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

$$P(|G|\le 1)+{\int }_0^{1}P(g_1^a\in dt)=1$$

From the equality $G^2\stackrel{\text{law}}{=}2e{g}_1$ where $e$ is exponentially distributed with parameter 1 and $G$ is a standard Gaussian variable (see $⟼$ Appendix A.4.2), prove that $P(|G|>a)=E\left(e^{-a^2/\left(2g_1\right)}\right)$ and conclude.

Proposition 4.3.3.3:

Let $g_1^a=sup\{t\le 1:B_t=a\}$, where $sup\left(\varnothing \right)=1$. The law of $g_1^a$ is

$$\begin{array}{cc}& P(g_1^a\in dt)=\exp (-\frac{a^2}{2t})\frac{dt}{\pi \sqrt{t(1-t)}}{1}_{\{0<t<1\}},& & P(g_1^a=1)=P(|G|\le a)\\ \text{(4.3.2)}\end{array}$$

where $G$ is a standard Gaussian random variable. The r.v.

$$d_1^a=inf\{u\ge 1:B_u=a\}$$

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