I am trying to get some intuition for the fact that a Black-Scholes price for an option is equal to the cost of replicating the option.

Say the interest is 0. The option is obviously still worth something, which must be the cost of hedging the option. From what I have read, this cost comes from the fact that you always rebalance your portfolio after the stock has moved, so you incur a small loss from "buying high, selling low".

But this is not completely clear when considering the fact that we are trading in continuous time and in a frictionless market. Does it mean that even in the limit, for infinitesimal time steps, the stock moves faster than the trading strategy?

In that case, which assumptions in the BS-model create this effect, and on the other hand what assumptions would make the cost of hedging equal 0?


1 Answer 1


In the BS model, with friction-free markets in continuous time, the cost of the hedging portfolio is the initial cost of setting up the portfolio. There are no costs over time as the hedging portfolio is self-financing: any purchase of the underlying is paid for by borrowing money and any selling of the underlying is invested at the risk-free rate. At maturity of the derivative, the hedging portfolio will have exactly the same value as the payoff of the derivative. Hence, the cost today of setting up the hedging portfolio has to be the same as the price of the derivative.


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