American put option. Exercise time is a random variable, calculation of expected payoff

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: $$E[V] = V\left(X_{\frac{1}{\lambda}}, \frac{1}{\lambda}\right) \text{ ?}$$

• 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 $X$... – James Spencer-Lavan Apr 4 at 18:08
• 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 – Makina Apr 4 at 18:23

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.