I am studying continuous stochastic processes and I would like to show that if my stochastic process $X$ satisfies the hypothesis of the well known Kolmogorov continuity theorem, I can find a set $A$ with $\mathbb{P}(A)=1$ on which $X_{t}(.)$ is $\delta$ Holdër continuous.
Here is the important part of the Kolmogorov continuity theorem that I use to show this (that is the existence of a modification\version $Y$ satisfying the following inequality) :
Given assumptions we assume here on a stochastic process $X$, there exists a version $Y$ of $X$ satisfying for all $\delta\in[0, \frac{\beta}{\alpha}[$
$$
\mathbb{E}\Big[\Big(\sup\Big\{\frac{\lvert Y_s – Y_t\rvert}{\lvert s -t\rvert^{\delta}} : s\neq t\in [0,1]\Big\}\Big)^{\alpha}\Big]<\infty.
$$
where $\alpha,\beta>0$.
Here is my attempt :
I denote $k_{\delta} := \sup\big\{\frac{\lvert Y_s – Y_t\rvert}{\lvert s -t\rvert^{\delta}} : s\neq t\in [0,1]\big\}$. This random variable is finite almost surely from the theorem. Thus, we have
$$
\mathbb{P}\Big(\omega \,\big\vert\, \forall\delta\in[0,\frac{\beta}{\alpha}[, \forall s\neq t\in[0,1] : \frac{\lvert Y_s(\omega)- Y_t(\omega)\rvert}{\lvert s -t\rvert^{\delta}}\leq k_{\delta}(\omega)\Big).
$$
By continuity of the process, we have
$$
\mathbb{P}\Big(\bigcap_{s\neq t\in [0,1]\cap\mathbb{Q}}\bigcap_{\delta\,\in\, [0,\frac{\beta}{\alpha}[\,\cap\,\mathbb{Q}}\big\{\omega :\lvert Y_s(\omega)- Y_t(\omega)\rvert\leq k_{\delta}(\omega)\lvert s -t\rvert^{\delta}\big\}\Big).
$$
From there, I don’t know how to proceed. I would like to take out the intersection and transform it as a limit, but I cannot justify this.
I would like to know if what I have done so far is okay, and, if not, how to improve it. Thank you a lot!
Best Answer
To clear up a misconception, Kolmgorov's continuity theorem does not guarantee that there exists a set $A$ with $\mathbb{P}(A) = 1$ on which $t \mapsto X_t$ is continuous. Rather, it guarantees that there exists a continuous process $Y$ such that, for all $t \ge 0$, $\mathbb{P}(Y_t = X_t) = 1$. The important distinction is that $\{\omega : Y_t(\omega) = X_t(\omega)\}$ may be different for each $t$.
From what I understand of your question, you've shown that there exists a version $Y$ of $X$ satisfying $$\mathbb{E}\left[\left( \sup_{s \ne t \in [0,1]} \frac{|Y_t-Y_s|}{|t-s|^{\delta}}\right)^{\alpha} \right] < \infty$$ for all $\delta \in [0,\frac{\beta}{\alpha}[$, and what remains is to show that this version is Holder-continuous almost surely.
As a reminder, that means we want to show that for almost every $\omega$, we have that there exists constants $p = p(\omega) > 0$ and $C = C(\omega) < \infty$ (possibly depending on $\omega$) such that $|Y_t(\omega)-Y_s(\omega)| \le C |t-s|^p$ for all $t,s \in [0,1]$.
Using your definition of $k_\delta(\omega) := \sup_{s \ne t \in [0,1]} \frac{|Y_t(\omega)-Y_s(\omega)|}{|t-s|^{\delta}}$, we have immediately that $|Y_t(\omega)-Y_s(\omega)| \le k_\delta(\omega) |t-s|^\delta$ for all $t,s \in [0,1]$ just from the definition of $k_\delta$. Since $\mathbb{P}(k_\delta<\infty) = 1$, we conclude that we do indeed have that $Y$ is Holder-continuous almost surely. Even better, we have that it is almost surely $\delta$-Holder continuous for all $\delta \in [0,\frac{\beta}{\alpha}[$.
As an example to show we only have continuity of a modification, consider $\Omega = [0,1]$ with $\mathbb{P}$ equal to the Lebesgue measure, and define $X_t(\omega) = t1_{t \ne \omega}$. Then, for all $t,s \in [0,1]$, we have $\mathbb{E}[|X_t-X_s|^\alpha] = |t-s|^{\alpha}$, so Kolmgorov's continuity theorem applies and we can find a Holder continuous modification (namely $Y_t(\omega) = t$). However, for every $\omega \in \Omega$, the path $t \mapsto X_t(\omega)$ has a discontinuity on $[0,1]$, so there is no chance of finding a set with probability $1$ on which $X$ is continuous (Holder or otherwise).