Why $\mathbb E\left[\sup\frac{|Y_t-Y_s|}{|t-s|^\alpha }\right]<\infty$ imply $(Y_t)_t$ continuous?

Question

Let $(Y_t)_t$ a stochastic process s.t. $$\mathbb E\left[\sup_{s,t\in [0,1], s\neq t}\frac{|Y_t-Y_s|}{|t-s|^\alpha }\right]<\infty,$$ with $\alpha >0$. Why does this implies that $(Y_t)_t$ is continuous a.s. ?

Does it come from the fact that if $\mathbb E[X]<\infty$ then $\mathbb P\{X<\infty\}=1$, and thus $$\mathbb P\left\{\sup_{s,t\in [0,1], s\neq t}\frac{|Y_t-Y_s|}{|t-s|^\alpha }<\infty\right\}=1.$$

Also $$\mathbb P\left\{\sup_{s,t\in [0,1], s\neq t}\frac{|Y_t-Y_s|}{|t-s|^\alpha }<\infty\right\}\leq \mathbb P\left\{\frac{|Y_t-Y_s|}{|t-s|^\alpha}<\infty\right\}=1.$$

1) How can I continue ? Does it implies that there is $C>0$ s.t. $$\mathbb P\{|Y_t-Y_s|<C|t-s|^\alpha \}=1,$$ or that $$\mathbb P\{\exists C>0: |Y_t-Y_s|\leq C|t-s|^\alpha \}=1 \ \ ?$$

2) And will it implies that $$\mathbb P\{\lim_{t\to s}|Y_t-Y_s|\}=1 \ \ ?$$ If yes, why ? I don't understand why I can put the limit inside.

Answer 1

• Assume that $Y = (Y_t)_{t\in[0, 1]}$ is realized over a complete probability space $(\Omega, \mathcal{F}, \mathbb{P})$ such that the function $C : \Omega \to [0, \infty]$ defined by

  $$ C(\omega) = \sup_{\substack{s, t \in [0, 1] \\ s \neq t}} \frac{|Y_s(\omega) - Y_t(\omega)|}{|s-t|^{\alpha}} $$

  is measurable, then

  1. If $\omega \in \Omega$ is such that $C(\omega) < \infty$, then $|Y_t(\omega) - Y_s(\omega)| \leq C(\omega)|t - s|^{\alpha}$ for all $s, t \in [0, 1]$ and hence $t \mapsto Y_t(\omega)$ is $\alpha$-Holder continuous. In particular, $$ E := \{ \omega : t \mapsto Y_t(\omega) \text{ is continuous} \}$$ satisfies $$ E^{c} \subseteq \{ \omega : C(\omega) = \infty \} $$

  2. If in addition that $\mathbb{E}[C] < \infty$, then $\{ C = \infty\}$ is measurable and $\mathbb{P}[C = \infty] = 0$. Since the probability space is complete, this implies that $E^c$ is also a measurable null-set. Therefore $E$ is also measurable and $\mathbb{P}[E] = 1$, i.e. the sample path is $\mathbb{P}$-a.s. continuous.

• It is worth to notice that $C$ and $E$ need not be measurable for any choice of the base probability space. For instance, under the naive choice $(\Omega, \mathcal{F}) = (\mathbb{R}^{[0,1]}, \mathcal{B}(\mathbb{R}^{[0,1]}))$, one can prove that

  $$ C([0, 1], \mathbb{R}) = \{ \omega \in \mathbb{R}^{[0,1]} : \omega \text{ is continuous} \} \notin \mathcal{B}(\mathbb{R}^{[0,1]}). $$

  (This is because, for any $A \in \mathcal{B}(\mathbb{R}^{[0,1]})$, there exists $I \subset [0, 1]$ such that $I$ is at most countable and $A$ is determined only by the information at times $t \in I$, i.e., $A \in \sigma( \pi_t : t \in I)$, where $\pi_t : \mathbb{R}^{[0,1]} \to \mathbb{R}$ is the projection $\pi_t(\omega) = \omega_t$.) This tells that, in order to discuss regularity of sample paths, one needs to choose a suitable base probability space to begin with.

  If one is given a stochastic process $S$, the usual workaround is to show that $X$ satisfies some weaker notion of stochastic continuity, and then show that this notion allows to construct a modification which has continuous sample paths. For instance, If $W = (W_t)_{t\geq 0}$ is the Brownian motion, then it satisfies $\mathbb{E}[|W_s - W_t|^2] = |s - t|$, hence by Kolmogorov-Chenstov theorem, $W$ has a modification whose sample paths are continuous. Then one may realize this modification on a nice probability space where sample-path continuity can be discussed as measurable event.