April 25, 2024
The theory developed by Fischer Black, Myron Scholes, and Robert Merton III is the foundation of modern mathematical finance. They showed an option can be replicated by executing a dynamic hedge using a money market account and the underlying specified in the option payoff. The value of the option is the cost of setting up the initial hedge. Their theory relies heavily on Ito processes.
Assume a money market account with price R_t at time t satisfying dR = Rρ\,dt, R_0 = 1. For constant ρ the solution is R_t = e^{ρ t}. Assume a stock with price S_t at time t paying no dividends satisfying dS = S\mu\,dt + Sσ\,dB, S_0 = s. For constant \mu and σ the solution is S_t = se^{\mu t}e^{σ B_t - σ^2 t/2}.
At time t assume amount M_t is held in the money market account and N_t shares are held in the stock where M_t and N_t are Ito diffusions. The evolution of the portfolio value V = MR + NS is \begin{aligned} dV_t &= d(M_t R_t + N_t S_t) \\ &= (M_t dR_t + R_t dM_t + dM_t dR_t) + (N_t dS_t + S_t dN_t + dN_t dS_t) \\ &= M_t dR_t + dM_t(R_t + dR_t) + N_t dS_t + dN_t(S_t + dS_t) \\ \end{aligned}
The cost of re-hedging the position (M,N) at t to (M + dM, N + dN) at the prices prevailing at t + dt is dM(R + dR) + dN(S + dS). The self-financing condition is 0 = dM(R + dR) + dN(S + dS). For a self-financing strategy (M,N) dV = M\,dR + N\,dS = (MRρ + NS\mu)\,dt + {\color{blue} NSσ\,dB}.
The sum and product of Ito diffusions are Ito diffusions so V_t is an Ito diffusion and V_t = v(t, R_t, S_t) for some function v\colon[0,\infty)\times\boldsymbol{R}^2\to\boldsymbol{R}. Using the Ito formula and (dR)^2 = 0. We have \begin{aligned} dV_t &= v_t\,dt + v_r\,dR + v_s\,dS + \frac{1}{2} v_{ss}\,(dS)^2 \\ &= v_t\,dt + v_r\,Rρ\,dt + v_s\,(S\mu\,dt + Sσ\,dB) + \frac{1}{2} v_{ss}S^2σ^2\,dt \\ &= (v_t + v_r Rρ + v_s S\mu + \frac{1}{2} v_{ss}S^2σ^2)\,dt + {\color{blue} v_s Sσ\,dB} \\ \end{aligned}
Equating dt terms we see MRρ + NS\mu = v_t + v_r Rρ + v_s S\mu + \frac{1}{2} v_{ss}S^2σ^2. Since V = MR + NS we have (V - NS)ρ + NS\mu = v_t + v_r Rρ + v_s S\mu + \frac{1}{2} v_{ss}S^2σ^2. Equating {\color{blue} dB} terms we have N = v_s so (V - v_s S)ρ = v_t + v_r Rρ + \frac{1}{2} v_{ss}S^2σ^2.
since the NS\mu = V_s S\mu terms, remarkably, cancel. The PDE does not involve \mu. This key fact was one of the reasons Scholes and Merton were awarded a Nobel prize. They were the first to show valuing an option on a stock does not require estimating the growth rate of the stock.
Note the B-S/M PDE is a necessary condition of a self-financing strategy but is it also sufficient?
Note that the PDE above involves a v_r Rρ term. This term does not appear in the original Black and Scholes paper (BlaSch73?). It does appear in Merton’s original treatment (Mer73?). He also assumes the bond has a non-zero diffusion term to derive a PDE and then takes a limit as the diffusion coefficient tends to zero. His later treatment (Mer75?) showed this was not necessary.
Exercise. Define \bar{v}(t, s) = v(t, R_t, s) where R_t = e^{ρ t}. Show (\bar{V} - \bar{v}_s S)ρ = \bar{v}_t + \frac{1}{2} \bar{v}_{ss}S^2σ^2. where \bar{V}_t = \bar{v}(t, S_t).
Hint: What is the partial derivative of \bar{v} with respect to t?
Note that when ρ = 0 we have 0 = \bar{v}_t + \frac{1}{2} \bar{v}_{ss}S^2σ^2.
Replacing \mu by ρ in the stock price process is related to change of measure and involves Girsanov’s Theorem. A fundamental problem is specifying models that are arbitrage-free. A clever trader will find ways to book a “profit” given a model that is not arbitrage-free.
A simple way to do this is to specify a model of stock price where S_t/R_t is a martingale.
Exercise. Show S/R is a martingale if and only if dS/S = dR/R + σ\,dB.
