5 The Black–Scholes formula

5.1 Asset price behaviour

We have seen how to price contingent claims using the binomial model in terms of the given parameters $u,d,r$, etc. This is of course only of practical use if the binomial model gives a reasonable approximation to real stock price behaviours. In this chapter we look at the how well the binomial model matches observed data and look at some particular choices of $u,d, p_u$ that give rise to the Black–Scholes formula for pricing a European call option.

We’d like for our binomial model to match the observed behaviour of stock prices, that the simple returns \[ \frac{S_{t+\delta t} - S_t}{S_t} \] over short time periods of length $\delta t$ are approximately Normally distributed, and have distribution $\cN(m \delta t, \sigma^2 \delta t)$, for some values $m$ and $\sigma^2$ that depend on the stock and other features of the market.

A common way to achieve this is to make the assumption that the price $S_t$ is distributed as a geometric Brownian motion with drift $\mu$ and volatility $\sigma > 0$. Specifically, \[\begin{equation} \log \frac{S_{u+t}}{S_u} \sim \cN\big((\mu - \sigma^2/2)t, \sigma^2 t\big) \tag{5.1} \end{equation}\] for all $u \geq 0, t > 0$ and additionally $S_{u+t}/S_u$ is independent of $S_v, v\leq u$.

Next term we’ll study the theory of Brownian motion and this continuous-time model of stock prices in more detail; here, we just try to make our discrete-time binomial model approximate the geometric Brownian motion as well as it can.

5.2 Tuning the binomial model

To do this, we start by breaking our time interval $[0,T]$ into $n$ steps, each of size $\Delta t = T/n$. We’ll be letting $n \to \infty$, so we should think of $n$ being large and $\Delta t$ being small. This is a slight difference from how we set up the binomial model before, where the time steps between nodes of the tree were all of size 1, but it does not really complicate matters very much, as long as we use the correct value for the interest rate per time step. We want the interest rate per unit time to be $r > 0$, so we take the interest rate over the small time step of size $\Delta t$ to be $r\Delta t$. This means the price $B_t$ of the risk-free asset is given by \[ B_{k\Delta t} = (1+r\Delta t)^k, \quad\text{for $k=0,\dots,n$}. \]

Other than this difference, the binomial model is exactly the same; there are constants $u > d$ and objective probabilities $p_u, p_d$ which specify how the price of the risky asset changes. Specifically, at each of the $n$ steps the share price is multiplied by either $u$ or $d$ (with probability $p_u$ or $p_d$), so that at time $T$ (after $n$ steps) the share price is \[ S_T = S_0 u^Yd^{n-Y},\] where $Y \sim \Bin(n,p_u)$. Note that this is a description of $S_T$ under the objective measure given by $p_u$ and $p_d$. We will make use of the martingale measure later.

To match the volatility, we choose $u = \e^{\sigma \sqrt{\Delta t}}$ and $d = \e^{-\sigma \sqrt{\Delta t}}$ for some $\sigma > 0$.

Remark. Note that for $\e^{-\sigma \sqrt{\Delta t}} < 1+r\Delta t<\e^{\sigma \sqrt{\Delta t}}$ to hold it is enough that $r\Delta t < \sigma \sqrt{\Delta t}$ and this holds if $n > T(r/\sigma)^2$. In other words, for large enough $n$ the market with these parameters is arbitrage-free and we can sensibly price contingent claims on this market.

How well do these parameters approximate geometric Brownian motion? We see that \[ \log \frac{S_T}{S_0} = Y \log{\frac{u}{d}} + n \log d = \sigma \sqrt{\Delta t} (2Y - n) = \sigma\sqrt{T}\frac{2Y-n}{\sqrt{n}}. \] Now, since $Y = \sum_{i=1}^n X_i$ is a sum of i.i.d. random variables, with $\Var{X_1}<\infty$ we can apply the Central Limit Theorem, which says that \[ \frac{\frac{1}{n}\sum_{i=1}^n X_i - \E(X_1)}{\sqrt{\frac{1}{n}\Var(X_1)}} = \frac{ Y - \E(Y) }{\sqrt{\Var(Y)}} \to \cN(0,1) \] in distribution as $n \to \infty$.

