## Martingale and its application to dynamical systems

In the last week of May I attended two lectures given by Professor Matthew Nicol.

Let $(\Omega,\mu)$ be a prob space with a $\sigma$-algebra $\mathcal{B}$. Let $\mathcal{F}\prec \mathcal{B}$ be a sub $\sigma$-algebra.

Example. $f(x)=2x (\text{mod} 1)$ on $\mathbb{T}$, and $\mathcal{B}$ be the Borel $\sigma$-algebra. Let $\mathcal{F}=f^{-1}\mathcal{B}$. Note that $(0.2,0.3)\notin\mathcal{F}$.

Let $Y$ be a $\mathcal{B}$-measurable r.v. and $Y\in L^1(\mu)$. The conditional expectation $E(Y|\mathcal{F})$ is the unique $\mathcal{F}$-measurable r.v. $Z$ satisfying $Z^{-1}(a,b)\in \mathcal{F}$ for all $(a,b)$, and $\int_F Z d\mu=\int _A Y d\mu$ for all $A\in \mathcal{F}$.

Note that $E(Y|\mathcal{F})=Y$ if and only if $Y$ is $\mathcal{F}$-measurable; and $E(Y|\mathcal{F})=E(Y)$ if $Y$ is independent of $\mathcal{F}$.

Let $(X_n)_{n\ge 0}$ be a stationary ergodic process with stationary initial distribution $\mu$. A basic problem is to find sufficient conditions on $(X_n)_{n\ge 0}$ and on functions $\phi\in L^2_0(\mu)$ such that $\displaystyle S_n(\phi)=\sum_{k=1}^n \phi(X_k)$ satisfies the central limit theorem (CLT) $\displaystyle \frac{1}{\sqrt{n}}S_n(\phi) \to N(0,\sigma^2)$, where the limit variance is given by $\displaystyle \sigma^2(\phi)=\lim_{n\to\infty}\frac{1}{n}E(S^2_n(\phi))$.

Let $f$ be a conservative diffeomorphism on $(M,m)$. There are two operators: $\phi\mapsto U\phi=\phi\circ f$, and $\phi\mapsto P\phi$ via $\int P\phi\cdot \psi=\int \phi\cdot \psi\circ f$ for all test function $\psi$.

Property. $PU(\phi)=\phi$ (vol-preserving) and $UP(\phi)=E(\phi|f^{-1}\mathcal{B})$.

Let $\mathcal{F}_n$ be an increasing sequence of $\sigma$-algebras. Then a sequence of r.v. $S_n$ is called a martingale w.r.t. $\mathcal{F}_n$, if $S_n$ is $\mathcal{F}_n$-measurable, $E(S_{n+1}|\mathcal{F}_n)=S_n$.

Let $\mathcal{F}_n$ be a decreasing sequence of $\sigma$-algebras. Then a sequence of r.v. $S_n$ is called a reverse martingale w.r.t. $\mathcal{F}_n$, if $S_n$ is $\mathcal{F}_n$-measurable, $E(S_{n}|\mathcal{F}_m)=S_m$ for any $n\le m$.

Theorem. Let $\{X_n:n\ge 1\}$ be a stationary ergodic sequence of (reverse) martingale differences w.r.t. $\{\mathcal{F}_n\}$. Suppose $E(X_n)=0$, and $\sigma^2=\text{Var}(X_i)>0$. Then $\displaystyle \frac{1}{\sigma\sqrt{n}}\sum_{i=1}^n X_i \to N(0,1)$ in distribution.

Gordin: Suppose $(f,m)$ is ergodic. Consider the Birkhoff sum $\displaystyle \sum_{i=1}^n \phi\circ f^i$ for some $\phi$ with $\int \phi=0$. The time series $\phi\circ f^i$ can be approximated by martingale differences provided the correlations decay quickly enough.

Suppose there exists $p(n)$ with $\sum p(n) < \infty$, such that $\|P^n\phi\|\le C\cdot p(n)\|\phi\|$. Then define $\displaystyle g=\sum_{n\ge 1}P^n\phi$, and let $X=\phi+g-g\circ f$.

