Hitting probabilities from an exponential martingale

Stopping identity at a bounded continuous time: fix $T>0$, put $\rho =\tau \wedge T$, and for $n\ge 1$ round $\rho$ upward to the grid $\{jT{2}^{-n}:0\le j\le 2^n\}$, obtaining ${\rho }_n$. At a grid point $u<T$, $\{{\rho }_n\le u\}=\{\rho \le u\}\in {\mathcal{F}}_u$, so ${\rho }_n$ is a bounded stopping index for the sampled discrete martingale. Discrete optional sampling gives $E_x{M}_{{\rho }_n}=E_x{M}_0=e^{-2\mu x}$. It also identifies $M_{{\rho }_n}$ as a conditional expectation of the fixed integrable variable $M_T$ at the grid stopped sigma-algebra, so [F4] makes $(M_{{\rho }_n}{)}_n$ uniformly integrable. Continuity gives $M_{{\rho }_n}\to M_{\rho }$ almost surely, hence in probability; [F4] upgrades this to $L^1$, proving $E_x{M}_{\tau \wedge T}=e^{-2\mu x}$. This uses the discrete theorem only on finite grids and proves the continuous bounded-time passage explicitly.

Define $X_{\tau }$ by evaluation on $\{\tau <\mathrm{\infty }\}$ and as $0$ otherwise, and define $M_{\tau }=e^{-2\mu X_{\tau }}$. These are measurable: bounded-time evaluations are limits of the finite-grid evaluations in step 1.1, and the finite-exit value is their eventual value as integer horizons increase. Since $\tau <\mathrm{\infty }$ almost surely by [F3], $M_{\tau \wedge T}\to M_{\tau }$ as $T\to \mathrm{\infty }$; the deterministic bound in [F5] gives $E_x{M}_{\tau }=e^{-2\mu x}$ by dominated convergence.

The value at the exit: $X_{\tau }\in \{-a,b\}$ almost surely by continuity and the definition of $\tau$ as the first exit, so $M_{\tau }=e^{-2\mu X_{\tau }}$ equals $e^{2\mu a}$ on $\{X_{\tau }=-a\}$ and $e^{-2\mu b}$ on $\{X_{\tau }=b\}$. Writing $p:=P_x(X_{\tau }=b)$, the identity of step 2.1 becomes $e^{-2\mu x}=p{e}^{-2\mu b}+(1-p)e^{2\mu a}$.

Solving: $p={\displaystyle \frac{e^{-2\mu x}-e^{2\mu a}}{e^{-2\mu b}-e^{2\mu a}}}={\displaystyle \frac{1-e^{-2\mu (x+a)}}{1-e^{-2\mu (a+b)}}}$, multiplying numerator and denominator by $e^{-2\mu a}$; the denominator is nonzero because $\mu \mathrm{\ne }0$ and $a+b>0$ make the two endpoint exponentials distinct.

The case $\mu =0$: then $X=Z$ is Brownian motion started at $x$, and Two-sided Brownian exit probability gives $P_x(X_{\tau }=b)=(x+a)\mathrm{/}(a+b)$ directly; this agrees with the limit of the formula of step 4.1 as $\mu \to 0$.

Boundary and consistency cases: for $x\to -a$ the probability tends to $0$ and for $x\to b$ it tends to $1$, consistent with the starting point being at the boundary; for $\mu \to 0$ step 4.1 has a removable singularity with limit $(x+a)\mathrm{/}(a+b)$; for $\mu <0$ the same computation applies with the sign carried through; the stopping time is not bounded, and the passage to the limit was justified by the uniform boundedness of $M_{\tau \wedge t}$ from [F5] rather than by assuming uniform integrability of an unbounded family; the exit time is finite almost surely by the law of the iterated logarithm; and AC supplies the conditional-expectation interface and the inherited Brownian and LIL choice requirements, as declared in the Given hypotheses.