At this point we specify $p_u$. What happens if we choose $p_u = 1/2$? Then $\E(Y) = n p_u = n/2$ and $\Var(Y) = np_u(1-p_u) = n/4$, which means that \[ \frac{ Y - \E(Y) }{\sqrt{\Var(Y)}} = \frac{1}{\sigma\sqrt{T}} \log \frac{S_T}{S_0} \to \cN(0,1). \] In other words, $\log (S_T/S_0)$ is approximately $\cN(0,\sigma^2 T)$. This is a good first attempt, as we’ve managed to emulate the volatility $\sigma$ using our binomial model, but this choice of $p_u$ doesn’t give us a general drift term. To get that we need to use a different value of $p_u$, which is a slight perturbation from $p_u=1/2$.

If we choose $p_u = 1/2 + (\mu - \sigma^2/2)\sqrt{\Delta t}/2\sigma$, then we have \[ \E(Y) = np_u = n/2 + \sqrt{n} (\mu-\sigma^2/2)\sqrt{T}/2\sigma, \quad \Var(Y) = n p_u (1-p_u) = n/4 + O(1), \] so for this choice of $p_u$ we find that \[ \frac{ Y - \E(Y) }{\sqrt{\Var(Y)}} \approx \frac{1}{\sigma\sqrt{T}} \log \frac{S_T}{S_0} - (\mu-\sigma^2/2)\sqrt{T}/\sigma \to \cN(0,1), \] meaning that $\log(S_T/S_0)$ is distributed Normally as $\cN\big((\mu-\sigma^2/2)T, \sigma^2 T\big)$, asymptotically as $n \to \infty$.

Aside. Actually, we haven’t quite shown that the price $S_t$ is distributed as a geometric Brownian motion; we need to show rather more, that $\log( S_{t+u} / S_u) \sim \cN\big( (\mu-\sigma^2/2)t, \sigma^2 t\big)$ for all times $u$ and $t>0$. This is in fact true but requires rather too much advanced probability for this course. The general scheme is (i) show convergence to a vector of suitable random variables $( S_{t_1}, S_{t_2}, \dots, S_{t_n})$ for any finite $n$; (ii) prove that Brownian motion paths are almost surely continuous (this is the hard bit); (iii) use path continuity and convergence at finite sets of times to show convergence of the trajectories to those of Brownian motion (this needs basic ideas about distances between functions).

5.3 Risk-neutral drift

We have chosen parameters \[\begin{equation} u = \e^{\sigma\sqrt{\Delta t}},\quad d=\e^{-\sigma\sqrt{\Delta t}}, \quad p_u = \frac12 + \frac{(\mu - \sigma^2/2)\sqrt{\Delta t}}{2\sigma} \tag{5.2} \end{equation}\] so that under the objective probabilities $p_u, p_d$, the random variable $\log (S_T/S_0)$ is approximately Normally distributed. However, to calculate the arbitrage-free price of a derivative, we need to use the martingale probabilities $q_u, q_d$, where \[ q_u = \frac{1+r\Delta t - d}{u-d} = \frac{1 + r\Delta t - \e^{-\sigma\sqrt{\Delta t}}}{\e^{\sigma\sqrt{\Delta t}} - \e^{-\sigma\sqrt{\Delta t}} }, \] (and $q_d = 1- q_u$).

