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American put option. Exercise time is a random variable, calculation of expected payoff
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<blockquote data-quote="Makina" data-source="post: 676867"><p>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.</p><p></p><p>I need to calculate the expectation $E[V]$ under the assumption that $\tau$ has exponential distribution with intensity $\lambda = 0.025$.</p><p></p><hr /><p></p><p>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: $$E[V] = V\left(X_{\frac{1}{\lambda}}, \frac{1}{\lambda}\right) \text{ ?}$$</p></blockquote><p></p>
[QUOTE="Makina, post: 676867"] 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 $\lambda = 0.025$. [HR][/HR] 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: $$E[V] = V\left(X_{\frac{1}{\lambda}}, \frac{1}{\lambda}\right) \text{ ?}$$ [/QUOTE]
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American put option. Exercise time is a random variable, calculation of expected payoff
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