# Brownian motion and conditional expectation

I'm looking through the proof on Brownian motion with drift carried out by my professor and I can't get my head around a passage.

Suppose I have a Brownian motion with drift $\mu$ and an absorbing barrier in $a$, that is

$$\bar{B}^\mu(t)=\left\{\begin{matrix}B^\mu(t) &\text{ for t} <T_a\\ a &\text{ for }t\geq T_a \end{matrix}\right.$$

where $T_a = \text{inf}\{t:B^\mu(t)\geq a\}$.

If I want to find the cdf of such process, we can see that, for $y<a,$

$P(\bar{B}^\mu(t) \leq y) = P(B^\mu(t) \leq y, T_a > t)=P(B^\mu(t) \leq y) - P(B^\mu(t) \leq y, T_a \leq t)$.

The first part of the last difference can be found with the transition probability of a free Brownian motion with a drift. For the second part, I can say that

$P(B^\mu(t) \leq y, T_a \leq t) = \mathbb{E}[I_{B^\mu(t) \leq y},I_{T_a \leq t}] = \mathbb{E}[\mathbb{E}[I_{B^\mu(t) \leq y},I_{T_a \leq t}|\mathcal{F}_{T_a}]]$

where $I$ is the indicator function and $\mathcal{F}_{T_a}$ is the sigma algebra stopped at the random time $T_a$. Since $T_a$ is a stopping time, the indicator of the event $T_a < t$ is $\mathcal{F}_{T_a}$-measurable and I can take the indicator random variable out of the conditional expected value.

$P(B^\mu(t) \leq y, T_a \leq t) = \mathbb{E}[I_{T_a \leq t}\mathbb{E}[I_{B^\mu(t) \leq y)}|\mathcal{F}_{T_a}]]$

So far so good. The problem comes with the remaining conditional expectation. If I take it, I can rewrite it as

$\mathbb{E}[I_{B^\mu(t) \leq y}|\mathcal{F}_{T_a}] = P(B^\mu(t) \leq y|\mathcal{F}_{T_a})$

and this is where the problem starts. The next passage would be to say that since Brownian motion (even with a drift) is a Markov process, then conditioning on the $\sigma$-algebra $\mathcal{F}_{T_a}$ is the same as conditioning on the position of a standard Brownian motion at random time $T_a$.

I guess that not understanding this passage shows I have not fully grasped what is contained inside $\mathcal{F}_{T_a}$ in this case.

I know that if $T_a$ is a stopping time with respect to the filtration $\{\mathcal{F}_t,t\geq 0\}$ on the probability space $(\Omega, \mathcal{F},P)$ then

$$\mathcal{F}_{T_a}=\{A\in\mathcal{F}:A \cap (T_a<t) \in \mathcal{F}_t \text{ }\forall t\}$$

So does $\mathcal{F}_{T_a}$ contain the positions of all the possible trajectories of the Brownian motion that have first passage time through $a$ before time $t$?

If this is the case I'm a happy man. Otherwise, could somebody clear up my confused mind? Thank you.