April 25, 2024
Let I be the set of instruments available over a time period and \Omega be the set of possible outcomes. The prices at the beginning of the period are a vector x\in\boldsymbol{R}^I. The prices at the end of the period are a function X\colon\Omega\to \boldsymbol{R}^I where X(\omega)\in\boldsymbol{R}^I are the instrument prices if \omega\in\Omega occurred.
For example, the binomial model has a bond and stock where x = (1, s), X(\omega) = (R, \omega), and \Omega = \{S^-, S^+\} where R > 0, and {S^- < S^+}. The bond has realized return R and the stock can go from price s to either S^- or S^+. A slightly more realistic model is {\Omega = [S^-, S^+]} where the final stock price can be any value in the interval.
A position \xi\in\boldsymbol{R}^I is the number of shares held in each instrument. The cost of acquiring the initial position is \xi\cdot x. The value of liquidating the final position is \xi\cdot X(\omega) if \omega\in\Omega occurred.
Arbitrage exists in a one-period model if there is a position \xi\in\boldsymbol{R}^I with \xi\cdot x < 0 and \xi\cdot X(\omega)\ge0 for all \omega\in\Omega: you make money acquiring the initial position and never lose money when unwinding it at the end.
Exercise. If sR\notin [S^-,S^+] in the binomial model there is an arbitrage.
Hint: If sR > S^+ the bond is always more valuable than the stock so short the the stock and buy bond with \xi = (S^+, -R). If Rs < S^- the bond is always less valuable than the stock so short the bond and buy the stock with with \xi = (-S^-, R).
In the first case, \xi\cdot x = S^+ - sR < 0, {\xi\cdot X(S^-) = S^+R - RS^- > 0}, and {\xi\cdot X(S^+) = S^+R - RS^+ = 0}, so \xi\cdot X\ge0 on \{S^-,S^+\}.
In the second case, \xi\cdot x = -S^- + sR < 0, {\xi\cdot X(S^-) = -S^-R + RS^- = 0}, and {\xi\cdot X(S^+) = -S^-R + RS^+ > 0}, so \xi\cdot X\ge0 on \{S^-,S^+\}.Suppose x = \sum_i X(\omega_i) \pi_i is a finite linear combination with \pi_i\in\boldsymbol{R} positive, \omega_i\in\Omega. If {\xi\cdot X(\omega)\ge0} for \omega\in\Omega then {\xi\cdot x\ge 0}, so no arbitrage exists.
A cone C is a subset of a vector space closed under positive scalar multiplication and vector addition.
Exercise. A cone is convex.
Hint: Show x,y\in C implies tx + (1-t)y\in C for 0 < t < 1.
Exercise. The set of arbitrage positions is a cone.
Exercise. If x belongs to the smallest cone containing the range of X then there is no arbitrage.
Hint: The range of X is \operatorname{ran}X = \{X(\omega)\mid\omega\in\Omega\}. The smallest cone containing the range of X is the set of finite linear combinations \sum_i X(\omega_i) \pi_i, \pi_i > 0, \omega_i\in\Omega.
Theorem. Arbitrage exists in the one-period model if and only if x does not belong to the smallest closed cone containing the range of X.
The previous exercise proves the “easy” direction. The contrapositive follows from
Lemma. If x\in\boldsymbol{R}^n and C is a closed cone in \boldsymbol{R}^n with x\not\in C then there exists {\xi\in\boldsymbol{R}^n} with {\xi\cdot x < 0} and {\xi\cdot y \ge0} for {y\in C}.
Proof. Let x^* be the point in C closest to x. It exists since C is closed and is unique since C is convex.
Since ty + x^*\in C for any t > 0 and y\in C we have \|x^* - x\| \le \|ty + x^* - x\|. Simplifying gives {t^2||y||^2 + 2t\xi\cdot y\ge 0}. Dividing by t > 0 and letting t decrease to 0 shows {\xi\cdot y\ge 0}.
We also have \|x^* - x\| \le \|tx^* + x^* - x\| for t + 1 > 0. Simplifying gives {t^2||x^*||^2 + 2t\xi\cdot x^*\ge 0} for t > -1. Dividing by t < 0 and letting t increase to 0 shows {\xi\cdot x^*\le 0} so {0 < ||\xi||^2 = \xi\cdot (x^* - x) \le -\xi\cdot x} hence {\xi\cdot x < 0}.
