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When trying to value an American option we often use grid-based methods (e.g. Monte Carlo in combination with Longstaff Schwartz; or Finite Difference Methods). As such, we are in fact estimating the value of a Bermudan option with discrete time points where we can exercise the option, i.e. $0=t_0 < t_1 < ... < t_n = T$.

However, as the time grid gets finer the value of the Bermudan option converges to the American option. I have heard that the rate of convergence is of order $O(\Delta t)$. How can this be shown analytically?

I am especially considering the following exercise strategy

  1. Find the optimal exercise time of the American option, $\tau^*$, where $t_{h-1}<\tau^* < t_{h}$,
  2. Exercise the Bermudan option at the first time point after $\tau^*$, i.e. at time $t_{h}$.

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My starting point would be to have a look at the price difference between the two options and continue from there to find an upper bound of order $O(\Delta t)$... but how?.

$$0\leq C^{A} - C^{B} \leq ... \approx O(\Delta t).$$

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2 Answers 2

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You do not have to get very fancy to find a convergence rate $O(\Delta t)$ for Bermudan to American option value. Consider a put option that is well below the early exercise boundary. We can then ignore optionality, and view it as one where we can exercise either now or in time $\Delta t$. Let's also assume zero carry costs. Exercising now results in present value $ P_0=K-S $, exercising later results in (future expected) value $R_1=K-S e^{r \Delta t}$, which has present value $P_1 = K e^{-r \Delta t} - S$.

So, at any given moment, (especially the in-the-money region of) Bermudan option prices are lower than American option prices by an amount proportional to $\Delta t$.

At this point it is worth noting that almost all options actually traded OTC or on exchanges are Bermudan with daily exercise or sparser, and not American.

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Is this paper useful? Discussed usage of Richardson extrapolation for such purposes http://www.fin.ntu.edu.tw/~conference2002/proceding/5-4.pdf

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  • $\begingroup$ It seems to be within the same topic from what I can tell about Schmidt's Inequality (if you are looking at that section?). However, I am not entirely sure if that result can be applied to obtain the order of convergence? $\endgroup$
    – Landscape
    Commented Jan 14, 2023 at 13:19
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    $\begingroup$ Because you put this as an answer, it would be much better if you can explain in sufficient detail how the question is answered in the referenced paper in case the link is removed or changed. $\endgroup$
    – Alper
    Commented Jan 14, 2023 at 23:14

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