# 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 $X$... Apr 4, 2019 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 Apr 4, 2019 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.

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.

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.

• Can you provide a bit more detail? Jun 12, 2020 at 10:11
• @DaneelOlivaw: I have added some details. Is it clear to you?
– Hans
Jun 15, 2020 at 18:59
• Sorry but, where did the random variable $Z$ go? There are 2 random variables involved, shouldn’t there be a double integral? Nov 11, 2020 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, 2020 at 21:00
• I am not the OP. OP though stated in a comment below his question that both variables are indeed independent. Nov 13, 2020 at 16:07

Let's recall why no analytical formula is known for the price of American Put option. This is due to the early exercise condition, which requires that options's price should be bigger or equal to its exercise payoff for any $$S_t$$ at any time $$t \le T$$ (in other words $$V(S_t,t) \ge max(K-S_t, 0)$$). This property is also known as path-dependence, as payoff at maturity $$T$$ doesn't depend only on $$S_T$$, but also on previous values of $$S_t$$ where it could be potentially exercised. An arbitrage opportunity exist when this condition is broken: exercise put and instantly buy it back with a net profit OR buy put and instantly exercise.

Early exercise condition is very hard to account for in analytical solution, hence numerical methods are used. Most popular are Finite Difference and Monte-Carlo. In both cases you explicitly ensure that put price doesn't fall below the exercise payoff. (Finite difference works well for 1D and 2D problems and is easier to implement, however Monte-Carlo is more stable in higher dimensions but is harder to code, again due to the early exercise condition.)

In your problem, a new exercise condition is introduced, which is not path-dependent, but random (distributed exponentially). It might be interpreted as

1. Replacement for the path-dependent condition. In this case exercise can be initiated only by some random chance and holder has no right to exercise. As path-dependent condition was lifted, we have a chance for a closed-form solution (assuming $$r=q=0$$):

$$E[V] = \int_0^T d\tau \ \lambda e^{-\lambda \tau} \int_0^{K} ds \ (K-S) \ \rho_s(S,\tau)$$

Here $$\rho_s(s,t)$$ is distribution density of $$S$$ at time $$t$$ in risk-neutral measure:

$$\rho_s(s,t) = \frac {s_0} {\sqrt{2\pi} \ s\ \sigma\sqrt{t}} \exp\left( - \frac{\left(\ln s / s_0 + 0.5 \sigma^2 t \right)^2}{2\sigma^2 t}\right)$$

The last integral is BS price of European Put with maturity $$\tau$$, hence you need to find

$$E[V] = \int_0^T dt \ \lambda e^{-\lambda t} \ V_{put}(S_0, t)$$

2. Addon to the path-dependent condition. In this case exercise can be initiated by some random chance or by the holder. This is a "loosing" strategy as by randomly exercising we always lose extrinsic value (which is non-negative for American options).

This is more challenging as path-dependence is still present. Not sure if you can incorporate exercise randomness into finite-difference approach, so what is left is Monte-Carlo. It will be 2D, as you need to simulate both $$S_t$$ and exercise time $$\tau$$ (if exercise happens). Otherwise. it should not be much different from standard American pricing a la Longstaff-Schwartz.

Hopefully, these thoughts will be helpful.