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


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?

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    $\begingroup$ Hello u123, and welcome to this place. Would you be able to give the exact location of the statement in Hull's book, or could you even quote verbatim? That would be helpful for any discussion! $\endgroup$ – Kermittfrog May 27 at 6:06
  • $\begingroup$ But if we adjust the number at a very small time interval Δt such that the buying/selling prices are almost equal - if this is true, then there is nothing to "hedge" and hence no cost. I suppose you are talking about a short gamma trade where you hedge delta. In this case, we need to hedge when the market moves, you accumulate delta opposite to market direction, hence buy high and sell low, and hence the cost of hedging $\endgroup$ – nimbus3000 May 27 at 6:37
  • $\begingroup$ @Kermittfrog Thanks for the advice! I have added the quote. $\endgroup$ – username123 May 27 at 15:03

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").

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  • $\begingroup$ I understand that the "manufacturing cost" of bringing an option should be equal to its price. But I am not fully convinced that this cost comes from "buy-high, sell-low". I have made some edits clarifying my point if you could take a look. Thank you! $\endgroup$ – username123 May 27 at 15:11
  • $\begingroup$ Hull actually has explained quite beautifully how hedging leads to “buy high, sell low”. I believe it’s page 2-3 into the Greeks chapter $\endgroup$ – Dhruv Mahajan May 27 at 16:19
  • $\begingroup$ Hull is saying that $c$ is equal to the Discounted Expected Value under the RN Measure (of which you wrote the integral) and also is equal to the Sell High Buy Low Loss. Which is far from obvious and requires a detailed proof. You would not add these 2 together, that would be double counting of the same thing. $\endgroup$ – noob2 May 29 at 10:22
  • $\begingroup$ “purchases must be made at a price K + e 􏰎 and sales must be made at a price K -e 􏰁 􏰎, for some small positive number e 􏰎. Thus, every purchase and subsequent sale involves a cost (apart from transaction costs) of 2􏰎e“ I was talking about this explanation in the Hull book for dynamic rehedging. He has explained how there are hidden costs in hedging and how number of trades need to be infinity for e to be 0. Detailed explained in “Section 19.3 : A stop loss strategy” $\endgroup$ – Dhruv Mahajan May 30 at 17:39

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.

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  • $\begingroup$ Thanks for your answer! I understand that the cost would be equal to the money made from selling the option, but just wonder where the cost really come from. I have made some edits if you could take a look. $\endgroup$ – username123 May 27 at 15:09

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