Aug 1, 2026
The one‑period model is the simplest framework for rigorously representing a financial market over a single period of time. The model defines the initial prices of tradeable instruments and their terminal prices contingent on the realized outcome. If there are no arbitrage opportunities available then initial prices are subject to geometric constraints determined by the final prices.
The one-period model specifies a finite set of tradeable instruments I and the set of possible outcomes \Omega representing what can happen over the period. The initial prices are given by a vector {x\in\boldsymbol{R}^I}1, indexed by the instruments. Terminal prices are defined by a vector-valued function {X\colon\Omega\to\boldsymbol{R}^I} where {X(\omega)\in\boldsymbol{R}^I} are the prices for each instrument corresponding to outcome {\omega\in\Omega}.
Arbitrage exists (in this very simple and unrealistic model) if we can purchase \xi\in\boldsymbol{R}^I shares at the beginning of the period with cost \xi\cdot x < 0 and sell them at the end of the period for profit {\xi\cdot X(\omega)\ge0} for all \omega\in\Omega. We make money putting on the position and never lose when unwound at the end.
It is not a definition that would pass muster with traders and risk managers. They will compare {|\sum_i \xi_i x_i|} with {\sum_i |\xi_i x_i|} as a measure of how much capital will be tied up. If the ratio is small they will take a pass on the mathematical “arbitrage.”
Exercise. Show if \xi is an arbitrage then t\xi is an arbitrage for any positive real number t.
This is a defect in the one-period model. There is only a finite amount of each instrument that can be traded.
Exercise. If \xi_0 and \xi_1 are arbitrages then so is \xi_0 + \xi_1.
A subset of a vector space closed under positive scalar multiplication and vector addition is a cone. The previous two exercises show the set of arbitrage portfolios is a cone.
There is a connection between cones and convex sets.
Exercise. Cones are convex.
Hint: A set C\subseteq\boldsymbol{R}^I is convex if and only if for x,y\in C we have {(1 - t)x + ty\in C} for 0 < t < 1.
Exercise. Show if C\subseteq\boldsymbol{R}^I is convex then \cup_{t>0} tC is a cone.
Hint: tC = \{tx\mid x\in C\} for t\in\boldsymbol{R}.
The smallest cone containing the range of X is {\operatorname{cone}X(\Omega) = \{\sum_j X(\omega_j) D_j\mid D_j > 0\}}. If {\xi\cdot X\ge0} on \Omega and x is in the cone then \xi\cdot x = \sum_j \xi\cdot X(\omega_j) D_j\ge0 so there is no arbitrage.
Exercise. Show this holds for every point in the closure of the cone.
Hint: If x_n\to x and \xi\cdot x_n\ge0 then \xi\cdot x\ge0.
This proves the “easy” direction of the FTAP.
Theorem. A one-period model is arbitrage-free if and only if x belongs to the smallest closed cone containing the range of X.
We prove the contrapositive of the easy direction by showing if x is not in the cone then arbitrage exists. One arbitrage is {\xi = x^* - x} where x^* is the point in the closed cone closest to x.
Lemma. If x\in\boldsymbol{R}^n and K\subseteq\boldsymbol{R}^N is a closed cone with x\not\in K then there exists {\xi\in\boldsymbol{R}^n} with {\xi\cdot x < 0} and {\xi\cdot y \ge0} for all {y\in K}.
Proof. Let x^* be the point in K closest to x. It exists because K is closed and unique since K is convex. Let \xi = x^* - x and note \xi\not=0.
We have ty + x^*\in K for any t > 0 and y\in K so \|x^* - x\| \le \|ty + x^* - x\|. Simplifying gives {t^2||y||^2 + 2ty\cdot\xi\ge0}. Dividing by t > 0 and letting t decrease to 0 shows {\xi\cdot y\ge0} for all y\in K.
We have (t + 1)x^*\in K for t + 1 > 0 so \|x^* - x\| \le \|tx^* + x^* - x\|. Simplifying gives {t^2||x^*||^2 + 2tx^*\cdot\xi^*\ge 0} for t > -1. Dividing by t < 0 and letting t increase to 0 shows {\xi\cdot x^*\le 0}.
Since {0 < ||\xi||^2 = \xi\cdot (x^* - x) \le -\xi\cdot x} we have {\xi\cdot x < 0}.
