# Category: Uncategorized

​​​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 $\mathrm{e}^b=(\pi/a)^{-n/2}$.

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.

Further reading. 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 of 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). It is based on the following 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.

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 martingales are either in discrete time or in continuous time with continuous paths.

The theorem. Let $M={(M_t)}_{t\geq0}$ be a continuous martingale bounded in $\mathrm{L}^1$. 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. 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 and $\sup_{t\geq0}|M_{t\wedge T_n}|\leq n$. Since it 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

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

About 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.

Final comments. 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.

Here are the Mathematical Citation Quotient (MCQ) for journals in probability, statistics, analysis, and general mathematics. The numbers were obtained using a home brewed Python script extracting data from MathSciNet. The graphics were obtained by using LibreOffice.

Recall that the MCQ is a ratio of two counts for a selected journal and a selected year.  The MCQ for year $Y$ and journal $J$ is given by the formula $\mathrm{MCQ}=m/n$ where

• $m$ is the total number of citations of papers published in jounal $J$ in years $Y-1$,…,$Y-5$ by papers published in year $Y$ in any journal known by MathSciNet;
• $n$ is the total number of papers published in journal $J$ in years $Y-1$,…,$Y-5$.

The Mathematical Reviews compute every year the MCQ for every indexed journal, and make it available on MathSciNet. This formula is very similar to the one of the five years impact factor, the main difference being the population of journals which is specifically mathematical for the MCQ (reference list journals) and the way the citations are extracted. Both biases are negative.

Let $X={(X_t)}_{t\geq0}$ be an irreducible Markov process with generator $G$ and unique invariant law $\mu$. What is the difference between irreversible and out of equilibrium? If you do not know what is a generator, they you may replace it by a transition kernel without loss of interpretation.

Trajectories. A Markov process is a way to transform an initial law into a law of trajectory, by mean of a generator. The law of the trajectory ${(X_t)}_{t\geq0}$ depends on the law of $X_0$ as well as on the generator $G$. The equilibrium or invariant law $\mu$ depends only on $G$.

Reversibility. We say that $X$ is reversible when $X_0\sim\mu$ implies that for all $t\geq0$ the trajectory $(X_s)_{0\leq s\leq t}$  and the reversed trajectory ${(X_{t-s})}_{0\leq s\leq t}$ have same law. This means that if we start the process from its equilibrium, the notion of orientation of time is not visible. Up to regularity considerations, this is equivalent to say that the generator is a symmetric operator in $L^2(\mu)$. When $X$ is a finite Markov chain, this corresponds to the detailed balance condition $\mu(x)G(x,y)=\mu(y)G(y,x)$ for all $x,y$ in the state space. A famous theorem by Kolmogorov states that this is equivalent to say that the weight of all cycles with respect to the transition kernel or generator does not depend on the orientation of the cycle, and the remarkable fact about this criterion is that it does not involve $\mu$.

Out of equilibrium. We say that the process $X$ is out of equilibrium when it is not started from the equilibrium $\mu$ meaning that $X_0\not\sim\mu$. Up to regularity assumptions, for all initial law, the process converges in law to its equilibrium in long time, in other words $\lim_{t\to\infty}X_t=\mu$ in law. On the contrary, if $X_0\sim\mu$, then the process satisfies $X_t\sim\mu$ for all $t\geq0$ since $\mu$ is invariant, and we say in this case that the process is at equilibrium.

Conclusion. Being out of equilibrium or not has nothing to do with being reversible or not. We may start a reversible process or not, out of equilibrium, or not. If we start a reversible process out of equilibrium, then typically it will converge exponentially fast to its equilibrium $\mu$ in $L^2(\mu)$, and the exponential speed of this convergence is given by the spectral gap of $G$.

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