38 — Two-Sided Hitting, Gambler’s Ruin for BM, and Exponential Martingales
1. Two-Sided Hitting Time
Fix two levels \(a<0<b,\) and define the stopping time \(T = T_b \wedge T_a, \qquad T_b = \inf\{t>0 : B_t = b\},\quad T_a = \inf\{t>0 : B_t = a\}.\)
Question: What is the probability that Brownian motion hits $a$ before $b$? \(P_0(T=T_a)=P_0(T_a<T_b).\)
2. Using the Martingale $B_t$
The stopped process ${B_{t\wedge T}}$ is bounded: \(|B_{t\wedge T}| \le \max(|a|,b),\) so it is uniformly integrable.
Thus \(B_{t\wedge T} \xrightarrow[t\to\infty]{a.s.} B_T,\) and \(E_0[B_{t\wedge T}] = E_0[B_0]=0.\)
Apply dominated convergence: \(0 = E_0[B_T] = P_0(T_a<T_b)\cdot a + P_0(T_b<T_a)\cdot b.\)
Since the two events partition the probability space, \(P_0(T_a<T_b)+P_0(T_b<T_a)=1.\)
Solving: \(\boxed{P_0(T_a<T_b)=\frac{b}{\,b-|a|\,}},\qquad \boxed{P_0(T_b<T_a)=\frac{|a|}{\,b-|a|\,}}.\)
3. Translation to the General Starting Point $x$
If the Brownian motion starts at $x\in(a,b)$, by spatial translation: \(\boxed{ P_x(T_a<T_b)=\frac{b-x}{b-a}},\qquad \boxed{ P_x(T_b<T_a)=\frac{x-a}{b-a}}.\)
(See the diagram on page 1 of the notes: the interval is shifted so that the left boundary becomes $0$.)
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4. Expected Value of the Minimum Exit Time
New question: \(E_x(T_a\wedge T_b).\)
Start with the martingale \(B_t^2 - t.\)
Since ${B_{t\wedge T}^2}$ is bounded (BM confined to $[a,b]$ until exit), it is uniformly integrable.
Then: \(E_x[B_{t\wedge T}^2 - (t\wedge T)] \xrightarrow[t\to\infty]{} E_x[B_T^2 - T].\)
Because $B_T\in{a,b}$, \(E_x[B_T^2] = a^2 P_x(T_a<T_b) + b^2 P_x(T_b<T_a).\)
Hence \(E_x(T) = E_x[B_T^2] = \frac{b-x}{b-a}a^2 + \frac{x-a}{b-a}b^2.\)
This is finite since $T$ has finite expectation when BM is confined to an interval.
5. An Exponential Martingale and the Probability of Ever Hitting $a$
We consider the exponential martingale \(M_t = e^{\theta B_t - \frac12 \theta^2 t}.\)
We study it stopped at $T\wedge t$: \(M_{t\wedge T} = e^{\theta B_{t\wedge T} - \frac12 \theta^2 (t\wedge T)}.\)
Take $\theta = 2b$.
From the bound in the notes (page 3)
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\(B_{t\wedge T} \le a + b(t\wedge T),\)
we get
\(M_{t\wedge T}
\le e^{\,2b[a+b(t\wedge T)] - 2b^2 (t\wedge T)}
= e^{2ab}.\)
Thus ${M_{t\wedge T}}$ is uniformly integrable and \(E_0[M_{t\wedge T}] = 1 \quad\Rightarrow\quad E_0[M_T]=1.\)
On ${T<\infty}$, \(M_T = e^{2b B_T - 2b^2 T}.\)
Since hitting $a$ is the only relevant event in this argument, \(B_T=a \quad\Rightarrow\quad M_T=e^{2ab}.\)
On ${T=\infty}$, one shows (via SLLN in the notes, page 3) that \(M_T = 0.\)
Thus \(1 = E_0[M_T] = e^{2ab} P_0(T<\infty) + 0\cdot P_0(T=\infty).\)
Solving: \(\boxed{ P_0(T<\infty) = e^{-2ab}}.\)
This is the classical one-sided non-return probability for Brownian motion with drift-like exponential weighting.
6. An Example of Optional Stopping with a Discrete Random Target
Consider a discrete random variable $X$ with distribution (page 4 of notes):
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| $x$ | $-\tfrac{7}{4}$ | $1$ | $2$ |
|---|---|---|---|
| $p(x)$ | $0.4$ | $0.5$ | $0.1$ |
Construct a stopping time $\tau$ such that \(B_\tau^0 \overset{d}{=} X.\)
Define \(\mathcal F_1 = \sigma\{X\in\{-\tfrac74\},\, X\in\{1,2\}\}.\)
Then \(E[X\mid \mathcal F_1] = \begin{cases} -\tfrac74, & X=-\tfrac74,\\ \frac{1\cdot0.5 + 2\cdot0.1}{0.6} = \tfrac76, & X\in\{1,2\}. \end{cases}\)
Since $E[X]=0$, Brownian motion can realize this by setting \(\tau = T_{7/6} \wedge T_{-7/4}.\)
Then \(B_\tau \overset{d}{=} X.\)
This is an application of optional stopping + gambler’s ruin probabilities returning a prescribed distribution.
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