Stochastic Processes and Financial Mathematics
(part two)

15.5 The Black-Scholes formula

In principle, Theorem 15.3.1 tells us the arbitrage free price, under the Black-Scholes model, of any contingent claim (such that \(\E ^\Q [\Phi (S_T)]\) exists). In many cases the only way to evaluate (15.13) is with numerics. However, in some cases it is possible to derive an explicit formulae.

In this section, we find explicit formulae for the case of European call options. The result is often referred to as the Black-Scholes formula.

We begin from the formula given to us by Theorem 15.3.1: the price of the contingent claim \(\Phi (S_T)\) at time \(0\) is

\begin{equation} \label {eq:call_rnv} F(t,S_t)=e^{-r(T-t)}\E ^\Q \l [\Phi (S_T)\|\mc {F}_t\r ]. \end{equation}

Here, since we take expectation in the risk-neutral world \(\Q \), the stock price process \(S_t\) has dynamics

\[dS_t=rS_t\,dt+\sigma S_t\,dB_t.\]

From (13.10) we have

\begin{align} S_T &=S_t\exp \l ((r-\tfrac 12\sigma ^2)(T-t)+\sigma (B_T-B_t)\r )\label {eq:gbm_tT}\\ &=S_te^Z\label {eq:gbm_Z} \end{align} where \(Z\sim N\l [(r-\tfrac 12\sigma ^2)(T-t),\sigma ^2(T-t)\r ]=N[u,v^2]\) is independent of \(\mc {F}_t\). Here, we use \(u\) and \(v\) to keep our notation manageable.

Before we attempt to price a call option, let us work through a simpler case.

Using the (15.19) in combination with either (15.20) or (15.21) is usually the best way to compute explicit pricing formulae. See exercises 15.2-15.5 for more examples in the same style.

We’ll give one more example in these notes. Consider the case of a call option with strike price \(K\) and exercise date \(T\). This case is not easy, and it will involve some hefty calculations because of the \(\max \) present in the contingent claim:

\[\Phi (S_T)=\max (S_T-K,0).\]

We are now looking to evaluate

\begin{align*} F(t,S_t) &=e^{-r(T-t)}\E ^\Q [\Phi (S_T)\|\mc {F}_t] \\ &=e^{-r(T-t)}\E ^\Q [\Phi (S_te^Z)\|\mc {F}_t], \end{align*} where we use (15.21). Writing \(s=S_t\in m\mc {F}_n\) and using the probability density function of \(Z\) gives us

\begin{align} F(t,s) &=e^{-r(T-t)}\int _{-\infty }^\infty \Phi (se^z)f_Z(z)\,dz\notag \\ &=e^{-r(T-t)}\l (0+\int _{\log (K/s)}^\infty (se^z-K)f_Z(z)\,dz\r )\notag \\ &=e^{-r(T-t)}\int _{\log (K/s)}^\infty (se^z-K)\frac {1}{\sqrt {2\pi }v}e^{-\frac {(z-u)^2}{2v^2}}\,dz\notag \\ &=\,\stackrel {\mc {A}}{\overbrace {e^{-r(T-t)}\int _{\log (K/s)}^\infty se^z\frac {1}{\sqrt {2\pi }v}e^{-\frac {(z-u)^2}{2v^2}}\,dz}} -\stackrel {\mc {B}}{\overbrace {e^{-r(T-t)}\int _{\log (K/s)}^\infty K\frac {1}{\sqrt {2\pi }v}e^{-\frac {(z-u)^2}{2v^2}}\,dz}}\label {eq:bs_form_pre} \end{align} Here, to deduce the second line, we split the integral into the cases \(se^Z< K\) and \(se^Z\geq K\). In the first case, \(\Phi (se^Z)=\max (se^Z-K,0)=0\).

The strategy now is to treat \(\mc {A}\) and \(\mc {B}\) separately. We plan to re-write each of \(\mc {A}\) and \(\mc {B}\) in terms of \(\mc {N}(z)=\frac {1}{\sqrt {2\pi }}\int _{-\infty }^z e^{-x^2}\,dx\), the cumulative distribution function of the \(N(0,1)\) distribution.