Hint. Let dS/S = \mu\,dt + σ\,dB. Show d(S/R) = dS/R - (S/R)\,dR/R since (dR)^2 = 0. Use this to show d(S/R) = (S/R)((\mu - ρ)\,dt + σ\,dB).
Exercise. If S/R is a martingale, V = MR + NS, and (M, N) is self-financing then V/R is a martingale.
Hint. For any Ito process V we have d(V/R) = dV/R - (V/R)\,dR/R when (dR)^2 = 0. Substitute dV = M\,dR + N\,dS and V/R = M + NS/R to show d(V/R) = -NS\,dR/R + (N/R)\,dS. Use dS = S\,dR/R + Sσ\,dB.
There is no need to solve a PDE when using the martigale method.
If an option pays \nu(S_T) at expiration T and there is a self-financing strategy (M_t, N_t) with V_T = \nu(S_T) then the value of the option is the cost of the initial hedge, V_0. Since V_t/R_t is a martingale the cost is V_0 = E[\nu(S_T)/R_T]. This is a subtle but important point. The option value can be computed without specifying the details of the hedging process. However, it assumes a perfect hedge exists. In reality, no hedge is perfect and risk management involves measuring how imperfect it is. Given a money market account it is trivial to construct a self-financing hedge, but V_T is almost never exactly equal to \nu(S_T).
In the B-S/M model with constant rate and volatility the stock price S_t = se^{ρ t}e^{σ B_t - σ^2 t/2} is log-normal. As Fischer Black noted, mathematical formulas for option valuation become simpler when working with forward values instead of spot values since the interest rate disappears.
Let F_t = fe^{σ B_t - σ^2 t/2} where f = se^{ρ t} is the forward value of the stock. We can replace σ B_t by any random variable with mean zero and variance σ^2 t. Let X have mean zero and variance one and set s = σ\sqrt{t} We can value any option with expiration price F = fe^{sX - κ(s)}, where κ^X(s) = \log E[e^{sX}] is the cumulant of X,
Exercise. Show E[F] = f and \operatorname{Var}(\log F) = s^2.
Exercise. If F is a positive random variable then there exists a random variable X having mean zero and variance one with F = fe^{sX - κ(s)}.
Exercise. Show κ^X(s) = s^2/2 if X is standard normal.
If h(F) is the payoff of an option in shares of F then Fh(F) is the dollar amount of the payoff when converted at price F. It is trivial that E[Fh(F)] = E[F] E[(F/E[F])h(F)]. We can define share measure E_* by E_*[Y] = E[(F/E[F])Y]. Since F/E[F] > 0 and E[F/E[F]] = 1 this is a probability measure.
If F = fe^{sX - κ(s)} then F/E[F] = e^{sX - κ(s)} so share measure is E_*[Y] = E[e^{sX - κ(s)}Y] and we write E_s for E_*. This is also referred to as the forward measure or the Esscher transform.
Exercise. If X is standard normal show E_s[g(X)] = E[g(X + s)].
Hint: E[e^N g(M)] = E[e^N]E[g(M + \operatorname{Cov}(N, M))] if N and M are jointly normal.
A put with strike k pays \max\{k - F,0\} = (k - F)^+ at expiration. Its forward value p is \begin{aligned} p &= E[\max\{k - F, 0\}] \\ &= E[(k - F)1(F\le k)] \\ &= k P(F\le k) - E[F1(F\le k)] \\ &= k P(F\le k) - fP_s(F\le k) \\ \end{aligned} Moneyness x is defined by F = k if and only if X = x. The value of a put option in terms of the distribution for X is p = kP(X\le x) - fP_s(X\le x).
Exercise. Show moneyness x = x(k) = (\log(k/f) + κ(s))/s.
Hint. Solve k = fe^{sx - κ(s)} for x.
Exercise. Show F\le k if and only if X\le x.
Hint: dx/dk = 1/ks > 0.
Exercise. If X is standard normal show x = \log(k/f)/s + s/2.
Black and Scholes defined d_2 = \log(f/k)/s - s/2 where s = σ\sqrt{t}.
Exercise. Show -d_2 = x.
Exercise. If X is standard normal show P_s(X\le x) = P(X\le x - s).
Black and Scholes defined d_1 = \log(f/k)/s + s/2 where s = σ\sqrt{t}.
Exercise. Show -d_1 = x - s.
This shows the forward put value is p = k\Phi(-d_2) - f\Phi(-d_1) where \Phi is the standard normal cumulative distribution.
Since \max\{F - k, 0\} - \max\{k - F, 0\} = F - k we have c - p = f - k where c = E[\max\{F - k, 0\}] is the call value.
Exercise. Show the forward value of a call is c = fP_s(F\ge k) - kP(F\ge k).
