Kolmogorov continuity criteria

THE KOLMOGOROV CONTINUITY CRITERIA

Let $X : \mathbb{R}^d_+ \to S$ be a process on $\mathbb{R}^d$ with values in a complete metric space $(S, \, \rho)$ such that

$$\begin{equation*}\mathbb{E} \left\{ \rho \left(X_s, X_t\right) \right\}^a \leq c|s-t|^{d+b}, \quad s, t \in \mathbb{R}^d,\end{equation*}$$

for some constants $a, b > 0$. Then there exists a modification $\tilde{X}$ of $X$ that is locally Hölder continuous of order $p$, for every $p \in(0, \frac{b}{a})$.

The Kolmogorov continuity criterion is a simple moment condition that ensures the existence of a Hölder-continuous version of a given process in a complete metric space. It has many applications in probability theory. For example, one can show that Brownian motion is almost surely locally Hölder continuous of order $p$, for every $p \in(0, \frac{1}{2})$.

Motivation

The fundamental motivation for Kolmogorov's continuity criterion is to provide a practical way to prove that random processes have continuous sample paths.

HÖLDER CONTINUOUS FUNCTION

A function $f$ on Euclidean space is called Hölder continuous if there are real constants $C > 0$ and $\alpha > 0$ such that for all $x$ and $y$,

$$\begin{equation*}|f(x) - f(y)| \leq C \|x - y\|^{\alpha}.\end{equation*}$$

Kolmogorov's insight was that instead of checking continuity directly, we could look at the moments of increments. For a process $X_t$, if we can show that for some $p > 0$ and $\alpha > 0$:

$$\begin{equation*}\mathbb{E} \left\{ \rho \left(X_s, X_t\right) \right\}^a \leq c|s-t|^{d+b}, \quad s, t \in \mathbb{R}^d,\end{equation*}$$

Then under mild additional conditions, the process has a continuous modification.

This criterion is particularly valuable for Brownian motion $B(t)$. We have:

$$\begin{equation*}\mathbb{E}[|B(t) - B(s)|^4] = 3|t-s|^2\end{equation*}$$

Taking $p=4$ and $\alpha=1$, the criterion immediately gives continuity.

Proof of Kolmogorov continuity criteria

We consider the restriction of $X$ to $[0,1]^d$ without the loss of generality and make proof for $[0,1]^d$. Then we build the approximation sets $G_n$ and define the mapping

$$\begin{equation*}\pi_n (u) : [0,1]^d \to G_n \quad \text{such that} \quad | \pi_n (u) - u | \leq 2^{-n}\end{equation*}$$

Construction approximation sequence

We can define the approximation dataset $G_n$ of points $[0, 1]^d$ such that the coordinates of $2^n x$ are positive integers and $| G_n | = (2^n + 1)^d$. The for each $u \in [0, 1]^d$ we choose $\pi_n (u) \in G_n$ as close to $u$ as possible, so that

$$\begin{equation*}| \pi_n (u) - u | \leq 2^{-n}\end{equation*}$$

and

$$\begin{equation*}|\pi_n (u) - \pi_{n-1} (u)| \leq | \pi_n (u) - u | + | u - \pi_{n-1} (u) | \leq \frac{3}{2^n}.\end{equation*}$$

Then consider the random variable

$$\begin{equation*}Y_n = \max \left\{ \rho(X_s, X_t); \quad |s-t| \leq \frac{3}{2^n} \right\}\end{equation*}$$

and since $G_{n-1} \subset G_n$, we get $\forall u \in [0,1]^d$

$$\begin{equation*}|\pi_n (u) - \pi_{n-1} (u)| \leq \frac{3}{2^n} \Rightarrow \rho(X_{\pi_{n}(u)}, X_{\pi_{n-1}(u)}) \leq Y_n\end{equation*}$$

For the set $G = \bigcup_{n\geq0} G_n$ we claim that

$$\begin{equation}\sup_{s, t \in G; \newline |s-t| \leq 2^{-k}} \rho(X_s, X_t) \leq 3 \sum_{n\geq k} Y_n\end{equation}$$

