Prove that, for (0 leq a b-a ight),end{aligned}] where (widehat{T}) is associated with the (mathrm{BM}left(widehat{W}_{t}=W_{t+a}-W_{a}, t geq

Prove that, for \(0 \leq a

\[\mathbb{P}\left(W_{s} eq 0, \forall t \in[a, b]\right)=\frac{2}{\pi} \arcsin \sqrt{\frac{a}{b}} .\]

From elementary properties of Brownian motion, we have

\[\begin{aligned}\mathbb{P}\left(W_{s} eq 0, \forall s \in[a, b]\right) & =\mathbb{P}\left(\forall s \in[a, b], W_{s}-W_{a} eq-W_{a}\right) \\& =\mathbb{P}\left(\forall s \in[a, b], W_{s}-W_{a} eq W_{a}\right)=\mathbb{P}\left(\widehat{T}_{W_{a}}>b-a\right),\end{aligned}\]

where \(\widehat{T}\) is associated with the \(\mathrm{BM}\left(\widehat{W}_{t}=W_{t+a}-W_{a}, t \geq 0\right)\). Using the scaling property, we compute the right-hand side of this equality

\[\begin{aligned}\mathbb{P}\left(W_{s} eq 0, \forall s \in[a, b]\right) & =\mathbb{P}\left(a W_{1}^{2} \widehat{T}_{1}>b-a\right)=\mathbb{P}\left(\frac{G^{2}}{\widehat{G}^{2}}>\frac{b}{a}-1\right) \\& =\mathbb{P}\left(\frac{1}{1+C^{2}}<\frac{a}{b}\right)=\frac{2}{\pi} \arcsin \left(\sqrt{\frac{a}{b}}\right)\end{aligned}\]

where \(G\) and \(\widehat{G}\) are two independent standard Gaussian variables and \(C\) a standard Cauchy variable.