April 25, 2024
The one period model for a set of instruments I is determined by a vector of initial prices x\in\boldsymbol{R}^I 1 and a vector-valued function X\colon\Omega\to\boldsymbol{R}^I where \Omega is the set of possible outcomes. The instrument prices at the end of the period are X(\omega)\in\boldsymbol{R}^I if \omega\in\Omega occurred. For example, we could take \Omega = \boldsymbol{R}^I and let X be the identity function. At the other extreme we could take \Omega to be any one element set to get a deterministic model. Binomial models correspond to \Omega having two elements. We can extend any one-period model to allow a catastrophic event \bot. Let \Omega^\bot = \Omega\cup\{\bot\} and define X^\bot(\omega) = X(\omega), \omega\in\Omega, and X^\bot(\bot) = 0 modeling the notion all instrument prices are 0 in the event of a catastrophe.
Note we do not assume a probability measure on \Omega.
The cost of obtaining portfolio \gamma\in\boldsymbol{R}^I at the beginning of the period is the inner product \sum_{i\in I} \gamma_i x_i = \gamma\cdot x = \gamma'x. The components of \gamma are the amount purchased in each instrument. At the end of the period it has value \gamma'X(\omega) given \omega\in\Omega occured. The model has arbitrage if there exists \gamma\in\boldsymbol{R}^I with \gamma'x < 0 and \gamma'X(\omega)\ge0 for all \omega\in\Omega. If the cost is negative we make money initially and if the final value is always non-negative we can liquidate the position and not lose money.
The Fundamental Theorem of Asset Pricing states that a model is arbitrage free if and only if x belongs to the smallest cone containing the range of X. Note this is a purely geometric condition. There is no need to use probability to define arbitrage.
Recall a cone is a subset C\subset V, where V is a vector space, such that x\in C implies px\in C for every scalar p > 0 and if x,y\in C then x + y\in C.
Exercise. Show the set of arbitrage portfolios is a cone.
The range of X is the set X(\Omega) = \{X(\omega)\mid \omega\in\Omega\}. Points of the form x = \sum_j p_j X(\omega_j) where p_j > 0 and \omega_j\in\Omega belong the the smallest cone containing the range of X.
Exercise. If X(\omega)\ge0, \omega\in\Omega, show \gamma'x\ge0 for all \gamma\in\boldsymbol{R}^I.
Hint: We are assuming x = \sum_j p_j X(\omega_j) where p_j > 0 and \omega_j\in\Omega.
This (almost) proves the easy direction: if x belongs to the cone determined by the range of X then there are no arbitrage portfolios. If x is not in the cone and x^* is the point in the cone closest to x, then \gamma = x^* - x is an arbitrage. However, proving that involves more machinery2.
Consider a bond, stock, and at-the-money call option where the bond has zero interest rate, the stock has initial price 100 and either 90 or 110 as the final prices, and the call option has price c. The one-period model for this is x = (1, 100, c) and X(\omega) = (1, \omega, \max\{\omega - 100, 0\}) where \omega\in\Omega = \{90, 110\}. The range of X is X(\Omega) = \{X(90), X(110)\} and the smallest cone containing the range of X consists of \{aX(90) + bX(110) where a,b\ge 0. Since x = X(90)/2 + X(110)/2 we must have c = 0/2 + 10/2 = 5 if the model is arbitrage-free.
More generally, assume the bond has realized return R, the stock has initial price s and final price \omega where \omega\in\{L,H\}. Assume an European option has price c and pays \nu(\omega) at the end of the period. The one-period model is x = (1, s, c) and X(\omega) = (R, \omega, \nu(\omega)). The range of X is X(\Omega) = \{X(L), X(H)\} and the smallest cone containing the range of X consists of \{aX(L) + bX(H)\mid a,b\ge 0\}. If x = aX(L) + bX(H) then 1 = aR + bR and s = aL + bH so a = (H - Rs)/R(H - L) and b = (Rs - L)/R(H - L). The no-arbitrage condition implies a,b\ge0 so L \le Rs \le H. In this case c = \nu(L)(H - Rs)/R(H - L) + \nu(H)(Rs - L)/R(H - L).