Property. Let $f:M\to M$ be such that $f^{-n}\mathcal{B}$ is decreasing. $\displaystyle S_n=\sum_{i=1}^n X\circ f^i$ is a reverse martingale with respect to $f^{-n}\mathcal{B}$.

Proof. Note that $PX=P\phi+Pg-PUg=0$. Then $E(X|f^{-1}\mathcal{B})=UP(X)=U0=0$.
Let $k < n$. It remains to show $E(X\circ f^k|f^{-n}\mathcal{B})=0$. To this end, we pick an element $A\in f^{-n}\mathcal{B}$ and write it as $A=f^{-k-1}C$ for some $C\in f^{k+1-n}\mathcal{B}$. Then $\displaystyle \int_A X\circ f^k dm=\int_{f^{-1}C}X dm =\int_{f^{-1}C} E(X|f^{-1}\mathcal{B}) dm=\int_{f^{-1}C}0 dm=0$. This completes the proof.

Three theorems of Gordin. Let $(\Omega,\mu,T)$ be an invertible $\mu$-preserving ergodic system, $X\in L^1(\mu)$ and $X_k(x)=X(T^kx)$ be a strictly stationary ergodic sequence.

(*) $\displaystyle \limsup_{n\to\infty}\frac{1}{\sqrt{n}}E|S_n| < \infty$

Theorem 1. Suppose there exists $\displaystyle \mathcal{F}_k\subset T^{-1}\mathcal{F}_k=\mathcal{F}_{k+1}$ such that $\displaystyle \sum_{k\ge 0} E|E(X_0|\mathcal{F}_{-k})|<\infty$, $\displaystyle \sum_{k\ge 0} E|X_0-E(X_0|\mathcal{F}_{k})| < \infty$. Then (*) implies $\displaystyle \lambda:=\lim_{n\to\infty}\frac{1}{\sqrt{n}}E|S_n|$ exists, and $\displaystyle \frac{1}{\sqrt{n}}S_n\to N(0,\lambda^2\pi/2)$ in distribution (degenerate if $\lambda=0$).

–Mixing condition. Let $\displaystyle \alpha(n):=\sup\{P(A\cap B)-P(A)P(B):A\in\mathcal{F}^0_{-\infty}, B\in\mathcal{F}^{\infty}_n\}$.

Theorem 2. Suppose for some $1/p+1/q=1$, $X\in L^p(\mu)$ and $\displaystyle \sum_{n\ge 1}\alpha(n)^{1/q} < \infty$. Then (*) implies the conclusion of Theorem 1.

–uniform mixing condition. Let $\displaystyle \phi(n):=\sup\{P(B|A)-P(B):A\in\mathcal{F}^0_{-\infty}, \mu(A) > 0, B\in\mathcal{F}^{\infty}_n\}$.

Theorem 3. Suppose $X\in L^1(\mu)$ and $\displaystyle \sum_{n\ge 1}\phi(n) < \infty$. Then (*) implies the conclusion of Theorem 1.

Cuny–Merlevede: not only the CLT, but also the ASIP holds under the above conditions.

Note that we started with an invariant measure $m$. The operator $U$ and $P$ can be defined for all non-conservative maps. To emphasize the difference, we use $\hat P$. Suppose $\hat P h=h$ for some $h\in L^1(m)$. Then $\mu=hm$ is an absolutely continuous invariant prob. measure:

$\displaystyle \int \phi\circ f d\mu=\int \phi\circ f h dm=\int \phi\cdot \hat P h dm=\int \phi hdm=\int\phi d\mu$.

Then we can rewrite $\displaystyle P\phi=\frac{1}{h}\hat P(h\phi)$, in the sense that $\displaystyle \int P(\phi)\cdot \psi d\mu=\int \phi\cdot \psi\circ f d\mu =\int \phi h\cdot \psi\circ f dm$
$\displaystyle =\int\hat P(\phi h)\cdot \psi dm \int \frac{1}{h}\hat P(\phi h)\cdot \psi d\mu$.