Probability of Position of Brownian motion at hitting time

Question

this might be a stupid question but I am a bit stuck here. let $B$ be a standard Brownian motion and $H_a$ the first hitting time of level $a$. I now want to find the probability $\mathbb{P}(B_{H_a} \in dw | H_a \in dt)$ Of course I know that $B_{H_a} = a$ but is the probability then $\frac{1}{\sqrt{2\pi t}} e^{-\frac{a^2}{2t}}$? what happens with the $dw$? or is it $\frac{1}{\sqrt{2\pi t}} e^{-\frac{w^2}{2t}}1_{w = a} dw$? I am very confused and would appreciate any help. thanks in advance!

Answer 1

For every $t$ the distribution of $B_{H_a}$ conditionally on $H_a=t$ is the Dirac measure at $a$. That is, for every Borel set $B$, $$ P(B_{H_a}\in B\mid H_a)=\mathbf 1_{a\in B}. $$

Answer 2

Note that, as you say, with $\mathbb{P} (B_{H_a} = a) = 1$. That is, $B_{H_a}$ is almost surely deterministic. It is thus indepedent of $T_a$. Therefore

$$\mathbb{P} (B_{H_a} \in A|H_a \in B) =\mathbb{P} (B_{H_a} \in A) = \mathbb{1}_a(A) $$

for any Borel subsets $A,B$ of $\mathbb{R}$.