Delta is the derivative of option value with respect to the forward. The delta of a put option is \begin{aligned} \partial_f p &= \partial_ f E[\max\{k - F, 0\}] \\ &= E[-1(F \le k)\partial_f F] \\ &= -E[1(F \le k)e^{sX - κ(s)}] \\ &= -P_s(F \le k) \\ \end{aligned} Where \partial_f p = \partial p/\partial f is the partial derivative of put value with respect to forward.
Exercise. Show \partial_ f E[\max\{k - F, 0\}] = -E[1(F \le k)]\partial_f F].
Hint Show d(\max\{k - x, 0\})/dx = -1(x\le k) and dF/df = e^{sX - κ(s)}.
Exercise. If X is standard normal then \partial_f v = -\Phi(-d_1).
The derivative of p = k\Phi(-d_2) - f\Phi(-d_1) with respect to f is \partial_f v = k\nu(-d_2)\partial_f(-d_2) - f\nu(-d_1)/\partial_f(-d_1) - \Phi(-d_1) where \nu = \Phi' is the standard normal density. The calculation showing the first two terms on the right hand side cancel is not trivial. Our more general approach avoids this.
Taking the derivative of c - p = f - k with respect to f we have \partial_f c - \partial_f p = 1 so call delta equals put delta plus one.
Gamma is the second derivative of option value with respect to the forward.
\begin{aligned} \partial_f^2 p &= -\partial_f P_s(F \le k) \\ &= -\partial_f P_s(X \le x) \\ &= -\partial_m P_s(X \le x)\partial_f x \\ &= -\partial_m P_s(X \le x)/\partial_x f \\ &= \partial_m P_s(X \le x)/f s\\ \end{aligned}
Vega is the derivative of option value with respect to volatility. The vega of a put option is \begin{aligned} \partial_s p &= \partial_s E[\max\{k - F, 0\}] \\ &= E[-1(F \le k)\partial_s F] \\ &= -E[1(F \le k)F(X - κ'(s))] \\ &= -\partial_s P_s(X \le x)f \\ \end{aligned}
Exercise. Show \partial_s P_s(X\le x) = E[1(X \le x)F(X - κ'(s))]/f.
Theta is the derivative of option value with respect to time. Using s = σ\sqrt{t} the theta of a put option is \begin{aligned} \partial_t p &= \partial_s p\,\partial_t s \\ &= \partial_s p\,σ/2\sqrt{t} \\ &= -\partial_s P_s(X \le x)fσ/2\sqrt{t}. \\ \end{aligned}
Traders think in terms of time decay of the option value so increasing time means shorter time to expiration and use the negative of this value, and sometimes divide that by 250 (business days per year) to approximate 1 day time decay.
Kappa is the derivative of option value with respect to strike \partial_k p = \partial_k E[\max\{k - F\}, 0\}] = P(F\le k), which is the cumulative distribution of the underlying. This implies the distribution of the underlying can be inferred from option prices (BreLin?).
Note value and all greeks can be expressed in terms of the share distribution function P_s(X\le x) and its derivatives with respect to x and s.
Market traded options have a price based on strike and expiration. Given a strike and expiration the Black-Sholes/Merton formula produces a value as a function of volatility. The volatility matching the market price is called the (Black-Scholes) implied volatility. It is used as a proxy for the price under the convention that the B-S/M formula is used for computation.
The implied volatility at a fixed expiration as a function of strike is the volatility curve for that expiration. If the B-S/M model of stock price dynamics were correct the volatility curve would be constant. It is a market fact that this is not the case.
Continuous time trading is physically impossible. The speed of light is 3\times 10^8 meters per second so it travels 0.3 meters, or approximately 1 foot, in a nanosecond. Electrons traveling over a wire are are much slower. In reality trades occur at a finite number of discrete times.
The B-S/M model assumes continuous time trading and stock price is geometric Brownian motion. These assumptions allow a mathematical proof that options can be perfectly hedged. In practice traders put on a hedge for some period of time before adjusting it based on market movements. One measure of risk is the difference between the actual value of the hedge at the end of the period and the theoretical model value.
For simplicity we assume zero funding rate so R_t = 1 and S_t is the price of a stock at time t. The market model is X_t = (R_t, S_t), C_t = 0, and the position is \Delta_t = (M_t, N_t).
A European option paying \nu(S_T) at expiration T has model value \bar{V}_t = E_t[\nu(S_T)], for t\le T. Let \bar{v}(t, s) = E[\bar{V}_t\mid S_t = s]. Note \bar{V}_t is a stochastic process and \bar{v}(t,s) is a function. The risk involved with discrete time hedging over the interval [t, t + \Delta t] is the difference V_{t + \Delta t} - \bar{V}_{t + \Delta t}.
