Black-Scholes option pricing formula and martingale approach.

By Hachemi Benyahia
On 2010-05-04

Abstract

Through the martingal approach we set out a detailed proof of the Black-Scholes-Merton valuation formula for a european vanilla option. This approach, wich can be interpretated as a Feynman-Kac's theorem's corollary, enables to solve quickly and elegantly problems related to derivatives valuation when analytical formula is available.

Mathematical notations

$S_{T}$ the underlying price at maturity
$\mathbb{Q}$ the risk neutral measure
$\mathbb{E}_{\mathbb{Q}}[.]$ the expectation operator under the risk neutral measure
$\Phi(.)$ the standard cumulative normal distribution
$\phi(.)$ the standard normal disctribution density function
$\mathbf{1}_\{expression\}$ the indicator function that is equal to $1$ if expression is true $0$ otherwise
$W_{t}$ the standard brownian motion
$f(T)$ the european option payoff
$f$ the european option value
$f_{c}$ the european call option value
$f_{p}$ the european put option value

Assumptions

There are no riskless arbitrage opportunities.
Security trading is continious.
The underlying asset follows the process: $dS_{t} = S_{t}(rdt + \sigma dW_{t})$ .
There are no dividends during the life of the derivative.
The risk free rate of interest $r$ and the volatility of the underlying asset $\sigma$ are constants.
The short selling of securities with full use of proceeds is permitted.
There are no transactions costs or taxes.
All securities are perfectly divisible.

Admitted propositions

Under the risk neutral probability measure the european option's price is a martingale : $f = e^{-rT}\mathbb{E_{\mathbb{Q}}}[f(T)]$
The solution of the stochastic differential equation follows by the underlying asset $dS_{t} = S_{t}(rdt + \sigma dW_{t})$ is given by:

$S_{T} = S_{0}\exp[{(r-\frac{\sigma^{2}}{2})T+\sigma W_{T}}]$

Proof

By the proposition 1 the call european value is given by:

$f_{c} = e^{-rT}\mathbb{E}_{\mathbb{Q}}[\max(S_{T}-K,0)]$

Taking into account the indicator function:

$1_{\{S_{T}\ge K\}}$

we obtain:

$f_{c} =e^{-rT}\mathbb{E}_{\mathbb{Q}}[(S_{T}-K)\mathbf{1}_{\{S_{T}\ge K\}}]$

By distributivity of the indicator function:

$f_{c} =e^{-rT}\mathbb{E}_{\mathbb{Q}}[S_{T}\mathbf{1}_{\{S_{T}\ge K\}} - K\mathbf{1}_{\{S_{T}\ge K\}}]$

By linearity of the expectation operator and distributivity of the discount factor:

$f_{c} =e^{-rT}\mathbb{E}_{\mathbb{Q}}[S_{T}\mathbf{1}_{\{S_{T}\ge K\}}]-e^{-rT}\mathbb{E}_{\mathbb{Q}}[K\mathbf{1}_{\{S_{T}\ge K\}}]$

The strike price is a constant :

$f_{c} =e^{-rT}\mathbb{E}_{\mathbb{Q}}[S_{T}\mathbf{1}_{\{S_{T}\ge K\}}]-Ke^{-rT}\mathbb{E}_{\mathbb{Q}}[\mathbf{1}_{\{S_{T}\ge K\}}]$

Finally, noting that:

$\mathbb{E}_{\mathbb{Q}}[\mathbf{1}_{\{S_{T}\ge K\}}] = \mathbb{Q}[S_{T}\ge K]$ ,

we can write:

$f_{c} =e^{-rT}\mathbb{E}_{\mathbb{Q}}[S_{T}\mathbf{1}_{\{S_{T}\ge K\}}]-Ke^{-rT}\mathbb{Q}[S_{T}\ge K]$

Finally we obtain an equality in wich the right hand side part is made out of two terms:

$e^{-rT}\mathbb{E}_{\mathbb{Q}}[S_{T}\mathbf{1}_{\{S_{T}\ge K\}}]$ and $Ke^{-rT}Q[S_{T}\ge K]$

Regarding the term:

