# 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.

• I do not think that by a delta hedge alone, you are able to replicate the option price; you should also take the money market account into consideration. – Gordon Aug 21 '17 at 14:44
• @Gordon Can we assume a simplifying assumption that interest rates are zero, just to make the math easier? – kasa Aug 21 '17 at 19:23
• Even if the interest rate is zero, this is still not possible: you need to borrow or deposit for self-financing purpose, though you won't earn any interest rate. – Gordon Aug 21 '17 at 19:27

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, you PnL will be: $$d\Pi_t - r_t\Pi_t = \frac{1}{2}S^2\partial^2_SP \left( \sigma_R^2 - \sigma_M^2\right) dt$$ the difference between realised variance $\sigma_R^2dt$ and pricing model variance $\sigma_M^2 dt$, weighted by the portfolio's Gamma.

• I agree with your answer. Can you please let me know if my understanding is correct: So the cost of replicating the option is the initial investment in delta*stock + money; from there on, the portfolio is self financing, hence no change in PnL. So, the cost of delta hedging is nothing but the price of the option at that initial moment. – kasa Aug 22 '17 at 6:18
• No, the P&L of holding the replicating portfolio (self-financing portfolio which is currently balanced with regards to the option delta) is exactly equal to the P&L of holding the option... which is the definition of the term 'replicating'. But indeed the cost of setting up this replicating portfolio = the initial investment since it is self-financing and if you assume zero market frictions. – Quantuple Aug 22 '17 at 8:23
• @Quantuple: I stand corrected. My new understanding: I start with a Portfolio P worth the price of the option C and having positions as (Deltastocks + Money). Here the Quantity of money = C(option value) - DeltaStocks. This is self-financing is because there is no exogenous capital flow into my portfolio P at any time during the life of the option. At the same time, P replicates the option payoffs at all times. But as a counter example, I start with only Delta*stocks, without any money component to begin with. Then my PnL from Option and from the new Portfolio will be the same. But, – kasa Aug 22 '17 at 13:43
• since the option and portfolio did not have the same value to begin with, so the payoffs from holding the option and the portfolio would be different. Hence, we are not exactly replicating the option. Is this understanding correct? – kasa Aug 22 '17 at 13:45
• Please read the answer pointed in AFK's answer. It contains everything you need IMO. Ps: I don't know why I wrote "currently rebalanced" I meant "dynamically rebalanced" of course. Basically you receive the option premium in cash if you are short the option i.e. $+C_t$ and use part of that to purchase $\Delta_t$ stocks worth $S_t$. Meaning you have at the end of the day $\Delta_t S_t$ worth of stocks and $C_t - \Delta_t S_t$ in the money market account. This is the replication portfolio at $t$. And this should be equal to holding $C_t$. – Quantuple Aug 22 '17 at 17:21

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