Analytical solutions to $E[f(X_\tau) e^{-\alpha\tau}]$

Question

Let $X_t$ be a geometric Brownian motion: $$dX_t=\alpha X_tdt+sX_tdB_t,$$ where $\alpha,s>0$ are constants. Let $\tau=\inf_{u\geq t}\{X_u=K\}$ be the first time $X_t$ reaches the threshold $K$ (from above). It's well-known that $$E_t[e^{-\alpha\tau}]=\left(\frac{X_t}{K}\right)^{-\frac{2\alpha}{s^2}}.$$

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A simple proof is that $E_t[e^{-\alpha\tau}]$ needs to satisfy the Cauchy-Euler ODE $$ \frac{1}{2}s^2x^2u_{xx}+\alpha xu_x -\alpha u=0$$ which implies $u(x)=ax^{1}+bx^{-\frac{2\alpha}{s^2}}$. From the boundary conditions $\lim_{x\to\infty}u(x)=0$ and $u(K)=1$, we find $a=0$ and $b=K^{\frac{2\alpha}{s^2}}$.

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Question:

I'd like to see how far we can push the availability of analytical solutions.$^1$

• Can we calculate $E_t[X_\tau e^{-\alpha\tau}]$ analytically?
• For what other non-constant functions $f$ can $E_t[f(X_\tau) e^{-\alpha\tau}]$ be computed analytically?

$^1$analytical solution = closed-form solution typically only including elementary functions. It may require special functions but otherwise should not contain remaining integrals, etc.

Answer 1

As discussed in the comments [big thanks to @Thomas], the answer is trivial. The stopping time $\tau$ is defined such that $X_\tau=K$. Thus, $$E_t[f(X_\tau)e^{-\alpha\tau}]=f(K)E_t[e^{-\alpha\tau}].$$