October 9, 2025
The One‑Period Model provides the simplest formal framework for describing a financial market over a single period where trading is allowed only at the start and end of the period. It specifies the prices of available instruments at the start of the period and their prices and cash flows at the end depending on what occurred. We make the usual unrealistic assumptions that prices are real numbers instead of integral multiples of each instrument’s minimum trading increment/tick size, and there is no bid-ask spread depending on the amount being bought or sold, much less any consideration of the credit rating of counterparties involved. The appendix proposes a model that can incorporate more realistic assumptions.
Quants turn math into software that traders use. If they roll out a model that is not arbitrage-free then a “clever” trader will find an “arbitrage” and any business that didn’t have risk controls in place to prevent this will most likely lose money.
The Fundamental Theorem of Asset Pricing identifies arbitrage-free models and provides an arbitrage if they are not. As (Ross 1978) showed, this is a purely geometric result having nothing to do with probability. Positive measures with mass one make an appearance, but they are not the probability of anything.
The One-Period Model consists of a finite set of market instruments, I, available for trading at the start and end of a period. The set of possible outcomes, \Omega, is what can happen over the period. The prices at the beginning of the period are a vector x\in\boldsymbol{R}^I1. 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. Instruments can have associated cash flows such as bond coupons or stock dividends. We use C\colon\Omega\to\boldsymbol{R}^I to denote the cash flows paid at the end of the period.
The binomial model (with no dividends) has a bond and stock where x = (1, s), X(\omega) = (R, \omega), and \Omega = \{L, H\} with {L < H}. The bond has realized return R and the stock can go from price s to either L or H. A somewhat more realistic model is {\Omega = [L, H]} where the final stock price can be any value in the interval.
The bond could be a cash deposit or Treasury bill that has a cash flow at the end of the period. These are typically quoted as a yield having a day count basis that can be used to calculate the realized return. T-bills use Actual/360 day count basis. The quoted yield y indicates the price now of unit notional received in t days is 1 - yt/360. The realized return R is the reciprocal of this.
A position \xi\in\boldsymbol{R}^I is the number of shares purchased in each instrument. The cost of acquiring the initial position is \xi\cdot x. The value of liquidating the position at the end of the period is {\xi\cdot(X(\omega) + C(\omega))} if \omega\in\Omega occurred. The one-period model implicitly assumes the position is liquidated and the prices and cash flows are received in proportion to the position at the end of the period.
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) + C(\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, and don’t need, 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 definite to say about how much they will make nor how likely it is they will make it.
Even our stronger definition is still not good enough for traders and risk managers. Even though {\xi\cdot x} is strictly negative they will compute {|\xi|\cdot|x|} as a measure of how much capital will be tied up putting on the position. No business would approve using 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 [L,H] in the binomial model then there is an arbitrage.
Hint: If Rs > H the bond is always more valuable than the stock so short the stock and buy bond. If Rs < L the bond is always less valuable than the stock so short the bond and buy the stock.
If Rs > H take \xi = (H, -R) so \xi\cdot x = H - Rs < 0, {\xi\cdot X(L) = HR - RL > 0}, and {\xi\cdot X(H) = HR - RH = 0}, so \xi\cdot X\ge0 on \{L,H\}.
If Rs < L take \xi = (-L, R) so \xi\cdot x = -L + Rs < 0, {\xi\cdot X(L) = -LR + RL = 0}, and {\xi\cdot X(H) = -LR + RH > 0}, so \xi\cdot X\ge0 on \{L,H\}.
This argument also holds for the somewhat more realistic model where the final stock price can be any value between the low and the high, i.e., \Omega = [L, H].A cone C is a subset of a vector space closed under positive scalar multiplication and vector addition: if x\in C then tx\in C for t > 0 and if x,y\in C then {x + y\in C}.
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.
The contrapositive is also true.
Theorem. Arbitrage exists in the one-period model if x does not belong to the smallest closed cone containing the range of X. If x^* is the closest point in the cone then \xi = x^* - x is an arbitrage.
In general, the arbitrage is not unique. We will establish the theorem using the purely geometric
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\ge0}. Dividing by t > 0 and letting t decrease to 0 shows {\xi\cdot y\ge0} 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 FTAP and that \xi = x^* - x implements an arbitrage.
