Calculate $\mathbb{E} W_{\tau + 1}$, $\mathbb{E}W^{2}_{\tau+1}$ and $\mathbb{E}W_{\tau \land 5}$. 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?
 A: 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, 
$E[W_{\tau \wedge 5}] = EW_5 = 0$.
