Two equivalent formulas for the volume of a random tetrahedron are given. Further on you can find an interesting conjecture for the expected volume that shall be proved.

Tetrahedron volume

Given are 12 independent standard normal distributed variables $$x_i=\mathcal{N}(0,1)_{i=1,...,12}$$ that define the 4 coordinates $$\vec{a}=(x_1,x_2,x_3),\;\; \vec{b}=(x_4,x_5,x_6),\;\; \vec{c}=(x_7,x_8,x_9),\;\; \vec{d}=(x_{10},x_{11},x_{12})$$ of a 3-simplex in $\mathbb{R}^3$. The first formula for the non-oriented simplex volume is $$V=\frac{1}{6}\left| (\vec{a}-\vec{d})\cdot \left((\vec{b}-\vec{d}) \times (\vec{c}-\vec{d})\right) \right|\tag{1}$$ $$=\frac{1}{6}\left| x_2 x_6 x_7 + x_3 x_4 x_8+ x_1 x_5 x_9+ x_3 x_5 x_{10} + x_6 x_8 x_{10} + x_2 x_9 x_{10}+ x_1 x_6 x_{11}+ x_3 x_7 x_{11}+ x_4 x_9 x_{11}+ x_2 x_4 x_{12}+ x_5 x_7 x_{12}+ x_1 x_8 x_{12}-x_3 x_5 x_7- x_2 x_6 x_{10}- x_3 x_8 x_{10} - x_1 x_6 x_8 - x_2 x_4 x_{9}- x_5 x_9 x_{10}- x_3 x_4 x_{11}- x_6 x_7 x_{11}- x_1 x_9 x_{11}- x_1 x_5 x_{12}- x_2 x_7 x_{12}- x_4 x_8 x_{12}\right|.$$

If the coordinate system is shifted $$\vec{p}=\vec{a}-\vec{d},\;\;\vec{q}=\vec{b}-\vec{d},\;\;\vec{r}=\vec{c}-\vec{d}$$ the new coordinates are $$\vec{p}=(y_1,y_2,y_3),\;\; \vec{q}=(y_4,y_5,y_6),\;\;\vec{r}=(y_7,y_8,y_9)$$ with new random variables $$y_i=\mathcal{N}(0,\sqrt{2})_{i=1,...,9}.$$

The shift reduces the number of random variables from 12 to 9 and increases the standard deviation from $1$ to $\sqrt{2}$ (this corresponds to a double variance $=\sqrt{2}^2)$. However the variables are not independent anymore. Their correlation $\rho=0.5$ is given by their covariance normalized by the standard deviation $$\rho=\frac{\mathbb{Cov}[y_i,y_j]}{\sqrt{\mathbb{Var}[y_i]}\sqrt{\mathbb{Var}[y_j]}}= \frac{\mathbb{Cov}[x_m-x_k,x_n-x_k]}{\sqrt{\mathbb{Var}[x_m-x_k]}\sqrt{\mathbb{Var}[x_n-x_k]}} =\frac{\mathbb{E}[x_k^2]}{\sqrt{\mathbb{Var}[x_m-x_k]}\sqrt{\mathbb{Var}[x_n-x_k]}}=\frac{\mathbb{E}[x_k]^2+\mathbb{Var}[x_k]}{\sqrt{\mathbb{Var}[x_m-x_k]}\sqrt{\mathbb{Var}[x_n-x_k]}} =\frac{1}{\sqrt{2}\sqrt{2}}=\frac{1}{2}\;\;\;\text{for}\;i\ne j \land n\ne m \ne k.$$

The second formula for the non-oriented volume as function of the dependent variables is $$V=\frac{1}{6}\left|\vec{p}\cdot (\vec{q} \times \vec{r}\right)|\tag{2}$$ $$=\frac{1}{6}\left| y_2y_6y_7+y_3y_4y_8+y_1y_5y_9-y_1y_6y_8-y_2y_4y_9-y_3y_5y_7\right|.$$

Equation (2) has only a quarter of summands of eq.(1) however the variables correlate with $\rho=0.5$.


What is the analytical expression for the expected volume $\mathbb{E}[V]$?

