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For standard Brownian motion started at zero and exit time $T$ from $(-b,a)$, $a,b>0$, stop the two exponential Brownian martingales with parameters $\lambda$ and $-\lambda$. Their stopped values are bounded by $e^{|\lambda |\max (a,b)}$. The optional stopping theorem and dominated convergence theorem give $e^{\lambda a}p+e^{-\lambda b}q=1$ and $e^{-\lambda a}p+e^{\lambda b}q=1$, where $p,q$ are the discounted upper and lower exit probabilities. Solving gives the displayed formula and $q=\sinh (\lambda a)/\sinh (\lambda (a+b))$. ThusAt $\lambda =0$, the first ratio has its removable value $b/(a+b)$, and the second equals one. Finiteness of $T$ follows from a uniformly positive one-unit-time chance to leave any fixed bounded interval.

The exit time $T$ is finite almost surely. Indeed, if the Brownian motion is still in $(-b,a)$ at an integer time, its next unit increment exceeds $a+b$ with a fixed probability $p>0$ independent of the past; on this event it must exit before the next integer time. Consequently $\mathbb{P}(T>n)\le (1-p{)}^n\to 0$. Path continuity gives $B_T\in \{a,-b\}$, and the two exit events partition the probability space up to null sets.

For each real $\lambda$, the exponential Brownian martingale $Z_t^{\lambda }=\exp (\lambda B_t-{\lambda }^{2}t/2)$ has expectation one. This follows directly from the moment-generating function of a normal distribution and independent increments. At $t\wedge T$ its value is bounded by $e^{|\lambda |\max (a,b)}$. Bounded-time optional stopping and then dominated convergence therefore giveWrite $p_{\lambda }=\mathbb{E}[e^{-{\lambda }^{2}T/2}{1}_{\{T=T_a\}}]$ and $q_{\lambda }=\mathbb{E}[e^{-{\lambda }^{2}T/2}{1}_{\{T=T_{-b}\}}]$. Applying the identity to $\lambda$ and $-\lambda$ yieldsFor $\lambda \ne 0$, multiply the first equation by $e^{\lambda b}$, the second by $e^{-\lambda b}$, and subtract. Solving the resulting system givesAdding and using the sum formula for the hyperbolic sine gives the asymmetric Brownian interval-exit transform:At $\lambda =0$, the printed sine ratio is $0/0$ and needs its removable extension. Dominated convergence as $\lambda \to 0$ givesThus the formulas hold for every real parameter with this endpoint convention. The expectation on the left of the first formula is legible in the original PDF; the converted TeX's nested exponential is an OCR error.

Use the positive exponential Brownian martingaleThe strong law for Brownian motion, $W_t/t\to 0$, makes its exponent tend to $-\infty$ and hence $M_t\to 0$. If $S={\sup }_{t\ge 0}(W_t-\lambda t)$, then $\sup M=e^{2\lambda S}$. Part (b) gives, for $x>0$,Thus the maximum is exponentially distributed with rate $2\lambda$:There is no atom at zero, by letting $x\downarrow 0$ in the tail. This is the infinite-horizon crossing probability for Brownian motion with negative drift.

The price process is a geometric Brownian motion withPutand define the equivalent measure $Q$ on ${\mathcal{F}}_T$ by the exponential Brownian martingaleThe Cameron-Martin-Girsanov theorem says that $W_t^Q=B_t+\vartheta t$ is Brownian motion under $Q$. Thereforeso $e^{-rt}{S}_t$ is a $Q$-martingale. This is the unique equivalent martingale measure, since cancellation of the single Brownian risk forces the displayed drift shift.

By risk-neutral valuation,Writing $W_T^Q=\sqrt{T}Y$ with $Y\sim N(0,1)$ gives

For $0\le s\le t$, the increment $W_t-W_s$ is independent of ${\mathcal{F}}_s$ and is $N(0,t-s)$. HenceAlso $\mathbb{E}{M}_t=1$, so $M$ is the exponential Brownian martingale.

For $0\le s\le t$, write $B_t=B_s+(B_t-B_s)$. The increment is independent of ${\mathcal{F}}_s$ and is normally distributed with variance $t-s$. Its moment generating function givesThe process is integrable for every real $\lambda$, so it is the exponential Brownian martingale.

By the Reflection invariance of Brownian motion, $-B$ is Brownian motion, and the first time $B_t-t$ reaches $b<0$ is the first time $-B_t+t$ reaches $-b>0$. Part b therefore gives

Now $L_t=\exp (B_t-t/2)$ is the exponential Brownian martingale. At $S_b$, the identity $B_{S_b}=b+S_b$ gives $L_{S_b}=e^{b+S_b/2}$, so $\mathbb{E}{L}_{S_b}=1$. The nonnegative stopped martingale $(L_{t\wedge S_b})$ therefore loses no mass at infinity and is uniformly integrable. The optional sampling theorem at any stopping time $R$ gives

Apply Itô formula to $Z_t^{(A)}=\exp (A{X}_t-A^{2}t/2)$. Its semimartingale decomposition isThe second term is a continuous finite-variation process. Since $Z^{(A)}$ is assumed to be a local martingale, uniqueness of the semimartingale decomposition makes this term identically zero. Both $A$ and $Z^{(A)}$ are nonzero, so the quadratic variation of $X$ is $[X{]}_t=t$.

The Lévy characterization of Brownian motion now says that $X_t-X_0$ is a Brownian motion. Consequentlyis a constant multiple of an exponential Brownian martingale. It is therefore a true martingale for every $a\in \mathbb{R}$.

For a continuous semimartingale $X$ and a $C^{1,2}$ function $f$, Itô's formula isFor Brownian motion $B$, $d[B{]}_t=dt$. With $f(t,x)=e^{-t/2}\cosh x$, the drift cancels:ConsequentlyThe stochastic-integral representation makes this a local martingale; localization handles unbounded $B$. In fact it is the average of the two positive exponential Brownian martingales $e^{B_t-t/2}$ and $e^{-B_t-t/2}$, so it is a true martingale on every finite time interval.