Formal (mathematical) proof of $\mathbb{P}(W_{t}>b)=\int_{0}^{t} \mathbb{P}\left( W_{t} > b \mid \tau_{b} =s \right) dF(s)$

Question

We begin with a probability space $(\Omega, \mathfrak{F}, \mathbb{P})$. I would like to know what is the formal way to prove: $$\mathbb{P}(W_{t}>b)=\int_{0}^{t} \mathbb{P}\left( W_{t} > b \mid \tau_{b} =s \right) dF(s),$$ where $W_{t}$ is a Standard Brownian Motion, $\tau_{b} = \min \left\{t \geq 0:W_{t} =b \right\}, b > 0$ and $F(t)=\mathbb{P}(\tau_{b}\leq t)$.

My attempt, starting from the left was: \begin{equation} \mathbb{P}(W_{t}>b)=\mathbb{P}(W_{t}>b, \tau_{b}\leq t) = \mathbb{P}(W_{t}>b \mid \tau_{b}\leq t) \mathbb{P}( \tau_{b}\leq t)=\mathbb{P}(W_{t}>b \mid \tau_{b}\leq t) F(t), \end{equation} and from here I'm not sure how to formally get the integral. Also, what confuses me a bit is the integrated term $\mathbb{P}\left( W_{t} > b \mid \tau_{b} =s \right)$. This is (by definition of conditional probability) equal to $\mathbb{P}(W_{t}>b, \tau_{b}= t)/\mathbb{P}(\tau_{b}=s)$, but isn't $\mathbb{P}(\tau_{b}=s)=0$?.

Thanks

Answer 1

I will write $Y$ for $\tau_b$. Let $\phi (Y)=P(W_t>b|Y)$. Then $\int_0^{\infty} P(W_t>b|Y=s)dF(s)$ is just a notation for $\int_0^{\infty} \phi (s)dF(s)$. Taking expectation on both sides of $\phi (Y)I_{Y\leq t}=P(W_t>b|Y)I_{Y\leq t}$ we get $P(W_t>b)=E\phi (Y)I_{Y\leq t}=\int_0^{t} \phi (s)dF(s)$. [I have used the fact that $(W_t>b,\tau_b \leq t)=(W_t>b)$].

Answer 2

@geetha290krm: Thanks for your reply. It pointed me in the right direction, so I think I can formalise the answer as follows:

Given that $\mathbb{P}(W_{t}>b \mid \tau_{b}) := \mathbb{P}(W_{t}>b \mid \sigma( \tau_{b} ) )$ is a version of $\mathbb{E}[ \mathbb{1}_{ \{ W_{t} > b \} } \mid \sigma(\tau_{b})]$ (where $\sigma(\tau_{b})$ is the $\sigma-$field generated by $\tau_{b}$), then for the event $A=\{ \tau_{b} \leq t \} \in \sigma(\tau_{b})$ we have: $$\int_{A} \mathbb{P}(W_{t}>b \mid \tau_{b} )(\omega) d\mathbb{P}(\omega) = \int_{A} \mathbb{E}[ \mathbb{1}_{ \{ W_{t} > b \} } \mid \sigma(\tau_{b})](\omega) d\mathbb{P}(\omega) = \int_{A} \mathbb{1}_{ \{ W_{t} > b \} }(\omega) d\mathbb{P}(\omega) = \mathbb{P}(W_{t}>b,\tau_{b}\leq t).$$

Now, since $ \mathbb{E}[ \mathbb{1}_{ \{ W_{t} > b \} } \mid \sigma(\tau_{b})]$ (i.e. $\mathbb{P}(W_{t}>b \mid \tau_{b})$ ) is $\sigma(\tau_{b})-$measureable, there exists a Borel function $\phi:[0,\infty] \to [0,\infty]$ such that $\mathbb{P}(W_{t}>b \mid \tau_{b}) = \phi(\tau_{b})$ and therefore (as you mentioned in your reply): $$ \int_{A} \mathbb{P}(W_{t}>b \mid \tau_{b} )(\omega) d\mathbb{P}(\omega) = \int_{A} \phi(\tau_{b}(\omega)) d\mathbb{P}(\omega) = \int_{0}^{t}\phi(x)dF(x).$$