If we assume prices are bounded, as they are in the real world, then every point in the closed convex cone generated by the range of X can be written \int_\Omega X(\omega)\,dD(\omega) for some positive, bounded, finitely additive measure D\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. Risk-neutral measures are not generally unique.
A zero coupon bond pays 1 unit at the end of the period on every outcome. A portfolio \zeta\in\boldsymbol{R}^I with {\zeta\cdot X(\omega) = 1} for all {\omega\in\Omega} is a zero coupon bond. The price, or discount, of a zero coupon bond is {\zeta\cdot x = \int_\Omega \zeta\cdot X(\omega)\,dD(\omega) = D(\Omega)}. The measure {D/D(\Omega)} is positive and has mass 1 so {x = E[X]D(\Omega)} if we pretend it is a probability measure.
Exercise. If \zeta is a zero coupon bond with only one non-zero component then that component is equal to the discount.
If \xi\cdot X = 0 then one instrument is a linear combination of the others and can be removed. This can be repeated until \xi\cdot X\not=0 for any \xi\in\boldsymbol{R}^I. If so the map \xi\mapsto\xi\cdot X is one-to-one. A model is complete if this map is onto. This cannot be the case if the cardinality of I is less than the cardinality of \Omega. In general the number of instruments is much smaller than the number of outcomes. Although complete markets are common in mathematical finance books, they almost never occur in models faithful to the real world.
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}.
This is actually a deleterious feature of the model. Going long or short typically affects the realized return. It also implies a portfolio stategy can be scaled to arbitrarily large positions. At some point you will run out of instruments to buy and sell.
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 so perhaps this should be called a risk-blind measure.
Exercise. If D is a risk-neutral measure then R = E[R_\xi] = 1/D(\Omega) is constant for any portfolio \xi\in\boldsymbol{R}^I.
Hint: The expectation is with respect to the “probability” measure DR = D/D(\Omega).
Let E be the set of points in space. The (Grassmann 1844) algebra \mathcal{G}(E) is the algebra over the real numbers generated by points in E with the condition PQ = 0 if and only if P = Q for P,Q\in E. In particular PP = 0. Multiplication is associative so (PQ)R = P(QR) and we can write PQR unambiguously.
Exercise. Show PQ = -QP.
Hint: 0 = (P + Q)(P + Q) = PP + PQ + QP + QQ.
Exercise. Show P(Q + R)(P + Q + R) = 0.
Congratulations! You have just proved the medians of a triangle intersect its centroid. In Grassmann Algebra the centroid of a triangle determined by points P, Q, and R is (P + Q + R)/3, the midpoint of Q and R is (Q + R)/2 and the median from P to the midpoint is P(Q + R)/2. Permuting P, Q, and R shows all three medians meet at the centroid. Grassmann used algebra instead of Euclid’s ruler and compass. It was the greatest invention in mathematics after René Descartes introduced coordinates to specify points in two and three dimensions. Grassmann got rid of coordinates and used points in space directly. His algebra works in any number of dimensions and predated Einstein in not assuming an absolute origin of points in space. Vectors come along for free as the difference of two points.
Descartes assumed an origin for his coordinate system. Given an origin O\in E and points P_1,\ldots,P_n\in E is is possible to write any point in their span as P = (1 - \sum_j t_j)O + \sum_j t_j P_j, as we will see shortly. The Cartesian coordinates of the P_j relative to origin O are the tuple (t_1,\ldots,t_n). Note the coefficient of O is determined by the tuple.
If PQ = tRS for some scalar t (with P\not=Q and R\not=S) then PQ and RS are congruent and we write t = \frac{PQ}{RS}. If R(t) = (1 - t)P + tQ = P + t(Q - P) we can think of R(t) as the point P plus a scalar multiple of the vector Q - P.
Exercise. Show R(t) = Q + (1 - t)(P - Q).
Since R(t)Q = (1 - t)PQ and PR(t) = tPQ. we have R(t) = \frac{RQ}{PQ}P + \frac{PR}{PQ}Q.
Exercise. Show PQR = 0.
Exercise. Show R(t) \not= Q - P for any t.
Note PQ(Q - P) = 0.
