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It is known that Brownian motion is almost surely locally Holder continuous, on a range that is random, i.e. depends on the particular path. This question explores the maximal range on which Brownian motion is Holder continuous.

Let $W$ be a standard Brownian motion, and let $C > 0$ and $0 < \alpha < \frac{1}{2}$ be constants.

Define the parametrized family of random variables $H_{C, \alpha}$ by

$$H_{C, \alpha} := \sup \big \{T > 0 \, \big | \, |W_t - W_s| \leq C |t - s|^\alpha \text{ for all } t, s \in [0, T] \big \}$$

Question: What is known about the probability distribution of $H_{C, \alpha}$? Does it admit a density, or an expression in terms of known distributions?

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    $\begingroup$ A trivial remark: this is equivalent to asking what is the distribution of the optimal Hölder constant $C_\alpha$ for the Brownian motion $W_t$, $t \in [0,1]$. No idea whether this has been studied, though. $\endgroup$ Commented Dec 10, 2021 at 8:28

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Claim: Let $0<\alpha<1/2$ and $0<C<\infty$. The random variable $H_{C,\alpha}$ has an absolutely continuous distribution.

Proof: Let $\{{\cal F}_t\}_{t \ge 0}$ be the standard Brownian Filtration. Given a non-negative rational $q$, define the ${\cal F}_q$-measurable random boundary functions $\psi_q^+:[q,\infty) \to (-\infty,\infty)$ and $\psi_q^-:[q,\infty) \to (-\infty,\infty)$ by $$\psi_q^+(t)=\inf_{s \le q} \; [W_s+C(t-s)^\alpha]$$ and $$\psi_q^-(t)=\sup_{s \le q} \; [W_s-C(t-s)^\alpha]\,.$$ Observe that both these functions are a.s. continuous, and $\psi_q^+$ is nondecreasing. For $$q \in G^+=\{q \in {\mathbb Q}: \psi_q^+(q)>W_q\} \,,$$ consider the stopping time $$H_{C, \alpha}^+(q) := \inf \big \{t \ge q \,: W_t = \psi_q^+(t)\} \,.$$ Similarly, for $$q \in G^-:=\{q \in {\mathbb Q}: \psi_q^-(q)<W_q\} \,,$$ consider the stopping time $$H_{C, \alpha}^-(q) := \inf \big \{t \ge q \,: W_t = \psi_q^-(t)\} \,.$$ Since $W$ is locally $\beta$-Holder continuous for $\beta \in (\alpha,1/2)$, we infer that with probability 1, we have $$H_{C,\alpha} \in \{H_{C, \alpha}^+(q) : q \in G^+\} \cup \{H_{C, \alpha}^-(q) : q \in G^-\} \,,$$ so by symmetry, it suffices to prove that for fixed $q \in G^+$, the stopping time $H_{C, \alpha}^+(q)$ has an absolutely continuous distribution. We deduce this from the next Lemma using the Markov property of $W$ at time $q$. QED

Lemma: Let $\psi:[0,\infty) \to [r,\infty)$ be a nondecreasing function, with $\psi(0)=r>0$. Denote $H:=\inf \{t \ge 0 : W_t=\psi(t)\},$ where the infimum of the empty set is $\infty$. Then for all $t,\epsilon>0$ and some absolute constant $A$, we have
$$P\Bigl(H \in [t,t+\epsilon]\Bigr) \le Ar^{-2}\epsilon \,.$$

Proof: For $b>0$, let $\tau_b:=\inf \{t \ge 0 : W_t=b\}.$ Then $\tau_1$ has a Levy distribution, with density bounded above by some absolute constant $A>0$. See e.g. https://en.wikipedia.org/wiki/First-hitting-time_model Since $\tau_b$ has the same law as $b^2 \tau_1$, the density of $\tau_b$ is bounded above by $Ab^{-2}$. Take $b=\psi(t) \ge r$. Then $$P\Bigl(H \in [t,t+\epsilon]\Bigr) \le P\Bigl(\tau_b \in [t,t+\epsilon]\Bigr) \le Ab^{-2}\epsilon \,,$$ and the lemma follows.

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  • $\begingroup$ Wow, very nice. $\endgroup$
    – Nate River
    Commented Dec 31, 2021 at 23:29

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