Applying Taylor’s theorem to $\e^x = 1+x+x^2/2+\dotsm$, we get \[ \begin{split} q_u %&= \frac{1 + r\Delta t - (1-\sigma\sqrt{\Delta t}+\sigma^2\Delta t/2 + \dotsm)}{(1+\sigma\sqrt{\Delta t}+\sigma^2\Delta t/2 + \dotsm)-(1-\sigma\sqrt{\Delta t}+\sigma^2\Delta t/2 + \dotsm)}\\ &= \frac{\sigma \sqrt{\Delta t}+ (r - \sigma^2/2)\Delta t + o(\Delta t)}{2\sigma \sqrt{\Delta t}+ o(\Delta t)} = \frac{1}{2} + \frac{(r-\sigma^2/2)\sqrt{\Delta t}}{2\sigma} + o(\sqrt{\Delta t}). \end{split} \] We see that (ignoring the error term) the expression for $q_u$ looks like the expression for $p_u$ but with $\mu$ replaced by $r$. Consequently, this means that under the martingale measure $\log (S_T/S_0)$ is also asymptotically Normally distributed, but as $\cN\big( (r-\sigma^2/2)T, \sigma^2 T\big)$. In other words, under the martingale measure the price $S_t$ is again a geometric Brownian motion with the same volatility $\sigma$, but with drift $r$, which is called the risk-neutral drift.

In summary, we have shown that it is possible to make the binomial model emulate observed stock price behaviour by choosing $u,d$ and $p_u$ according to (5.2). Furthermore if we use the risk-neutral probabilities $q_u,q_d$ in the binomial model then there is a corresponding risk-neutral drift for the geometric Brownian motion model. This means that if we calculate the price $\Pi_0(\Phi)$ of a contingent claim $\Phi(S_T)$, as given by Theorem 3.2 (using interest rate $r\Delta t$ per step), \[ \Pi_0(\Phi) = \frac{1}{(1+r\Delta t)^n} \E_\Q [ \Phi( S_T) ] \] and let $n \to \infty$, we get the limiting expression \[ \e^{-rT} \E[ \Phi(S_0 \e^W) ] \] where $W \sim \cN\big((r-\sigma^2/2)T, \sigma^2 T\big)$.

To price a European call option, we use this expression with $\Phi(S_T) = (S_T-K)^+$, and we find an expression in terms of an expectation under the risk-neutral measure for the cost $C(K,T,\sigma,S,r)$ of a European call option, which depends on the strike price $K$, expiry date $T$, the volatility $\sigma$ of the stock, the current share price $S$ and the interest rate $r$: \[\begin{equation} C(K,T,\sigma,S,r) = \e^{-rT} \E[(S\e^W-K)^+], \tag{5.3} \end{equation}\] where $S = S_0$ and $W = \log(S_T/S) \sim \cN\big((r-\sigma^2/2)T, \sigma^2 T\big)$.

Remark. In fact, there is a slight gap in our reasoning above. The formula (5.3) should contain a term of the form $\lim_{n \to \infty} \E[X_n]$ but we are claiming it equals $\E[ \lim_{n\to\infty} X_n]$. We know that for arbitrary random variables these need not be the same (for example, the sequence of random variables $X_n$ on state space $\Omega = (0,1)$ with the uniform measure defined by \[ X_n(\omega) = \begin{cases} n &\text{if $\omega < 1/n$},\\ 0 &\text{otherwise}, \end{cases} \] has $\E[X_n] = 1$ for all $n$, but $\lim_{n\to\infty} X_n(\omega) = 0$ for all $\omega \in (0,1)$ so $\E[\lim_{n\to\infty} X_n] = 0$), but in our case the random variables in question are well-behaved enough to mean that we can push this limit inside our expectation.

The expectation formula for the call option price can be evaluated with a straightforward computation involving the Normal distribution. Let’s start with the answer and how to use it and then see how to find the answer.

