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Why Monte Carlo Estimates Give a Lower Bound for Bermudan Options

Article Quant Q&A · Author: Peter
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Summary

The discussion addresses why a Monte Carlo estimate of a Bermudan option value can fall below the true value when continuation values and exercise decisions are estimated from simulated paths. The central explanation is the supremum-estimation effect: optimizing over a finite sample tends to produce an estimate no greater than the true supremum, so the resulting policy gives a lower bound on the optimal stopping value.

For an upper bound, the answer points to the Andersen–Broadie approach, which uses an optimization over discrete martingales and has the opposite bound direction. The document offers a conceptual explanation and names an alternative bounding method, but it does not give implementation details, convergence evidence, or a numerical example. Its brief reference to convexity and call options is not developed in the accepted explanation.

Key ideas

  • Optimizing exercise decisions on a Monte Carlo sample tends to underestimate the true supremum.
  • A policy derived from estimated continuation values therefore yields a lower bound on the optimal stopping price.
  • The Andersen–Broadie method is cited as a way to estimate an upper bound using discrete martingales.
  • The discussion is conceptual and does not provide implementation guidance or numerical validation.

Tags

OptionsDerivatives pricingBacktesting
Full text
# Lower bound for Bermudan Option Price


# Lower bound for Bermudan Option Price












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

## Answer by Valometrics.com (score 1)

https://quant.stackexchange.com/a/51020

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

## Answer by siou0107 (score 0)

https://quant.stackexchange.com/a/51022

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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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.