The lemma proves the “hard” direction of the FTAP and \xi = x^* - x implements an arbitrage.
Note the lemma is a purely geometric fact. It is similar to Farkas’ lemma and is a special case of the Hahn-Banach theorem in finite dimensional space.
For the binomial model the smallest closed cone containing the range of X is \{X(S^-)\pi^- + X(S^+)\pi^+\mid \pi^-,\pi^+\ge0\}. If (1, s) is in the cone then {(1,s) = (R, S^-)\pi^- + (R, S^+)\pi^+} for some \pi^-,\pi^+\ge0. Solving for \pi^- and \pi^+ gives the no arbitrage condition S^- \le Rs \le R^+.
Exercise. Show \pi^- = (S^+ - Rs)/R(S^+ - S^-) and \pi^+ = (Rs - S^-)/R(S^+ - S^-).
If Rs > S^+ then (R, s) is above the line through the origin with slope R/S^+ so \xi should be proportional to (S^+, -R). If Rs < S^- then (R, s) is below the line through the origin with slope R/S^- so \xi should be proportional to (-S^-, R).
If we add an option with payoff \nu to the binomial model then x = (1, s, c) and X(\omega) = (R, \omega, \nu(\omega)), \omega\in\{S^-,S^+\}. There is no arbitrage if and only if (1, s, c) = X(S^-)\pi^- + X(S^+)\pi^+ for some \pi^-,\pi^+\ge0. The first two equations determine \pi^- and \pi^+ as above so v = \nu(S^-)\pi^- + \nu(S^+)\pi^+. Every option payoff is linear in the binomial model.
A slightly more interesting model is a bond with zero interest rate, a stock that can go from 100 to 90, 100, or 110, and an at-the-money call option with price v. The model is x = (1, 100, v), X(\omega) = (1, \omega, \max\{\omega - 100, 0\}), \omega\in\{90, 100, 110\}. The smallest closed cone containing the range of X is \{X(90) a + X(100) b + X(110) c\mid a,b,c\ge0\}. If x belongs to the cone then \begin{aligned} 1 &= a + b + c \\ 100 &= 90a + 100b + 110 c \\ v &= 10c \\ \end{aligned} for some a,b,c\ge0 so \begin{aligned} 1 - c &= a + b \\ 100 - 110c &= 90a + 100b \\ \end{aligned}
Exercise. Show a = c and b = 1 - 2c.
Hint: Multiply the first equation by 90 and subtract the second.
The FTAP proves the model is arbitrage-free if and only if the option value is between 0 and 5 since b\ge0 implies 0\le c\le 1/2 and v = 10c.
Exercise. Show the model without the bond is arbitrage-free if and only the option value is between 0 and 100/11 < 10.
Hint: Consider \{(90, 0)a + (100, 0)b + (110, 10)c\mid a,b,c\ge0\}. The smallest closed convex cone is bounded below by y = 0 and above by y = (10/110)x.
If you are willing to assume prices are bounded, as they are in the real world, then the one-period model should require X\colon\Omega\to\boldsymbol{R}^I is bounded. This is not the case in the Black-Scholes/Merton continuous time model of lognormal stock prices. Scholes and Merton won a Nobel prize “for a new method to determine the value of derivatives.” Their assumptions eliminated the need to measure the real-world return on a stock. There is still a Nobel prize to be won by someone who can come up with a generally accepted way to determine the volatility of a stock. And perhaps multiple future prizes for those who develop theories not founded on the mathematical fictions of continuous time trading and unbounded prices.
If x\in\boldsymbol{R}^I and X\colon\Omega\to\boldsymbol{R}^I is bounded then there is no arbitrage if and only if x = \int_\Omega X(\omega)\,d\Pi(\omega) for some positive, bounded, finitely additive measure on \Omega, but this requires a bit more math to establish.
If there exists a zero coupon bond \zeta\in\boldsymbol{R}^I with \zeta\cdot X(\omega) = 1 for all {\omega\in\Omega} then {D = \zeta\cdot x = \int_\Omega \zeta\cdot X(\omega)\,d\Pi(\omega) = \Pi(\Omega)} is the cost of the bond, aka discount. The measure {P = \Pi/D} is positive and has mass 1 so {x = E[XD]} if we pretend it is a probability measure.