This lemma proves the FTAP and that \xi = x^* - x implements an arbitrage.
We now give an alternate proof that will generalize to multi-period models. See (Dunford and Schwartz 1958) Volume I, Chapter IV for the mathematical details. We assume X is bounded, because it is, and write X\in B(\Omega,\boldsymbol{R}^I) where B(\Omega,\boldsymbol{R}^I) = \{X\colon\Omega\to\boldsymbol{R}^I\mid \|X\| = \sup_{\omega\in\Omega}\|X(\omega)\| < \infty\}. Define {A\colon\boldsymbol{R}^I\to\boldsymbol{R}\oplus B(\Omega)} by A\xi = -\xi\cdot x\oplus \xi\cdot X. The components of the right-hand side are the amounts associated with buying \xi at the beginning of the period and selling \xi at the end.
No arbitrage is equivalent to \operatorname{ran}A\cap\mathcal{P}^+ = \emptyset where \mathcal{P}^+ = \{p\oplus P\mid p > 0, P(\omega)\ge 0, \omega\in\Omega\}. Since \mathcal{P}^+ has an interior point the Hahn-Banach theorem implies there is a hyperplane {H\supseteq\operatorname{ran}A} with {H\cap\mathcal{P}^+ = \emptyset}. Every hyperplane is the preannihilator of an element in the dual {(\boldsymbol{R}\oplus B(\Omega))^*\cong\boldsymbol{R}\oplus ba(\Omega)} where ba(\Omega) is the set of finitely additive measures on \Omega.
There exists {d\oplus D} in the dual with {H = {}^\perp\{d\oplus D\}}. We can and do assume d = 1. Since {0 = \langle -\xi\cdot x\oplus \xi\cdot X, 1\oplus D\rangle = -\xi\cdot x + \langle \xi\cdot X, D\rangle} for all \xi\in\boldsymbol{R}^I we have x = \langle X,D\rangle. This shows x belongs to the closed cone containing the range of X. It does not show how to find an arbitrage if one exists.
Let’s apply the FTAP to particular one-period models.
Consider a model with a single bond having realized return R. In this case x = 1 and X(\omega) = R for all \omega no matter the sample space. If R \le 0 then \xi = -1 is an arbitrage since \xi x = -1 and \xi R = -R \ge 0.
Exercise. Show this model is arbitrage-free if and only if R > 0.
Hint: x = X D for some D > 0 so D = 1/R.
Note that negative interest rates (R < 1) do not imply arbitrage.
Consider a model with a bond having realized return R and a stock with price s at the beginning of the period that can go to either a low L or a high H at the end. This is modeled by {x = (1, s)} and {X(\omega) = (R, \omega)} where {\omega\in\Omega = \{L,H\}}. If {x = X(L)D_L + X(H)D_H} then 1 = R D_L + R D_H and s = L D_L + H D_H so D_L = (H - Rs)/R(H - L) and D_H = (Rs - L)/R(H - L).
Exercise. Show the model is arbitrage-free if and only if L/R \le s \le H/R and R > 0.
Hint: We have already shown R > 0 is necessary. Use D_L,D_H\ge 0.
This is the simplest example of how end of period prices constrain initial prices. This is a geometric result that has nothing to do with probability.
Exercise. Show the same result holds for the slightly more realistic model with \Omega = [L,H].
Hint: The smallest cone containing X(L) and X(H) is the same as the smallest cone containing \{X(\omega)\mid L\le\omega\le H\} because cones are convex.
We can also establish this result using Grassmann algebra. The computation is more tedious but entirely mechanical and generalizes to any number of instruments.
Grassmann starts with the Euclidean (no absolute origin) space of points E and considers the associative algebra \mathcal{G}(E) generated by points in E with the rule PQ = 0 if and only if P = Q for P,Q\in E. 2
Exercise. Show PQ = -QP for P,Q\in E.
Hint: 0 = (P + Q)(P + Q).
Given points P_0, P_1,\cdots,P_k\in E let \Pi = P_0 \cdots P_k be their product. Replacing P_j by P\in E in the product define \Pi_j(P) = (\prod_{i=0}^{j-1} P_i) P (\prod_{i = j+1}^k P_i).
