Black Scholes — Stochastic Calculus Notes

Chapter 3 Black Scholes

In these notes we price an option contract using the no-arbitrage pricing theory.

3.1 Black Scholes Option Pricing

Definition 3.1 (Black Scholes Model).

Black Scholes model is model for the market where the risk-free interest rate r is constant and tradable asset price process follows Geometric Brownian motion i,e.

	$\displaystyle D_{t}$	$\displaystyle\sim e^{-rt}$
	$\displaystyle\frac{dS_{t}}{S_{t}}$	$\displaystyle=\mu dt+\sigma dW_{t}$

We assume that the no-arbitrage condition holds in the market and that there is a risk neutral probability measure $\mathbb{P}$ such that discounted asset process is a local martingale under this measure.

	$\displaystyle d(D_{t}S_{t})_{\text{drift}}$	$\displaystyle=0$
	$\displaystyle\big{(}-rD_{t}S_{t}dt+D_{t}S_{t}(\mu dt+\sigma dW_{t}^{\mathbb{P}% })+0\big{)}_{\text{drift}}$	$\displaystyle=0$
	$\displaystyle(\mu-r)D_{t}S_{t}$	$\displaystyle=0$

Therefore we have $\mu=r$ .

We know the solution to the Geometric Brownian motion

S_{t}=S_{0}e^{(r-\frac{1}{2}\sigma^{2})t+\sigma W_{t}^{\mathbb{P}}}

Let’s consider a call option on the stock S with expiration at time T and strike K. The value of this option at time T is

V_{T}=\max(S_{T}-K,0)

From the no-arbitrage pricing theory we have $D_{t}V_{t}$ is a local martinage and under bounded regularity conditions we have

D_{t}V_{t}=\mathbb{E}^{\mathbb{P}}[D_{T}V_{T}|\mathcal{F}_{t}]

(3.1)

i.e. $D_{t}V_{t}$ is a martingale.

Solving we have,

	$\displaystyle V_{t}$	$\displaystyle=\frac{1}{D_{t}}\mathbb{E}^{\mathbb{P}}[D_{T}V_{T}\|\mathcal{F}_{t}]$
		$\displaystyle=e^{rt}\mathbb{E}^{\mathbb{P}}[e^{-rT}\max(S_{T}-K,0)\|\mathcal{F}% _{t}]$
		$\displaystyle=e^{-r(T-t)}\mathbb{E}^{\mathbb{P}}[\max(S_{t}e^{(r-\frac{1}{2}% \sigma^{2})(T-t)+(\sigma W_{T}^{\mathbb{P}}-\sigma W_{t}^{\mathbb{P}}}-K,0)\|% \mathcal{F}_{t}]$
		$\displaystyle=e^{-r(T-t)}\int_{-\infty}^{\infty}\max(S_{t}e^{(r-\frac{1}{2}% \sigma^{2})(T-t)+\sigma\sqrt{T-t}x}-K,0)\frac{1}{\sqrt{2\pi}}e^{-x^{2}}dx$

Denoting $\tau=T-t$ and $d_{1,2}=\frac{1}{\sigma\sqrt{\tau}}\big{[}\log(\frac{S_{t}}{K})+(r\pm\frac{1}{% 2}\sigma^{2})\tau\big{]}$ , we have

	$\displaystyle V_{t}$	$\displaystyle=e^{-r\tau}\int_{-d_{2}}^{\infty}\big{(}S_{t}e^{(r-\frac{1}{2}% \sigma^{2})\tau+\sigma\sqrt{\tau}x}-S_{t}e^{(r-\frac{1}{2}\sigma^{2})\tau+% \sigma\sqrt{\tau}(-d_{2})}\big{)}\frac{1}{\sqrt{2\pi}}e^{\frac{-x^{2}}{2}}dx$
		$\displaystyle=S_{t}e^{-\frac{1}{2}\sigma^{2}\tau}\int_{-d_{2}}^{\infty}\big{(}% e^{\sigma\sqrt{\tau}x}-e^{\sigma\sqrt{\tau}(-d_{2})}\big{)}\frac{1}{\sqrt{2\pi% }}e^{\frac{-x^{2}}{2}}dx$
		$\displaystyle=S_{t}e^{-\frac{1}{2}\sigma^{2}\tau}\big{[}\int^{-d_{2}}_{\infty}% e^{-\sigma\sqrt{\tau}x}\frac{1}{\sqrt{2\pi}}e^{\frac{-x^{2}}{2}}dx-\int_{-% \infty}^{d_{2}}e^{\sigma\sqrt{\tau}(-d_{2})}\frac{1}{\sqrt{2\pi}}e^{\frac{-x^{% 2}}{2}}dx\big{]}$
		$\displaystyle=S_{t}e^{-\frac{1}{2}\sigma^{2}\tau}\big{[}e^{\frac{1}{2}\sigma^{% 2}\tau}\int_{-d_{2}}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-(x+\sigma\sqrt{% \tau})^{2}}{2}}dx-e^{\sigma\sqrt{\tau}(-d_{2})}\int_{-\infty}^{d_{2}}\frac{1}{% \sqrt{2\pi}}e^{\frac{-x^{2}}{2}}dx\big{]}$
		$\displaystyle=S_{t}\int_{-d_{2}-\sigma\sqrt{\tau}}^{\infty}e^{-y^{2}}dy-Ke^{-r% \tau}\int_{-\infty}^{d_{2}}\frac{1}{\sqrt{2\pi}}e^{\frac{-x^{2}}{2}}dx$
		$\displaystyle=S_{t}\int_{-\infty}^{d_{2}+\sigma\sqrt{\tau}}e^{-y^{2}}dy-Ke^{-r% \tau}\int_{-\infty}^{d_{2}}\frac{1}{\sqrt{2\pi}}e^{\frac{-x^{2}}{2}}dx$
		$\displaystyle=S_{t}N(d_{1})-Ke^{-r\tau}N(d_{2})$

where N is the cumulative distribution function of the standard normal distribution.

	$\displaystyle V_{t}$	$\displaystyle=\frac{1}{D_{t}}\mathbb{E}^{\mathbb{P}}[D_{T}V_{T}\|\mathcal{F}_{t}]$
		$\displaystyle=e^{rt}\mathbb{E}^{\mathbb{P}}[e^{-rT}\max(S_{T}-K,0)\|\mathcal{F}% _{t}]$
		$\displaystyle=e^{-r(T-t)}\mathbb{E}^{\mathbb{P}}[\max(S_{t}e^{(r-\frac{1}{2}% \sigma^{2})(T-t)+(\sigma W_{T}^{\mathbb{P}}-\sigma W_{t}^{\mathbb{P}}}-K,0)\|% \mathcal{F}_{t}]$
		$\displaystyle=e^{-r(T-t)}\int_{-\infty}^{\infty}\max(S_{t}e^{(r-\frac{1}{2}% \sigma^{2})(T-t)+\sigma\sqrt{T-t}x}-K,0)\frac{1}{\sqrt{2\pi}}e^{-x^{2}}dx$