$e^{-rT}\mathbb{E}_{\mathbb{Q}}[S_{T}\mathbf{1}_{\{S_{T}\ge K\}}]$

the assumption 3 and the proposition 2 imply that the event:

$S_{T} \ge K$

can also be written:

$S_{0}\exp[{(r-\frac{\sigma^{2}}{2})T+\sigma W_{T}}] \ge K$

This last equation is equivalent to:

$W_{T} \ge \frac{1}{\sigma}[\ln(\frac{K}{S_{0}})-(r-\frac{\sigma^{2}}{2})T]$

$W_{T}$ follows a centered normal distribution with a variance $T$ , thus the event:

$S_{T}\ge K$ ,

is equivalent to:

$U \ge \frac{1}{\sigma\sqrt{T}}[\ln(\frac{K}{S_{0}})-(r-\frac{\sigma^{2}}{2})T]$

Therefore:

$\mathbb{Q}[S_{T}\ge K] =\mathbb{Q}[U \ge \frac{1}{\sigma\sqrt{T}}[\ln(\frac{K}{S_{0}})-(r-\frac{\sigma^{2}}{2})T]]$

By a formal calculus we have:

$\mathbb{Q}[S_{T}\ge K] = 1- \mathbb{Q}[U \le \frac{1}{\sigma\sqrt{T}}[\ln(\frac{K}{S_{0}})-(r-\frac{\sigma^{2}}{2})T]]$

$\mathbb{Q}[S_{T}\ge K] = 1- \Phi[\frac{1}{\sigma\sqrt{T}}[\ln(\frac{K}{S_{0}})-(r-\frac{\sigma^{2}}{2})T]]$

$\mathbb{Q}[S_{T}\ge K] = \Phi[-\frac{1}{\sigma\sqrt{T}}[\ln(\frac{K}{S_{0}})-(r-\frac{\sigma^{2}}{2})T]]$

$\mathbb{Q}[S_{T}\ge K] = \Phi[\frac{1}{\sigma\sqrt{T}}[\ln(\frac{S_{0}}{K})+(r-\frac{\sigma^{2}}{2})T]]$

$\mathbb{Q}[S_{T}\ge K] = \Phi(d_{2})$

Thus :

$Ke^{-rT}\mathbb{Q}[S_{T}\ge K] = Ke^{-rT}\Phi(d_{2})$

with:

$\Phi(d_{2}) = \frac{1}{\sigma\sqrt{T}}[\ln(\frac{S_{0}}{K})+(r-\frac{\sigma^{2}}{2})T]$

As for the second term we have:

$\mathbb{E}_{\mathbb{Q}}[S_{T}\mathbf{1}_{\{S_{T}\ge K\}}]=\int_{S_{T}\ge K}S_{T}\phi(U)dU$

From what we have establish for the event :

$S_{T}\ge K$ ,

we can write:

$\mathbb{E}_{\mathbb{Q}}[S_{T}\mathbf{1}_{\{S_{T}\ge K\}}]=\int_{-d_{2}}^{\infty}S_{0}\exp[{(r-\frac{\sigma^{2}}{2}})T+\sigma\sqrt{T}U]\frac{1}{\sqrt{2\pi}}\exp({-\frac{U^{2}}{2}})dU$

$\mathbb{E}_{\mathbb{Q}}[S_{T}\mathbf{1}_{\{S_{T}\ge K\}}]=S_{0}e^{-rT}\int_{-d_{2}}^{\infty}\frac{1}{\sqrt{2\pi}}\exp({\sigma\sqrt{T}U}-\frac{U^{2}}{2}-\frac{\sigma^{2}}{2}T)dU$

Taking into account the change of variable:

$V = U-\sigma\sqrt{T}$ et $dV=dU$

and by a formal and rudimentary calculus, we obtain:

$\mathbb{E}_{\mathbb{Q}}[S_{T}\mathbf{1}_{\{S_{T}\ge K\}}]=S_{0}e^{-rT}\int_{-d_{2}-\sigma\sqrt{T}}^{\infty}\frac{1}{\sqrt{2\pi}}\exp({-\frac{V^{2}}{2}})dV$