5.4 The formula

The formula $C = \e^{-rT}\E[(S\e^W - K)^+]$ stated in (5.3) can be written the form \[\begin{equation} C = S \PhiN(d_1) - Ke^{-rT} \PhiN\bigl(d_1 - \sigma\sqrt{T} \bigr) \quad \mbox{where} \quad d_1 = \frac{(r + \sigma^2/2)T + \log (S/K)}{\sigma\sqrt{T}} \tag{5.4} \end{equation}\] and $\PhiN$ denotes the standard Normal cdf given by \[ \PhiN(x) = \int_{-\infty}^x \phi(z) \ud z = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^x \e^{-z^2/2} \ud z \] (This is also commonly denoted by $\Phi$ but we’ve already used $\Phi$ as the notation for a general contract function, so we’ll use $\PhiN$ for the Normal cdf to avoid confusion.) Some text books define $d_2 = d_1 - \sigma\sqrt{T}$, for the input to $\PhiN$ in the last term of the formula.

Example 5.1 The current price of a stock is $S = 20$, its volatility is $\sigma = 0.1$, the nominal interest rate is $r = 0.06$ (or 6% per year). The no-arbitrage price of a call option with strike price $K = 21$ which expires in 3 months is found as follows.

With $T = 0.25$, we have \[ d_1 = \frac{0.015 + 0.00125 + \log 20/21}{0.1 \times 0.5} \approx \frac{-0.032540}{0.05} = -0.65080 \] (remember that we use natural logs) and hence \[ C \approx 20\PhiN(-0.65080) - 21e^{-0.015}\PhiN(-0.70080) \approx 5.152 - 5.000 = 0.152 \] i.e. just about $15.2$ pence (if the stock value is pounds).

Remark. There are a few important points to note.

If we wish to buy/sell a $(K, T)$ call option at time $t \in (0, T)$ when the stock price is $S_t$ then the appropriate price is $C(K, T-t, \sigma, S_t, r)$. This form is commonly encountered as option expiry dates are concentrated into just a few dates in the year so generally we look at dependence upon the variable $t$ rather than $T$. Modify $d_1$ from (5.4) by replacing $T$ with $T-t$.
Using put-call option parity the no-arbitrage price of a European $(K, T)$ put option satisfies \[ P = C(K, T, \sigma, S, r) + Ke^{-rT} - S = Ke^{-rT}\PhiN(\sigma \sqrt{T} - d_1) - S \PhiN(-d_1) \]
The underlying mean drift $\mu$ of the stock price does not appear in the Black–Scholes formula as we use the risk-neutral drift but the volatility $\sigma$ does appear and we have to find a value for this somehow! More on this later.
We have derived this formula via a limit using risk-neutral probabilities on an artificial tree model which we ‘tuned.’ In fact the original argument was rather different using self-financing portfolios and delta hedging.

5.5 Calculating the expectation

We start by introducing the indicator function that the option is worth something at time $T$, i.e., \[ I_K = \begin{cases} 1 & \text{if $S_T > K$},\\ 0 & \text{otherwise}. \end{cases} \] Also suppose that $t = 0$ for the moment. In what follows, we write $\E_r$ to remind ourselves that we are taking the expected value using the risk-neutral drift. Then, recalling that $S_T = Se^W$ we have \[ \E_r[(S\e^W - K)^+] = \E_r[I_K(S\e^W - K)] = S\E_r[I_K\e^W] - K\E_r[I_K] \]

Lemma 5.1 We have \[ \E_r[(S\e^W - K)^+] = \e^{rT}S\PhiN(d_1) - K\PhiN\bigl(d_1 - \sigma\sqrt{T} \bigr) \] where $d_1 = [(r + \sigma^2/2)T + \log S/K ]/\sigma\sqrt{T}$ and $\PhiN$ is the standard Normal cdf.

