# Compute $\mathbb P(\sup_{t\in [a,b]}B_t>x)$, what's wrong with Markov property here?

I would like to compute $$\mathbb P\left(\sup_{t\in [a,b]}B_t>x\right)$$ where $$(B_t)$$ is a Brownian motion and $$0. What I would say using Markov property is $$\mathbb P\left(\sup_{t\in [a,b]}B_t>x \right)=\mathbb P\left(\sup_{t\in [0,b-a]}B_t>x\mid B_0=B_a\right),$$ but something looks strange since the LHS is a number whereas the RHS is a random variable. Can someone tel me how to manage ?

• They're both numbers...
– Ian
Commented May 22, 2021 at 15:44
• @Ian: Are you sure ? For me if $f(u)=\mathbb P(Y\in A\mid X=u)$, then if $Z$ is a random variable, then so is $f(Z)$. So, why do you think that the RHS is not a random variable but really a number ?
– Surb
Commented May 22, 2021 at 16:46
• @Surb Maybe if $B_0$ is also random then that makes sense, but that would be somewhat unusual for Brownian motion.
– Ian
Commented May 22, 2021 at 16:48
• @Ian: How do you interpret $\mathbb P\left\{\sup_{t\in [0,b-a]}B_t>x\mid B_0=B_a\right\}$ ? I do it as : there is $\Omega '$ s.t. $\mathbb P(\Omega ')=1$ and for all $\omega '\in \Omega '$, $$\mathbb P\left\{\sup_{t\in [0,b-a]}B_t>x\mid B_0=B_a\right\}(\omega ')=\mathbb P\left\{\sup_{t\in [0,b-a]}B_t>x\mid B_0=B_a(\omega ')\right\}.$$ In other words, it's a "brownian motion" that start at $B_a$ a.s. The OP is right at this point, the RHS of his last equality is a random variable, whereas the LHS is a number...
– Surb
Commented May 22, 2021 at 19:14
• @joshua : your equality is indeed not correct (for the reason you mentionned). Maybe something as $$\mathbb P\left\{\sup_{t\in [a,b]}B_t>x\right\}=\int_{\mathbb R}\mathbb P\left\{\sup_{t\in [0,b-a]}(y+B_t)>x\mid B_0=y\right\}\mathbb P\left\{B_a\in \,\mathrm d y\right\},$$ should be better...
– Surb
Commented May 22, 2021 at 19:22