Brownian motion: exit time from symmetric strip

Question

I'm trying to figure out why my way of thinking is not valid.

Consider a Wiener process $W_t$ and the strip $S_a = [-a/2,a/2]$ with $a$ positive. I'm wandering what is the probability $P_a(t)$ that the process has never left the strip?

So I thought that:

$$ P_a(t) = Prob\{|x(\tau)|\le a/2, \forall \tau \in [0,t]\} $$

could be re-written as:

$$ P_a(t) =Prob\{ M_t \le a/2 \cap m_t \ge -a/2, \forall \tau \in [0,t]\} $$

where: $M_t=max\{W_s | 0 \le s \le t\} $ and: $m_t=min\{W_s | 0 \le s \le t\} $. Now I see those two events (the one involving the maximum and the one involving the minimum) as independent events, so i can write:

$$ P_a(t) =Prob\{ M_t \le a/2, \forall \tau \in [0,t]\} * Prob\{m_t \ge -a/2, \forall \tau \in [0,t]\} $$

I think that something stated above is already wrong, because, when I compute it, it turns out to be something like:

$$ P_a(t) \sim erf(1/\sqrt{t}) - erf^2 (1/\sqrt{t}) $$

where $erf$ is the error function. From the last one (with all the factors) it turns out that the expected value of the exit time is infinite, nor $a^2/4$ as I would expect (I'm referring to: An informal introduction to Stochastic Calculus with Applications - Ovidiu Calin, pg. $83$, exercise $4.3.9$).

Thanks for your help.

Answer 1

The probability of not walking over max includes paths that walked below the min. By symmetry the probability of walking below the min includes paths that walked above the max.

(Consider rolling a 6 sided dice. The probability of rolling a 3 or 4 is 1/3. If you multiply the probability of rolling a 1,2,3 or 4 by that of rolling a 3,4,5 or 6, you will get 2/3*2/3 which is 4/9, which is different.)

I think you should compute

$$ P_a(t) =1-((1-Prob\{ M_t \le a/2, \forall \tau \in [0,t]\}) + (1-Prob\{m_t \ge -a/2, \forall \tau \in [0,t]\}) $$

this is the probability that's left cutting away the probabilities of exceeding each limit.

Answer 2

I think I have spotted the flaw in your argument. The problematic piece is $\forall \tau\in [0,t]$. For any $\tau \in [0,t],M_t\leq a/2$ and $M_t\geq -a/2$ are not independent events. This is because given that $M_t\leq a/2$ is true (say), we now know that $Prob(M_t> a/2)=0$, and so all the unconditional probability (for $M_t> a/2$) gets reassigned over the interval $M_t\leq a/2$, which then also changes the probability for $-a/2\leq M_t\leq a/2$ in general.

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I don't know stochastic calculus, so the supposed mistakes I have spotted in your argument may not actually be.

$|x(\tau)|<a/2$ is not logically equivalent to $(M_t\leq a/2~\cap~m_t\geq a/2)$. I am not sure what a Wiener process is, but first condition is about a range of values for $x$, while the latter seems to be only about two particular values. Probability measure over those conditions, as far as I can see, cannot be the same.