​​​This tiny post is about a basic characterization of Gaussian distributions.​

The theorem. A random vector of dimension two or more has independent components and is rotationally invariant if and only if its components are Gaussian, centered, with same variances.

In other words, for all $n\geq2$, a probability measure on $\mathbb{R}^n$ is in the same time product and rotationally invariant if and only if it is a Gaussian distribution $\mathcal{N}(0,\sigma^2I_n)$ for some $\sigma\geq0$.

Note that this does not work for $n=1$. In a sense it is a purely multivariate phenomenon.

A proof. For all $\sigma\geq0$, the Gaussian distribution $\mathcal{N}(0,\sigma^2I_n)$ is product and is rotationally invariant, and if $\sigma>0$, its density is, denoting $|x|:=\sqrt{x_1^2+\cdots+x_n^2}$, $$x\in\mathbb{R}^n\mapsto\mathrm{exp}\Bigr(-\frac{|x|^2}{2\sigma^2}-n\log\sqrt{2\pi\sigma^2}\Bigr).$$ Conversely, suppose that $\mu$ is a rotationally invariant product probability distribution on $\mathbb{R}^n$. We can assume without loss of generality that it has a smooth positive density $f:\mathbb{R}^n\to(0,\infty)$, since otherwise we can consider the probability measure $\mu*\mathcal{N}(0,\varepsilon I_n)$ for $\varepsilon>0$, which is also product and rotationally invariant. By rotational invariance, $\log f(x)=g(|x|^2)$, and thus $$\partial_i\log f(x)=2g'(|x|^2)x_i.$$ On the other hand, since $\mu$ is product, we have $\log f (x)=h(x_1)+\cdots+h(x_n)$ and thus $$\partial_i\log f (x)=h'(x_i).$$ Hence $\partial_i\log f(x)$, which depends on $|x|$ via $g'(|x|)$, depend only on $x_i$. Since $n\geq2$, it follows that $g’$ is constant. Therefore there exist $a,b\in\mathbb{R}$ such that $g(u)=au+b$ for all $u$, and thus $f(x)=\mathrm{e}^{a|x|^2+b}$ for all $x\in\mathbb{R}^n$. Since $f$ is a density, $a<0$ and $\mathrm{e}^b=(\pi/a)^{-n/2}$.

Another proof (for the converse). Suppose that $X_1,\ldots,X_n$ are $n\geq2$ independent random variables, such that the random vector $(X_1,\ldots,X_n)$ is rotationally invariant. Since for all $1\leq i\neq j\leq n$, there exists a rotation $R$ such that $Re_i=\pm e_j$, it follows that $X_1,\ldots,X_n$ are independent and identically distributed of say law $\nu$ which is symmetric. This reduces the problem to show that $\nu$ is Gaussian. This also means that it suffices to solve the problem for $n=2$. Now, for all $\theta\in\mathbb{R}$, the rotation of angle $\theta$ in $\mathrm{span}\{e_1,e_2\}$ gives that $\cos(\theta)X_1-\sin(\theta)X_2$ has the law of $X_1$. This indicates that $\nu$ is (symmetric) stable. But denoting $\varphi$ its characteristic function, and using the independence, we obtain $\varphi(\cos(\theta)t)\varphi(-\sin(\theta)t)=\varphi(t)$ for all $t\in\mathbb{R}$. Using the expression of the characteristic function of symmetric stable distributions, this leads to the Gaussianity of $\nu$. Note that without using stability, this shows also that all the cumulants or order $\neq2$ are all zero when $\nu$ has all its moments finite. This alternative proof without regularization is inspired by the idea of reduction to $n=2$ due to Dinh-Toan Nguyen, PhD student, communicated by my colleague Laure Dumaz.

History. ​This was probably known before Maxwell, maybe by Carl Friedrich Gauss (1777 – 1855) himself. The proof above is roughly the reasoning followed by James Clerk Maxwell (1831 – 1879) to derive the distribution of velocities in an ideal gas at equilibrium. In his case $n=3$, and the distribution is known in statistical physics as the Maxwellian distribution. This was a source of inspiration for Ludwig Boltzmann (1844 – 1906) for the derivation of his kinetic evolution equation and his H-theorem about entropy.

Characterizations. This characterization of Gaussian laws among product distributions using invariance by the action of transformations (rotations) leads to the same characterization for the heat semi-group and for the Laplacian operator. There are of course other remarkable characterizations of the Gaussian, for instance as being an eigenvector of the Fourier transform, and also, following Boltzmann, as being the maximum entropy distribution at fixed variance.

• Robert Robson, Timon Mehrling, and Jens Osterhoff
Great moments in kinetic theory: 150 years of Maxwell’s (other) equations
European Journal of Physics 38(6) 2017 (PDF)

Maxwell characterization for unitary invariant random matrices. A random $n\times n$ Hermitian matrix has in the same time independent entries and a law invariant by conjugacy with respect to unitary matrices if and only if it has a Gaussian law with density of the form $$H\mapsto\exp(a\mathrm{Tr}(H^2)+b\mathrm{Tr}(H)+c).$$ Note that the unitary invariance implies that the density depends only on the spectrum and is actually a symmetric function of the eigenvalues. A complete solution can be found for instance in Madan Lal Mehta book on Random matrices (Theorem 2.6.3), who attributes the result to Charles E. Porter and Norbert Rosenzweig (~1960). It is related to a lemma due to Hermann Weyl: all the invariants of an $n\times n$ matrix $H$ under non-singular similarity transformations $H\mapsto UHU^*$ can be expressed in terms of traces of the first $n$ powers of $H$. The assumption about the independence of entries kills all powers above $2$.

Complement. It is not difficult to show that if $X$ is a random vector of $\mathbb{R}^n$, $n\geq1$ with independent Gaussian and centered components of positive variance then $\mathbb{P}(X=0)=0$ and $X/|X|$ is uniformly distributed on the sphere. Conversely, it was shown by my former teacher and colleague Gérard Letac in The Annals of Statistics (1981) that if a random vector $X$ of $\mathbb{R}^n$, $n\geq3$, has independent components and is such that $\mathbb{P}(X=0)=0$ and $X/|X|$ is uniformly distributed on the sphere, then $X$ is Gaussian and in particular its components are Gaussian with zero mean and same positive variance. Moreover there are counter examples for $n=1$ and $n=2$. When $n\geq3$, this result of Letac implies the Maxwell theorem.

Last Updated on 2022-03-19

1. amic 2018-12-15

n⩾2…
En tout cas je viens d’apprendre un truc que je ne connaissais pas (ou que j’avais sciemment bien oublié…), merci.

2. Djalil Chafaï 2018-12-16

Merci Amic, c’est corrigé. Il y avait même un $n\geq0$ dans la première version 🙂

3. Gerard Letac 2018-12-20

On peut completer ainsi: si $X=(X_1,\ldots,X_n)$ est forme de va independantes et si $X/\|X\|$ est uniforme sur la sphere unite, alors les $X_i$ sont normaux et de meme loi, a condition que $n\geq 3$ C’est dans Annals of Statistics (1981) pages 408-417.

4. Djalil Chafaï 2018-12-28

Merci Gérard, j’ai mis à jour le billet pour mentionner ce résultat.

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