## May 4, 2009

### Black Scholes using Hedging and the Partial Differential Equation

Using the Black Scholes model assumptions, we assume that the underlying stock follows Geometric Brownian Motion.

$\mathrm{d}S = \mu S \mathrm{d}t + \sigma S \mathrm{d}X.$

Use $\Pi$ to denote the value of a portfolio of one long option position and a short position in some quantity $\Delta$ of the underlying:

$\Pi = V(S, t) - \Delta S$

As the time $t$ to $t + \mathrm{d}t$ the value of the portfolio will change. This change in value is partly due to the change in the option value and partly to the change in the underlying

$\mathrm{d} \Pi = \mathrm{d} V - \Delta \mathrm{d} S$

From Ito's Lemma we have

$\mathrm{d} V = \frac{\partial V}{\partial t} \mathrm{d} t + \frac{\partial V}{\partial S} \mathrm{d} S + \frac{1}{2} \sigma^2 S^2 \frac{\partial ^2 V}{\partial S^2} \mathrm{d} t$

Substituting, the value of $\mathrm{d} V$ in the previous equation and grouping for $\mathrm{d} t$ and $\mathrm{d} S$ terms.

$\mathrm{d} \Pi = \left(\frac{\partial V}{\partial t} + \frac{1}{2} \sigma^2 S^2 \frac{\partial ^2 V}{\partial S^2} \right) \mathrm{d} t + \left( \frac{\partial V}{\partial S} - \Delta \right) \mathrm{d} S$

$\mathrm{d} t$, terms are deterministic terms and $\mathrm{d} S$ terms are random.

We can eliminate the random terms by choosing -

$\Delta = \frac{ \partial V}{\partial S}$,

Thus by choosing a suitable value of $\Delta$, our portfolio change is deterministic and completely riskless. This riskfree change must be equal to the growth we would get by putting the equivalent amout in a risk-free asset.

$\mathrm{d} \Pi = r \Pi \mathrm{d} t$

Putting it all together and elimiating $\Delta$ and $\Pi$, we get

$\frac{\partial V}{\partial t} + \frac{1}{2} \sigma^2 S^2 \frac{\partial ^2 V}{\partial S^2} + rS\frac{\partial V}{\partial S} - rV = 0$

The Black-Scholes PDE.

#### 1 comment:

1. Can you talk about the equivalence of BSM to the Heat Equation?