In this question Anthony Quas asks about the expected absolute value of the determinant of an $n\times n$ row stochastic matrix $A$, where the rows are independently selected from the uniform distribution on the unit $(n-1)$-dimensional simplex $x_1+\cdots+x_n=1$, $x_i\geq 0$. I can show by a messy computation that the integral over all such matrices of $(\det A)^2$ is $1/(n+1)!^{n-1}$. Is there a noncomputational reason for such a simple value?

  • $\begingroup$ Interesting. Is this the $(n-1)$st power of the factorial? This looks a lot smaller than I was expecting; or maybe I just misunderstood the random matrix result. $\endgroup$ Commented Dec 3, 2016 at 15:53
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    $\begingroup$ This is $(n+1)!$ to the $(n-1)$st power. However, this is not the average value of $(\det A)^2$. To get the average value one must divide by the volume, which is $1/(n-1)!^n$. $\endgroup$ Commented Dec 4, 2016 at 1:11
  • $\begingroup$ Great - so this gives the sort of size that I was expecting. Thanks for taking an interest in my original question. $\endgroup$ Commented Dec 4, 2016 at 3:12
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    $\begingroup$ @RichardStanley: that's not an answer to the "reason" question, but I think I see how to make a perhaps nicer computation: by computing the expectation of the $det^2$ for the i.i.d. exponential variables case (see mathoverflow.net/a/13040/31371, though I would do the handling of $Fix$ summation part differently). Then, divide it by $(n(n+1))^n$ that corresponds to the normalization (expectation of $(a_{i1}+\dots+a_{in})^2$ per row), that sends it to the simplex. It provides $\frac{(n+1)!}{(n(n+1))^n}= \frac{(n-1)!}{(n(n+1))^{n-1}}$ for the expectation; perhaps this way is less messy? $\endgroup$ Commented Dec 5, 2016 at 15:40
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    $\begingroup$ @Hans: The technique is that of Exercise 5.64 in my book Enumerative Combinatorics, vol. 2. $\endgroup$ Commented Jan 30, 2018 at 23:31

2 Answers 2


Hidden in the comment by Victor Kleptsyn is a really nice argument. Since nobody upvoted that comment yet (I just did) it's probably worth expanding it.

From a probabilistic approach it's more natural to show the equivalent statement that the expectation of $(\det A)^2$ equals $\frac{(n+1)!}{(n(n+1))^n}$. The trick is to realize the uniform measure on the simplex as the distribution of $$ (X_1,\dots,X_n) := \frac{(Y_1,\dots,Y_n)}{Y_1 + \cdots +Y_n} $$ where $(Y_i)$ are i.i.d. exponential random variables (say, of parameter 1). All you need to know is that $\mathbf{E} [Y_i] =1$ and $\mathbf{E} [Y_i^2] =2$. Moreover in this representation $(X_i)$ are independent from $S:=Y_1+\cdots+Y_n$. It follows that we have the following identity in distribution $$ B = D A $$ where $B$ is a matrix with i.i.d. exponential entries and $D$ a diagonal matrix whose entries are independent copies of $S$, with $A$ and $D$ moreover independent. Therefore $$ \mathbf{E} [\det B^2] = \mathbf{E} [\det D^2] \cdot \mathbf{E} [\det A^2] .$$ It is very easy to check that $\mathbf{E} [\det D^2] = (n(n+1))^n$, and $\mathbf{E} [\det B^2]$ can be expanded as $$ \sum_{\sigma,\tau \in \mathfrak{S}_n} \varepsilon(\sigma) \varepsilon(\tau) 2^{Fix(\sigma \tau^{-1})} = n! \sum_{\pi \in \mathfrak{S}_n} \varepsilon(\pi) 2^{Fix(\pi)} = n! \det \begin{pmatrix} 2 & 1 & \cdots & 1 \\ 1 & 2 & \ddots & \vdots \\ \vdots & \ddots & \ddots & \vdots \\ 1 & \cdots & \cdots & 2 \end{pmatrix} = (n+1)!$$ (Here $Fix$ denotes the number of fixed points.)

  • $\begingroup$ I haven't understood why $(X_1,\ldots ,X_n)$ is independent from $S = Y_1 + \cdots + Y_n$ ? $\endgroup$
    – Machinato
    Commented Jun 21, 2022 at 11:59
  • $\begingroup$ This is because the density of $(Y_1,\dots,Y_n)$ at a point $(y_1,\dots,y_n)$ depends only on $y_1+\dots+y_n$. It is similar to the following situation, which is maybe more standard: if $Z=(Z_1,\dots,Z_n)$ are i.i.d. $N(0,1)$ random variables, then $Z/||Z||_2$ is independent from $||Z||_2$ because the density of $Z$ depends only on $\|Z\|_2$ (the standard Euclidean norm). $\endgroup$ Commented Jun 22, 2022 at 4:31
  • $\begingroup$ Does it imply that since the density of $Z$ depends actually on any strictly monotone function $ g(\left \| Z \right \|_2)$, so the pair of random variables $Z/g(\left \| Z \right \|_2)$ and $g(\left \| Z \right \|_2)$ are independent as well? $\endgroup$
    – Machinato
    Commented Jun 22, 2022 at 6:20
  • $\begingroup$ No (If for example $g(x)=\sqrt{x}$, then $g(||Z||_2)$ is a function of $Z/g(||Z||_2)$). You also need to use the fact that the polar decomposition induces a bijection between $\mathbf{R}^n \setminus \{0\}$ and $(0,\infty) \times S^{n-1}$; after passing to polar coordinates the density has a product form, hence independence. In the case of exponential variables, the "polar" decomposition induces a bijection between $(0,\infty)^n$ and $(0,\infty) \times \Delta_{n-1}$ where $\Delta_{n-1}$ is the standard simplex. $\endgroup$ Commented Jun 22, 2022 at 6:56

Not a full answer.

The determinant of an $n\times n$ matrix $A$ is the volume of the parallelotope spanned by its rows. That volume is equal to $n!$ times the volume of the $n$-simplex whose vertices are at the origin and at the vectors which are the rows of $A$.

When those vectors are on the standard $(n-1)$-simplex, we can compute the volume of the $n$-simplex by the formula for the volume of a pyramid: the volume of the base times the height, divided by $n$.

In this case, the base is the $(n-1)$-simplex formed by the rows of $A$, and the height is $\frac{1}{\sqrt{n}}$.

Thus the determinant of $A$ is $\frac{n! V}{n\sqrt{n}}$, where $V$ is the volume of the $(n-1)$-simplex spanned by the rows of $A$.

The square of the determinant is $(\det A)^2 = \frac{n!^2 V^2}{n^3}$.

Therefore the average value over all such $A$ of the square of the determinant is $\frac{n!^2}{n^3}E(V^2)$.

"All" that remains is to compute the expected value $E(V^2)$ of the squared volume of a subsimplex of a standard simplex. Note that the analogous question for the expected volume $E(V)$ is apparently quite difficult, see http://mathworld.wolfram.com/SimplexSimplexPicking.html


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