Exercise. Show \Pi_j(P_i) = 0 if i\not=j and \Pi_j(P_j) = \Pi.
Hint: PQ = 0 if and only if P = Q.
This shows if P = \sum_i x_i P_i then \Pi_j(P) = x_j \Pi. We write x_j = \Pi_j(P)/\Pi so P = \sum_j \frac{\Pi_j(P)}{\Pi} P_j.
This provides a coordinate-free equation for expressing points in space.
Exercise. Show \sum_j \Pi_j(P) = \Pi.
Exercise. Show \sum_j x_j = 1.
P belongs to the convex hull of \{P_j\}_{j=0}^k if and only if x_j\ge0 for all 0\le j\le k.
Since \sum_j x_j = 1 we have P = (1 - \sum_{j=1}^k x_j)P_0 + \sum_{j=1}^k x_j P_j = P_0 + \sum_{j=1}^k x_j (P_j - P_0).
Given an _origin P_0 this shows how \boldsymbol{R}^n fits into Grassmann space. The vector x = (x_1,\ldots,x_n) corresponds to the point P(x) = P_0 + \sum_j x_j \pi_j where \pi_j = P_j - P_0 is the vector from the origin to P_j.
P belongs to the cone with origin P_0 containing \{P_j\}_{j=1}^k if and only if x_j\ge0 for 1\le j\le k.
To apply this to the bond and stock model we let O be the origin, \rho a vector in the bond direction, and \sigma a vector in the stock direction. We have x = O + \rho + s\sigma and X(\omega) = O + R\rho + \omega\sigma.
Using Grassmann algebra
x = \frac{xX(L)X(H)}{\Pi} O + \frac{OxX(H)}{\Pi} X(L) + \frac{OX(L)x}{\Pi} X(H)
Since OO = 0 we have
\begin{aligned} OX(L)X(H) &= O(R\rho + L\sigma)(R\rho + H\sigma) \\ &= (R O\rho + L O\sigma)(R\rho + H\sigma) \\ &= RH \rho\sigma + LR O\sigma\rho \\ &= (RH - LR) O\rho\sigma \\ &= R(H - L) O\rho\sigma \\ \end{aligned} \begin{aligned} OxX(H) &= O(O + \rho + s\sigma)(O + R\rho + H\sigma) \\ &= (O\rho + s O\sigma)(O + R\rho + H\sigma) \\ &= H\,O\rho\sigma + sR\,O\sigma\rho \\ &= (H - Rs)\,O\rho\sigma \\ \end{aligned} \begin{aligned} OX(L)x &= O(O + R\rho + L\sigma)(O + \rho + s\sigma) \\ &= (R O\rho + L O\sigma)(O + \rho + s\sigma) \\ &= Rs\,O\rho\sigma + L\,O\sigma\rho \\ &= (Rs - L)\,O\rho\sigma \\ \end{aligned} There is no arbitrage if and only if x belongs to the smallest closed cone containing the range of X. This is equivalent to the coefficients of \rho and \sigma being non-negative. Since R(H - L) is positive we get H - Rs\ge0 and Rs - L\ge0 as before.
We model adding a call option with strike K to the bond and stock by {x = (1,v,s)} and {X(\omega) = (R, \max\{\omega - K, 0\}, \omega)}.
TODO: 90-100-110 problem Making the rounds at conferences a number of years ago.
If \kappa is a vector in the option direction then x = O + \rho + s\sigma + v\kappa and {X(\omega) = O + R\rho + \omega\sigma + \max\{\omega - K, 0\}\kappa}.
By Grassmann we know x is a linear combination of O, X(L), X(K), and X(H) and belongs to the smallest cone containing the range of X if and only if the coefficients of X(L), X(H), and X(K) are non-negative.
