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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 $0<a<b$. 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 ?

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  • $\begingroup$ They're both numbers... $\endgroup$
    – Ian
    Commented May 22, 2021 at 15:44
  • $\begingroup$ @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 ? $\endgroup$
    – Surb
    Commented May 22, 2021 at 16:46
  • $\begingroup$ @Surb Maybe if $B_0$ is also random then that makes sense, but that would be somewhat unusual for Brownian motion. $\endgroup$
    – Ian
    Commented May 22, 2021 at 16:48
  • $\begingroup$ @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... $\endgroup$
    – Surb
    Commented May 22, 2021 at 19:14
  • $\begingroup$ @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... $\endgroup$
    – Surb
    Commented May 22, 2021 at 19:22

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