# Month: December 2018

​​​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

This tiny post is devoted to a proof of the almost sure convergence of martingales bounded in $\mathrm{L}^1$. This proof that we give below relies on the almost sure convergence of martingales bounded in $\mathrm{L}^2$, after a truncation step. In order to keep the martingale property after truncation, we truncate with a stopping time. The boundedness in $\mathrm{L}^1$ is used to show via the maximal inequality that the martingale is almost surely bounded. Note that this proof differs from the classical and historical proof from scratch which is based on up-crossing or oscillations.

The theorem. Let $M={(M_t)}_{t\geq0}$ be a continuous time martingale, with continuous trajectories, bounded in $\mathrm{L}^1$ in the sense that $\sup_{t\geq0}\mathbb{E}(|M_t|)<\infty$. Then there exists $M_\infty\in\mathrm{L}^1$ such that $\lim_{t\to\infty}M_t=M_\infty$ almost surely. Moreover the convergence holds in $\mathrm{L}^1$ if and only if $M$ is uniformly integrable.

A proof. The fact that $M_\infty\in\mathrm{L}^1$ follows without effort from the almost sure convergence, the boundedness in $\mathrm{L}^1$, and the Fatou lemma, namely
$\mathbb{E}(|M_\infty|) =\mathbb{E}(\varliminf_{t\to\infty}|M_t|) \leq\varliminf_{t\to\infty}\mathbb{E}(|M_t|) \leq C<\infty.$ Moreover, it is a general fact that a sequence of random variables that converges almost surely to a limit belonging to $\mathrm{L}^1$ does converge in $\mathrm{L}^1$ if and only if it is uniformly integrable.

It remains to prove a.s. convergence. We can assume that $M_0=0$, otherwise consider the martingale $M-M_0={(M_t-M_0)}_{t\geq0}$ which is also bounded in $\mathrm{L}^1$, making $M$ converge a.s. to $M_0+(M-M_0)_\infty$. We proceed by truncation and reduction to the square integrable case.

By the Doob maximal inequality with $p=1$, and $r>0$,

$$\mathbb{P}\Bigr(\sup_{s\in[0,t]}|M_s|\geq r\Bigr) \leq\frac{\mathbb{E}(|M_t|)}{r}.$$
By monotone convergence, with $C:=\sup_{t\geq0}\mathbb{E}(|M_t|)<\infty$, for all $r>0$,
$\mathbb{P}\Bigr(\sup_{t\geq0}|M_t|\geq r\Bigr) \leq\frac{C}{r}.$
It follows that $\mathbb{P}\Bigr(\sup_{t\geq0}|M_t|=\infty\Bigr)\leq\lim_{r\to\infty}\mathbb{P}\Bigr(\sup_{t\geq0}|M_t|\geq r\Bigr)=0.$
In other words almost surely ${(M_t)}_{t\geq0}$ is bounded.
As a consequence, on an almost sure event, say $\Omega’$, for large enough $n$,
$T_n:=\inf\{t\geq0:|M_t|\geq n\}=\infty.$

On the other hand, by the Doob stopping theorem, for all $n\geq0$, ${(M_{t\wedge T_n})}_{t\geq0}$ is a martingale. Moreover, since $M_0=0$, we have $\sup_{t\geq0}|M_{t\wedge T_n}|\leq n$. Since ${(M_{t\wedge T_n})}_{t\geq0}$ is bounded in $\mathrm{L}^2$, there exists $M^{(n)}_\infty\in\mathrm{L}^2$ such that $\lim_{t\to\infty}M_{t\wedge T_n}=M^{(n)}_\infty$ almost surely (and in $\mathrm{L}^2$ but this is useless here). Let us denote by $\Omega_n$ the almost sure event on which this convergence holds. Then, on the almost sure event $\Omega’\cap(\cap_n\Omega_n)$, we have, for all $m,n$, $M^{(n)}_\infty=M^{(m)}_\infty=:M_\infty$, and

$\lim_{t\to\infty}M_t=M_\infty.$

Truncation. Truncation is very natural to increase integrability. It is for instance used in the proof of the strong law of large numbers for independent random variables in $\mathrm{L}^1$ in order to reduce the problem to variables in $\mathrm{L}^p$ with $p>1$, the case $p=4$ being particularly simple.

Prerequisites. The ingredients should be established before and without using this theorem namely maximal inequalities for martingales, almost sure convergence of martingales bounded in $\mathrm{L}^2$, and stopping theorem for martingales and arbitrary stopping times.

Note. How to adapt this proof to discrete time martingales ? In this case, $\sup_{t\geq0}|M_{t\wedge T_n}|$ is not necessarily $\leq n$ but it can be bounded by the random variable $n+M_{T_n}\mathbf{1}_{T_n<\infty}$. Note that it is not the discreteness of time that poses a problem but rather the discontinuity of the trajectories. Continuous martingales are more convenient to handle for some aspects. See also this more recent post devoted to the convergence of discrete time martingales.

Acknowledgements. Thanks to my friends and colleagues Arnaud Guyader and Nathaël Gozlan who pointed out the fact that the argument does not work for discrete time martingales and the fact that it requires a preprocessing to ensure that $M_0=0$.

Last Updated on 2020-10-02

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