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I got an American put option, where the payoff is $V_\tau = \max(K - X_{\tau}, 0)$ and $X_{\tau}$ is the price of an underlying at the stopping time $\tau < T$. The underlying follows a standard GBM with $r = q = 0$; $X_0$ is given.

I need to calculate the expectation $E[V]$ under the assumption that $\tau$ has exponential distribution with intensity $\lambda = 0.025$.


I tried transforming this equation into: $$\int_0^\infty (K - X_0e^{-\frac{1}{2}\sigma^2 \tau + \sigma \sqrt{\tau}Z})^+\lambda e^{-\lambda \tau}d\tau$$ but then I'm just completely lost with how to proceed with the square root. I know that by definition $E[\tau] = \frac{1}{\lambda}$ but can I use this as an answer? As in, can I claim that: $$E[V] = V\left(X_{\frac{1}{\lambda}}, \frac{1}{\lambda}\right) \text{ ?}$$

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  • $\begingroup$ What's the probability it calls at time t1, and how much is a t1-maturity call worth? What about t2? I assume here that $\tau$ is not correlated with $X$... $\endgroup$ – James Spencer-Lavan Apr 4 at 18:08
  • $\begingroup$ originally i was doing monte carlo where i could simulate the stopping time, but here i have to avoid the simulation and somehow calculate the final result without any stopping time simulation. i got values of stock at each time for all paths and intensity. This is it. And yes, it is not correlated $\endgroup$ – Makina Apr 4 at 18:23
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You cannot make that claim. $v(T,s_0,k...)$ increases with approximately of $\sqrt t$. It is not linear with respect to $t$. By Jenson's inequality $E[v(T,s_0,k...)] < v(E[T],s_0,k...)$ when $v''(T)<0$ and $T$ is not a constant.

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