In the martingale approach to derivative pricing, we show that there exists a replicating strategy $(\phi_t, \psi_t)$ which mimics the derivative payoff. My textbook then goes on to state that it is even possible to know what exactly the value of $\phi_t$ is: it's equal to $\Delta_t$, the mathematical derivative of the derivative price with respect to the price of the underlying.

I would like to know the proof and/or intuition for this—something that my textbook fails to provide.


Recall firstly the way how Black and Scholes derive their famous result:

Let $\mathrm{d}S_t=rS_t\mathrm{d}t+\sigma S_t\mathrm{d}W_t$ and $\pi(t,x)=-C(t,x)+\delta x$ such that $\pi(t,S_t)$ is the time $t$ price of a portfolio which is short one European-style call option and is long $\delta$ units of the stock. Here, $\delta\in\mathbb{R}$ is just some real constant and not the sensitivity of the option price. Since the portfolio is self-financing,

\begin{align*} \mathrm{d}\pi(t,S_t) &=-\mathrm{d}C(t,S_t)+\delta \mathrm{d}S_t \\ &=-\mathrm{d}C(t,S_t)+\delta rS_t\mathrm{d}t + \delta \sigma S_t\mathrm{d}W_t. \end{align*}

Furthermore, by Ito's Lemma,

$$ \mathrm{d}C(t,S_t)=\left(\frac{\partial C}{\partial t}(t,S_t)+rS_t\frac{\partial C}{\partial x}(t,S_t)+\frac{1}{2}\sigma^2S_t^2\frac{\partial^2 C}{\partial x^2}(t,S_t)\right)\mathrm{d}t+\left(\sigma S_t\frac{\partial C}{\partial x}(t,S_t) \right)\mathrm{d}W_t.$$

Recall that $\delta$ is any arbitrary real constant. If you now set $\delta=\frac{\partial C}{\partial x}(t,S_t)=\Delta_t$ (i.e. your option delta), then, the Brownian motion cancels and you get (by no-arbitrage)

\begin{align*} \mathrm{d}\pi(t,S_t) &= \left(-\frac{\partial C}{\partial t}(t,S_t)-\frac{1}{2}\sigma^2S_t^2\frac{\partial^2 C}{\partial x^2}(t,S_t)\right)\mathrm{d}t \\ &= r\pi(t,S_t)\mathrm{d}t \\ &= r\left(-C(t,S_t)+\frac{\partial C}{\partial x}S_t\right)\mathrm{d}t. \end{align*} Thus, you of course arrive at the famous Black Scholes PDE.

This derivation however demonstrates that the call option can be hedged in a risk-free manner and hence, you can replicate its payoff by investing in the stock and the bond. As $\Delta$ gives you the sensitivity for changes in the stock price, it is not surprising that $\Delta_t$ tells you how much you need to invest in the stock.

Alternatively, recall that $$ C(t,S_t) = S_te^{-q(T-t)}\Pi_1-Ke^{-r(T-t)}\Pi_2.$$ This is the price of a call option and $\Pi_1$ and $\Pi_2$ are some probabilities. This formula is quite general and holds for much more models then the Black-Scholes model. The terms $\Pi_1$ and $\Pi_2$ again tell you how much you need to invest in the stock and the bond in oder to hedge the call option and again, $e^{-q(T-t)}\Pi_1$ is your Delta which tells you how much you need to invest in the stock.


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