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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.