Example

The first step in solving a math problem is writing down the problem. This note shows how to use SUM to express the problem of finding arbitrage constraints on initial European option values paying A_i at t_i, i = 1,2. The model is a replacement for Black-Scholes/Merton and extends Ross 1978 to explicitly include cash flows. Unlike earlier models, every market instrument can be incorporated, including futures and limit orders.

As a warm-up, let’s start with a one period model of bond and stock: bond has constant realized return R, stock has initial price s and final price any bounded function S\in B(\Omega).

SUM notation: initial market prices X_0 = x = (1,s), final market prices X_1 = X = (R, S), where S(\omega) = \omega, \omega\in\Omega.

By FTAP, no arbitrage if there exists a non-negative finitely additive measure D\in ba(\Omega) with x = XD(\Omega) = \int_\Omega X(\omega)\,dD(\omega).

This is equivalent to x belonging to the smallest closed cone containing the range of X in \boldsymbol{R}^2.

Extend the model with an option paying {A} at expiration. Let x = (1,s,v), X = (R,S,0), C = (0, 0, {A}).

No arbitrage if x = (X + C)D(\Omega) so (1,s,v) = \int_\Omega (R,S,{A})\,dD.

Example: \Omega = {L,H} (binomial model). We must have (1,s,v) = (R, L, A) D_L + (R, H, A) D_H for some D_L,D_H\ge 0. Bond and stock components determine D_L and D_H. Non-negativity of D_L and D_H imply L\le Rs\le H. These determine a unique option value v.

Example: \Omega = [L,H]. Option value is no longer unique, the only constraint is (1, s, v) belongs to the cone determined by the range.

If A(\omega) = \max\{\omega - K,0\}, L < K < H, then X(L), X(K), and X(H) are extreme points and (1, s, v) = X(L) D_L + X(K) D_K + X(H) D_H for some D_L,D_K,D_H\ge 0.

Note X(H) = (0,0,H-K) so the third component is v = (H - K)D_H and D_H = v/(H - K). Substitute that in the first two components and solve for D_L and D_K to get D_L = A + Bv > 0 and D_K = C + Dv > 0 for the constraints on option value v.

Exercise. Compute A, B, C, and D.

Exercise. What are the limits as L\to 0 and H\to\infty?

In general, let \Omega = B([0,\infty)). Any bounded path is a possible trajectory. This is a very general model. Let the information at time t, \mathcal{A}_t, be the smallest algebra of sets for which \omega \mapsto \omega(s) are measurable for all s\le t. The atom in \mathcal{A}_t containing \omega\in\Omega is the set A_t(\omega) = \{\omega'\in\Omega\mid \omega(s) = \omega'(s), 0\le s \le t\}, i.e., all paths agreeing with \omega up to time t.

Consider a model with two European options paying A_1 at t_1 and A_2 at t_2. We have X_0 = (1, s, v_1, v_2), C_{t_1} = (0, 0, A_1, 0), C_{t_2} = (0, 0, 0, A_2), and C_t = (0,0,0,0) otherwise. Note the third component of X_t = 0 for t > t_1 and the fourth component of X_t = 0 for t > t_2 since the value of an option is zero after expiration.

By the FTAP, the model is arbitrage free if there exist non-negative finitely additive measure D_t\in ba(\mathcal{A}_t) with X_t D_t = (X_u D_u + \sum_{t < s \le u} C_s D_s)|_{\mathcal{A}_t}, t\le u.

This in not a complete solution of the problem, it is only a rigorous way of writing it down.