The problem is as follows:
Let $(W_t)$ be a Brownian Motion, $\alpha>0$, and $\tau = \inf\{t>0 : W_t \geq \alpha\}$ be the First Exit Time.
Compute $\mathbb{E}(\tau)$.
I am aware that the result is not finite, however I am having trouble showing that the integral does not converge.
This problem has answers here. However they use results I haven't seen or understand, such as Wald's Identities.
What I have done so far:
$$
\mathbb{P}(\tau \leq t)
= \mathbb{P}(W_t^* \geq \alpha)
= 2 \cdot \mathbb{P}(W_t \geq \alpha)
= 2(1- \Phi(\alpha / \sqrt{t}))
$$
Where $W_t^* = \underset{0\leq s \leq t}{\sup} W_s$.
$$
f(t)
= \frac{\partial}{\partial t} \mathbb{P}(\tau \leq t)
= 2\phi\left(\frac{\alpha}{\sqrt{t}}\right)\left(\frac{\alpha}{2 \sqrt{t^3}}\right)
= \frac{\alpha}{\sqrt{2\pi t^3}} e^{-\frac{1}{2t}\alpha^2}
$$
I am stuck here:
$$
\mathbb{E}(\tau)
= \int_0^{\infty} t \cdot \frac{\alpha}{\sqrt{2\pi t^3}} e^{-\frac{1}{2t}\alpha^2} dt
\overset{?}{=} \infty
$$
Any help would be appreciated.
Best Answer
I do not know whether your expression is correct, but $$\begin{align} &\int_0^{\infty} t \cdot \frac{\alpha}{\sqrt{2\pi t^3}} e^{-\frac{1}{2t}\alpha^2} dt \\ \ge &\int_1^{\infty} t \cdot \frac{\alpha}{\sqrt{2\pi t^3}} e^{-\frac{1}{2t}\alpha^2} dt \qquad\text{ shortening the integral}\\ \ge &\int_1^{\infty} t \cdot \frac{\alpha}{\sqrt{2\pi t^3}} e^{-\frac{1}{2}\alpha^2} dt \qquad\text{ since } e^{-\frac{1}{2t}\alpha^2} \text{ is an increasing function of } t\\ = &\frac{\alpha}{\sqrt{2\pi}} e^{-\frac{1}{2}\alpha^2}\int_1^{\infty} \frac{1}{\sqrt{t}} dt\\ = &+\infty \end{align}$$