Brownian Motion inequality (related to Dvoretzky-Erdoes test)

Question

i have the following question: Let $B(t)$ be a d-dimeansional Brownian motion $d\ge 3$, and $f$ be a monoton increasing function from the positive reals to the positive reals. Let $A_n=(\exists t\in [2^n,2^{n+1}]:|B(t)|\le f(t))$ and $P_x$ be a probabiity measure, such that $B(t)$ is a B.M. starting at point $x$. Let $m<n-1$. Now why does: $$P(A_n|A_m)\le \sup_{x\in R^d} P_x(\exists t>2^{n-1} s.t. |B(t)|\le f(2^{n+1}))$$ hold? I know one has to use the Markov property at time $2^{n-1}$ but I am completely stuck.

Answer 1

Denote by $\mathcal{F}_t$ the (canonical) $\sigma$-algebra generated by $(B_t)_{t \geq 0}$. By the monotonicity of $f$, we have

$$A_n \subseteq \big\{ \exists t \in [2^n,2^{n+1}]: |B_t| \leq f(2^{n+1}) \big\} =: \tilde{A}_n.$$

Therefore, the Markov property implies

$$\begin{align*} \mathbb{P}(A_n \mid \mathcal{F}_{2^{n-1}}) &\leq \mathbb{P}(\tilde{A}_n \mid \mathcal{F}_{2^{n-1}}) \\ &= \mathbb{P}^{B_{2^{n-1}}}\left( \exists t \in [2^n-2^{n-1},2^{n+1}-2^{n-1}]: |B_t| \leq f(2^{n+1}) \right) \\ &\leq \sup_{x \in \mathbb{R}^d} \mathbb{P}^x(\exists t \geq 2^{n-1}: |B_t| \leq f(2^{n+1}))=: C_n. \end{align*}$$

In the last step, we have used that $t \in [2^n-2^{n-1},2^{n+1}-2^{n-1}]$ implies $t \geq 2^{n-1}$. Consequently, since $A_m \in \mathcal{F}_{2^{n-1}}$, the tower property shows

$$\mathbb{P}(A_n \mid A_m) = \mathbb{E} \bigg[ \underbrace{\mathbb{P}(A_n \mid \mathcal{F}_{2^{n-1}})}_{\leq C_n}) \mid A_m\bigg] \leq C_n \mathbb{E}(1 \mid A_m) \leq C_n.$$