Let’s look at the the \(\mc {B}\) term first. Setting \(y=-z\), we have

\begin{align*} \mc {B} &=\frac {K}{\sqrt {2\pi } v} e^{-r(T-t)}\int _{-\log (K/s)}^{-\infty } e^{-\frac {(-y-u)^2}{2v^2}}(-1)\,dy\\ &=\frac {K}{\sqrt {2\pi } v} e^{-r(T-t)}\int ^{\log (s/K)}_{-\infty }e^{-\frac {(y+u)^2}{2v^2}}\,dy \end{align*} and setting \(x-\frac {y+u}{v}\) we have

\begin{align*} \mc {B} &=\frac {K}{\sqrt {2\pi } v} e^{-r(T-t)}\int ^{\frac {1}{v}\l (\log (s/K)+u\r )}_{-\infty }e^{-\frac {x^2}{2}}v\,dx\\ &=Ke^{-r(T-t)}\int ^{\frac {1}{v}\l (\log (s/K)+u\r )}_{-\infty }\frac {1}{\sqrt {2\pi }}e^{-\frac {x^2}{2}}\,dx\\ &=Ke^{-r(T-t)}\,\mc {N}\l [\frac {1}{v}\l \{\log \l (\frac {s}{K}\r )+u\r \}\r ]. \end{align*}

For the \(\mc {A}\) term, we have an extra factor \(e^{z}\). To handle this factor we will have to complete the square inside the exponential term before we make the \(x\) substitution (i.e. the same technique as in exercise 11.5). Again setting \(y=-z\) we have

\begin{align*} \mc {A} &=\frac {s}{\sqrt {2\pi } v}e^{-r(T-t)}\int _{-\log (K/s)}^{-\infty } e^{-y}e^{-\frac {(-y-u)^2}{2v^2}}(-1)\,dy\\ &=\frac {s}{\sqrt {2\pi } v}e^{-r(T-t)}\int ^{\log (s/K)}_{-\infty } e^{-y}e^{-\frac {(y+u)^2}{2v^2}}\,dy\\ &=\frac {s}{\sqrt {2\pi } v}e^{-r(T-t)}\int ^{\log (s/K)}_{-\infty } \exp \l (\frac {-1}{2v^2}\l [y^2+(2u+2v^2)y+u^2\r ]\r )\,dy\\ &=\frac {s}{\sqrt {2\pi } v}e^{-r(T-t)}\int ^{\log (s/K)}_{-\infty } \exp \l (\frac {-1}{2v^2}\l [\l (y+(u+v^2)\r )^2-(u+v^2)^2+u^2\r ]\r )\,dy\\ &=\frac {s}{\sqrt {2\pi } v}e^{-r(T-t)+\frac {(u+v^2)^2-u^2}{2v^2}}\int ^{\log (s/K)}_{-\infty }\exp \l (-\frac {(y+(u+v^2)^2)}{2v^2}\r )\,dy \end{align*} Noting that \(r(T-t)=u+\frac 12v^2\), it is easily seen that \(-r(T-t)+\frac {(u+v^2)^2-u^2}{2v^2}=0\), so the first exponential term is simply equal to \(1\). Setting \(x=\frac {y+(u+v^2)}{2v^2}\) we obtain

\begin{align*} \mc {A} &=\frac {s}{\sqrt {2\pi } v}\int _{-\infty }^{\frac {1}{v}\l (\log (s/K)+u+v^2\r )}\exp \l (-\frac {x^2}{2}\r )v\,dx\\ &=s\int _{-\infty }^{\frac {1}{v}\l (\log (s/K)+u+v^2\r )}\frac {1}{\sqrt {2\pi }}\exp \l (-\frac {x^2}{2}\r )\,dx\\ &=s\,\mc {N}\l [\frac {1}{v}\l \{\log \l (\frac {s}{K}\r )+u+v^2\r \}\r ]. \end{align*}

Substituting back in for \(u\) and \(v\), and recalling that we use the shorthand \(s=S_t\), we obtain the following formula for price, at time \(t\in [0,T]\), of a European call option with strike price \(K\) and exercise date \(T\).

\begin{equation} \label {eq:bs_formula} F(t,S_t)= S_t\,\mc {N}[d_1]-Ke^{-r(T-t)}\,\mc {N}[d_2] \end{equation}

where