Derivation of the Black-Scholes equation

A neutral hedge equity is constructed by selling $\frac{1}{\partial w/\partial x} =\Delta x/\Delta w$ call options at price w(x,t), so that the net equity invested is

$$x - w \frac{\Delta x}{\Delta w}.$$ (1)

A change in x by $\Delta x$ accompanied by a change in w by $\Delta w$ then gives no change in the equity

$$\Delta x - \Delta w \frac{\Delta x}{\Delta w} = 0.$$ (2)

In writing the Black-Scholes equation, we will find the value of the price of the call option w(x,t) necessary to allow the hedge equity to grow at the same rate as investing the equity value in an interest account or instrument at the fixed interest rate r per day so that

$$\Delta x - \Delta w \frac{1}{\partial w /\partial x} =\left(x - w \frac{1}{\partial w /\partial x} \right) r \Delta t.$$ (3)

The Taylor expansion for the the change $\Delta w$ is

$$\Delta w = w(x+\Delta x, t + \Delta t)-w(x,t) =\frac{\partial... ...tial x^2}(\Delta x)^2 + \frac{\partial w}{\partial t}\Delta t.$$ (4)

It is now assumed that the variance $<(\Delta x)^2>$ comes from a random walk in the fractional price, and is therefore proportional to $\Delta t$, giving

$$(\Delta x)^2 = v^2 x^2 \Delta t.$$ (5)

v² is the variance per unit time, or the variance rate.

Putting Eq. 4 and Eq. 5 into Eq. 3, cancelling the $\Delta x$, dividing by $\Delta t$ and multiplying by $\partial w/\partial x$ gives the Black-Scholes equation

$$\frac{\partial w}{\partial t}= r w - r x \frac{\partial w}{\partial x} -\frac{1}{2} v^2 x^2 \frac{\partial^2 w}{\partial x^2}.$$ (6)