Calculate $\mathbb{E} W_{\tau + 1}$, $\mathbb{E}W^{2}_{\tau+1}$ and $\mathbb{E}W_{\tau \land 5}$.

Question

Let $W$ be a standard Wiener process. Let $\tau = \inf \{t \geqslant 0 : W_{t} = 4\}$.

Calculate:

a) $\mathbb{E} W_{\tau + 1}$

b) $\mathbb{E}W^{2}_{\tau+1}$

c) $\mathbb{E}W_{\tau \land 5}$

I guess that a) should be just $\mathbb{E} (W_{\tau + 1} - W_{\tau} + W_{\tau})$ = $\mathbb{E} W_1 + \mathbb{E}W_\tau = \mathbb{E}W_\tau$ , so I have to calculate $ \mathbb{E}W_\tau$. Although I'm not sure how - probably it won't be just an expected value of four increments?

Answer 1

Interesting question. Just want to add more details on calculating $E(W_{\tau + 1} - W_\tau)$ and $E[(W_{\tau + 1} - W_\tau)^2]$.

By iterative expectation law: \begin{align} & E(W_{\tau + 1} - W_\tau) = E\{E[W_{\tau + 1} - W_\tau | \tau]\} \\ = & \int_0^\infty E[W_{\tau + 1} - W_\tau | \tau = t] f_\tau(t) dt \\ = & \int_0^\infty E[W_{t + 1} - W_t | \tau = t] f_\tau(t) dt \\ = & \int_0^\infty E[W_{t + 1} - W_t] f_\tau(t) dt \quad \text{because } W_{t + 1} - W_t \text{ is independent of } \{\tau = t\} \\ = & 0. \end{align}

Similarly, \begin{align} & E(W_{\tau + 1} - W_\tau)^2 = E\{E[(W_{\tau + 1} - W_\tau)^2 | \tau]\} \\ = & \int_0^\infty E[(W_{\tau + 1} - W_\tau)^2 | \tau = t] f_\tau(t) dt \\ = & \int_0^\infty E[(W_{t + 1} - W_t)^2 | \tau = t] f_\tau(t) dt \\ = & \int_0^\infty E[(W_{t + 1} - W_t)^2] f_\tau(t) dt \quad \text{because } W_{t + 1} - W_t \text{ is independent of } \{\tau = t\} \\ = & \int_0^\infty f_\tau(t) dt = 1. \quad \text{Use } W_{t + 1} - W_t \sim N(0, 1). \end{align}

Finally, because $\{W_t\}$ is a martingale, by the optimal sampling theorem, $E[W_{\tau \wedge 5}] = EW_5 = 0$.