Say $B$ is a Brownian motion and $B^n$ is the function that is equal to $B$ at all numbers of the form $2^{-n}\mathbb{N}$ and linear on the intervals $[k2^{-n},(k+1)2^{-n}]$. I am trying to show that $\lim_{n\to\infty} P(\sup_{[0,1]}|B-B^n|\leq\epsilon )=1$. I was thinking that $|B-B^n|$ can be bounded by $|B_{\frac{k}{2^n}}-B_{\frac{k+1}{2^n}}|$ which is $\mathcal{N}(0,\frac{1}{2^n})$ distributed. Is this correct? Any hint would be appreciated
Brownian motion dyadic construction convergence
brownian motionprobabilityprobability theory
Related Solutions
Fix $\delta>0$. By the very definition of the supremum, there exists for any $n \in \mathbb{N}$ some $s_n \in (0,s_0]$ such that
$$\sup_{0<s \leq s_0} \left| \frac{W(t_n(s))-W(s)}{t_n(s)^{1/2-\epsilon}} \right| \leq \left| \frac{W(t_n(s_n))-W(s_n)}{t_n(s_n)^{1/2-\epsilon}} \right| + \delta.$$
If we can show that
$$\limsup_{n \to \infty} \left| \frac{W(t_n(s_n))-W(s_n)}{t_n(s_n)^{1/2-\epsilon}} \right| =0 \tag{4}$$
then this proves the assertion.
Proof of $(4)$: There exists a subsequence $(s_n')_{n \in \mathbb{N}}$ of $(s_n)_{n \in \mathbb{N}}$ such that the $\limsup$ in $(4)$ is attained, i.e.
$$\lim_{n \to \infty} \left| \frac{W(t_n(s_n'))-W(s_n')}{t_n(s_n')^{1/2-\epsilon}} \right| = \limsup_{n \to \infty} \left| \frac{W(t_n(s_n))-W(s_n)}{t_n(s_n)^{1/2-\epsilon}} \right| \tag{5}$$
Since $(s_n')_{n \in \mathbb{N}}$ is contained in the compact interval $[0,s_0]$, we may assume without loss of generality that $u := \lim_{n \to \infty} s_n'$ exists (otherwise we take another subsequence). In order to prove $(4)$, we will show that the left-hand side of $(5)$ is zero, and to this end we consider two cases separately.
Case 1: $u>0$. Since $s_n' \to 0$ implies, by $(1)$, $t_n(s_n') \to u$ it follows from the continuity of the sample paths of Brownian motion that
$$\frac{W(t_n(s_n'))-W(s_n')}{t_n(s_n')^{1/2-\epsilon}} \xrightarrow[]{n \to \infty} \frac{W(u)-W(u)}{u^{1/2-\epsilon}} = 0.$$
Case 2: $u=0$. Fix some $\gamma>0$. Because of $(2)$, there exists $r>0$ such that $$\sup_{0<s \leq r} \left| \frac{W(s)}{s^{1/2-\epsilon}} \right| \leq \gamma.$$
As in Case 1 we have $t_n(s_n') \to u$ and so $t_n(s_n') \to 0$. In particular, we can choose $N \in \mathbb{N}$ such that $$|t_n(s_n')| \leq r \quad \text{and} \quad |s_n'| \leq r $$ for all $n \geq N$. Hence,
$$\begin{align*} \left|\frac{W(t_n(s_n'))-W(s_n')}{t_n(s_n')^{1/2-\epsilon}}\right| &\leq \left| \frac{W(t_n(s_n'))}{t_n(s_n')^{1/2-\epsilon}} \right| + \frac{|s_n'|^{1/2-\epsilon}}{t_n(s_n')^{1/2-\epsilon}} \left| \frac{W(s_n')}{|s_n'|^{1/2-\epsilon}} \right| \\ &\leq \gamma + \frac{|s_n'|^{1/2-\epsilon}}{t_n(s_n')^{1/2-\epsilon}} \gamma \end{align*}$$
for all $n \geq N$. It is immediate from $(1)$ that
$$\frac{|s_n'|^{1/2-\epsilon}}{t_n(s_n')^{1/2-\epsilon}} \leq 1+ \gamma$$
for $n$ sufficiently large, and therefore we conclude that
$$\left|\frac{W(t_n(s_n'))-W(s_n')}{t_n(s_n')^{1/2-\epsilon}}\right| \leq \gamma + (1+\gamma) \gamma$$
for all $n$ sufficiently large. As $\gamma>0$ was arbitrary, this proves
$$\lim_{n \to \infty} \left|\frac{W(t_n(s_n'))-W(s_n')}{t_n(s_n')^{1/2-\epsilon}}\right| =0.$$
Remark: Note that we have actually shown that $(4)$ holds for any sequence $(s_n)_{n \in \mathbb{N}} \subseteq (0,s_0]$ (... and not only for the sequence which we picked at the very beginning of this answer).
All you need is $\lim \sup P(E_n) \leq P( \lim \sup E_n)$ which follows by applying Fatou's Lemma to then sequence $(I_{E_n^{c}})$.
Best Answer
First, the process $W_t^{n,k}:=(B_t-B_t^n)$, $t\in[k2^{-n},(k+1)2^{-n}]$ is a Brownian bridge. Therefore, using for example this result, \begin{align*} &\mathsf{P}\!\left(\sup_{t\in [k2^{-n}, (k+1)2^{-n}]}|W_t^{n,k}|>\epsilon\right) \\ &\qquad\le 2\mathsf{P}\!\left(\sup_{t\in [k2^{-n}, (k+1)2^{-n}]}W_t^{n,k}>\epsilon\right)=2e^{-2^{n+1}\epsilon^2}, \end{align*} and so, \begin{align*} \mathsf{P}\!\left(\sup_{t\in[0,1]}|B_t-B_t^n|> \epsilon\right)&= \mathsf{P}\!\left(\max_{0\le k\le 2^n-1}\sup_{t\in [k2^{-n}, (k+1)2^{-n}]}|W_t^{n,k}|> \epsilon\right) \\ &\le \sum_{k=0}^{2^n-1}\mathsf{P}\!\left(\sup_{t\in [k2^{-n}, (k+1)2^{-n}]}|W_t^{n,k}|> \epsilon\right) \\ &\le 2^{n+1}e^{-2^{n+1}\epsilon^2}\to 0 \quad\text{as}\quad n\to\infty. \end{align*}