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If a trader shorts an option and dynamically delta hedges to ensure the delta is equal to 0 if that option expires out of the money does the trader profit that options extrinsic value at the time of selling it?
To me this makes sense as even if Vol rises or the option moves ITM, at expiration all extrinsic value goes to 0 and since the delta is hedged the trader will profit the extrinsic value that they sold. But I feel like I must be missing something, where am I mistaken?
Consider your question in the idealized case of zero transaction cost and where the underlying stock price follows geometric Brownian motion with constant volatility -- identical to the implied volatility used to price the option.
If the delta hedge is rebalanced over time short time intervals of length $\Delta t,$ then the cost of hedging is a random variable that converges almost surely to the option price in the limit as $\Delta t \rightarrow 0$. In practice, of course, it is impossible to continuously delta hedge. Furthermore, in the presence of transaction costs the hedging cost with continuous rebalancing diverges to infinity -- Brownian motion has unbounded variation.
Hence, the hedging cost for any realistic strategy has some distribution around the option price with non-zero variance for $\Delta t > 0$. The actual cost will depend on the path of the underlying price. Even if the option expires OTM, there are paths where the cost can be very low -- underlying price runs steadily in the OTM direction, and there are paths where the cost can be very high -- underlying price oscillates frequently through the strike price close to expiration and finishes OTM.
In summary, with delta-hedging over non-zero time steps, the hedging cost (conditioned on the option expiring OTM) is not less than the option price with certainty.
