Inequality of an expectation (here: perpetual put of an american option)

Question

for a given function $u(x):=\sup_{\tau \in T_{0,\infty}}E[(Ke^{-r\tau}-xe^{\sigma B_{\tau}-(\sigma^{2}\tau)/2})_{+}1_{\tau <\infty}]$ and $x \in [0,\infty)$, K a positive real number, $(B_{t})$ a standard brownian, $T_{0,\infty}$ the set of all stopping times of the filtration of $(B_{t})_{t\geq 0}$, i want to show that $u(x)\geq (K-x)_{+}$. In the book it is also said that for all $T>0$ it holds that $u(x)\geq E[(Ke^{-rT}-xe^{\sigma B_{T}-(\sigma^{2}T)/2)})_{+})]$ which implies $u(x)>0$. I could apply Jensen inequality here, but why do we get strictly $>0$?

Answer 1

Define the stopping time $$ T = \inf\left\{ t : \sigma B_t > \left(\frac{\sigma^2}2 - r\right) t + C \right\} $$ It is finite with non zero probability.

Now define $\tau = \inf (T,1)$ (say), so that $\tau <\infty$. Hence $$ u(x) \ge E[(Ke^{-r\tau}-xe^{\sigma B_{\tau}-(\sigma^{2}\tau)/2})_{+}1_{\tau <\infty}] \\ \ge P(\tau = T) \inf_{\omega: \tau\le 1} (Ke^{-r\tau}-xe^{\sigma B_{\tau}-(\sigma^{2}\tau)/2})_{+} $$

Choose $C$ such as $$ \inf_{\omega: \tau\le 1} (Ke^{-r\tau}-xe^{\sigma B_{\tau}-(\sigma^{2}\tau)/2})_{+} >0 $$ and you will get $u(x)>0$.