What is known?


It is conjectured that $\mathbb{E}[V]=\frac{2}{3}\sqrt{\frac{2}{\pi}}$ or $\mathbb{E}[V]=\frac{21}{4\pi^2}$. Assuming the first conjecture is true please note the relation to a standard half-normal distribution in $\mathbb{R^1}$ that has expectation $\sqrt{\frac{2}{\pi}}$.


All even moments are precisely known and the odd moments are approximately known. The first moments are

\begin{array}{|l|l|}\hline \text{odd moments} & \text{even moments} \\ \text{(simulation)} & \text{(analytic)} \\ \hline m_1\approx 0.532 & m_2=\frac{2}{3}\\ \hline m_3\approx\sqrt{2} &m_4=\frac{40}{9} \\ \hline m_5\approx18.9 &m_6=\frac{2800}{27} \\ \hline \end{array}

(more moments on demand).

Solution strategies

One could try to integrate over a subvolume where the sign of the volume is constant. Due to symmetry every subvolume should have equal size. The challenge is therefore to find the right suitable integration borders.

A related question about the expected area of a triangle with standard normal distributed coordinates in $\mathbb{R}^3$ was proven to be $\sqrt{3}$. If these methods would be applied to the tetrahedron case then according to the answerer "ultimately it comes down to the product of independent chi-distributed variables and a variable for the spherical angle they determine: finding the expectation of the latter is the crux of the question."

Other equations for the volume

There are other methods to calculate the volume however they include at least 1 square root, an unwanted property for such problems.

Expected oriented volume

The expression for the volume is a sum of triple products of random variables. As the expectations of the independent $x_i$ in eq.(1) are $\mathbb{E}[x_i]=0$ it holds
$$\mathbb{E}[x_i x_j x_k\pm x_l x_m x_n]=0\cdot 0 \cdot 0\pm 0\cdot 0 \cdot 0=0\;\;\;\text{for}\; 1\le i,j,k,l,m,n \le 12$$ The expected oriented volume is therefore $0$.

  • 1
    $\begingroup$ Please, avoid making several edits. $\endgroup$
    – Aloizio Macedo
    Feb 14, 2020 at 15:29

1 Answer 1


Let $X_0, X_1, \dots, X_n$ be i.i.d. standard normal vectors in $\mathbb{R}^n$ (so each $X_i \sim \mathcal{N}(0, I_n)$). Writing $Y_i = X_i - X_0$ for $i = 1, \dots, n$, we have that the $n$-volume of the $n$-simplex with vertices $X_0, X_1, \dots, X_n$ is equal to $$\frac{1}{n!} |\det(Y_1, \dots, Y_n)|$$ where we consider $Y_1, \dots, Y_n$ as column vectors.

Define $(W_1, \dots, W_n) = (Y_1, \dots, Y_n)^T$, i.e. $W_{i, j} = X_{j, i} - X_{0, i}$, so $W_1, \dots, W_n$ are independent, and $W_i \sim \mathcal{N}(0, \Sigma)$, where the covariance matrix $\Sigma$ has $2$'s on the diagonal and $1$'s off the diagonal. Note that $J_n$ (the matrix of ones) has eigenvalues $n, 0, \dots, 0$, hence since $\Sigma = I_n + J_n$, $\Sigma$ has eigenvalues $n+1, 1, \dots, 1$ and thus $\det \Sigma = n+1$. Now, defining $Z_i = \Sigma^{-1/2} W_i$ for $i = 1, \dots, n$, we have that $Z_1, \dots, Z_n$ are independent with each $Z_i \sim \mathcal{N}(0, I_n)$, and also that $$\det(Y_1, \dots, Y_n) = \det(W_1, \dots, W_n) = \det(\Sigma^{1/2}Z_1, \dots, \Sigma^{1/2}Z_n) = \det \Sigma^{1/2} \cdot \det(Z_1, \dots, Z_n).$$ It follows that the desired expected volume is $$\frac{\sqrt{n+1}}{n!} \mathbb{E}[|\det(Z_1, \dots, Z_n)|]$$ for independent $Z_1, \dots, Z_n \sim \mathcal{N}(0, I_n)$. To finish, we compute $\mathbb{E}[|\det(Z_1, \dots, Z_n)|]$.

