Pathwise linearization of diffusion processes

Question

Let $W$ be a standard $n$-dimensional Brownian motion, and $X$ the diffusion process given by the solution to the SDE

$$dX_t = \mu(X_t) \, dt + \sigma(X_t) \, dW_t,$$

with $\mu: \mathbb R^n \to \mathbb R$, $\sigma: \mathbb R^n \to \mathbb R^{n \times n}$ Lipschitz continuous, and $\sigma$ uniformly elliptic.

Then is it true that for all $t \geq 0$, almost surely we have

$$\limsup_{h \to 0^+} \left | \frac{X_{t + h} - X_t}{h^{1/2}} - \sigma (X_t) \left ( \frac{W_{t + h} - W_t}{h^{1/2}} \right ) \right | = 0?$$

Remark: The bound requested is quite sharp. To demonstrate its power, note that we would obtain as an immediate corollary the following:

Corollary: (Law of iterated logarithm for diffusion processes)

Almost surely, for all $t \geq 0$, we have

$$\limsup_{h \to 0_+} \frac{X_{t+h} - X_t}{\sqrt{2h \log \log \frac{1}{h}}} = -\liminf_{h \to 0_+} \frac{X_{t+h} - X_t}{\sqrt{2h \log \log \frac{1}{h}}} = |\sigma(X_t)|.$$

Answer 1

This is true. We need the following result, essentially proven in this post.

Proposition: (Linearization of diffusion process)

With $X$ as in the problem statement, we have $$ \mathbb E \left [\frac{\sup_{0 \leq t \leq T} |X_t - x - \sigma(X_0) W_t|}{\sqrt T} \right ] = O(\sqrt T)$$ as $T \to 0^+$.

Proof of main theorem: We aim to deduce this from the above proposition. The idea will be to apply the Borel-Cantelli lemma along a sparse enough subsequence, but not too sparse so that we do not lose control of the difference quotients.

To this end, let $\varepsilon > 0$ be arbitrary. Consider the sequence $t_n := \frac{1}{n^3}$. By the proposition and the Markov inequality, there exists some $C, N> 0$ such that for all $n \geq N$,

$$\mathbb P \left ( \frac{\sup_{0 \leq t \leq {t_n}} |X_{t_n} - x - \sigma(X_0) W_{t_n}|}{\sqrt {t_{n+1}}} \geq \varepsilon \right )$$ $$= \mathbb P \left (\frac{\sup_{0 \leq t \leq {t_n}} |X_{t_n} - x - \sigma(X_0) W_{t_n}|}{\sqrt {t_{n}}} \geq \varepsilon \sqrt \frac{t_{n+1}}{t_n}\right )$$ $$ \leq \frac{C}{\varepsilon}\sqrt \frac{t^2_{n}}{t_{n+1}}.$$ Since $\frac{t_n}{t_{n+1}} \to 1$ as $n \to \infty$, we conclude that for all large enough $n$,

$$\mathbb P \left (\frac{\sup_{0 \leq t \leq {t_n}} |X_{t_n} - x - \sigma(X_0) W_{t_n}|}{\sqrt {t_{n+1}}} \geq \varepsilon \right ) \leq \frac{C}{\varepsilon} \sqrt t_n,$$

with a different constant $C$. Recalling that $t_n := \frac{1}{n^3}$, we see that the RHS is summable in $n$, and so $\mathbb P (\limsup_{n} E_{n, \varepsilon}) = 0$, where $E_{n, \varepsilon}$ denotes the event that $\frac{\sup_{0 \leq t \leq {t_n}} |X_{t_n} - x - \sigma(X_0) W_{t_n}|}{\sqrt {t_{n+1}}} \geq \varepsilon$.

To conclude, we note that the event $ \{ \limsup_{h \to 0^+}| \frac{X_h - x - \sigma(X_0) W_h}{h^{1/2}} | > 0 \}$ is contained within the union $\bigcup_{m = 1}^\infty \limsup_n E_{n, 1/m}$ which has probability $0$. Thus $$\limsup_{h \to 0^+} \left | \frac{X_h - x - \sigma(X_0) W_h}{h^{1/2}} \right | = 0$$ almost surely, as claimed.