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i have the following question. The price of an Bermudan option is given by \begin{align*} V_{0} = \sup_{\tau \in \mathcal{T}(0,\dots, T)} \mathbb{E}[f_{\tau}(X_{\tau})]. \end{align*}

It is possible to approximate this price using Monte-Carlo and the continuation values defined as \begin{align*} q_{t}(x) = \sup_{\tau \in \mathcal{T}(t+1, \dots, T)}\mathbb{E}[f_{\tau}(X_{\tau})\mid X_{t} = x]. \end{align*}

My question is now, why do I get a lower bound for the Bermudan option price when calculating the continuation values recursively via \begin{align*} q_{t}(x) = \mathbb{E}[\max\{f_{t+1}(X_{t+1}), q_{t+1}(X_{t+1})\} \mid X_{t} = x]? \end{align*}

Is it because the $supremum$ of the continuation values is always smaller than the $supremum$ of the actual stopping problem, because the range of stopping times is a subset of the other?

Best regards,

Peter

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  • $\begingroup$ I guess you can prove by induction starting from the terminal nodes rolling back. $\endgroup$
    – Vim
    Feb 5 '20 at 17:10
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It depends of the convexity of the function f. I guess you already heard about the fact that american call price is the same as european call price when there is no dividends. It is still valid for bermudan call as its price is between american call price and european call price.
Please have a look on this document for more details:
http://www.stat.uchicago.edu/~lalley/Courses/391/Lecture15.pdf

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You guessed the reason correctly. When you try to estimate the supremum across a sample, your estimator is indeed less than (or equal if you are lucky) the true supremum.

Another method to get an upper bound for the price is the Andersen-Broadie algorithm, which estimates an infimum over a set of (discrete) martingales, which for the opposite reason is always greater than (or equal if you are lucky) the true infimum :)

There are many good references on the topic, I personally enjoyed Guyon and Henry-Labordère’s Nonlinear Option Pricing

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