Let \bar{v}(S, T) = E[\nu(S e^{σ B_T - σ^2T/2})]. Note \bar{V} is a stochastic process and \bar{v} is a function.
Exercise. Show E_t[\nu(S_T)\mid S_t = S] = \bar{v}(S, T - t).
Note \partial_t E_t[\nu(S_T)\mid S_t = S] = -\partial_t \bar{v}(S, T - t).
Suppose you are given V_0 in cash and it is now your job to replicate the option payoff. At each time t_j you can use the money market account to put on a position of N_j shares in the stock. The value of your hedge at time t_j is V_j = M_j + N_j S_j. Your initial position N_0 in the stock will be deducted from V_0 and result in a money market position M_0 = V_0 - N_0 S_0.
At time t_j, 0 < j < n, you change the stock position from N_{j-1} to N_j. Using the money market account to finance this results in M_j = M_{j-1} - (N_j - N_{j-1})S_j; the cost of the shares purchased are deducted based on the prevailing stock price.
At time t_n = T the position in the stock is closed out so the accumulated shares N_{n-1} are sold and the money market position becomes M_n = M_{n-1} + N_{n-1}S_n.
A successful hedge would result in V_T = V_n = M_n R_n = \nu(S_T). How should you choose N_0,\ldots,N_{n-1}?
The progress of a hedge over time can be tracked by comparing the hedge value V_t = M_t + N_t S_t to the model value \bar{V}_t = \bar{v}(S_t, T - t).
The difference between the hedge value and the model value is the profit and loss. The change in the hedge value over the interval [t_{j-1},t_j] is the backward difference \nabla V_j = V_j - V_{j-1}.
Exercise. Show \nabla V_j = N_{j-1}\nabla S_j.
The change in model value over the interval is \nabla\bar{V}_j = \bar{V}_j - \bar{V}_{j-1}. Since σ and T are fixed this can be computed using Taylor’s formula \begin{aligned} \nabla \bar{V}_j &= -\partial_t\bar{v}_j\,\nabla t_j + \partial_s\bar{v}_j\,\nabla S_j \\ &\quad + \frac{1}{2}\partial_t^2\bar{v}_j\,(\nabla t_j)^2 - \partial_t\partial_s \bar{v}_j(\nabla t_j)(\nabla S_j) + \frac{1}{2}\partial_s^2\bar{v}_j(\nabla S_j)^2 \\ &\qquad + \mathrm{higher\ order\ terms} \end{aligned} where the partial derivatives are evaluated at t = t_{j-1} and S = S_{j-1}. Ignoring (\nabla t_j)(\nabla S_j) = 0, (\nabla t_j)^2 = 0, and higher order terms we have
\begin{aligned} \nabla V_j - \nabla \bar{V}_j &= N_{j-1}\nabla S_j - \nabla \bar{V}_j \\ &\approx N_{j-1}\nabla S_j - (-\partial_t\bar{v}_j\,\nabla t_j + \partial_s\bar{v}_j\,\nabla S_j + \frac{1}{2}\partial_s^2\bar{v}_j(\nabla S_j)^2) \\ &= \partial_t\bar{v}_j\,\nabla t_j - \frac{1}{2}\partial_s^2\bar{v}_j(\nabla S_j)^2 \\ \end{aligned} if we choose N_{j-1} = \partial_s\bar{V}_j. The P&L on a delta hedged trade can be approximated using theta and gamma. The term (1/2)\partial_s^2\bar{v}_j (\nabla S)^2 is called dollar gamma.
Exercise. Show \nabla V_j = N_{j-1}\nabla S_j.
Exercise. Show if N, M, and K are jointly normal then E[e^N 1(M \ge K)] = E[e^N] P(M + \operatorname{Cov}(N,M) \ge K + \operatorname{Cov}(N,K)).
Hint. Use E[e^N g(N_1,\ldots)] = E[e^N] E[g(N_1 + \operatorname{Cov}(N, N_1), \ldots)].
Exercise. (Margrabe) Show if N and M are jointly normal then E[\max\{e^N - e^M, 0\}] is E[e^N] P(N - M \ge - \operatorname{Var}(N) + \operatorname{Cov}(N,M)) - E[e^M] P(N - M \ge - \operatorname{Cov}(M, N) + \operatorname{Var}(M)).
Hint. E[\max\{e^N - e^M, 0\}] = E[(e^N - e^M) 1(N \ge M)].
Exercise. Show E[\max\{S_T - S_t, 0\}] = s (Φ(σ\sqrt(T - t)/2) - Φ(-σ\sqrt(T - t)/2)) where Φ is the standard normal cumulative distribution.
Hint. Recall S_t = se^{σB_t - σ^2 t/2} and t < T then use the previous exercise.