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Abstract

This article shows that the volatility smile is not necessarily inconsistent with the Black–Scholes analysis. Specifically, when transaction costs are present, the absence of arbitrage opportunities does not dictate that there exists a unique price for an option. Rather, there exists a range of prices within which the option's price may fall and still be consistent with the Black–Scholes arbitrage pricing argument. This article uses a linear program (LP) cast in a binomial framework to determine the smallest possible range of prices for Standard & Poor's 500 Index options that are consistent with no arbitrage in the presence of transaction costs. The LP method employs dynamic trading in the underlying and risk-free assets as well as fixed positions in other options that trade on the same underlying security. One-way transaction-cost levels on the index, inclusive of the bid–ask spread, would have to be below six basis points for deviations from Black–Scholes pricing to present an arbitrage opportunity. Monte Carlo simulations are employed to assess the hedging error induced with a 12-period binomial model to approximate a continuous-time geometric Brownian motion. Once the risk caused by the hedging error is accounted for, transaction costs have to be well below three basis points for the arbitrage opportunity to be profitable two times out of five. This analysis indicates that market prices that deviate from those given by a constant-volatility option model, such as the Black–Scholes model, can be consistent with the absence of arbitrage in the presence of transaction costs. © 2001 John Wiley & Sons, Inc. Jrl Fut Mark 21:1151–1179, 2001