Find distribution of $\sup_{t \leq \tau} W_t$ for a stopping time $\tau$ Let $\tau=\inf(t>0:W_{t}=-1)$. How to find distribution $\sup_{t\leq \tau}W_{t}$ ?
 A: 
Lemma: Let $(W_t)_{t \geq 0}$ be a one-dimensional Brownian motion, and define
$$T_a := \inf\{t \geq 0; W_t = a\}, \qquad a \in \mathbb{R}.$$ If $a,b>0$ then $$\mathbb{P}(T_{-a} < T_b) = \frac{b}{a+b} \qquad \mathbb{P}(T_b<T_{-a}) = \frac{a}{a+b}. \tag{1}$$

Proof: For fixed $a,b>0$ denote by $T = \min\{T_{-a},T_b\}$ the first exit time from the interval $(-a,b)$. It is known that $$\mathbb{P}(W_{T} = a) = \frac{b}{a+b} \qquad \mathbb{P}(W_{T}=b) = \frac{a}{a+b}$$ (this follows e.g. by an application of Wald's identities, see here for a detailed proof). Noting that $\{W_{T}=a\} = \{T_{-a}<T_b\}$ and $\{W_T=b\} = \{T_b<T_{-a}\}$, this proves the assertion.

Note that $\tau$, defined in your problem, satisfies $\tau=T_{-1}$. If $x>0$, then
$$\mathbb{P} \left( \sup_{s \leq \tau} W_s \geq x \right) = \mathbb{P}(T_x < T_{-1}) \stackrel{(1)}{=} \frac{1}{1+x}.$$
Consequently, the distribution of $\sup_{s \leq \tau} W_s$ is absolutely continuous with respect to Lebesgue measure; its density is given by $$p(x) = \frac{1}{(1+x)^2} 1_{\{x>0\}}.$$
A: Let $M_t:=W_{t\wedge\tau} +1$. Then $M$ is a nonnegative continuous local martingale such that $M_t \to 0$ almost surely as $n \to \infty$. By this question (note that the argument is easily adapted to local martingales) it holds that $$\mathbb{P}\left(\sup_{t\geq 0} M_t > x\right) = 1 \wedge \frac{1}{x}, \quad x>0.$$
We obtain for $x > -1$
$$\mathbb{P}\left(\sup_{s\leq \tau}W_{s} > x\right) = \mathbb{P}\left(\sup_{t\geq 0} M_t > 1+x\right)= 1 \wedge \frac{1}{1+x}.$$
