Consider a vanilla European call option C, with underlying asset S, strike price K and time to maturity T. Assume that S follows a geometric Brownian motion with mean growth rate of μ and volatility σ. r represents the continuously compounding risk free interest rate.
I sold one call option, and I decided to hedge my risks, using Delta-Neutral strategy. So, I make sure that I always have $$ S\frac{\partial C}{\partial S}$$ worth of stock with me at any moment.
In John Hull, it is mentioned that the cost of such a strategy is always equal to BSM price, irrespective of the actual path that the stock price follows. I am trying to prove this statement mathematically.
Here is how I am trying to prove: The total cost of delta hedging should be $$ \int_0^T SN(d1) $$
$$ as \frac{\partial C}{\partial S} = N(d1) $$ ( If possible, we can assume r = 0 for simplicity )
Can you please guide me on how to proceed further or any other method
Thank you.
P.S. (edit): Intuitively, I know that cost of delta-hedging strategy is always equal to the price of the option. To prove this, let us assume that I sold a call option. Now, I want to hedge myself against downward movement in stock prices. The strategy I would follow is to maintain Delta * Stocks at every point in time, therefore my payoff at the end of maturity would be zero. Essentially, the cost of such a strategy has to be always equal to the price of the call option because only then, No-arbitrage holds.