Let $(Y_t)_{t\leq 0}$ be a continuous uniformly integrable martingale. It can be shown that for any $p\geq 1$, the following two properties are equivalent :
- there is a constant $C$ such that for any stopping time $T$ $$\mathbb{E}(|Y_\infty -Y_T|^p)\leq C\mathbb{P}(T<\infty ) $$
- there is a constant $C$ such that for any stopping time $T$ $$\mathbb{E}(|Y_\infty -Y_T|^p|\mathcal{F}_T)\leq C .$$
If one of these properties if satisfied we say that $(Y_t)$ is a $BMO_p$ martingale. For example, any uniformly integrable martingale with $Y_\infty$ bounded a.s. is a $BMO_p$ martingale, $(B_{t\wedge 1})_{t\geq 0}$ where $(B_t)$ is the standard brownian motion is also a $BMO$ martingale.
Proposition : If $(Y_t)$ is a $BMO$ martingale then for any stopping time $S$, $Y^S=(Y_{t\wedge S})$ is also a $BMO$ martingale.
Assume now that $C=1$. Let $a>0$ be a fixed real and define the sequence of stopping time $(R_n)_{n\geq 0}$ by $$ R_0=0 \text{ and }R_{n+1}=\inf \{ t\geq R_n : |Y_t-Y_{R_n}|>a\}.$$
By applying $1.$ to the martingale $Y^{R_{n+1}}$ with the stopping time $T=R_n$ it can be shown that
Proposition : $a^p\mathbb{P}(R_{n+1}<\infty )<\mathbb{P}(R_{n}<\infty )$
I have proved all the previous statements but I didn't succeed to prove that the last proposition implies that for all $x>0$ $$ \mathbb{P}(\sup _{t\geq 0}|Y_t|>x)\leq 2^p2^{-px/2}.$$
First define the upcrossing stopping time as $$T^+_{n + 1} = \inf\{t \geq T^+_n : Y_t - Y_{T^+_n} > a \}$$ Since $Y_t$ is a Martingale, by the Strong Markov Property, the probabilities that $T^+_n$ and $R_n$ are finite are equal: $$P(T^+_n < \infty) = P(R_n < \infty)$$ So $T^+_n$ satisfies the same scaling relationship as $R_n$: $$a^p\mathbb{P}(T^+_{n+1} < \infty) < \mathbb{P}(T^+_n < \infty)$$
For a given $a$ and since probabilities are at most $1$ it holds from the scaling relationship that: $$P(T^+_{n} < \infty) < \frac{1}{a^{(n-1)p}}$$
Let $x > 2$ (the inequality holds trivially otherwise) and define $n$ to be the first integer greater than $x / 2$. If we experience at least $n$ upcrossings in finite time, we can be sure the supremum of $Y_t$ is greater than $x$, i.e. $$P(\sup Y_t > x) \leq P(T^+_{n} < \infty)$$ Therefore $$P(\sup Y_t > x) \leq \frac{1}{2^{(n-1)p}} \leq \frac{1}{2^{(x/2 -1)p}}$$