To prove this, consider $n \geq k$ such that $s, t \in G_n$, so that $s = \pi_n (s)$ and $t = \pi_n (t)$. Assuming $| s - t| \leq 2^{-k}$, we have

$$\begin{equation*}|\pi_k (s) - \pi_{k} (t)| \leq | \pi_k (u) - s | + |s - t| + | t - \pi_k (t) | \leq \frac{3}{2^k}\end{equation*}$$

and thus

$$\begin{equation*}\rho(X_{\pi_{k}(s)}, X_{\pi_k(t)}) \leq Y_k\end{equation*}$$

Next, for $u \in \{ s, t \}$,

$$\begin{equation*}\rho(X_u, X_{\pi_k(u)}) = \rho(X_{\pi_{n}(u)}, X_{\pi_k(u)}) = \sum_{k \leq l \leq n} \rho(X_{\pi_{l+1}(u)}, X_{\pi_l(u)}),\end{equation*}$$

and since $\rho(X_{\pi_{l+1}(u)}, X_{\pi_l(u)}) \leq Y_{l+1}$, we obtain

$$\begin{equation*}\rho(X_u, X_{\pi_k(u)}) \leq \sum_{ l \geq k} Y_{l+1}\end{equation*}$$

We then use the previous inequalities and get

$$\begin{equation*}\rho(X_s, X_t) \leq \rho(X_s, X_{\pi_k(s)}) + \rho(X_{\pi_k(s)}, X_{\pi_k(t)}) + \rho(X_{\pi_k(t)}, X_t) \leq 3 \sum_{ l \geq k} Y_{l+1}\end{equation*}$$

The inequality for maximum

The number of points on which we consider the maximum process $Y_n$ can be bounded as follows:

$$\begin{equation*}\left| \left\{(s,t) \in G_n \colon |s-t| \leq \frac{3}{2^n} \right\} \right| \leq 3^d 2^{nd}\end{equation*}$$

Then using Chebyshev inequality and condition (1):

$$\begin{align*}\mathbb{P} \left\{ Y_n > \frac{1}{2^{pn}}\right\}&= \mathbb{P} \left\{\bigcup_{\substack{(s,t) \in G_n \\ |s-t| \leq 3 2^{-n}}}\left\{\rho(X_s, X_t) > \frac{1}{2^{pn}}\right\}\right\} \\&\leq \sum_{\substack{(s,t) \in G_n \\ |s-t| \leq 3 2^{-n}}}\mathbb{P} \left\{\rho(X_s, X_t) > \frac{1}{2^{pn}}\right\} \\&= \sum_{\substack{(s,t) \in G_n \\ |s-t| \leq 3 2^{-n}}}\mathbb{P} \left\{\rho(X_s, X_t)^a > \frac{1}{2^{pna}}\right\} \leq C(d) 2^{n(pa - b)}.\end{align*}$$

where the constant $C(d)$ depends only on $d$.

If we choose $p \in (0, \frac{b}{a})$

$$\begin{equation*}\sum_{n=1}^{\infty} \mathbb{P} \left\{ Y_n \geq \frac{1}{2^{pn}} \right\} < \infty\end{equation*}$$

By Borel-Cantelli Lemma, we get that $\exists N_0 \, \forall n \geq N_0 \colon Y_n \leq 2^{-p n}$ a.s. This implies convergence of the series $\sum_{n\geq0} Y_n$ a.s.

Hölder continuous of $X$

Thus

$$\begin{equation}\sup \left\{ \rho(X_s, X_t) \colon s, t \in G, |s-t| \leq 2^{-m} \right\} \leq 3 \sum_{n\geq m} Y_n \leq 3 \sum_{n\geq m} 2^{-pn} \leq \frac{2^{-pm}}{1 - 2^{-p}}\end{equation}$$

showing that $X$ is a.s. Hölder continuous on $G$ of order $p$.
In particular, $X$ agrees a.s. on $\bigcup_n \mathcal{D}_n$ with a continuous process $\tilde{X}$ on $[0,1]^d$, and we note that the Hölder continuity of $\tilde{X}$ on $G$ extends with the same order $p$ to the entire cube $[0,1]^d$. \newline To show that $\tilde{X}$ is a version of $X$, fix any $t \in[0,1]^d$, and choose $t_1, t_2, \ldots \in G$ with $t_n \rightarrow t$. Then $X_{t_n}=\tilde{X}_{t_n}$ a.s. for each $n$. Since also $X_{t_n} \stackrel{\mathbb{P}}{\rightarrow} X_t$ by (1) and $\tilde{X}_{t_n} \rightarrow \tilde{X}_t$ a.s. by continuity, we get $X_t=\tilde{X}_t$ a.s.

References

• [Durrett2019]Durrett, Rick. Probability—theory and examples. Fifth edition. Cambridge University Press, Cambridge, 2019
• [Billingsley2012]Billingsley, Patrick. Probability and measure. Anniversary edition. John Wiley Sons, Inc., Hoboken, NJ, 2012