Let $B$ be a Brownian motion (in fact, I have in mind a Levy process). Let $I_s := [a_s,b_s]$ be an interval for all $s \in [0,2t]$.
How do you prove the following statement rigorously, just using the stationary and independent increments of $B$?
$$P(B_s \in I_s \forall s \in [0,2t]) \leq P(B_s \in I_s \forall s \in [0,t]) \cdot \sup_{x \in I_t} P(x + B_s \in I_{s+t} \forall s \in [0,t]). $$
It's clear 'by drawing a picture' and using intuitive understanding of independent stationary increments. But I'd like to see a very formal proof if possible.
Let $\{\mathcal F_s\}$ be the filtration associated with Brownian motion. Condition on $\mathcal F_t$, and use the Markov property and tower property of conditional expectation: \begin{align*} P(B_s\in I_s\text{ for all }s\in[0,2t]) &= E\Big[\mathbf1_{\{B_s\in I_s\text{ for }s\in[0,t]\}}P\big(B_s\in I_s\text{ for }s\in(t,2t]\,\big|\,\mathcal F_t\big)\Big]\\ &=E\Big[\mathbf1_{\{B_s\in I_s\text{ for }s\in[0,t]\}}P\big(B_t+(B_s-B_t)\in I_s\text{ for }s\in(t,2t]\,\big|\,B_t\big)\Big]\\ &\le E\Big[\mathbf1_{\{B_s\in I_s\text{ for }s\in[0,t]\}}\sup_{x\in I_t}P\big(x+(B_s-B_t)\in I_s\text{ for }s\in(t,2t]\,\big|\,B_t\big)\Big]\\ &=E\Big[\mathbf1_{\{B_s\in I_s\text{ for }s\in[0,t]\}}\sup_{x\in I_t}P\big(x+B_u\in I_{t+u}\text{ for }u\in(0,t]\big)\Big]\\ &=P(B_s\in I_s\text{ for }s\in[0,t])\sup_{x\in I_t}P(x+B_s\in I_{s+t}\text{ for }s\in[0,t]). \end{align*}