# 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 '19 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 '19 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.
Substitute $$x=\sqrt{\tau}$$. There may be two intervals of $$[0,\infty)$$ for the integral over $$\tau$$ that is effective under the function $$(\cdot)^+$$. For convenience, we take the right most interval starting from $$x^2_1$$. \begin{align} &\frac12\int_{x^2_1}^\infty (K - X_0e^{-\frac{1}{2}\sigma^2 \tau + \sigma \sqrt{\tau}Z}) e^{-\lambda \tau}d\tau \\ =& \int_{x_1}^\infty (K - X_0e^{-\frac{1}{2}\sigma^2 x^2 + \sigma xZ}) e^{-\lambda x^2}xdx \\ =& \frac{K}{2\lambda}e^{-\lambda x_1^2}-X_0\Big(\int_{x_1}^\infty e^{-\frac12a(x -b)^2 +c}(x-b) dx +b\int_{x_1}^\infty e^{-\frac12a(x -b)^2 +c}dx \Big) \\ =& \frac{K}{2\lambda}e^{-\lambda x_1^2}-X_0 \Big(\frac12 e^{-\frac12a(x_1 -b)^2 +c}+bF(x_1)\Big) \end{align} where $$a, b, c$$ are appropriate constants and $$F(x)$$ is essentially the complementary error function.
• Sorry but, where did the random variable $Z$ go? There are 2 random variables involved, shouldn’t there be a double integral? – Daneel Olivaw Nov 11 '20 at 9:07
• @DaneelOlivaw: Are you assuming $\tau$ is independent of $Z$? If so, you should specify it in your question. My $(a,b,c)$ is a function of $Z$. Yes, you need to integrate with respect to $Z$. That integral may not be clean. I will get the details out later. – Hans Nov 11 '20 at 21:00