Does $B_{\tau\wedge t}\sim C\mathbb 1_{[-h,h]}(x)e^{-\frac{x^2}{2t}}$ where $\tau=\inf\{t>0\mid |B_t|\geq h\}$

Question

Let $(B_t)$ a Brownian motion and $$\tau=\inf\{t>0\mid |B_t|\geq h\}.$$

Does $B_{t\wedge \tau}$ has density $$f(x,t)=C\mathbb 1_{[-h,h]}(x)e^{-\frac{-x^2}{2t}}$$ where $C$ is s.t. $\int_{\mathbb R}f_X(x,t)dx=1$ ? Indeed, $(B_{t\wedge \tau})$ is a Brownian motion until it reach $h$, and after the process is just killed.

Answer 1

No, $B_{t \wedge \tau}$ does not have a density with respect to Lebesgue measure. Note that

$$\mathbb{P}(|B_{t \wedge \tau}| = h) = \mathbb{P}(\tau \leq t)>0$$

for any $t>0$, i.e. (the distribution of) $B_{t \wedge \tau}$ has an atom of positive measure; in particular, it is not absolutely continuous w.r.t. Lebesgue measure.