One-Period Model

Binomial Model

The binomial model has a bond and stock where x = (1, s), X(\omega) = (R, \omega), and \Omega = \{S^-, S^+\} with {S^- < S^+}. The bond has realized return R and the stock can go from price s to either S^- or S^+. A somewhat more realistic model is {\Omega = [S^-, S^+]} where the final stock price can be any value in the interval.

Arbitrage

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 don’t lose money when unwinding it at the end of the period.

Some authors define arbitrage as a portfolio with \xi\cdot x = 0 and \xi\cdot X is non-negative and strictly positive on some set having positive probability. We haven’t specified a probability measure so we can’t use this definition. No trader would consider that to be an arbitrage anyway. Even though the position costs nothing to put on, that definition has nothing to say about how much they will make nor how likely it is they will make it.

In reality, even our stronger definition is still not good enough for traders. Even though \xi\cdot x is strictly negative, they will consider how much capital will be tied up putting on that position and compute |\xi|\cdot |x|. If the ratio |\xi\cdot x|/|\xi|\cdot |x| is small, they will take a pass on that “arbitrage” opportunity. No trader would use a million dollars from their funding account to make at a penny even though that satisfies the mathematical definition of arbitrage.

Exercise. If Rs\notin [S^-,S^+] in the binomial model then there is an arbitrage.

Hint: If Rs > S^+ the bond is always more valuable than the stock so short the the stock and buy bond. If Rs < S^- the bond is always less valuable than the stock so short the bond and buy the stock.

Cone

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.

The smallest cone containing the range of X, \operatorname{ran}X = X(\Omega) = \{X(\omega)\mid \omega\in\Omega\}, is the set of finite linear combinations \sum_i X(\omega_i) \pi_i, \pi_i > 0, \omega_i\in\Omega. If x = \sum_i X(\omega_i) \pi_i is in the cone and \xi\cdot X is non-negative on \Omega then {\xi\cdot x\ge 0} so no arbitrage exists.

Exercise. If x belongs to the smallest closed cone containing the range of X then there is no arbitrage.

Fundamental Theorem of Asset Pricing

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 contra-positive follows from the

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} for all y\in C.

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 that \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.

Application

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.

Exercise. Find a,b\in\boldsymbol{R} with aR + b\omega = \nu(\omega) for \omega\in\{S^-,S^+\}.

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.

The FTAP proves the model is arbitrage-free if and only if the option value is between 0 and 5 since v = 10c\ge0 and b\ge0 implies 0\le c\le 1/2 so v = 10c\le 5.

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 cone is bounded below by y = 0 and above by y = (10/110)x.

Reality

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.

FTAP

We will assume X\colon\Omega\to\boldsymbol{R}^I is bounded so \|X\| = \sup_{\omega\in\Omega}|X(\omega)| is finite, just as it is in the real world. We write this {X\in B(\Omega,\boldsymbol{R}^I)}, the normed vector space of bounded \boldsymbol{R}^I-valued functions. 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 \Pi\in ba(\Omega) on \Omega, but this requires a bit more math to establish. See (Dunford and Schwartz 1958). We call any such measure risk-neutral, although risk-blind might be more appropriate. In general, risk-neutral measures are not unique.

If we let Q = \Pi/\Pi(\Omega) then Q is a positive measure with mass 1 and x = E[X]\Pi(\Omega) if we pretend it is a probability measure.

If a zero coupon bond exists then \Pi(\Omega) is its price. A zero coupon bond is a portfolio \zeta\in\boldsymbol{R}^I with {\zeta\cdot X(\omega) = 1} for all {\omega\in\Omega}. The price, or discount, of a zero coupon bond is {\zeta\cdot x = \int_\Omega \zeta\cdot X(\omega)\,d\Pi(\omega) = \Pi(\Omega) = D}. The measure {Q = \Pi/D} is positive and has mass 1 so {x = E[X]D} if we pretend it is a probability measure.

Define the map M_X\colon\boldsymbol{R}^I\to B(\Omega) by M_X(\xi) = \xi\cdot X. We can and do assume M_X is one-to-one, otherwise there would be redundant market instruments. We say prices X form a complete market if M_X is onto.

Exercise. Show the market is complete implies the cardinality of \Omega is less than or equal to the cardinality of I.

Hint: If T\colon V\to W is a linear operator between vector spaces that is onto, then the dimension of V is greater than or equal to the dimension of W.

Although complete markets are common in mathematical finance books, they almost never occur in models faithful to the real world.

Return

The realized return on a portfolio \xi\in\boldsymbol{R}^I is {R_\xi = \xi\cdot X/\xi\cdot x} whenever {\xi\cdot x\not=0}.

Exercise. Show R_\xi = R_{t\xi} for any non-zero t\in\boldsymbol{R}.

If \xi\cdot x = 1 then R_\xi = \xi\cdot X. One unit invested at the beginning of the period results in {R_\xi\colon\Omega\to\boldsymbol{R}} at the end of the period. Every portfolio has the same expected realized return under a risk-neutral measure.

Exercise. If \Pi is a risk-neutral measure then E[R_\xi] = 1/\Pi(\Omega) for any portfolio \xi\in\boldsymbol{R}^I.

Hint: The expectation is with respect to the “probability” measure \Pi/\Pi(\Omega).

Probability

We now assume there is a probability measure P on \Omega representing the real-world event probabilities.

Let U\colon B(\Omega)\to\boldsymbol{R} be a utility function. A common choice is U(Y) = E[Y] - \alpha \operatorname{Var}(Y) for some positive risk parameter \alpha\in\boldsymbol{R}. Note if Y\in B(\Omega) then moments of all orders exist.

To find a portfolio that maximizes the utility of the realized return we use Lagrange multipliers to solve \max_\xi U(\xi\cdot X) - \lambda(\xi\cdot x - 1).

We need to compute the Frechet Derivatives of D(UM_X) where M_X\xi = \xi\cdot X is the market map as above. The chain rule gives D(UM_X)\xi = DU(M_x\xi)DM_X\xi = DU(M_x\xi)M_X\xi since DM_X = M_X. Note DU(M_x\xi) is in B(\Omega)^* and M_X\xi\in B(\Omega).

The first order condition is 0 = DU(\xi\cdot X)M_X\xi - \lambda x^* for all \xi\in\boldsymbol{R}^I.

References