The following is a theorem in the book Brownian motion by Yuval Peres and Peter Morters. I have one difficulty in understanding the proof:
Theorem 2.11 For a linear Brownian motion $\{B(t): 0 \leqslant t \leqslant 1\}$, almost surely,
(a) every local maximum is a strict local maximum;
(b) the set of times where the local maxima are attained is countable and dense;
(c) the global maximum is attained at a unique time.
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. 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 .
What exactly have we done here in the last step? How do we get that the probability of the event $$
B\left(a_{2}\right)-B\left(b_{1}\right)=B\left(a_{2}\right)-B\left(b_{1}\right)-\left(m_{2}-B\left(a_{2}\right)\right) .
$$
Is zero, from the given argument? The only information that we have is that the random variables $B\left(a_{2}\right)-B\left(b_{1}\right)$, $B\left(a_{2}\right)-B\left(b_{1}\right)$ and $\left(m_{2}-B\left(a_{2}\right)\right)$ are independent.
Can someone elaborate the argument a bit. Thank you very much!
Best Answer
Suppose $X,Y,Z$ are independent r.v.'s. Let $U=Y-Z$. Then $P(X=U)=\int P(X=u)dF_U(u)$ (by conditioning on $U$) where $F_U$ is the distribution function of $U$. If $X$ has continuous distribution then $P(X=u)=0$ for all $u$ so $\int P(X=u)dF_U(u)=0$.