Derive Black-Scholes Formula

Mathematics

Basic Stochastic Calculus

I. Brownian Motion

1. Standard Brownian Motion = Wiener Process

— Requirement

  • $W_0=0$

  • $W_{t_n}-W_{t_k} \sim N(0,t_n-t_k)$

  • Independent increments

  • $W_t$ is continous

— Properties

  • $\mathbb{E}(dW_t)=0 $

  • $\mathbb{E}(dW_t^2)=dt $

  • $\text{Var}(dW_t)=dt$

  • $\text{Var}(dW_t^2)=2dt^2$

2. Geometric Brownian Motion

$$dS_t = S_{t_n}-S_{t_{n-1}} = \mu S_t dt + \sigma S_t \sqrt{dt}$$

Since $dW_t=\sqrt{dt}$, plug in:

$$dS_t=\mu S_t dt + \sigma S_t dW_t$$

II. Stochastic Calculus

1. Ito Lemma

Assume vanilla option is a function of underlying stock price $S_t$ and time $t$: $V(S_t,t)$, where $S_t$ contains a stochastic process.

Apply Ito Lemma:

$$dV=\frac{\partial V}{\partial t}dt + \frac{\partial V}{\partial S}dS + \frac{1}{2}\frac{\partial^2 V}{\partial S^2} dS^2$$

From $dS_t$ to $dS_t^2$:

$$dS_t^2=\mu^2S_t^2dt^2+\sigma^2S_t^2dW_t^2+2\mu\sigma S_tdW_tdt$$

Since $dW_t^2=dt$:

$$dS_t^2=\mu^2S_t^2dt^2+\sigma^2S_t^2dt+2\mu\sigma S_tdW_tdt$$

Ignore $dt^2$ and $dW_tdt$ because they are too small (order of $dt>1$):

$$dS_t^2=\sigma^2S_t^2dt$$

Plug $dS_t$ and $dS_t^2$ into Ito Lemma:

$$ dV=(\frac{\partial V}{\partial t}\mu S_t+\frac{\partial V}{\partial S}\mu S_t+\frac{1}{2}\frac{\partial^2 V}{\partial S^2}\sigma^2 S_t^2)dt + \frac{\partial V}{\partial S}\sigma S_t dW_t $$

2. Self-Financing

Above formula contains a stochastic term $dW_t$, in order to eliminlate it, we build a portfolio $\Pi$ with long some of underlying stocks and short some of options:

Set the portion of stock to be $-\frac{\partial V}{\partial S}$, and option to be $1$, portfolio should be:

$$\Pi = V - \frac{\partial V}{\partial S}S_t$$

$$d\Pi = dV-\frac{\partial V}{\partial S}dS_t$$

$$d\Pi = (\frac{\partial V}{\partial S}+\frac{1}{2}\frac{\partial^2 V}{\partial S^2}\mu^2 S_t^2)dt$$

3. Black-Schole Differential Equation

Since this portfolio contains no stochastic term, its value is certain, and from no-arbitragy theory, it must be equal to the same amount of money you gain from riskless account.

$$d\Pi = r \Pi dt$$

$$(\frac{\partial V}{\partial S}+\frac{1}{2}\frac{\partial^2 V}{\partial S^2}\mu^2 S_t^2)dt = r(V-\frac{\partial V}{\partial S}S_t) dt$$

$$\frac{\partial V}{\partial S}+\frac{1}{2}\frac{\partial^2 V}{\partial S^2}\mu^2 S_t^2 = rV-\frac{\partial V}{\partial S} r S_t$$

$$\frac{\partial V}{\partial S}+\frac{1}{2}\frac{\partial^2 V}{\partial S^2}\mu^2 S_t^2 + \frac{\partial V}{\partial S} r S_t - rV=0$$

This is the Black-Schole Differential Equation.

4. Feynman-Kac Theorem

Now we introducing Feynman-Kac Theorem:

If

$$dx_t = \mu(x_t,t)dt+\sigma(x_t)dW_t^{\mathbb{Q}}$$

and

$$\frac{\partial V}{\partial t} + \mu(x_t,t) \frac{\partial V}{\partial x} + \frac{1}{2} \sigma^2(x_t,t) \frac{\partial^2 V}{\partial x^2} -r(x,t) V(x_t,t)= 0$$

with boundary $V(x_T,T)$, then:

$$V(x_t,t)=E^{\mathbb{Q}}[e^{-\int_t^T r(x_u,u)du}V(x_T,T)| \mathcal{F}_t]$$

5. Solve Differential Equation

Applying Feynman-Kac:

$$V(S_t,t) = E^{\mathbb{Q}}[e^{-\int_t^T r(S_u,u)du}V(S_T,T)|\mathcal{F}_t]$$

If we look at the integration, since $r(S_t,t)$ is a constant riskfree rate, this becomes:

$$\int_t^T rdu = r\int_t^T du = ru|_t^T=r(T-t)$$

Black-Schole Differential Equation becomes:

$$V(S_t,t)=E^{\mathbb{Q}}[e^{-r(T-t)}V(S_T,T)|\mathcal{F}_t]=e^{-r(T-t)}E^{\mathbb{Q}}[V(S_T,T)|\mathcal{F}_t]$$

Our boundary condition for call option is $V(S_T,T)=\text{max}(S_T-K,0)$:

$$V(S_t,t)=e^{-r(T-t)}E^{\mathbb{Q}}[\text{max}(S_T-K,0)|\mathcal{F}_t]$$

$$=e^{-r(T-t)} [E^{\mathbb{Q}} [S_T-K] P^* (S_T>K)+0] $$

$$=e^{-r(T-t)}E^{\mathbb{Q}}[S_T-K]P^*(S_T>K)$$

Recall that the expected value of a continous distribution:

$$E(X)=\int_a^b x f(x)dx$$

Here we have

$$V(S_t,t)=e^{-r(T-t)}\int_K^{\infty}(S_T-K) f(S_T)dS_T$$

$$=e^{-r(T-t)}[\int_K^{\infty}S_T f(S_T)dS_T-K\int_K^{\infty}f(S_T)dS_T]$$

6. Lognormal Distribution

Since $S_t$ has a lognormal distribution, some of its properties are:

  • $$\int_K^{\infty}f(x)dx=\Phi(\frac{\ln x-\mu}{\sigma})$$
  • $$\int_K^{\infty}xf(x)dx=e^{(\mu+\frac{1}{2}\sigma^2)}\Phi(\frac{\mu+\sigma^2-\ln K}{\sigma})$$

Also, since $S_t$ is following geometric brownian motion, it also has the property that:

  • $$E(S_T)=S_t e^{r(T-t)}$$

7. Call Option Pricing

Apply lognormal properties to our previous equation:

$$\ln S_T \sim N(\ln S_t + (r-\frac{\sigma^2}{2})(T-t),\sigma^2(T-t))$$

After substitution, we have:

$$V(S_t,t)=S_0 \Phi(d_1) -e^{-r(T-t)}K \Phi(d_2)$$

Where:

$$d_1=\frac{1}{\sigma \sqrt{T-t}}[\ln\frac{S_0}{K}+(r+\frac{1}{2}\sigma^2)(T-t)]$$

$$d_2 = d_1 - \sigma \sqrt{T-t}$$