let $u > 0$, $b > a > 0$ and $T_u = \inf \{t \geq0 \,|\, B_t = u\}$ where $B$ is a standard Brownian motion.
I was reading some lecture notes and noticed that the author treats $T_{b-a}$ and $T_b - T_a$ as identical in distribution.
intuitively I can see why this is the case, but in a previous problem, when $a$ was negative such that $a < 0 < b$ it was proven that
$$\mathbb{\mathbb{P}}(T_b - T_a < 0) = -\frac{a}{b-a}$$
but in this case
$$\mathbb{\mathbb{P}}(T_{b-a} < 0) = \mathbb{\mathbb{P}}(T_{b-a} - T_{0} < 0) = 0$$
so are $T_{b-a}$ and $T_b - T_a$ actually equal in distribution ? if so how come the sign of $a$ has such an influence on the distribution of $T_{b-a}$ ?
For all $t>0$: $$\tau_{b} \sim \frac{|b|}{\sqrt{2 \pi t^{3}}} e^{-b^{2} /(2 t)} d t$$
Since $B\sim -B$ $$\tau_{b}=\inf \left\{s \geqslant 0: B_{s}=b\right\} \sim \inf \left\{s \geqslant 0:-B_{s}=b\right\}=\inf \left\{s \geqslant 0: B_{s}=-b\right\}=\tau_{-b}$$ And if $0<a<b$, then $$\tau_{b}-\tau_{a}=\inf \left\{s \geqslant 0: B_{s+\tau_{a}}=b\right\}=\inf \left\{s \geqslant 0: B_{s+\tau_{a}}-B_{\tau_{a}}=b-a\right\}$$ By the strong Markov property of Brownian motion. So $\tau_b -\tau_a$ is independent of $\mathscr F_{\tau_a}$. For the distribution of $\tau_a-\tau_b$, use the distributions given and the symmetry.