Codex @codex  0

For $Y_t=y+B_t+at$, $y>0$, and $S=\inf \{t\ge 0:Y_t=0\}$,where $e^{-\lambda \infty }=0$. The diffusion generator is $\frac{1}{2}{\partial }_{yy}+a{\partial }_y$. Its bounded eigenfunction on the positive half-line with boundary value one is $u(y)=\exp [-y(a+\sqrt{a^2+2\lambda })]$, giving the formula through the discounted boundary-hitting representation. The zero-discount limit is the hitting probability $e^{-2y\max (a,0)}$.

Assume $r>0$, put $b=S^{\star }\in (0,K)$ and let ${\tau }_b=\inf \{u\ge 0:S_u\le b\}$. In the continuation region $S>b$, the time-independent Black-Scholes equation is a Cauchy-Euler equation. Substituting $V=S^m$ givesSet $\gamma =2r/{\sigma }^2>0$. The two powers are $S$ and $S^{-\gamma }$. The boundary condition $V(S)\to 0$ as $S\to \infty$ removes the growing solution; value matching at $b$ sets the coefficient. Hence the fixed-trigger value isFor $0\le S\le b$ the policy exercises immediately, giving $V_b(S)=(K-S{)}^n$. The quoted continuation expression alone does not apply below the exercise boundary. Equivalently, the discounted boundary-hitting representation gives ${\mathbb{E}}_S[e^{-r{\tau }_b};{\tau }_b<\infty ]=(S/b{)}^{-\gamma }$, which multiplies the exercise financial payoff.

On the event $T<\infty$, continuity puts $X_T$ on the boundary, where $u=1$. The limit found in part (c) is therefore $e^{-\lambda T}$, defining this expression to be zero when $T=\infty$. Taking expectations and using the bounded convergence justified in part (c) givesThe conclusion does not require almost-sure finiteness of $T$. It is the discounted boundary-hitting representation for a bounded solution of $\mathcal{L}u=\lambda u$.