If we let \Omega = [L,H] be an interval we get the same answer.
Exercise. Show X(\omega) is a linear combination of X(L) and X(H) with positive coefficients if \omega\in[L, H].
This shows the smallest cone generated by X([L,H]) equals the smallest cone generated by X(L) and X(H).
A puzzle making the rounds a few years ago was given a bond with zero interest rate, a stock with initial price 100 and possible final prices of 90, 100, and 110, what are the allowable arbitrage-free prices of an at-the-money call? The call can expire at either 0 or 10, however the arbitrage-free price must be between 0 and 5. The one period model for this is x = (1, 100, c) where c is the call value, and X(\omega) = (1, \omega, \max\{\omega - 100,0\}) for \omega\in\Omega = \{90, 100, 110\}. The set of all points in the smallest cone containg the range of X is \{aX(90) + bX(100) + cX(110)\mid a,b,c \ge 0\}. Since x = X(90)c/10 + X(100)(1 - c/5) + X(110)c/10 we must have c\ge0 and 1 - c/5\ge0, so 0 \le c \le 5.
Exercise. Assume only the stock and call are available. What is the no arbitrage condition for c?
Hint: It is not 0\le c\le 10.
As above, the answer does not change if we take \Omega = [90, 110].
If the call has strike k…
If we assume X is bounded on \Omega then there exists a positive measure \pi on \Omega with x = \int_\Omega X\,d\pi. The mass of \pi is \|\pi\| = \int_\Omega 1\, d\pi. Since Q = \pi/\|pi\| is a probability measure we can write this as x = E[X]/R where R = 1/\|\pi\|. Of course Q is not the probability of anything, it is just a positive measure with mass 1. We call Q a risk-neutral measure.
Exercise. Show the risk-neutral measure is unique for a binomial model.
In general, there are many risk-neutral measures for a one period model having more than two outcomes.
Given any portfolio \gamma\in\boldsymbol{R}^I define R_\gamma = \gamma'X/\gamma'x to be the realized return of the portfolio whenever \gamma'x\neq 0.
Exercise Show R_\gamma = R_{t\gamma} for any scalar t\in\boldsymbol{R}, t\neq 0.
Exercise If x = E[X]/R show E[R_\gamma] = R for all \gamma\in\boldsymbol{R}^I with \gamma'x\neq 0.
This puts a crimp in portfolio optimization when using risk-neutral measures.
We call \zeta\in\boldsymbol{R}^I a zero coupon bond if \zeta'X(\omega) = 1 for all \omega\in\Omega. The discount is D = \zeta'x = \int_\Omega 1\,d\pi = 1/R so we have x = E[X]D.
An n-period model specifies trading times t_0 < t_1 < \cdots < t_n.
A partition \mathcal{A} of a set \Omega is used to represent partial information…
The multi-period model for a set of instruments i is determined by vector-valued functions for price and cash flows, X_t,C_t\colon\mathcal{A}_t\to\boldsymbol{R}^I.
Prices…
Cash flows…
A trading strategy consists of vector-valued functions \Gamma_t\colon\mathcal{A}_t\to\boldsymbol{R}^I indicating how many shares of each instrument are traded at time t. A strategy is closed out if \sum_t \Gamma_t = 0. The position at time t is \Delta_t = \sum_{s < t} \Gamma_j. Note the strict inequality…
Value V_t = (\Delta_t + \Gamma_t)\cdot X_t…
Account A_t = \Delta_t\cdot C_t - \Gamma_t\cdot X_t…
The set of all trading strategies is \mathcal{G} = \{(\Gamma_t)_{t\in T}\mid \Gamma_t\colon\mathcal{A}_t\to\boldsymbol{R}^I\}. Let \mathcal{G}_0 be the collection of closed out trading strategies. Arbitrage exists if there is a closed out trading strategy (\Gamma_t)_{t\in T} with A_{t_0} > 0 and A_{t_j}\ge0 for j > 0.
