Brownian motion Convergence If $X$ is a standard 1d brownian motion and $M_t$ $= \mbox{max}\{X_S: 0 \le s \le t\} $, what can we say about $M_t/t$ as $t \rightarrow \infty$?
Mainly, what can we say about the behavior of this martingale?
My attempt: P(Msubt > a) = 2P(B(t) >a), but what integral does this fit in with?
 A: Since $B_t$ is symmetric, the result recalled in the post reads $\mathrm P(M_t\gt at)=\mathrm P(|B_t|\gt at)$ $(\ast)$. 
Since $B_t$ equals $\sqrt{t}B_1$ in distribution, $(\ast)$ is $\mathrm P(|B_1|\gt a\sqrt{t})$. Since $|B_1|$ is almost surely finite and $a\sqrt{t}\to+\infty$ for every $a\gt0$, the last probability goes to zero, hence $\mathrm P(M_t/t\gt a)\to0$ for every $a\gt0$. That is, $M_t/t\to0$ in probability.
One can strengthen this convergence, noting that the series $\sum\limits_n\mathrm P(|B_1|\gt a\sqrt{n})$ converges since $\mathrm P(|B_1|\gt a\sqrt{n})$ is $\mathrm e^{-\frac12a^2n+o(n)}$. By the first Borel-Cantelli lemma, $M_n/n\to0$ almost surely when the integer $n$ goes to infinity. For every $t\geqslant1$ there exists an integer $n\geqslant1$ such that $n\leqslant t\lt n+1$, and $(M_t)_t$ is nondecreasing hence $M_t/t\leqslant M_{n+1}/n\leqslant 2M_{n+1}/(n+1)$. This proves that $M_t/t\to0$ almost surely when the real number $t$ goes to infinity.
The same identity $(\ast)$ shows the convergence in $L^p$ for every finite $p\geqslant1$.
