I am reading Morter's book on Brownian motion, and am confusing about a portion of Theorem 2.11. The lemma being proved is that for two non-overlapping intervals, the maxima are almost surely distinct.
Proof. We first show that, given two nonoverlapping closed time intervals, i.e. such that their interiors are disjoint, the maxima of Brownian motion on them are different almost surely, see Figure 2.2 for an illustration. Let $\left[a_1, b_1\right]$ and $\left[a_2, b_2\right]$ be two fixed intervals with $b_1 \leqslant a_2$. Denote by $m_1$ and $m_2$, the maxima of Brownian motion on these two intervals. Note first that, by the Markov property together with Theorem 2.8, almost surely $B\left(a_2\right)<$ $m_2$. Hence this maximum agrees with maximum in the interval $\left[a_2-\frac{1}{n}, b_2\right]$, for some $n \in \mathbb{N}$, and we may therefore assume in the proof that $b_1<a_2$. Applying the Markov property at time $b_1$ we see that the random variable $B\left(a_2\right)-B\left(b_1\right)$ is independent of $m_1-B\left(b_1\right)$. Using the Markov property at time $a_2$ we see that $m_2-B\left(a_2\right)$ is also independent of both these variables. The event $m_1=m_2$ can be written as
$$
B\left(a_2\right)-B\left(b_1\right)=m_1-B\left(b_1\right)-\left(m_2-B\left(a_2\right)\right)
$$
Conditioning on the values of the random variables $m_1-B\left(b_1\right)$ and $m_2-B\left(a_2\right)$, the left hand side is a continuous random variable and the right hand side a constant, hence this event has probability 0 .
Note that Theorem 2.8 being referred to is
Theorem 2.8 Suppose $\{B(t): t \geqslant 0\}$ is a linear Brownian motion. Define $\tau=\inf \{t>$ $0: B(t)>0\}$ and $\sigma=\inf \{t>0: B(t)=0\}$. Then
$$
\mathbb{P}_0(\{\tau=0\})=\mathbb{P}_0(\{\sigma=0\})=1
$$
My question is about the lines "note first that by Theorem 2.8, …, we may therefore assume." How does the Markov property and Theorem 2.8 impliy $B(a_2) < m_2$ almost surely? And why are we taking $a_2 – 1/n$, that is making the second interval bigger, surely we want $a_2 + 1/n$ so they are disjoint now, rather than merely non-overlapping, since $b_1 < a_2$ is given. Those lines in particular are giving me trouble. Thank you!
Best Answer
The Markov property implies $B(t+a_2) - B(a_2)$ is a Brownian motion. Applying 2.8 to this Brownian motion, we see that $\inf\{t \geq 0 \mid B(t+a_2) - B(a_2) > 0\}$ is almost surely $0$. This in particular means that there necessarily is some $t \in (0, b_2-a_2)$ (depending on the path) such that $m_2 \geq B(t+a_2) > B(a_2).$
The simple answer is that this is an error in the text, and it should be $a_2 + 1/n$.
One issue that the author seems to skirt around is the fact $n$ depends on the path, so we haven't actually reduced to $b_1 < a_2$ (as the $a_2$ in the problem statement necessarily is non-random). You should see if you can fix that issue as well.