Exercise. Show \frac{(Q - P)Q}{PQ} = -1 and \frac{P(Q - P)}{PQ} = 1.
In general if P_0\cdots P_k \not=0 and P_0\cdots P_kR = 0 then R = \sum_{j=0}^k \frac{P_0\cdots R \cdots P_k}{P_0\cdots P_k} P_j where R takes the place of P_j in the numerator of each congruence ratio.
Exercise. Show {P_1\cdots R \cdots P_k = (-1)^jR\prod_{i\not=j}P_i}.
Hint: PQ = -QP
The convex hull of points P_0,\ldots,P_k in E is \operatorname{co}(P_0,\ldots,P_k) = \{\sum_{j=0}^k t_j P_j\mid t_j \ge 0, \sum_j t_j = 1\}. Clearly P_j is in the convex hull for all j.
Exercise. Show if Q and R are in the hull then every convex combination of Q and R is also in the hull.
This gives a simple way to detect if a point is in the convex hull of a set of points whose product is not 0. Given a candidate point P\in E calculate P_0\cdots P\cdots P_k where P replaces P_j, 0\le j\le k in the product. The point P\in E is in the convex hull if and only if all the congruent ratios are non-negative.
Since there is no origin in Euclidean space we have to define a cone relative to some point O\in E. We say C\subseteq\mathcal{G}(E) is a cone with origin O if C is convex and for every P\in C we have the ray from O through P, O + t(P - O) for t\ge0, is in C.
Exercise. Show the smallest cone with origin O containing P_1, \ldots, P_k is \begin{aligned} \operatorname{cone}_O(P_1,\ldots,P_k) &= \{O + \sum_{j=1}^k t_j (P_j - O)\mid t_j\ge0\} \\ &= \{(1 - \sum_{j=1}^n t_j) O + \sum_{j=0}^k t_j P_j\mid t_j\ge0\}. \\ \end{aligned} Note the similarity to the convex hull, however the coefficient of O might be negative.
In the binomial model we have x = (1, s) and X(\omega) = \omega for \omega\in\{L,H\}. Let P_b be the point in the bond dimension and P_s be the point in the stock dimension so x = (1 - 1 - s)O + P_b + s\,P_s, X(L) = (1 - R - L )O + R\,P_b + L\,P_s, and X(H) = (1 - R - H )O + R\,P_b + H\,P_s. Since x = \frac{xX(L)X(H)}{OX(L)X(H)} O + \frac{OxX(H)}{OX(L)X(H)} X(L) + \frac{OX(L)x}{OX(L)X(H)} X(H) we can find the no arbitrage conditions by a simple, if somewhat tedious, calculation. \begin{aligned} OX(L)X(H) &= (R\,OP_b + L\,OP_s)X(H) \\ &= (R\,OP_b + L\,OP_s)( (1 - R - H )O + R\,P_b + H\,P_s) \\ &= RH\,OP_bP_s + LR\,OP_sP_b \\ &= R(H-L)\,OP_bP_s \\ \end{aligned} The coefficient of O is determined by the other two coefficient so we write \_O to streamline calculations. For the bond coefficient we compute \begin{aligned} OxX(H) &= (OP_b + s\,OP_s)X(H) \\ &= (OP_b + s\,OP_s)(\_O + R\,P_b + H\,P_s) \\ &= H\,OP_bP_s + sR\,OP_sP_b \\ & = (H - sR)\,OP_bP_s \\ \end{aligned} For the stock coefficient we compute \begin{aligned} OX(L)x &= (R\,OP_b + L\,OP_s)x \\ &= (R\,OP_b + L\,OP_s)(\_O + P_b + s\,P_s) \\ &= Rs\,OP_bP_s + L\,OP_sP_b \\ &= (Rs - L)OP_bP_s \\ \end{aligned} Since R and H - L are positive the no arbitrage condition are {H - sR\ge0} and {Rs - L\ge0} so L \le Rs \le H. We have already established this without using Grassmann algebra but generalizations to higher dimensional cases can be reduced to a calculation.
Exercise. What is the cone with origin O generated by \{X(\omega)\mid L\le\omega\le H\}.
Hint: It is the same as the cone with origin O generated by X(L) and X(H).
