(1) Deduce from Proposition 2.5.4.1 (mathbb{P}left(A_{t}^{-, 0} in d u mid W_{t}=x ight)). (2) Recover the formula...

(1) Deduce from Proposition 2.5.4.1 \(\mathbb{P}\left(A_{t}^{-, 0} \in d u \mid W_{t}=x\right)\).

 (2) Recover the formula (2.5.5) for \(\mathbb{P}\left(A_{t}^{-, 0} \in d u\right)\).

Proposition 2.5.4.1:

The density of the pair \(\left(A_{t}^{-, 0}, W_{t}\right)\) is

\[\mathbb{P}\left(A_{t}^{-, 0} \in d u, W_{t} \in d x\right)=d u d x \frac{|x|}{\sqrt{2 \pi}} \int_{u}^{t} \frac{1}{\sqrt{s^{3}(t-s)^{3}}} e^{-x^{2} /(2(t-s))} d s \mathbb{1}_{\{u

\[\begin{align*}
& \mathbb{P}\left(A_{t}^{-, 0, u} \in d u\right) / d u=\left[\sqrt{\frac{2}{\pi u}} \exp \left(-\frac{u^{2}}{2} u\right)-2 u \Theta(u \sqrt{u})\right] \\
& \quad \times\left[u+\frac{1}{\sqrt{2 \pi(t-u)}} \exp \left(-\frac{u^{2}}{2}(t-u)\right)-u \Theta(u \sqrt{t-u})\right] \tag{2.5.5}
\end{align*}\]