The Black-Scholes formula

So far we have established that if a risk neutral measure is constructed, then the prices of all derivatives can be computed using discounted expectations under this measure, ruling out arbitrage. In addition we admitted that this risk neutral measure is not unique, therefore different risk neutral measures would yield different prices. All these measures are consistent, in the sense that they do not allow arbitrage opportunities, it is just that they trading days are not dense enough for the derivative securities to be perfectly replicated, since there is an infinite number of possibilities .

One [of these infinite] risk neutral measures can be constructed using the ideas of Girsanov [together with Cameron and Martin for the discrete case] that utilizes the Radon-Nikodym derivative of the market measure with respect to the risk neutral one. Once again, one has to stress that this is not a unique one, and most probably it is not the true one. It nevertheless serves as a trackable approximation.

We fix a constant $\theta$ and define the random variable

$$M_{n}=\exp \left\{ \sum_{k=1}^{n}\left( -\theta Y_{k}-\frac{1}{2}\theta ^{2}\right) \right\}$$   .

We observe the following properties

• $M_{n}$ is always nonnegative, $M_{n}\geq 0$; and

• $\mathbf{E}M_{n}=\exp \left\{ -\frac{n}{2}\theta ^{2}\right\} \mathbf{%% E}\exp \left\{ -\sum_{k=1}^{n}\theta Y_{k}\right\} =1$.

• If we define the function $Q:\Omega \rightarrow \left[ 0,1\right]$ with $Q\left( A\right) =\int_{A}M\left( \omega \right) dP\left( \omega \right)$ [or in the Radon-Nikodym notation $\frac{dQ}{dP}=M_{n}$] we will have of course $Q\left( \Omega \right) =\mathbf{E}\left[ M_{n}\vert\mathscr{F}%% _{0}\right] =1$, making $Q$ a probability measure.

• On the other hand, expectations under $Q$ are related with expectations under $P$ in the following way
  $$\mathbf{E}_{Q}X=\mathbf{E}_{P}\left[ M_{n}X\right]$$    for any random variable $$X$$.

• If $\left( Y_{1},\ldots ,Y_{n}\right)$ are independent standardized normals under $P$, then $\left( Y_{1}^{Q},\ldots ,Y_{n}^{Q}\right) =\left( Y_{1}+\theta ,\ldots ,Y_{n}+\theta \right)$ are independent standardized normals under $Q$.

The last observation can be verified from the moment generating function [under $Q$] of the vector $\left( Y_{1}^{Q},\ldots ,Y_{n}^{Q}\right)$, namely

$$\mathbf{E}_{Q}\exp \left\{ \sum_{k=1}^{n}u_{k}Y_{k}^{Q}\right\} = \mathbf{E}%% _{P}\exp \left\{ \sum_{k=1}^{n}u_{k}\left( Y_{k}+\theta \right) -\theta Y_{k}-\frac{1}{2}\theta ^{2}\right\}$$

$$= \exp \left\{ \sum_{k=1}^{n}\frac{1}{2}u_{k}^{2}\right\}$$

What we want is such a value of $\theta$ that will make $Q$ a risk neutral probability measure, or equivalently a value of $\theta$ that would make the discounted prices of the stock under $Q$ martingales, namely

$$\mathbf{E}_{Q}\left[ \frac{S_{k}}{R_{k}}\vert\mathscr{F}_{k-1}\right] =\frac{%% S_{k-1}}{R_{k-1}}\text{.}$$

We can write

$$\frac{S_{k}}{R_{k}}=\frac{S_{k-1}}{R_{k-1}}\exp \left\{ \sigma Y_{k}^{Q}-%% \frac{1}{2}\sigma ^{2}\right\}$$   , where $$Y_{k}^{Q}=Y_{k}+\frac{\mu -r}{%% \sigma }\text{.}$$

The quantity $\frac{\mu -r}{\sigma }$ is the familiar market price of volatility risk. What we really want is to find a measure under which the random variables $\left\{ Y_{k}^{Q}\right\} _{k=1}^{n}$ are independent standard normals. Apparently such a measure is produced from Girsanov's theorem if $\theta =\frac{\mu -r}{\sigma }$. In addition one can observe that under the risk neutral measure the stock price evolves according to

$$S_{k}=S_{k-1}\exp \left\{ \sigma Y_{k}^{Q}+\left( r-\frac{1}{2}\sigma ^{2}\right) \right\}$$

and does not depend on the mean return $\mu$.

Say now that we want to price a European call option that matures at time $n$, with strike price $K$. The [ $\mathscr{F}_{n}$-measurable] payoff will be $%% \left( S_{n}-K\right) ^{+}$. According to our previous discussion the price of the call today will be equal to

$$C=\mathbf{E}_{Q}\left[ \frac{\left( S_{n}-K\right) ^{+}}{R_{n}}\vert\mathscr{F}%% _{0}\right]$$   .

One can solve the above relationship

$$C = e^{-rn}\mathbf{E}_{Q}\left[ \left( S_{0}\exp \left\{ \sigma B_{n}... ...{1}{2}\sigma ^{2}\right) n\right\} -K\right) ^{+}\vert%% \mathscr{F}_{0}\right]$$

$$= \frac{e^{-rn}}{\sqrt{2\pi n}}\int_{-\infty }^{+\infty }\left( S_{... ...a ^{2}\right) n\right\} -K\right) ^{+}\exp \left\{ -\frac{x^{2}}{2n}\right\} dx$$

which gives as a result the Black-Scholes formula

$$C=S_{0}\Phi \left[ \frac{\ln \left\{ \frac{S_{0}}{K}\right\} +\le... ...\} +\left( r-\frac{1}{2}\sigma ^{2}\right) n}{\sigma \sqrt{n}}\right] \text{.}$$