Let $(W\_t)\_{t\ge 0}$ be a standard Brownian motion and $\tau$ be a stopping time lying in $[1,2]$. For $x, y>0$, can we show


$$\mathbb E\big[{\bf 1}\_{\{x+\inf\_{0\le t\le 2}W\_t>0\}}(W\_{\tau}-y)^+\big]>0?$$



Is this expectation $\mathbb E\big[{\bf 1}_{\{x+\inf_{0\le t\le 2}W_t>0\}}(W_{\tau}-y)^+\big]$ strictly positive?
Let $A= \{W\_1>y+2, \, \inf\_{0 \le t \le 1} W\_t >-x\}.$
By the reflection principle,
$$P(A)=P(W\_1>y+2)-P(W\_1<-2x-y-2)$$ $$=P(W\_1>y+2)-P(W\_1>2x+y+2)>0\,.$$
Let $D= \{ \inf\_{1 \le t \le 2} W\_t >y\}.$ Then by the Markov property and the reflection principle,
$$P(D|A) \ge P(\inf\_{0 \le t \le 1} W\_t >-1)=P(W\_1>-1)-P(W\_1<-1)=P(W\_1 \in (-1,1]\:)>0\,.$$
Thus $P(D)>0$. We deduce that
$$\mathbb E\big[{\bf 1}\_{\{x+\inf\_{0\le t\le 2}W\_t>0\}}(W\_{\tau}-y)^+\big]>0$$
Since the integrand is strictly positive on the event $D$.


