I am a beginner in Brownian motion and not a pro of probability theory and I would like to check my solution of Exercise 2.26 of the book by LeGall, stating:
Let $B$ be a Brownian motion, for each $a\ge0$ set $T_a=\inf\{t\ge0: B_t=a\}$. Show that for $0\le a\le b$ the random variable $T_b - T_a$ is independent of $\sigma(T_c, 0\le c\le a)$ and it has the same distribution as $T_{b-a}$.
My idea is to use the strong Markov property of BM and to look at $B_t^{(T_a)} = 1_{T_a<\infty}(B_{T_a + t} - B_{T_a})$, the BM "rebooted" at time $T_a$. The strong Markov property states that this is again a BM and that it is independent of $$\mathscr{F}_{T_a} = \{A\in\mathscr{F}_\infty:\forall t\ge0,\ A\cap\{T_a\le t\}\in\mathscr{F}_t\}\ ,$$ where $\mathscr{F}_t$ is our filtration. Now the $\sigma$-algebra $\sigma(T_c, 0\le c\le a)$ is generated by $\{T_c\le s\}\in\mathscr{F}_\infty$ for $0\le c\le a$ and $s\ge 0$. We notice that since $c\le a$ and the sample paths of the BM are continuous, if $T_c>t$ then we need have $T_a>t$. Therefore, it follows that $$\{T_c\le s\}\cap\{T_a\le t\} = \{T_c\le s\wedge t\}\cap\{T_a\le t\}$$ and since both sets are in $\mathscr{F}_t$, so is their intersection. It follows that $\{T_c\le s\}\in\mathscr{F}_{T_a}$, and thus that $\sigma(T_c, 0\le c\le a)\subseteq\mathscr{F}_{T_a}$. In particular, we have obtained that $B_t^{(T_a)}$ is independent of $\sigma(T_c, 0\le c\le a)$.
Now we can conclude. Denote by $T_d^{(T_a)} = \inf\{t\ge0: B_t^{(T_a)}=a\}$. For $A\in\sigma(T_c, 0\le c\le a)$ we have \begin{align} P(T_b - T_a\mid A) ={}&P(T_{b-a}^{(T_a)} - T_0^{(T_a)}\mid A)\\ ={}&P(T_{b-a}^{(T_a)} - T_0^{(T_a)}) \end{align} where the first line is quite obvious from the definitions, and in the second line we used the fact that the "rebooted" stopping times are defined only in terms of the "rebooted" BM, which is independent of $A$. Now on the one hand the last line equals $P(T_b - T_a)$ by the same reasoning we already made in the first line of the above. On the other hand, we have that $T_0^{(T_a)} = 0$ a.s. and that $T_{b-a}^{(T_a)}$ has the same distribution as $T_{b-a}$ since the "rebooted" BM is again a BM.
I think this fully proves the statement we wanted. Did I miss anything or was I imprecise at any step?
Any help or comment is greatly appreciated. Thanks in advance!
I think your solution is fine, but your conclusion was a little bit hard for me to follow. I think it would be more clear to just argue that $T_b - T_a = T_{b-a}^{(T_a)}$ is measurable with respect to $\sigma(B_t^{(T_a)}, 0\le t)$ in the same way that you did and use the fact that $\sigma(B_t^{(T_a)}, 0 \le t)$ is independent of $\sigma(T_c, 0 \le c \le a)$ again. You use this fact anyway, but this avoids needing to use the conditional probabilities and streamlines the argument.