Convergence rate of Ito process path

Question

Let $|t-t_0|=O(\alpha_n)$ and $dX_t = \mu_t dt + \sigma_t dB_t$ with $X_0=0$. Then what is the convergence rate of the Ito process for a given realized path $\omega$, i.e., $$ |X_t(\omega) - X_{t_0}(\omega)| = O(?). $$

What I have tried.

Using the mean value theorem, we have $|X_t(\omega)-X_{t_0}(\omega)|=\left|X'_{t^*}(\omega)(t-t_0) \right|$. So If $X'_{t_0}(\omega)$ is well-defined, due to $|t^*-t_0|\le|t-t_0|=O(\alpha_n)$, we have $O(\alpha_n)$ unless $X'_{t_0}(\omega)=0$. However, it is well known that the Brownian motion is nowhere differentiable, so $X'_{t_0}(\omega)$ is not defined for all $t_0$. The only thing that I can know is that since $X_t(\omega)$ is continuous, $X_t(\omega)\rightarrow X_{t_0}(\omega)$ anyway. So is it possible to obtain the convergence rate of the above Ito process?

Thanks,

Answer 1

Under the bounded condition for absolute moments of $\mu_t$ and $\sigma_t$, it's $$ |X_t(\omega) - X_{t_0}(\omega)| = O_p(\sqrt{\alpha_n \log(1/\alpha_n)}). $$ It's kind of Lévy's modulus of continuity theorem.

For the detail, see this.