Skip to content
All library documents

Why Dynamic Delta Hedging Costs the Black–Scholes Premium

Article Quant Q&A · Author: username123

Summary

The document explains the theoretical cost of hedging a written call by repeatedly adjusting a stock position to match the option’s delta. Under the Black–Scholes assumptions, dynamic hedging replicates the option payoff, but the hedge trades generate a stochastic profit and loss. The adjustments tend to require buying stock after it rises and selling after it falls; even when hedging is continuous, the resulting expected hedge loss is not eliminated by making each time step smaller.

The answers interpret the option premium as the expected cost of carrying out this replicating hedge, which offsets the short option’s payoff risk. The document distinguishes that theoretical result from real trading: it assumes idealized conditions, including no transaction costs and no hedging error. Financing costs, discrete rebalancing, and other departures from the model can change actual outcomes. It also cautions against treating the payoff’s expected loss alone as the explanation for the premium; the hedge’s expected P&L provides the theoretical pricing rationale.

Key ideas

  • Dynamic delta hedging can replicate a written option’s payoff under idealized assumptions.
  • The hedge generates stochastic P&L as stock positions are adjusted over the option’s life.
  • The expected cost of the replicating hedge equals the Black–Scholes option value in the theoretical setup.
  • Continuous rebalancing does not make the expected hedge cost vanish.
  • Transaction costs, financing, and discrete hedging can create differences from the theoretical result.

Tags

Full text
# where does the cost of delta hedging come from?


# where does the cost of delta hedging come from?












I am reading John Hull's book, and am a bit confused about the explanation regarding the cost of delta hedging.

Here is the background: a financial institute is selling call options with strike price $K$, and it is applying delta hedging by adjusting number of purchased stocks to hedge the risk (of stock price going above $K$). The cost of the hedging is expected to be the price of the call option computed by Black-Scholes model. The explanation by author is that it is due to "buy-high, sell-low" when doing the adjustment (as quoted below, in Section 19.4 "Delta hedging" of 10th edition).

> The delta-hedging procedure in Tables 19.2 and 19.3 creates the equivalent of a long position in the option. This neutralizes the short position the financial institution created by writing the option. As the tables illustrate, delta hedging a short position generally involves selling stock just after the price has gone down and buying stock just after the price has gone up. It might be termed a buy-high, sell-low trading strategy! The average cost of $240,000 comes from the present value of the difference between the price at which stock is purchased and the price at which it is sold.

But if we adjust the number at a very small time interval $\Delta t$ such that the buying/selling prices are almost equal, and further we assume that risk-free interest rate is 0, would that imply that there is almost no cost associated with "buy-high, sell-low"?

My understanding is that the real cost comes from the probability that the final stock price $S_T$ is above $K$, in which case there will be inevitable loss for the financial institute. I am not sure if I misunderstand something, as this is not consistent with the explanation by the author.

Let me know what you think.

Edit: Thanks for all answers so far! Let me explain my idea in more formal way: we know that there will be inevitable expected loss of selling an call option being

$$\int_K^{\infty}(S_T-K)p(S_T)dS_T$$

which is exactly the basis for the Black-Scholes price. This loss is associated with the probability that $S_T$ goes above $K$. If we have additional loss related to "buy-high, sell-low" (due to finite-time interval when hedging), then the total cost would be larger than the Black-Scholes price. I wonder if there is any issue with this reasoning?

## Answer by nbbo2 (score 3)

https://quant.stackexchange.com/a/54447

In this statement Hull provides a theoretical justification for the initial value $c$ of the option. Why is $c$ equal to a specific number and not some other number? Where does $c$ come from?

The option by itself as you say is risky because its value depends on the chance that the final stock price $S_T$ is above $K$. (The Idiot says: therefore we cannot put a definite value on $c$, it depends on the Utility Function (risk aversion) of the buyer and seller. But the Idiot is wrong).

As a first step Hull shows that this risk can be eliminated by performing a Dynamic Hedging strategy, of which he provides all the details. Under some strict assumptions this hedging is perfect and all risk is eliminated (we are of course working in the realm of pure theory and in the real world the assumptions may not be fulfilled, resulting in some hedging error).

As a second step Hull asks if this hedging is free or if it has a cost. The answer is it has a cost, which is due to "buying high and selling low while doing the adjustment". He calculates this cost mathematically and comes to a remarkable conclusion: the expected value of the cost is precisely equal to the Black Scholes value of the option $c$.

The implications are:

(1) We now understand where $c$ comes from. It is the expected cost for the financial institute to undertake the dynamic hedging of the option, no more no less (again this is theory: in real life the institute will charge a bit more to buyers, and give a bit less to sellers in order to make a profit, but we are neglecting these trading costs by assumption).

(2) We can justify $c$ in an intellectual sense, as being the "manufacturing cost" of bringing an option (which previously did not exist) into existence through the dynamic hedging process. This also provides a justification why financial intermediaries such as option hedgers exist. They take in the amount $c$ from the option buyer and are able to use spend this amount (on average) to produce the required payoff to the customer. The Black Scholes formula, which at first appears to be the obscure result of some strange new calculus invented by a Japanese mathematician is seen to have a interesting intuitive interpretation. (At least interesting to me! Practical people don't care about the intellectual justification, they just want to memorize the Black Scholes formula to pass the exam, if asked to explain it they will say "It is derived from Ito's Calculus").

## Answer by Pontus Hultkrantz (score 2)

https://quant.stackexchange.com/a/54446

Disregarding the cost of capital to borrow money to buy the hedges, and assuming continuous hedging (no hedging error), the cost comes from your realised PnL during the liftime of your hedges ("buy-high, sell-low"). So the Pnl of your hedges is stochastic, as expected since you own a stock. If you sell the option, your expected PnL from the hedges after delivering the stock to the option holder is negative and equal to what you made from the option premium. So the money you made from selling the option, is what you expect to lose on the hedges.

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

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