My goal is to prove that $B(t)/t \longrightarrow 0$ as $t\to \infty$ for a Brownian motion $(B(t))_{t\geq0}$. First I am told to prove that $(X_k^{(n)})_{1\leq k \leq 2^n}$ defined by $$X_k^{(n)}(t):= \lvert B(t+\frac{k}{2^n}) – B(t) \rvert$$
is a submartingale for arbitrary $t\geq 0$ (I assume wrt to the generated filtration). Can someone give me a hint on how to do this? I am not very familiar with Brownian motion yet. I tried writing
$$\mathbb E[X_{k+1}^{(n)}(t) \mid \mathcal F_n]- X_k^{(n)}(t) = \mathbb E[X_{k+1}^{(n)}(t)- X_k^{(n)}(t) \mid \mathcal F_n]$$ but don't know how I can get a lower bound…
Prove that this these increments of Brownian motion form a submartingale
brownian motionmartingalesprobabilityprobability theorystochastic-processes
Best Answer
The process
$$M_s := B(t+s)-B(t), \qquad s \geq 0,$$
is for each fixed $t \geq 0$ a martingale. Since the mapping $f(x) := |x|$ is convex, this implies, by Jensen's inequality, that $(|M_s|)_{s \geq 0}$ is a submartingale. In particular, $(|M_{k/2^n}|)_{k=1,\ldots,2^n}$ is a submartingale.