Recall that the maximum $X_{0}^{t}:=\operatorname{Max}_{s \in[0, t]} W_{s}$ over $[0, t]$ of standard Brownian motion $\left(W_{s}ight)_{s \in[0, t]}$ has the probability density function

$$
\varphi_{X_{0}^{t}}(x)=\sqrt{\frac{2}{\pi t}} \mathrm{e}^{-x^{2} /(2 t)}, \quad x \geqslant 0 .
$$

a) Let $\tau_{a}=\inf \left\{s \geqslant 0: W_{s}=aight\}$ denote the first hitting time of $a>0$ by $\left(W_{s}ight)_{s \in \mathbb{R}_{+}}$. Using the relation between $\left\{\tau_{a} \leqslant tight\}$ and $\left\{X_{0}^{t} \geqslant aight\}$, write down the probability $\mathbb{P}\left(\tau_{a} \leqslant tight)$ as an integral from $a$ to $\infty$.

b) Using integration by parts on $[a, \infty)$, compute the probability density function of $\tau_{a}$.

The derivative of $\mathrm{e}^{-x^{2} /(2 t)}$ with respect to $x$ is $-x \mathrm{e}^{-x^{2} /(2 t)} / t$.

c) Compute the mean value $\mathbb{E}^{*}\left[\left(\tau_{a}ight)^{-2}ight]$ of $1 / \tau_{a}^{2}$.