Sample path of Brownian Motion within epsilon distance of continuous function

Question

Given a continuous function $f:[0,1]\rightarrow\mathbb{R}$, $f(0)=0$, how can one show that $P(\underset{0\leq t\leq1}{\sup}\left|B_{t}-f(t)\right|<\varepsilon)>0$, where $P$ is the probability measure under which $(B_{t})_{t\geq0}$ is a standard Brownian Motion.

Any help would be much appreciated.

Answer 1

Let $f: [0,1] \to \mathbb{R}$ be a continuous function. Since $[0,1]$ is compact, $f$ is uniformly continuous on $[0,1]$, i.e. we can choose $n \in \mathbb{N}$ such that

$$|f(s)-f(t)| < \frac{\varepsilon}{2} \quad \text{for all $|s-t| \leq \frac{1}{n}$.}$$

If we set $t_j := j/n$ for $j=0,\ldots,n$, then

$$\begin{align*} \mathbb{P} \left( \sup_{0 \leq t \leq 1} |B_t-f(t)| <\varepsilon\right) &\geq \mathbb{P}\left( \forall j=0,\ldots,n-1: \sup_{t \in [t_j,t_{j+1}]} |B_t-f(t_j)| < \frac{\varepsilon}{2 (n-j)} \right) \\ &= \mathbb{P}\left( \prod_{j=0}^{n-1} 1_{A_j} \right) \end{align*}$$

for

$$A_j := \left\{\sup_{t \in [t_j,t_{j+1}]} |B_t-f(t_j)| < \frac{\varepsilon}{2(n-j)} \right\} \in \mathcal{F}_{t_{j+1}}.$$

It follows from the Markov property (of Brownian motion) and tower property (of conditional expectation) that

$$\begin{align*} \mathbb{P}\left( \prod_{j=0}^{n-1} 1_{A_j} \right) &= \mathbb{P} \left[ \left(\prod_{j=0}^{n-2} 1_{A_j} \right) \mathbb{P}^{B_{t_{n-1}}} \left(\sup_{t \in [0,1/n]} |B_t-f(t_{n-1})| < \frac{\varepsilon}{2} \right) \right]. \end{align*}$$

It suffices to show that

$$\mathbb{P}^x \left( \sup_{t \in [0,1/n]} |B_t-f(t_{n-1})| < \frac{\varepsilon}{2} \right)>c>0 \tag{1}$$

for all $x \in B(f(t_{n-1}),\varepsilon/4)$. (Then we can iterate the procedure and obtain the desired lower bound.) To this end, we note that

$$\begin{align*} \mathbb{P}^x \left( \sup_{t \leq 1/n} |B_t-f(t_{n-1})| < \frac{\varepsilon}{2} \right) &= \mathbb{P} \left( \sup_{t \leq 1/n} |B_t+x-f(t_{n-1})| < \frac{\varepsilon}{2} \right) \\ &\geq \mathbb{P} \left( \sup_{t \leq 1/n} |B_t| < \frac{\varepsilon}{4} \right) \end{align*}$$

for all $x \in B(f(t_{n-1}),\varepsilon/4)$. As $M_{1/n} := \sup_{t \leq 1/n} B_t \sim |B_{1/n}|$ (by the reflection principle), $(1)$ follows.

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Remark: The asymptotics of the probability $\mathbb{P} \left( \sup_{t \in [0,1]} |B_t-f(t)| < \varepsilon \right)$ as $\varepsilon \to 0$ is subject of so-called small ball estimates.

Answer 2

Sorry I cant leave this as a comment. As a first idea, you can approximate the continuous function $f$ with a piece-wise linear continuous function (discretise the interval $[0,1]$. And on each subinterval $[t_k, t_{k-1}]$, you can use a brownian bridge to calculate the probability and show it's positive.

Answer 3

Too long for a comment.

I don't think that Saz's proof works as it is written. In the induction you don't want to approximate the constant function $f(t_j)$ but the piecewise linear function $s\mapsto sf(f_{j+1})+(1-s)f_j$ (in fact, you can just assume that $f$ itself is piecewise linear). More specifically, you must define $A_j:=\{\sup_{[0,1/n]}|B_{t_j+s}-(sf(t_{j+1})+(1-s)f(t_j))|<\epsilon/2(n-j)\}$ and then show that given a line $s\mapsto ms+q$ with $|q|<\epsilon/2(n-j+1)$, then $\mathbb{P}(\sup_{[0,1/n]}|B_s-(ms+q)|<\epsilon/2(n-j))>0$. This can be done for example using Girsanov and that you already know that for all $\sigma>0$ you have $\mathbb{P}(\sup_{[0,1/n]}|B_s|<\sigma)>0$ (as shown here How to show that $P( \sup_{0 \leq s \leq 1} |B_s| \leq \epsilon) > 0$ for any $\epsilon > 0$?). In this case $\sigma=\epsilon/2(n-j)-\epsilon/2(n-j+1)$.