I would like to ask about intuitively interpreting the results of Doob's Optiona Stopping Theorem applied to standard Brownian motion.
The theorem provides three conditions, under which a stopped process is a martingale. One of these conditions is that the stopping time $T_A$ (associated with an event "$A$" that results in stopping the process) is finite in expectation, i.e.: $\mathbb{E}\left[ T_A \right] < \infty$.
Case 1: Let's define the stopping time $T_A$ associated with $W_t$ as the first time the Brownian motion hits level "$a$". Since we have $\mathbb{E}[W_{T_A}]=a\neq\mathbb{E}[W_{T_A}|W_0]$, the stopped Brownian motion cannot be a martingale. Therefore it must be that none of the three conditions in Doob's theorem are true: therefore we must have that $\mathbb{E}[T_A]=\infty$.
Case 2: Let's define the stopping time $T_A$ associated with $W_t$ as the first time the Brownian motion hits either level "$a$" or "$-a$". It can be shown that in this case the probability $\mathbb{P}\left(T_a<t \right)\rightarrow1$ as $t\rightarrow\infty$. Therefore, we must have that $W_{T_A}$ is a martingale, so we must have that $\mathbb{E}\left[W_{T_A}|W_0\right]=0$.
How do these two cases fit together and how to interpret these intuitively?
Question Case 1: The first case tells us that the expected hitting time of a specific level is infinite: if we interpret expectation as the "average" over many paths, would it be correct to deduce that the result tells us that at least over some paths, Brownian motion tends to "diverge and never return"? I.e. it shoots off towards either positive infinity or negative infinity for at least some paths? Since even just one path that never hits the level "$a$" is enough for the expectation to diverge, this would intuitively explain why the stopping time is not integrable.
Question Case 2: Here, the stopping time is associated with $W_t$ breaking out of a "band" defined by the two boundaries $\{-a,a\}$: since this stopping time is finite, would it be correct to conclude that Brownian motion tends not to stay "confined to a band", not even for a single path out of infinite number of paths? I.e. can we conclude that Brownian motion always eventually breaks out of a range?
Best Answer
Q1: No, that's not a correct interpretation. In fact, it is well known that Brownian motion is recurrent, which is to say that $T_A < \infty$ with probability 1. So actually there is no chance that the Brownian motion will diverge and never return. One can also show that with probability 1 we have $\limsup_{t \to \infty} W_t = +\infty$ and $\liminf_{t \to \infty} W_t = -\infty$, which says that Brownian motion makes wider and wider swings from positive to negative values, recrossing all the numbers in between every time, so that it eventually makes infinitely many widely separated visits to every number.
You can think of this as saying that Brownian motion is guaranteed to hit $a$, but takes on average an extremely long time to do so. Simple random walk has the same phenomenon: if you take a fair coin and flip it until the number of heads exceeds the number of tails (i.e. until the simple random walk hits +1), with probability 1 you will eventually finish, but the expected number of flips needed is infinite. Basically, there is a possibility that you start off with several tails in a row, which will tend to make it take an extremely long time before you get a corresponding run of heads.
Q2: Well, from your argument alone we can only conclude that there is a nonzero probability that the Brownian motion breaks out of the band $[-a,a]$. But it is in fact the case that this happens with probability 1. It follows from the facts mentioned above, since we know that the Brownian motion will hit the values $a+1$ and $-a-1$ with probability 1.