Law of large numbers for Brownian Motion (Direct proof using L2-convergence)

Question

In “Brownian Motion” by Schilling and Partzsc, they give a HINT to prove the Law of Large Numbers for Brownian Motion (not in their solutions, fyi) by

(1) Noting that $\left|\frac{B_{t}}{t}\right|\leq\left|\frac{B_{t}-B_{\left\lfloor t\right\rfloor +1}}{\left\lfloor t\right\rfloor }\right|+\frac{\left\lfloor t\right\rfloor +1}{\left\lfloor t\right\rfloor }\left|\frac{B_{\left\lfloor t\right\rfloor +1}}{\left\lfloor t\right\rfloor +1}\right|\leq\max_{\left\lfloor t\right\rfloor \leq s\leq\left\lfloor t\right\rfloor +1}\left|\frac{B_{s}-B_{\left\lfloor t\right\rfloor +1}}{\left\lfloor t\right\rfloor }\right|+\frac{\left\lfloor t\right\rfloor +1}{\left\lfloor t\right\rfloor }\left|\frac{B_{\left\lfloor t\right\rfloor +1}}{\left\lfloor t\right\rfloor +1}\right|$

and (2) showing that the first term on RHS converges to 0 in $L^{2}$ using the Reflection Principle and then making a subsequence argument.

I've made the following observations, but other than that, I'm stuck. First, the usual (strong) law of large numbers gives us $\frac{\left\lfloor t\right\rfloor +1}{\left\lfloor t\right\rfloor }\left|\frac{B_{\left\lfloor t\right\rfloor +1}}{\left\lfloor t\right\rfloor +1}\right|\rightarrow0\;\mbox{a.s.}\; t\rightarrow\infty.$ Hence it suffices to show that $\max_{\left\lfloor t\right\rfloor \leq s\leq\left\lfloor t\right\rfloor +1}\left|\frac{B_{s}-B_{\left\lfloor t\right\rfloor +1}}{\left\lfloor t\right\rfloor }\right|$ converges a.s. to 0 (or maybe something weaker is enough??).

If we can show $L^{2}$ convergence to 0 , then we have a.s. convergence for a subsequence. I've been able to show the $L^{2}$ convergence (see below), but I cannot get my head around the subsequence argument. Diverging from the Hint, I've proved the a.s. convergence directly by Borel-Cantelli and Doob's Inequality. But I'd really like to understand the $L^{2}$ approach as well.

$L^{2}$ convergence to 0: By $\left(B_{s+n}\right)_{s\geq0}\overset{D}{=}\left(B_{s}+B_{n}\right)_{s\geq0}$ and Doob's Inequality we get $\mathbb{E}\left(\left(\max_{\left\lfloor t\right\rfloor \leq s\leq\left\lfloor t\right\rfloor +1}\left|\frac{B_{s}-B_{\left\lfloor t\right\rfloor +1}}{\left\lfloor t\right\rfloor }\right|\right)^{2}\right) = \frac{1}{\left\lfloor t\right\rfloor ^{2}}\mathbb{E}\left(\left(\max_{\left\lfloor t\right\rfloor \leq s\leq\left\lfloor t\right\rfloor +1}\left|B_{s}-B_{\left\lfloor t\right\rfloor +1}\right|\right)^{2}\right) = \frac{1}{\left\lfloor t\right\rfloor ^{2}}\mathbb{E}\left(\left(\max_{s\leq1}\left|B_{s+\left\lfloor t\right\rfloor }-B_{1+\left\lfloor t\right\rfloor }\right|\right)^{2}\right) = \frac{1}{\left\lfloor t\right\rfloor ^{2}}\mathbb{E}\left(\left(\max_{s\leq1}\left|B_{s}-B_{1}\right|\right)^{2}\right) \leq \frac{1}{\left\lfloor t\right\rfloor ^{2}}(2^{2}\mathbb{E}\left(\left|B_{1}\right|^{2}\right)+constants)$ On a sidenote, I didn't seem to use the Reflection Principle?

Answer 1

Hence it suffices to show that $\max_{\left\lfloor t\right\rfloor \leq s\leq\left\lfloor t\right\rfloor +1}\left|\frac{B_{s}-B_{\left\lfloor t\right\rfloor +1}}{\left\lfloor t\right\rfloor }\right|$ converges a.s. to $0$ [when $t\to\infty$.]

Here is a direct approach. For every $n\geqslant1$, let $$X_n=\max_{n \leq s\leqslant n +1}\left|\frac{B_{s}-B_{n +1}}{n}\right|,$$ then $$X_n\stackrel{d}{=}\frac1{n}X_1$$ and $E(X_1^2)$ is finite, hence $$E\left(\sum_{n\geqslant1}X_n^2\right)=E(X_1^2)\sum_{n\geqslant1}\frac1{n^2}\ \text{is finite},$$ in particular, the series $$\sum_{n\geqslant1}X_n^2\ \text{converges almost surely,}$$ in particular, $$X_n\to0\ \text{almost surely.}$$ (The book probably mentions the reflection principle for the proof that $X_1$ is square integrable. No idea about the subsequence argument.)

Answer 2

As I suggested in the question, diverging from the hint, one could also proceed as follows:

Note that $\mathbb{P}(\limsup_{t\rightarrow\infty}\frac{\left|B_{t}\right|}{t}>\frac{1}{M})\leq\mathbb{P}(\bigl\{\sup_{t\in\left[n,n+1\right]}\frac{\left|B_{t}\right|}{n}>\frac{1}{M}\bigr\}\;\mbox{i.o.})=0$ for all $M\in\mathbb{N}$ by applying Doobs Inequality to get $\mathbb{P}\left(\sup_{t\in\left[n,n+1\right]}\left|B_{t}\right|>\frac{n}{M}\right)\leq\frac{M^{4}}{n^{4}}\mathbb{E}\bigl|B_{n+1}\bigr|^{4}\asymp\frac{1}{n^{2}}$ since $B_{n+1}\thicksim\mathcal{N}(0,n+1)$ and then using Borel Cantelli.

Hence $\mathbb{P}(\limsup_{t}\frac{\left|B_{t}\right|}{t}>0)=\mathbb{P}\left(\bigcup_{M\in\mathbb{N}}\left\{ \limsup_{t\rightarrow\infty}\frac{\left|B_{t}\right|}{t}\right\} >\frac{1}{M}\right)=0$ by subadditivity and thus $\mathbb{P}\left(\lim_{t}\frac{\left|B_{t}\right|}{t}=0\right)=\mathbb{P}(\limsup_{t}\frac{\left|B_{t}\right|}{t}=0)=1$ .