# American Perpetual Put Option

I want to compute the expected payoff of a (classical) perpetual American put option in the Black-Scholes-Merton (BSM) framework with an optimal strategy of exercising the option at time $$\tau=\inf\{t:S_{t}=S^{*}\}$$. In here $$K,r>0,\sigma>0$$ and $$W_{t}$$ are the strike price, risk-free rate, implied volatility and a Brownian motion according to the BSM framework. $$S^{*} has to be determined by maximizing the expected payoff using this optimal strategy. The expected payoff is as following $$\mathbb{E}[e^{-r\tau}(K-S^{*})^{+}1_{\tau<\infty}]$$ My attempt was as following:

$$e^{-r\tau}=\frac{S_{0}}{S^{*}}e^{\sigma W_{t}-\frac{\sigma^{2}t}{2}}$$ and $$(K-S^{*})^{+}=(K-S^{*})$$, thus $$\mathbb{E}[e^{-r\tau}(K-S^{*})^{+}1_{\tau<\infty}]=(K-S^{*})\frac{S_{0}}{S^{*}}\mathbb{E}[e^{\sigma W_{t}-\frac{\sigma^{2}t}{2}}1_{\tau<\infty}]$$. Now, we have a Radon-Nikodym derivative changing or tilting the measure with $$\sigma$$ according to the Cameron-Martin theorem $$\frac{dP_{\sigma}}{dP}=Z_{\sigma}(t)=e^{\sigma W_{t}-\frac{\sigma^{2}t}{2}}$$ If we interpret the function $$1_{\tau<\infty}$$ as $$t$$ will ever hit $$\tau$$ we can write the expected payoff as following $$P_{\sigma}(\sigma W_{t}-\frac{\sigma^{2}t}{2}=\log(\frac{S^{*}}{S_{0}}))$$ wherein standard Brownian motion has the same law as Brownian motion with a drift $$\sigma$$ under the risk-neutral measure. The following equation is given $$P(B_{t}+\gamma t=\eta)=e^{2\eta\gamma}$$ and given the last equation the expected pay off should be set equal to $$\frac{S^{*}}{S_{0}}^{2r/\sigma^{2}}$$.

However, I fail to interpret the last step correctly and cannot find the right solution. Can you help me with the interpretation and equation?