Consider someone that writes a call, and wishes to delta-hedge against it to remain delta neutral. For this to be profitable, the price they sell this option for should be greater than or equal to the cost of delta-hedging it. (From this question answer: "An option price is equal to the cost of delta-hedging")

I understand the option writer has to use capital to delta-hedge, which would imply some risk-free cost to borrow. However, I do not understand how volatility increases the overall expense of delta-hedging.

The option writer could do the following, continuously: For each theoretical price point, compute what delta would be, and place a limit buy or sell order at that theoretical price that would result in their position matching that delta. This would result in them having a ladder of orders up and down the order book, which would slowly change over time. Notice: this is entirely agnostic of the actual volatility.

What am I missing here? Given that options' IVs correlate with historical volatilies there must be some real cost or risk associated with delta-hedging more volatile stocks. However, that cost and/or risk eludes me.


1 Answer 1


The key here is to observe that the volatility at the time the option is written is not exactly equal to the volatility that the markets actually experience during the option's lifetime.

The seller will price the option according to her best estimate of future volatility over its lifetime, but will always prove to have been too high or too low. More volatility increases the hedging cost (and the price our seller will have wished she had charged).

As to why extra volatility increases the hedging costs and option prices, the easiest way to see that is to note that a hedger who is short an option will be selling when the underlying drops in value, and buying when it increases. Buy high/sell low is well-known to be ruinous as an investment strategy. :-)

  • $\begingroup$ I don't have a better answer but isn't the hedging price increasing in vol also in the BS model with constant vol? In this model an increase in stock price also does not predict low returns going forward. $\endgroup$
    – fes
    Jul 21, 2021 at 11:14
  • $\begingroup$ Regarding buy high/sell low: When delta-hedging, do you not buy and sell the same amounts at the same price? That is, if "on the way up" you bought 1 share at \$1.00, and 1 share at \$1.01, if the price goes back down, you'll also sell 1 share at \$1.01, and 1 share at \$1.00. Where's the loss? $\endgroup$
    – JS_Riddler
    Jul 21, 2021 at 12:58
  • $\begingroup$ @JS_Riddler In the BS world, with a continuous stochastic process, only after the price of $1.00 has been observed does it enter the filtration (i.e. it is in the past) and therefore cannot be traded again until the next crossing. So the hedge happens at a price 1.00 plus or minus a little epsilon. And those add up. In the real world, of course, our hedger cannot always get her price either. $\endgroup$
    – Brian B
    Jul 21, 2021 at 15:23
  • $\begingroup$ @BrianB I think I get it... Is the following accurate: Let our hedger rehedge every period T. During each period, the required delta position will change, and as a result our hedger will be buying slightly higher (on upswings) or slightly lower (on downswings) than if they were to rehedge with a smaller T. Furthermore, the expected movement, and thus inaccuracy, in each T is greater with higher volatility. Our hedger will hedge more perfectly as T approaches zero, or as volatility approaches zero. $\endgroup$
    – JS_Riddler
    Jul 23, 2021 at 0:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.