Delta of options : A mathematical explanation

Question

Consider the Black-Scholes equation $$\begin{equation}\label{eq3} \frac{\partial{V}}{\partial{t}}+\frac{1}{2}\sigma^2S^2\frac{\partial^2{V}}{\partial{S}^2}+(r-D)S \frac{\partial{V}}{\partial{S}}-rV=0,~~~~S\in (0,\infty),~~~t\in(0,T) \end{equation}$$ where $D$ is the dividend yield, $\sigma$ is the market volatility, $r$ is the interest rate.

Doubt : The delta $\triangle =\frac{\partial{V}}{\partial{S}}$ of an option mathematically equivalent to the rate of change of option price with respect to the change in asset price. How this is connected with the amount of shares need to be short selled to hedge the position changes?

Answer 1

Assume we are long a call option, $C$. If the price of the asset, $S$, underlying it declines, the value of $C$ decreases and the long position loses money. To counter the decline in price of the underlying we want to short $\Delta$ units of the underlying asset. What this looks like is if $\Pi$ is the value of the portfolio of long one call option and short $\Delta$ units we have

$$\Pi = C - \Delta S$$

We want $\Pi$ to be insensitive to small changes in the price of the underlying, $S$. That is we want

$$\frac{\partial \Pi}{\partial S} = \frac{\partial C}{\partial S} - \Delta\frac{\partial S}{\partial S} = 0.$$

Solving for $\Delta$ gives

$$\Delta = \frac{\partial C}{\partial S}$$

We can also see this from a Taylor series approximation. If we let $V(S, t)$ be the value at time $t$ of a European option on an underlying with spot price $S$, and no dividend. $V(S, t)$ is infinity many times differentiable in both $S$ and $t$. We now expand $V$ around $(S, t)$

$$V(S + dS, t + dt) = V(S, t) + dS\frac{\partial V}{\partial S} + dt\frac{\partial V}{\partial t} + \frac{(dS)^2}{2}\frac{\partial^2 V}{\partial S^2} + \frac{(dt)^2}{2}\frac{\partial^2 V}{\partial t^2} + dSdt\frac{\partial^2 V}{\partial S \partial t} $$

We approximate $(dS)^2 \approx \sigma^2S^2dt$ and say that $dV = V(S + ds, t + dt) - V(S, t)$ we have

$$dV \approx \frac{\partial V}{\partial S}dS + \frac{\partial V}{\partial t}dt + \frac{\sigma^2S^2}{2}\frac{\partial^2 V}{\partial S^2}dt $$

We can recognize the terms on the right as the options Delta, Theta, and Gamma. Now using this in a portfolio $\Pi$ we have

$$\Pi = V - \Delta S$$ $$d\Pi = dV - \Delta dS$$

Which is approximatly

$$d\Pi \approx \frac{\partial V}{\partial S}dS + \frac{\partial V}{\partial t}dt + \frac{\sigma^2S^2}{2}\frac{\partial^2 V}{\partial S^2}dt - \Delta dS $$

So to remove the sensitivity to price we have $$\Delta = \frac{\partial V}{\partial S}$$

Leaving only Theta and Gamma sensitivity

$$d\Pi \approx \frac{\partial V}{\partial t}dt + \frac{\sigma^2S^2}{2}\frac{\partial^2 V}{\partial S^2}dt$$

Answer 2

The basic idea is to create a portfolio that (to first order) does not change in value when the share price changes. If you have a portfolio of put options and shares it has a certain value. If the stock price goes up, the shares are more valuable and the put options are less valuable. Let us say the put options decline in value by $0.50$ when the the share price increases by $1$, which is the delta you define. If you own twice as many puts as shares, the first derivative of your portfolio value with respect to share price is $0$ so you don't care whether the stock goes up or down.