For a one-period model \mathcal{G}_0 = \{(\gamma,-\gamma)\mid \gamma\in\boldsymbol{R}^I\}. In this case A_0 = -\gamma\cdot X_0 and A_1 = \gamma\cdot C_1 + \gamma\cdot X_1. As a warm-up for the proof of the multi-period case define A\colon\boldsymbol{R}^I\to\boldsymbol{R}^I\oplus B(\Omega) by A\gamma = (A_0, A_1). The no arbitrage condition is the range of the operator A does not intersect the cone \mathcal{P} = \{(p, P)\mid p > 0, P\ge 0\}
The Markowitz optimization problem is to maximize E[R_\gamma] - (\tau/2)\operatorname{Var}(R_\gamma) under the constraint \gamma'x = 1, where \tau is a risk aversion parameter. In general, higher returns and smaller variance are preferred. In what follows, we assume x = (1,\ldots,1) = \boldsymbol{1} by dividing the components of X by corresponding components of x. Note E[R_\gamma] = \gamma'E[X] and \operatorname{Var}(R_\gamma) = \gamma'V\gamma where V = E[XX'] - E[X]E[X'].
When \tau = 0, as we have seen above, E[R_\gamma] does not depend on \gamma if x = \int_\Omega X\,d\pi for some positive measure \pi.
Exercise. In a model with arbitrage show E[R_\gamma], \gamma'x = 1, is unbounded above and below.
When \tau > 0 we have the optimum \gamma = (\tau V)^{-1}(E[X] - \lambda\boldsymbol{1}) where \lambda = (\tau - B)/A, B = \boldsymbol{1}'V^{-1}E[X], and A = \boldsymbol{1}'V^{-1}\boldsymbol{1}. Note every optimal portfolio belongs to the two dimensional space spanned by V^{-1}E[X] and V^{-1}\boldsymbol{1}.
If x = \boldsymbol{1}= E[X]/R then B = AR, \lambda = \tau/A - R, E[X] - \lambda\boldsymbol{1}= R\boldsymbol{1}- (\tau/A - R)\boldsymbol{1}= (R - (\tau/A - R))\boldsymbol{1},
Since V is positive definite the spectal theorem implies V = \sum_i \lambda_i e_i where \lambda_i > 0 and \{e_i\} are orthonormal. Its inverse is V^{-1} = \sum_i \lambda_i^{-1} e_i. It is common for the covariance matrix of a large number of instruments have most of the eigenvalues, \lambda_i, be small. This implies V^{-1}x is largely determined by the eigenvectors corresponding to small eigenvalues.
Assume X = \boldsymbol{1}+ \sigma\epsilon where \epsilon is a vector of independent standard normally distributed random variables, so E[X] = \boldsymbol{1} and \operatorname{Var}(X) = \sigma^2 I. Clearly this is not amenable to any sort of portfolio optimization. Since (\tau V)^{-1}E[X] = \boldsymbol{1}/\tau\sigma^2 We have B = n/\tau\sigma^2. Since \boldsymbol{1}= E[X], A = B so \lambda = (\tau/(n/\tau\sigma^2) - 1) = 1/n\sigma^2 - 1.
Recall A^B = \{f\colon B\to A\} is the set of all functions from the set B to the set A. We prefer the notation x\in\boldsymbol{R}^I to the circumloqution x\in\boldsymbol{R}^n where I = {i_1,\ldots,i_n} and x = (x_1,\ldots,x_n)\in\boldsymbol{R}^n corresponds to x(i_j) = x_j in \boldsymbol{R}^I.↩︎
The machinery being the Hahn-Banach theorem or its finite dimensional version due to Farkas.↩︎