The aim of this exercise is to provide an explanation of the fact, obtained in Proposition 4.3.3.3, that

\[\mathbb{P}(|G| \leq 1)+\int_{0}^{1} \mathbb{P}\left(g_{1}^{a} \in d t\right)=1\]

From the equality \(G^{2} \stackrel{\text { law }}{=} 2 \mathbf{e} g_{1}\) where \(\mathbf{e}\) is exponentially distributed with parameter 1 and \(G\) is a standard Gaussian variable, prove that \(\mathbb{P}(|G|>a)=\mathbb{E}\left(e^{-a^{2} /\left(2 g_{1}\right)}\right)\) and conclude.

Proposition 4.3.3.3:

Let \(g_{1}^{a}=\sup \left\{t \leq 1: B_{t}=a\right\}\), where \(\sup (\emptyset)=1\). The law of \(g_{1}^{a}\) is

\[\begin{align*}
& \mathbb{P}\left(g_{1}^{a} \in d t\right)=\exp \left(-\frac{a^{2}}{2 t}\right) \frac{d t}{\pi \sqrt{t(1-t)}} \mathbb{1}_{\{0& \mathbb{P}\left(g_{1}^{a}=1\right)=\mathbb{P}(|G| \leq a)
\end{align*}\]

where \(G\) is a standard Gaussian random variable. The r.v.

\[d_{1}^{a}=\inf \left\{u \geq 1: B_{u}=a\right\}\]

has the same law as \(1+\frac{(a-G)^{2}}{\widetilde{G}^{2}}\) where \(G\) and \(\widetilde{G}\) are independent standard Gaussian random variables.