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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 $X$... $\endgroup$ Apr 4 '19 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 '19 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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The best way to price this kind of options is to use monte carlo: 1. You generate the stopping time. 2. you generate the underlying value at this generated stopping time and you store it in a vector. 3. The price will be the mean of your vector as r and q are equal to zero. That's all.

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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.

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  • $\begingroup$ Can you provide a bit more detail? $\endgroup$ Jun 12 '20 at 10:11
  • $\begingroup$ @DaneelOlivaw: I have added some details. Is it clear to you? $\endgroup$
    – Hans
    Jun 15 '20 at 18:59
  • $\begingroup$ Sorry but, where did the random variable $Z$ go? There are 2 random variables involved, shouldn’t there be a double integral? $\endgroup$ Nov 11 '20 at 9:07
  • $\begingroup$ @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. $\endgroup$
    – Hans
    Nov 11 '20 at 21:00
  • $\begingroup$ I am not the OP. OP though stated in a comment below his question that both variables are indeed independent. $\endgroup$ Nov 13 '20 at 16:07

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