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