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Andersen--Broadie converts an exercise strategy into an upper bound. The better the exercise strategy the better the upper bound. You can get the exercise strategy by using regression to approximate the continuation value and this is pretty standard -- the LS Method is widely used but does have defects. Once you have an exercise strategy you need the value ...


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Just to add an intuitive argument to @MJ73550's already very nice answer: When holding an American option - or any option callable by the holder for that matter -, the question you ask yourself before exercising it is whether the proceeds from early exercise (i.e. exercise now to get the option's intrinsic value) are greater than what you could expect to ...


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I would not say there is no link to what you say but here would be my view. Intuitive explanation If you wait for a delay $h$ before exercising, you lose your exercise right between $t$ and $t+h$, this leads to a loss in value. Supermartingale property proof (to apply it in your case : $\phi_t=e^{-rt}(L-S_t)^+$) If we denote $\phi$ the obstacle, and ...


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Ok, so I have been thinking about it, and may have found the solution, but please correct me if I'm wrong. I guess the discounted process goes down, because when the holder of the option doesn't exercise it, as long as the price $S(t)$ is less than the optimal exercise price $L^*$ he's loosing cash from not investing into money market?


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There is a logical fallacy in your argument. The price of a European call expiring 1 day before a dividend payment may well be greater than that of a call expiring after it. In other words, claiming that $$ C_E (S_0,K,t_D-1\text {day}; D, t_D) < C_E (S_0,K,T; D, t_D) $$ is not necessarily true. Try the above inequality with a huge dividend (e.g. $D ...


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I assume $r>0$. Let look at a) Let $v$ be the solution. $v$ is increasing (easy to prove, take $x<y$ and show that $v(x)<v(y)$ due $(S^x_t-K)^++a<(S^y_t-K)^++a$ on the continuity region $C$, i.e $x:v(x)>(x-K)^++a$, you have : $$\text{Black Scholes PDE perpetual case : }\frac{1}{2}\sigma^2x^2v''(x)+rxv'(x)-rv(x)=0$$ solutions are of the ...


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Use Dynkin's formula to write the expectation: $\mathbb{E}[e^{-r\tau} \phi(S_\tau)]= g(S_0)+\mathbb{E}[\int_ 0 ^ \tau (A g -rg) dt]$ where $\phi$ is the payoff. Use the infinitismal generator $A$ to derive an ODE which describes the solution Use the fact that American options must be equal to or greater than their intrinsic value to derive boundary ...


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there is a C++ implemented version in the gold part of the Kooderive open source project.



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