Proof. Under the risk-neutral drift, $\log (S_T/S) = W \sim \cN\big((r - \sigma^2/2)T, \sigma^2 T\big)$ so (key trick!) we can write it in the form \[\begin{equation} W = (r - \sigma^2/2)T + \sigma\sqrt{T} \cdot Z \tag{5.5} \end{equation}\] where $Z \sim N(0, 1)$. Now \[ I_K = 1 \Leftrightarrow S\e^W > K \Leftrightarrow (r - \sigma^2/2)T + \sigma\sqrt{T} Z > \log (K/S) \Leftrightarrow Z > \sigma\sqrt{T} - d_1. \] From this equivalence and the symmetry of the Normal density we get \[\begin{equation} \E_r[I_K] = \P_r(S\e^W > K) = \P(Z < d_1 - \sigma\sqrt{T}) = \PhiN(d_1 - \sigma\sqrt{T}). \tag{5.6} \end{equation}\] It remains to evaluate $\E_r(I_K \e^W)$. Recalling that the standard Normal density is $\phi(z) = \e^{-z^2/2}/\sqrt{2\pi}$ we have \[\begin{eqnarray} \E_r[I_K \e^W] & = &\int_{\sigma\sqrt{T} - d_1}^\infty \exp \bigl[ (r - \sigma^2/2)T + z\sigma\sqrt{T} \bigr] \phi(z)\,\ud z \nonumber \\ & = &\e^{rT} \frac{1}{\sqrt{2\pi}} \int_{\sigma\sqrt{T} - d_1}^\infty \exp \bigl[-(z^2 - 2z\sigma\sqrt{T} + \sigma^2T)/2 \bigr] \ud z \nonumber \\ & = &\e^{rT} \frac{1}{\sqrt{2\pi}} \int_{-d_1}^\infty \e^{-y^2/2} \,\ud y \hspace{3cm} (\mbox{set } y = z - \sigma\sqrt{T}) \nonumber \\ & = &\e^{rT} \P(Z > -d_1) = \e^{rT} \PhiN(d_1) \tag{5.7} \end{eqnarray}\] Combining these two parts together completes the proof.

Now, multiplying the statement of Lemma 5.1 by $\e^{-rT}$ gives us (5.4), the Black–Scholes formula.

Exercise: check that we can just replace $T$ by $T-t$ throughout if the option is purchased at time $t$ rather than 0.

5.6 Properties of $C$

We finish this chapter by stating some properties of $C$. We describe the behaviour of $C(K,T,\sigma, S,r)$ as the various inputs change, some of which we can see from the form of the Black–Scholes formula. We can show that $C$ is

decreasing and convex in $K$;
increasing in $T$ and hence $C(K, T-t, \sigma, S, r)$ is decreasing in $t$;
increasing and convex in $S$;
increasing in $\sigma$ and increasing in $r$.

In fact, the behaviour of $C$ as $K$ or $T$ changes does not depend upon the geometric Brownian motion model, and can be deduced by comparing the appropriate portfolios. The properties of $P$, the European put price follow directly from those of $C$ and the put-call parity formula.

These properties can also be deduced by calculating various partial derivatives of $C(K,T-t,\sigma,S,r)$. Some of these derivatives have special names given by Greek letters, and they are collectively known as “the Greeks.” We’ll look more at these next term.

Exercise: calculate $\pa C/\pa x$ with respect to $x = S$, $K$, $t$, $r$, $\sigma$ directly from (5.4). Their expressions involve the standard Normal cdf $N$ and pdf $\phi$, e.g., \[ \frac{\pa C}{\pa r} = K(T-t)e^{-r(T-t)}N(d_1 - \sigma\sqrt{T-t}) > 0. \]

It is interesting to observe that the price $C=C(K,T-t,\sigma,S,r)$ satisfies the following partial differential equation \[ \frac{\pa C}{\pa t} + rS\frac{\pa C}{\pa S} + \frac12\sigma^2S^2\frac{\pa^2 C}{\pa S^2} - rC = 0. \] We shall see this PDE again next term, when we study Itô calculus.

Exercise: show that $C(K, T-t,\sigma, S, r) = \alpha C(K/\alpha, T-t, \sigma,S/\alpha, r)$ for $\alpha > 0$ — this scaling property is entirely natural as we should find equivalent prices whether we choose to pay in units of pence, pounds or 100 pounds.

Mathematical Finance 2022/23 (Michaelmas)