Doob-Kolmogorov Inequality

Question

Denote by $(X(t),t\ge 0)$ a standard Brownian motion, i.e random variables with the following properties:

1. $X(0)=0$.

2. With probability 1, the function $t\mapsto X(t)$ is continuous on $[0,\infty)$.

3. For $0\le r<t$, we have $X(t)-X(r)$ is normally distributed with mean 0 and variance $t-r$.

Define $X(s)=\sup_{0\le t\le s} |X(t)|$. Prove that \begin{equation} P(X(s)\ge \delta) \le \frac{s}{\delta^2}. \end{equation}

My attempt is the following:

By no. 3 $E[X(s)]=s$. Thus, the problem can be reduced to show that \begin{equation} P(X(s)\ge \delta) \le \frac{E[X(s)]}{\delta^2}. \end{equation} Thanks a lot.

Answer 1

Hint: Use Doob's maximal inequality

$$\mathbb{P} \left( \sup_{s \leq t} |M(s)| \geq \delta \right) \leq \frac{1}{\delta^p} \mathbb{E}(|M_t|^p), \qquad \delta>0,$$

which holds for any continuous martingale $(M_t)_{t \geq 0}$ such that $M_t \in L^p$, $p \geq 1$.