Why Discrete Delta Hedging Can Produce Gains or Losses
Summary
The document presents a Black–Scholes call example in which a trader sells an at-the-money option and hedges its delta by buying stock and borrowing funds. The assumed stock follows geometric Brownian motion, and the hedge is adjusted at regular intervals rather than continuously. In the illustrated path, the stock's realized volatility is below the assumed implied volatility, and the position appears profitable after a small move. The explanation attributes the difference from the theoretical hedge cost to discrete rebalancing and the particular random price move.
It also raises a challenge about the bookkeeping and asks how a one-standard-deviation move would yield a zero-profit result. The text does not answer those questions or provide a full P&L derivation, so its numerical account should be treated as an illustration rather than a verified calculation. It conveys that discrete hedging leaves path-dependent gains and losses, while offering limited evidence about their size or expected value.
Key ideas
- Continuous-time replication assumptions do not exactly describe a hedge rebalanced at intervals.
- Discrete delta hedging can produce path-dependent gains and losses.
- The example compares implied volatility used for pricing and hedging with lower realized volatility on one path.
- The document leaves its book-value arithmetic and the proposed zero-profit move unresolved.
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Full text
# Representing a continuous time hedge discretely # Representing a continuous time hedge discretely I recently came across an example (Section 5.2.9 here) which does a simple delta hedging experiment. Below are the details: Market conditions Interest rates are at a 2% level --> r = 0.02 The underlying stock pays no dividend and follows a Geometric Brownian Motion (GBM) with µ = 10% and σ = 20% The initial spot price is 100€ --> S0 = 100 The call option is ATM with a maturity of 0.1 years --> T = 0.1 We are going to delta hedge at regular time step with dt = 0.005 so that there are 20 steps. We will generate sample paths of the stocks by generating standard normal deviates and using the known formula. From the time series, we can calculate back the realized volatility after the experiment --> realized volatility = 17%. First Day The trader sells the option at a price of 2.62€, corresponding to an implied volatility of 20%. He will also use this implied volatility for hedging purposes, which prescribed a delta of 52.52% so that he buys 0.5252 stocks. He does this by borrowing (0.5252 ∗ 100€ − 2.62€) = 49.90€. After the first day, the trader has the following instruments in his book: A sold call option at 2.62€ A loan with a present value of 49.90€ A stock portfolio with a value of 52.52€ Second Day The outstanding loan is increased because of the interest effect. In this case, the effect is about 1 cent. The stock has decreased by 10 cents. Because of the delta of his position, his stock portfolio loses about 5 cents. The sold option loses more than just 5 cents. All of this brings the trader's book to a positive value. The theory of Black-Scholes is prescribing that the hedging cost leads to the option price. However, we notice a profit coming into the book after just one day of hedging! Why? The reason is that the theory assumes that you are hedging continuously through time, rather than once a day. So the error is induced by the discrete nature of the hedging strategy. Another way of looking at this is by analyzing the random number that was used to generate the value of the second day. This number is a draw from the standard normal distribution. This one is a small number. One could say that the volatility of the first day, because of this small draw, is smaller than actually predicted by the distribution. It is smaller than the "typical" number in a standard normal distribution. Which random number one has to use in order to see no profit and no loss on this first day? Z = -1 or +1 --> one standard deviation out of the mean." My question is how is the book value positive? The loan which is a liability increased by a cent. The short call is fixed and was used to purchase the portfolio while the portfolio itself is down by 5 cents. Doesn't this add up to a net negative value? Also, how was Z = 1 derived for a net zero position?
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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.