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The Black-Scholes formula entails market completeness, so the price of an option is only the cost associated with dynamically hedging the option.

Where does this cost come from? I don't see how sustaining a replicating portfolio in a frictionless market can cost anything.

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The instruments you use for the dynamic hedge cost money. –  Bob Jansen Jun 14 at 15:09

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I think you misinterpreted what you read.

The whole point of the frictionless market assumption is that you can forget about any cost or any bound on volumes or latency associated with transactions used to rebalance a self-financed portfolio. So you are right when you say that sustaining a replicating portfolio doesn't cost anything.

This implies that the price of the option has to be equal to the initial value of any replicating portfolio (and by absence of arbitrage all replicating portoflios have the same initial value).

From a mathematical point of vue, your mistake comes from assuming that any martingale (like the actualized value of a self-financing portfolio $\widetilde{\Pi}_t := e^{-\int_0^t r_s ds} \Pi_t$, for $t\leq T$, under the risk neutral probability) has zero expected value when actually it just has constant expectated value. So if $\Pi$ is a replicating portfolio, the value of your derivative is equal to $\Pi_0$ and it can be non zero.

Take any simple example. If your terminal payoff is almost surely positive (like a call or put option), you will need some money (or to borrow some) to start your replication strategy. The replication of a bond with fixed coupons rates or the floating leg of a swap is quite enligntening imho.

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