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?