Let $Z_1', \dots, Z_n'$ be the result of performing the Gram-Schmidt process to $Z_1, \dots, Z_n$ without normalizing, so for each $k$, we have $\mathrm{span}(Z_1', \dots, Z_k') = \mathrm{span}(Z_1, \dots, Z_k)$, and we inductively define $Z_k' = Z_k - P_kZ_k$ (with $Z_1' = Z_1$), where $P_k$ is the orthogonal projection onto $\mathrm{span}(Z_1', \dots, Z_{k-1}')$. Notably, these are all elementary column operations, so $\det(Z_1', \dots, Z_n') = \det(Z_1, \dots, Z_n)$, and $Z_1', \dots, Z_n'$ are orthogonal, so $|\det(Z_1', \dots, Z_n')| = \prod_{k=1}^n |Z_k'|$. Equivalently, we have $Z_k' = P_k' Z_k$, where $P_k'$ is the orthogonal projection onto the orthogonal complement of $\mathrm{span}(Z_1', \dots, Z_{k-1}')$, so $Z_k'$ can be seen as a standard normal vector on this $(n-k+1)$-dimensional space. This means that conditioning on $Z_1', \dots, Z_{k-1}'$, $|Z_k'|$ has the chi distribution with $n-k+1$ degrees of freedom, so in fact $|Z_k'|$ is independent of $Z_1', \dots, Z_{k-1}'$ with $$\mathbb{E}[|Z_k'|] = \sqrt{2} \frac{\Gamma((n-k+2)/2)}{\Gamma((n-k+1)/2)}.$$ It follows that all $|Z_k'|$ are independent, giving \begin{align*} \mathbb{E}[|\det(Z_1, \dots, Z_n)|] &= \prod_{k=1}^n \mathbb{E}[|Z_k'|]\\ &= \prod_{k=1}^n \sqrt{2} \frac{\Gamma((n-k+2)/2)}{\Gamma((n-k+1)/2)} \\ &= \prod_{k=1}^n \sqrt{2} \frac{\Gamma((k+1)/2)}{\Gamma(k/2)} \\ &= 2^{n/2} \frac{\Gamma((n+1)/2)}{\Gamma(1/2)} \end{align*} so the expected volume is $2^{n/2} \frac{\Gamma((n+1)/2) \sqrt{n+1} }{\Gamma(1/2) n!}$. At $n = 3$ (the given case), this is $\frac{2}{3} \sqrt{\frac{2}{\pi}}$.

Higher moments can be computed in the same way, using the corresponding higher moments of the chi distribution.

  • $\begingroup$ Some more interesting results on Gaussian polytopes can be found here: arxiv.org/abs/1706.08092; my result above appears as a formula on page 8 $\endgroup$
    – user125932
    Feb 17, 2020 at 5:57
  • $\begingroup$ Great derivation. Which textbooks are recommended to understand this? I want to add that dhe derivation even dates back to a paper from 1971. jstor.org/stable/1426176 (eq.77). Is there literature for the case of different expectations of the coordinates? $\endgroup$ Feb 17, 2020 at 18:50
  • $\begingroup$ IMO these are all sort of standard tricks from linear algebra and probability/statistics, the main machinery I use here is the property of multivariate normal distributions under affine transformations (see e.g. here). For some good textbooks on linear algebra and probability, I'd probably recommend Axler and Durrett, respectively. I'm not sure if there are any good textbooks that would cover all the material in this answer specifically. I'd also be happy to clear up any parts that are confusing. $\endgroup$
    – user125932
    Feb 17, 2020 at 20:16
  • $\begingroup$ For your last question, it's not clear to me how you'd want to alter the problem -- do you want the points to be chosen again independently with normal distributions, just with different means? $\endgroup$
    – user125932
    Feb 17, 2020 at 20:19
  • $\begingroup$ yes, 12 different means in R3 for the tetrahedron and 9 different means in R3 for the triangle (the earlier post) $\endgroup$ Feb 17, 2020 at 20:50

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