For the following Brownian motion with drift $X_t = X_0 + \mu t + \sigma B_t$ where $\mu \in \mathbb{R}$, $ \sigma > 0$ and $X_0 \in (a,b)$ which is a solution to the stochastic differential equation: \begin{align*} dX_t = \mu dt + \sigma dB_t \ . \end{align*} We define the stopping time: \begin{align*} \gamma = \min\{\tau,T\} \end{align*} Where $\tau = \inf\{t: X_t \notin (a,b)\}$ and $T>0$ is a constant. There is three ways this process can be stopped: If $X_t$ hits the upper or the lower barrior, that is $X_t = b$ or $X_t = a$, or the process will be stopped by the time constrained t = T. I want to find these three probabilities $P(X_\gamma = a)$, $P(X_\gamma = b)$ and $P(a < X_\gamma < b)$ does anyone have a strategy to solve this problem?
I can solve the problem in the case where $T = \infty$, such that we only have $\gamma = \tau = \inf\{t: X_t \notin (a,b)\}$ by using Dynkin's formula, but I cannot see if it is possible to solve the problem using a similar approach?
Here is my solution to the simplified problem ($T=\infty$) :
We define a reward function $c:\{a,b\} \to \mathbb{R}$, such that $c(a) = 0$ and $c(b) = 1$. Now given that the process starts at $x$, that is $X_0 = x$, the expectation of the reward function at $X_{\tau}$, must be equal to the probability that $X_{\tau}$ equals $b$ thus: \begin{align*} \textbf{E}^x c(X_{\tau}) = c(a)\textbf{P}^x\{X_{\tau} = a \} + c(b)\textbf{P}^x\{X_{\tau} = b \} = \textbf{P}^x\{X_{\tau} = b \} \end{align*} By Dynkin's formula we have: \begin{align*} \textbf{E}^x h(X_{\tau}) = h(x) + \textbf{E}^x \int_0^\tau (Lh)(X_{t}) dt \ . \end{align*} Where $L$ is the forward Kolmogorov operator: \begin{align*} (Lh)(x) = \mu \partial_x h(x) + \frac{\sigma^2}{2} \partial_x^2 h(x) \ . \end{align*} We see that if we solve the Dirichlet problem: \begin{align*} \begin{cases} (Lh)(x) = 0 \ , \ x \in (a,b) \ , \\ h(x) = c(x) \ , \ x \in \{a,b\} \ . \end{cases} \end{align*} Then by Dynkin's formular we get: \begin{align*} \textbf{E}^x h(X_{\tau}) = h(x) + \textbf{E}^x \int_0^\tau (Lh)(X_{t}) dt = h(x) \end{align*} and since $h(X_{\tau}) = c(X_{\tau})$ we have: \begin{align*} h(x) = \textbf{E}^x h(X_{\tau}) = \textbf{E}^x c(X_{\tau}) = \textbf{P}^x\{X_{\tau} = b \} \ . \end{align*} In words the solution $h(x)$ to the Dirichlet problem is the probability that the process $X_{\tau}$ at time $\tau$ equals $b$ when the process starts at $x$.
We will now solve the Dirichlet problem. We have the following differential equation we have to solve: \begin{align*} D \partial_x^2 h + \mu \partial_x h = 0 \ . \end{align*} We define $k = \mu/D$ such that we have: \begin{align*} \partial_x^2 h + k \partial_x h = 0 \ . \end{align*} By multiplication by $\exp(kx)$ we see that: \begin{align*} (\partial_x^2 h) \exp(kx) + (\partial_x h) k \exp(kx) = 0 \end{align*} Then we have: \begin{align*} \partial_x ((\partial_x h) \exp(kx)) = 0 \ . \end{align*} We integrate both sides: \begin{align*} (\partial_x h) \exp(kx) = - C_1 \ . \end{align*} By multiplication of $\exp(-kx)$ and integration we get: \begin{align*} h(x) - C_2 = -\int C_1 \exp(-kx) dx = \frac{C_1}{k} \exp(-kx) \ . \end{align*} The final solution is therefore: \begin{align*} h(x) = \frac{C_1}{k} \exp(-kx) + C_2 \ . \end{align*} By the boundary conditions we have: \begin{align*} h(a) = \frac{C_1}{k} \exp(-ka) + C_2 = 0 \ , \end{align*} thus \begin{align*} C_2 = - \frac{C_1}{k} \exp(-ka) \ . \end{align*} For $x = b$ we get: \begin{align*} h(b) = \frac{C_1}{k} \exp(-kb) - \frac{C_1}{k} \exp(-ka) = 1 \ , \end{align*} then we have: \begin{align*} C_1 = \frac{k}{\exp(-kb) - \exp(-ka)} \ . \end{align*} The solution is therefore: \begin{align*} h(x) = \frac{\exp(-ka) - \exp(-kx)}{\exp(-ka) - \exp(-kb)} \ . \end{align*} Now by replacing $h(x)$ with $ \textbf{P}^x\{X_{\tau} = b \} $ and $k$ with $2\mu/\sigma^2$ we have: \begin{align*} \textbf{P}^x\{X_{\tau} = b \} = \frac{\exp(\frac{-2\mu a}{\sigma^2}) - \exp(\frac{-2\mu x}{\sigma^2})}{\exp(\frac{-2\mu a}{\sigma^2}) - \exp(\frac{-2\mu b}{\sigma^2})} = \frac{1 - \exp(\frac{-2\mu}{\sigma^2}(x-a))}{1 - \exp(\frac{-2\mu }{\sigma^2}(b-a))} \ . \end{align*} And then $\textbf{P}^x\{X_{\tau} = a \} = 1 - \textbf{P}^x\{X_{\tau} = b \}$.