Why Continuous Delta Hedging Replicates a Black–Scholes Option

# A simple question: Cost of delta hedging when a call option is sold


# A simple question: Cost of delta hedging when a call option is sold












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.

## Answer by AFK (score 5)

https://quant.stackexchange.com/a/35727

You should go back to the derivation of the Black-Scholes equation (see this answer for example). The main point is that you can cancel the risk of the derivative over an infinitesimal time period $dt$ by holding a certain amount $\Delta$ of the asset. When applying this hedging strategy, in this continuous limit, the variance of your PnL is zero. So regardless of the actual path followed by the underlying, the cost of the replication strategy is always the same.

Now this assumes that the rates are deterministic and moreover volatility is known and constant so the volatility we input to price is always going to be the realised volatility on any path. If the volatility is unknown, your PnL will be given by: $$ d\Pi_t - r_t\Pi_t = \frac{1}{2}S^2\partial^2_SP \left( \sigma_R^2 - \sigma_M^2\right) dt $$ the difference between realised variance $\sigma_R^2dt$ and pricing model variance $\sigma_M^2 dt$, weighted by the portfolio's Gamma.

## Answer by honeybadger (score 4)

https://quant.stackexchange.com/a/35733

Based on the inputs from other users, this is another non-rigorous proof of why the cost of delta-hedging is equal to the option price. This approach might be useful for students who use John Hull for reference and may not be familiar with self-financing strategy.

Context: E.g. 19.2 in Options, Futures and Other Derivatives by Hull.

We sold a plain European option and we want to hedge our position by replicating the option payoff.

$$ \text{Assume that at} \quad t = 0, \quad C_0, \, \Delta_0, \, \text{are the price & the delta of the option and } S_0 \text{to be the price of the stock}$$

We know invest a portfolio worth $ C_0 $that consists of $$ (\Delta_0 * S_0) \quad\text{stocks} + \quad (C_0 - \Delta_0 * S_0) \quad \text{money} $$

Now, $$ \text{At} \quad t = 1, \quad C_1, \, \Delta_1, \, \text{are the price & the delta of the option and } S_1 \text{to be the price of the stock}$$ So, we borrow $$B_1 = (\Delta_1*S_1 - \Delta_0*S_1) \text{worth of money from the money market}\\$$ After which our portfolio positions are

$$ (\Delta_1 * S_1) \quad\text{stocks} + \quad (C_0 - \Delta_0 * S_0 -(\Delta_1*S_1 - \Delta_0*S_1)) \quad \text{money} $$

Similarly, $$ \text{At} \quad t = 2, \quad C_2, \, \Delta_2, \, \text{are the price & the delta of the option and } S_2 \text{to be the price of the stock}$$ So, we borrow $$B_2 = (\Delta_2*S_2 - \Delta_1*S_2) \text{worth of money from the money market}$$ After which our portfolio positions are

$$ (\Delta_2 * S_2) \quad\text{stocks} + \quad (C_0 - \Delta_0 * S_0 -(\Delta_1*S_1 - \Delta_0*S_1)-(\Delta_2*S_2 - \Delta_1*S_2) ) \quad \text{money} $$

Assume that by end of $ t= 2 $, the option reaches its maturity and payoff to us, i.e. the seller of the option is $ \Delta_2*S_2 - C_2$

In Hull book, this money market concept is not explained clearly. The statement that Hull makes is that the cost of setting up the initial portfolio + borrowing + final payoff is equal to the option price: in other words,

$$ (\Delta_0 * S_0) \quad (\text{initial cost of the stock}) + \\ (\Delta_1*S_1 - \Delta_0*S_1) + (\Delta_2*S_2 - \Delta_1*S_2) \quad (borrorwings) + \\ (\Delta_2*S_2 - C_2) \quad (final payoff) \\ \text{is equal to } \\ C_0 \text{ -the initial price of the option} $$

Solving the above equation of Hull, we get, $$ \Delta_0*(S_0-S_1) + \Delta_1*(S_1-S_2) = C_0 - C_2$$

From the definition of $\Delta$, this should always hold true for small time intervals.

Hence, Hull's statement that cost of setting up delta-hedged portfolio is equal to the price of the option holds true