Note, the set I describe and the simplex set are not isometric/congruent, you nead some skew transformation which preserves volume. Basically, you need to know that the transformation $(x_1,\dots,x_n)\to (x_1,x_1+x_2,\cdots,\sum_{i=1}^n x_n)$ is a volume-preserving map. Commented Feb 26, 2013 at 22:14

First suppose that we want to find the volume of $T_n(a)$, then by change of variables $X=aU$, you can see that we have $V(T_n(a))=\color{red}{a^n}V(T_n(1))$.

Since $x_1+\cdots+x_n\le1$ if and only if $x_n\le1$ and $x_1+\cdots+x_{n-1}\le 1-x_n$, we have \begin{align}V(T_n(1))&=\int_{x_n\le 1}\left(\int_{x_1+\cdots+x_{n-1}\le1-x_n}dx_1\cdots dx_{n-1}\right)dx_n\\ &=V(T_{n-1}(1))\int_{x_n\le1}\color{red}{(1-x_n)^{n-1}}dx_n=\frac1 nV(T_{n-1}(1))\end{align} The numbers $V(T_n(1))$ satisfy in the above recursion formula, so $$V(T_n(1))=\frac1{n!}.$$

One another way is to consider the following integral

$$I=\int_{T_n(a)}e^{-(x_1+\cdots+x_n)}dx_1\cdots dx_n$$ Since $V(T_n(a))=a^nV(T_n(1))$, $$I=\int_0^\infty e^{-a}dV(T_n(a))=V(T_n(1))\int_0^\infty n a^{n-1}e^{-a}da=n!V(T_n(1))$$ But also we have $$I=\left(\int_0^\infty e^{-x}dx\right)^n=1$$ Hence, $$V(T_n(1))=\frac1{n!}.$$

• I got the answer to the other question :). plus the other question is easy. This one is harder. I have one question: earlier today I was thinking this problem can be solved maybe by using a change of variables? What do you think about using $y_i = \sum_{j=1}^i x_j$ ?
– user145801
Commented May 1, 2014 at 9:13
• ِDid you calculate Jacobian of this change of variable? Commented May 1, 2014 at 9:23
• I did it for the case $n=3$ which is $1$
– user145801
Commented May 1, 2014 at 9:24
• and $n=2$ also .
– user145801
Commented May 1, 2014 at 9:30
• Jacobian for arbitrary $n$ is also $1$, but I suggest you to use following change of variable $$y_1=x_n,y_2=x_{n-1}+x_{n},\cdots,y_n=x_1+\cdots+x_n$$ Then it's Jacobian is also $1$ and we have $$V(T_n(1))=\int_0^1\int_{y_1}^1\int_{y_2}^1\cdots\int_{y_{n-1}}^11~dy_n~dy_{n-1}~dy_{n-2}\cdots~dy_1.$$ Commented May 1, 2014 at 10:43

Hint: The general rule is that the $n$-volume of a simplex is $\frac 1n$ times the $(n-1)$ volume of the base times the height, which you can prove by integration. The height is $1$, so you have a recurrence.

Try using induction. $T_{1}$ is just in the interval $[0,1]$, which has volume (length) $1$. $T_{2}$ is a right triangle with volume (area) $1/2$. Now imagine the case for $n=3$. When we slice the tetrahedron $T_{3}$ at some height $z\in[0,1]$, we get a cross section that looks like $T_{2}$. But as $z$ gets bigger, the cross section gets smaller. Try to convince yourself that in general we have $$v(T_{n})=\int_{0}^{1}(1-x)^{n-1}v(T_{n-1})\,\mathrm{d}x=\frac{1}{n}v(T_{n-1})$$ Then, using the fact that $v(T_{1})=1$, we have $v(T_{n})=1/n!$. Note that we scale by $(1-x)^{n-1}$ instead of by $1-x$ because it is the linear dimensions of the $T_{n-1}$ slice that scale by $1-x$, which translates to the $\mathbb{R}^{n-1}$ volume of the slice scaling by $(1-x)^{n-1}$.

• Thanks Ben! I have one question. Is there a way to solve this problem using triple integrals? and maybe a change of variables?
– user145801
Commented Apr 26, 2014 at 1:03
• Unfortunately I answered too hastily and this approach is wrong! I apologize. I believe Ross' answer is correct. For $T_{3}$ the integral would be $\int_{0}^{1}\int_{0}^{1-z}\int_{0}^{1-(y+z)}1 \, \mathrm{d} x \, \mathrm{d} y \, \mathrm{d} z$. Commented Apr 26, 2014 at 1:14
• I have fixed my response, which is now a more fleshed out version of Ross'. Commented Apr 26, 2014 at 1:26