Let B(t) be one dimensional BM. How can I prove that τ=inf{t≥0:B(t)=2-4t} is a bounded stopping time (with respect to the filtration generated by BM)?
In the solution of the exercise, the lecturer used Wald in order to find the expectation of τ but in my opinion, there is no difference between τ and for example the first hitting time of 1, for which Wald cannot be used.
Thank you for your help.
This is not a bounded stopping time. In fact, one can show
$$ \lim_{t\to\infty}\frac1t\log\mathbb P(\tau > t) = -8, $$
so $\mathbb P(\tau > t) \approx e^{-8t}$, not zero. (More generally, for $\tau = \inf\{t\ge0: B(t) = a - bt\}$, where $a,b>0$, the above limit is equal to $-b^2/2$, which you can prove using Mogulskii's theorem.) However, there is a way to evaluate $\mathbb E[\tau]$ using the Optional Stopping Theorem (from which Wald's identities for Brownian motion are derived). Fix $\lambda>0$ and consider the process
$$ M_t := \exp\bigg( \lambda B_t - \lambda^2t/2 \bigg), $$
which is a martingale. Since the stopped martingale $\{M_{\tau\wedge t}\}$ is bounded (since $B_{\tau\wedge t} \le 2$ and hence $0 < M_{\tau\wedge t} \le e^{2\lambda}$), the Optional Stopping Theorem implies $\mathbb E[M_\tau] = \mathbb E[M_0] = 1$, or
$$ \mathbb E\left[ \exp\Big(\lambda(2-4\tau) - \lambda^2\tau/2\Big) \right] = 1 $$
for all $\lambda>0$. Differentiating with respect to $\lambda$ at $\lambda=0$, we deduce
$$\mathbb E[2-4\tau] = 0,$$
or $\mathbb E[\tau] = \frac12$. Note that this is the same answer you would have obtained by naively applying Wald's identity; roughly speaking, since $\tau$ has exponential tails, it is almost bounded, and so it behaves similarly to a bounded stopping time.