Brownian motion with 2 absorbing boundaries Consider the stochastic process
\begin{equation*}
X_t = \mu t + \sigma W_t
\end{equation*}
where $W_t$ is the standard Brownian motion. Suppose that $X_0 = x \in (0,1)$ and that $0$ and $1$ are absorbing boundaries. I would like to know how to tackle the following questions:

*

*What is the probability that $X_t$ will be absorbed at the $x=1$ boundary? i.e. What is the probability that $X_t$ hits $x = 1$ without ever hitting $x = 0$.

*What is the probability that $X_t$ will never be absorbed?

*What is the probability that $X_t$ will remain unabsorbed after $t > T$?

My Thoughts
I know that if there is only one boundary at $x = 1$ then the probability of hitting $x = 1$ at some time $t > T$ is given by
\begin{equation}
Prob(T(1) > T) = \int_{-\infty}^{1-x}e^{-\frac{1}{2}\frac{(s-\mu T)^2}{\sigma^2 T}}ds - e^{\frac{2\mu(1-x)}{\sigma^2}}\int_{-\infty}^{x-1}e^{-\frac{1}{2}\frac{(s-\mu T)^2}{\sigma^2 T}}ds.
\end{equation}
The formula can be derived using the reflection principle and some change of measure. We can see that the probability approaches zero as $T\rightarrow \infty$. Therefore, in the one boundary case, $X_t$ would be absorbed a.s.. However, I'm not sure how to get a similar formula if we also have a boundary at $x = 0$ since I don't think the reflection principle technique will apply. However, adding a boundary at $x = 0$ will not change to answer to 2, that is there is a zero probability that $X_t$ will never be absorbed.
 A: Though I guess that this already has been answered on MSE somewhere I will present what I know about it.
First let me make some notations. For a process $Y$ let
$T_a^Y := \inf \{t:Y_t =a\}$ and $T^Y_{a,b} = \min (T^Y_a , T^Y_b)$
For symplicity let $\sigma = 1, \mu \geq 0$.
First of all the answer to 2. is clear by $\Bbb P (T^X_{0,1} < \infty)\geq \Bbb P (T_{1}^X < \infty) \geq \Bbb P (T_1^W < \infty) =1$. The last inequality follows by the fact that the drift is positive. That the last probability is $1$ is well-known from the theory of Brownian motion.
At this point let me just note that parts of the problem can be reformulated as a two-sided so called first passage time problem over affine linear time dependent boundaries, i.e. for a (continuous) function $b: [0,\infty ) \to \Bbb R$ denote the hitting time (or first passage time) as $\tau_b^Y := \inf\{ t: Y_t = b(t) \}$. Then for $b_+(t) = 1 - \mu t$ and $b_-(t) = -\mu t$, we have that
$$\Bbb P (T^X_{0,1} > t) = \Bbb P (\tau^W_{b_+ , b_-} > t)$$
I found that A modification of the sequential probability ratio test to reduce the sample size by Anderson (1960) seems to cover this situation.
(Old text of this part: As far as I know for affine linear functions $b_i (t) = \alpha_i + \beta_i t$ hitting probabilties $\Bbb P (\tau_{b_1 ,b_2}^W > t)$ have only been established yet for $\alpha_1 < \alpha_2$ and $\beta_1 < 0 < \beta_2$ (for example see Some conditional crossing results of Brownian motion over a piecewise-linear boundary by Mario Abundo (2002) in Statistics & Probability Letters 58), which does not cover your case. )
But your problem as special case is even simpler. By the Girsanov theorem $(X_s)_{s\in [0,t]}$ is a Brownian motion under the measure $\Bbb Q_t =L_t d \Bbb P$ on $\mathcal F_t = \sigma (X_s , s\in [0,t])$, where $L_t = \exp (-\mu W_t - \frac 1 2 \mu^2 t )= \exp (-\mu X_t  + \frac 1 2 \mu^2 t)$.
For the purpose of addressing 2. let $A^X = \{T^X_{0,1} > t\} \in \mathcal F_t$ which depends in some particular way on the path $(X_s)_{s\in[0,t]}$.
Speaking in densities we can make the following informal computations (which can be made rigorous)
$$\Bbb P(X_t = z , A^X) = \Bbb E_{\Bbb P} [ 1_{A^X}1_{\{X_t = z\}}] = \Bbb E_{\Bbb P} [ 1_{A^X}1_{\{X_t = z\}} \frac 1 {L_t} L_t]\\ = \Bbb E_{\Bbb Q_t} [ 1_{A^X}1_{\{X_t = z\}} \exp (\mu X_t  - \frac 1 2 \mu^2 t)] = \Bbb E_{\Bbb P} [ 1_{A^W}1_{\{W_t = z\}} \exp (\mu W_t  - \frac 1 2 \mu^2 t)]\\ = \Bbb E_{\Bbb P} [ 1_{A^W}1_{\{W_t = z\}} \exp (\mu z  - \frac 1 2 \mu^2 t)] = \Bbb P (W_t = z , A^W)  \exp (\mu z  - \frac 1 2 \mu^2 t)$$
where by $A^W = \{T^W_{0,1} > t\}$. Thus the event $A^W$ depends on the path $(W_s)_{s\in[0,t]}$ in the same way $A^X$ does on $(X_s)_{s\in[0,t]}$.
Now the advantage is: The density
$$\Bbb P ( W_t = z , T^W_{0,1} > t)$$
for $z\in (0,1)$ is known and there are essentially two represantations (one is good for large $t$, the other for small $t$) (see for example Brownian motion and classical potential theory by Port and Stone at Proposition 8.2 in Chapter 2 or Brownian motion and Stochastic Calculus by Karatzas and Shreve at Proposition 8.10 in Chapter 2 ).
Thus we have
$$\Bbb P (T^X_{0,1} > t) = \int_0^1 \Bbb P(X_t = z , T^x_{0,1} > t) dz = \int_0^1 \Bbb P (W_t = z , T^W_{0,1} > t)  \exp (\mu z  - \frac 1 2 \mu^2 t) dz$$
which can be computed further.
In fact one could try to address 1. in the same way: By setting $A^X = \{T^X_1 < T^X_0 , T^X_{0,1} \leq t\}$ we get
$$\Bbb P (T^X_1 < T^X_0) = \lim_{t\to \infty} \Bbb P (T^X_1 < T^X_0 , T^X_{0,1} \leq t)\\= \lim_{t\to \infty} \int_{\Bbb R} \Bbb P (W_t = z,T^W_1 < T^W_0 , T^W_{0,1} \leq t) \exp (\mu z - \frac 1 2 \mu^2 t) d z \\= \lim_{t\to \infty} \int_{\Bbb R} \Bbb P (W_t = z,W_{T^W_{0,1}} =1, T^W_{0,1} \leq t) \exp (\mu z - \frac 1 2 \mu^2 t) d z$$
By the strong Markov property
$$\Bbb P (W_t = z,W_{T^W_{0,1}} =1, T^W_{0,1} \leq t) = \int_0^t \Bbb P (W_{t-s} = z | W_0 =1) p(s)ds$$
where $p(s) = -\frac{d}{ds} \Bbb P (T^W_{0,1} > s)$ is the density of $T^W_{0,1}$.
