Compute the expectation of $X_{\tau}$ and related probability given its SDE $dX_t = X_t\,dt + \sigma \, dB_t$

Question

Suppose there is a stochastic process $X_t$ satisfying the SDE $dX_t = X_t\,dt + \sigma \,dB_t$ where $B_t$ denotes the standard Brownian motion starting from $0$. Let $a<b$ being two real numbers. Define $\tau = \inf\{t:X_t\notin [a,b]\}$ and I want to find $E[X_\tau\mid X_0=x]$.

I know one method using Feynman-Kac theorem. Denote $u(t,X_t) = E[X_\tau\mid X_t]$ exploiting its markovian property. Then $u$ is a martingale by Tower property. By Ito's lemma: $$ du(t,X_t)=u_t\ dt+u_x\ dX_t+\frac{1}{2}u_{xx}\ d[X]_t=(u_t + u_xX_t + \frac{\sigma^2}{2} u_{xx})\ dt + \sigma u_x\ dB_t $$ So we get the PDE: \begin{cases} u_t + xu_x + \frac{\sigma^2}{2} u_{xx}=0\\ u(t,a)=a\\ u(t,b)=b \end{cases} Notice $u_t=0$, we simply swich $u(t,x)=u(x)=E[X_\tau\mid X_t=x]$ to get \begin{cases} xu_x + \frac{\sigma^2}{2} u_{xx}=0\\ u(a)=a\\ u(b)=b \end{cases} Solve this and we get $$ u(x)=a\frac{\int_x^{b}e^{-\frac{x^2}{\sigma^2}}\ dx}{\int_a^b e^{-\frac{x^2}{\sigma^2}}\ dx} + b\frac{\int_a^x e^{-\frac{x^2}{\sigma^2}}\ dx}{\int_a^b e^{-\frac{x^2}{\sigma^2}}\ dx} $$ Since $u(x)$ should have a form like $$ u(x)=aP(X_\tau=a) + bP(X_\tau=b) $$ Is it safe to guess $P(X_\tau=a \mid X_t=x)=\frac{\int_x^b e^{-\frac{x^2}{\sigma^2}}\ dx}{\int_a^b e^{-\frac{x^2}{\sigma^2}}\ dx}$ and $P(X_\tau=b \mid X_t=x)=\frac{\int_a^x e^{-\frac{x^2}{\sigma^2}}\ dx}{\int_a^b e^{-\frac{x^2}{\sigma^2}}\ dx}$?

My reason is that it seems very reasonable if $x$ increases from $a$ to $b$, $X_\tau$ varies from $a$ and more and more likely becomes $b$ when $x$ increases. Besides, $\frac{\int_a^x e^{-\frac{x^2}{\sigma^2}}\ dx}{\int_a^b e^{-\frac{x^2}{\sigma^2}}\ dx} + \frac{\int_x^b e^{-\frac{x^2}{\sigma^2}}\ dx}{\int_a^b e^{-\frac{x^2}{\sigma^2}}\ dx}=1$.

I generally have three questions:

1. How to justify $P(X_\tau = a) + P(X_\tau = b) = 1$, namely $P(\tau = \infty) = 0$?
2. How to justify $P(X_\tau = a)$ and $P(X_\tau = b)$ happen to be the coefficient of $a$ and $b$ in the expression of $u(x)$?
3. I also learned a technique to compute this type of problem by constructing martingales like $B_t^2-t$ when we compute the exit time of Brownian motion. Is there similar way to solve this problem? I notice $e^{-t}X_t$ is a martingale but don't know how to proceed.

Thank you for any help!

Answer 1

1. We can prove a more general result. For a bounded region $G$, let's assume the diffusion process $X_t$ starts from $X_0=x$ and the exit time is $\tau$. Suppose we can prove the following: $$ \forall x\in G, P(X_1 \in G^c) \geq \alpha>0 $$ namely $P(\tau \leq 1) \geq \alpha$, then from the Markov property, $P(\tau \geq 2) \leq (1-\alpha)^2$. Thus: $$\sum_{n=1}^{\infty}P(\tau \geq n) \leq \sum_{n=1}^{\infty}(1-\alpha)^n = \frac{1}{\alpha}-1 < \infty$$ By Borel-Cantelli lemma, $P(\limsup_{n\to \infty} (\tau > n))=0$, so $P(\tau<\infty)=1$. Since $X_t$ is Gaussian, to prove $\forall x \in [a,b]$, $\exists \alpha > 0$ such that $P(X_1\notin[a,b])>\alpha > 0$ is easy.

2. We can solve the same PDE with the boundary condition $u(a)=1$ and $u(b)=0$ to compute $P(X_\tau=a)$.

3. The generator of the process is $\mathscr{L}(f)=xf'(x)+\frac{\sigma^2}{2} f''(x)$, so $X_t-\int_{0}^{t}\mathscr{L}(f)(X_s)ds$ is a martingale where $f(x)=x$. Thus $$ E[X_\tau] = E[\int_{0}^{\tau}X_s\ ds] $$