Let $B_t$ be a one dimensional Brownian motion such that $B_0=a$. Here $t \in [0,T]$. Define the first exit time $\tau : =\inf\{ s \in [0,T] : B_s=b\}$ with $b<a$. I would like to find the conditional probability density function $f(B_t=x|\tau=s)$ with $x>b$, $s\in (0,T]$ and $t<s $. I am trying to write it in a derivative $$ f(B_t=x|\tau=s) = \dfrac{\partial}{\partial x} \mathbb{P}(B_t \leq x | \tau =s). $$ Let $W_{0,a}^{s,b}(t)$ be the Brownian bridge such that $W_{0,a}^{s,b}(0) = a$ and $W_{0,a}^{s,b}(s) =b$. I am thinking if we can do it by $$ f(B_t=x|\tau=s) = \dfrac{\partial}{\partial x} \mathbb{P}(W_{0,a}^{s,b}(t) \leq x | W_{0,a}^{s,b}(u)> b \text{ for all $u \in [0,s)$}). $$ But I do not know how to find this expression. And I think it is weired.
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