Using Grassmann algebra
x = \frac{xX(L)X(K)(H)}{\Pi} O + \frac{OxX(K)X(H)}{\Pi} X(L) + \frac{OX(L)xX(H)}{\Pi} X(K) + \frac{OX(L)X(K)x}{\Pi} X(H)
First we show \begin{aligned} OX(L)X(H)X(K) &= O(R\,\rho + L\,\sigma)(R\,\rho + H\,\sigma + (H - K)\,\kappa)X(K) \\ &= (RH\,O\rho\sigma + R(H - K)\,O\rho\kappa + LR\,O\sigma\rho + L(H - K)\,O\sigma\kappa)X(K) \\ &= ((RH - LR)\,O\rho\sigma + R(H - K)\,O\rho\kappa + L(H - K)\,O\sigma\kappa)X(K) \\ &= ((RH - LR)\,O\rho\sigma + R(H - K)\,O\rho\kappa + L(H - K)\,O\sigma\kappa)(R\,\rho + K\,\sigma) &= R(H - K)K\,O\rho\kappa\sigma + L(H - K)R\,O\sigma\kappa\rho \\ &= (-R(H - K)K + L(H - K)R)\,O\rho\sigma\kappa \\ &= R(H - K)(L - K)\,O\rho\sigma\kappa \\ \end{aligned}
\begin{aligned} OxX(H)X(K) &= O(\rho + s\,\sigma + v\,\kappa)(R\,\rho + H\,\sigma + (H - K)\,\kappa)X(K) \\ &= H\,O\rho\sigma + (H-K)\,O\rho\kappa + sR\,O\sigma\rho + s(H - K)\,O\sigma\kappa + vR\,O\kappa\rho + vH\,O\kappa\sigma)X(K) \\ &= ((H - sR)\,O\rho\sigma + (H - K - vR)\,O\rho\kappa + (s(H - K) - vH)\,O\sigma\kappa)X(K) \\ &= ((H - sR)\,O\rho\sigma + (H - K - vR)\,O\rho\kappa + (s(H - K) - vH)\,O\sigma\kappa) (R\,\rho + K\,\sigma) \\ &= (s(H - K) - vH)R - (H - K - vR)\,O\rho\sigma\kappa \\ \end{aligned}
So v\le s(H - K)/H. 100(10)/100 = 10. \begin{aligned} OX(L)xX(K) &= O(R\,\rho + L\,\sigma)(\rho + s\,\sigma + v\,\kappa)X(K) \\ &= (Rs\,O\rho\sigma + Rv\,O\rho\kappa + L\,O\sigma\rho + Lv\,O\sigma\kappa)X(K) \\ &= (Rs - L)\,O\rho\sigma + Rv\,O\rho\kappa + Lv\,O\sigma\kappa) (R\,\rho + K\,\sigma) \\ &= (LvR - RvK)\,O\rho\sigma\kappa \\ \end{aligned} So v\ge 0. TODO: check sign \begin{aligned} OX(L)X(H)x &= O(R\,\rho + L\,\sigma)(R\,\rho + H\,\sigma + (H - K)\,\kappa)x \\ &= RH\,O\rho\sigma + R(H - K)\,O\rho\kappa + LR\,O\sigma\rho + L(H - K)\,O\sigma\kappa \\ &= ((RH - LR)\,O\rho\sigma + R(H - K)\,O\rho\kappa + L(H - K)\,O\sigma\kappa) (\rho + s\sigma + v\kappa) \\ &= R(H - L)v - R(H - K)s + L(H - K)\,O\rho\sigma\kappa \\ &= R(H - L)v - (H - K)(Rs - L) \\ \end{aligned} so v\ge (H - K)(Rs - L)/R(H - L). v\ge 10(10)/20 = 100/20 = 10/2 = 5.
Prices must be integral multiples of minimum price increments of each instrument. Likewise, positions must be integral multiples of minimum trading increments. This easy to incorporate into the model.
This model also ignores the bid/ask spread in pricing. This not only depends on the sign of the amount being traded, but also the size of the trade. It can also depend on the credit relationship of the counterparties involved in the transaction, but that is beyond the scope of this short write-up.
To model bid/ask spread we can define X\colon\Omega\times\boldsymbol{R}\to\boldsymbol{R}^I where X(\omega,a) is the price of amount a being traded. A simple example is X(\omega,a) = X(\omega) + \delta\operatorname{sgn}(a) for some fixed transaction cost \delta per share.
If we are being intellectually honest then we must admit X(\omega) = 0 for \omega\in\Omega. At the end of the period there is no further economic activity so all prices must be zero. We should replace prices X with cash flows C that accrue to the investor in proportion to the position. There is no final trade equal to the negative of the position. This will be clarified when we consider multi-period models.