We can add a call option with price o and strike K\in(L,H). The one period model is now x = (1, s, o) and {X(\omega) = (R, \omega, \max\{\omega - K, 0\})} for {\omega\in[L,H]}.
Exercise. Show the cone with origin O generated by the range of X is the same as that generated by X(L), X(K), and X(H).
Hint: Every X(\omega) is a convex combination of either X(L) and X(K) if L\le\omega\le K or X(K) and X(H) if K\le\omega\le H.
If P_o is a point in the option dimension then x = \_O + P_b + s\,P_s + o\,P_o and X(\omega) = \_O + R\,P_b + \omega\,P_s + (\omega - K)^+\,P_o.
The arbitrage free conditions are reduced to calculating congruence ratios as above. \begin{aligned} OX(L)X(K)X(H) &= (R\,OP_b + L\,OP_s)X(K)X(H) \\ &= (R\,OP_b + L\,OP_s)(\_O + R\,P_b + K\,P_s)X(H) \\ &= (RK\,OP_b P_s + LR\,OP_s P_b)(\_O + R\,P_b + H\,P_s + (H - K)\,P_o) \\ &= RK(H - K)\,OP_b P_s P_o + LR(H - K)\,OP_s P_b P_o \\ &= R(H - K)(K - L)\,OP_b P_s P_o \\ \end{aligned} The coefficient of X(L) is \begin{aligned} OxX(K)X(H) &= (OP_b + s\,OP_s + o\,OP_o)X(K)X(H) \\ &= (OP_b + s\,OP_s + v\,OP_o)(\_O + R\,P_b + K\,P_s)X(H) \\ &= (RK\,OP_b P_s + sR\,OP_s P_b)(\_O + R\,P_b + H\,P_s + (H - K)\,P_o) \\ &= RK(H - K)\,OP_b P_s P_o + sR(H - K)\,OP_s P_b P_o \\ &= R(H - K)(K - s)\,OP_b P_s P_o \\ \end{aligned} The coefficient of X(K) is \begin{aligned} OX(L)xX(H) &= (R\,OP_b + L\,OP_s)xX(H) \\ &= (R\,OP_b + L\,OP_s)(P_b + s\,P_s + o\,P_o)X(H) \\ &= (Rs\,OP_b P_s + Ro\,OP_b P_o + L\,OP_s P_b + Lo\,OP_s P_o)X(H) \\ &= (Rs - L)OP_b P_s + Ro\,OP_b P_o + Lo\,OP_s P_o)( R\,P_b + H\,P_s + (H - K)\,P_o) \\ &= ((Rs - L)(H - K) - RoH + LoR) OP_b P_s P_o \\ &= ((Rs - L)(H - K) - Ro(H - L)) OP_b P_s P_o \\ \end{aligned} The coefficient of X(H) is \begin{aligned} OX(L)X(K)x &= (R\,OP_b + L\,OP_s)X(K)x \\ &= (R\,OP_b + L\,OP_s)(R\,P_b + K\,P_s)x \\ &= (RK\,OP_b P_s + LR\,OP_sP_b)x \\ &= R(K - L)OP_b P_s(P_b + s\,P_s + o\,P_o) \\ &= (R(K - L)o\,OP_b P_s P_o \\ \end{aligned} The no arbitrage conditions are s\le K and 0\le o\le\frac{(Rs - L)(H - K)}{R(H - L)}. Of course the conditions L\le Rs \le H hold from the model not containing the option.
Exercise. Consider a model with a bond having zero interest, and a stock that starts at 100 and can go to either 90, 100, or 110. What are the arbitrage-free values of an at-the-money call?
Hint. The model is x = (1, 100, o), X(\omega) = (1, \omega, \max\{\omega - 100, 0\}) where \omega\in\{90,100,110\}. Show 0\le o\le 5.
Exercise. Remove the bond from the above model and answer the same question.
Hint: The answer is 0\le o\le 100/11 = 9.090\ldots.
Grassmann algebra can also be used to detect and find an arbitrage when it exists. If any of the coefficients are negative then arbitrage exists. If only one coefficient is negative the arbitrage is perpendicular the the hyperplane determined by the points having positive coefficients.
We can also extend this to an arbitrary number of call options with strikes between the low and high values for the underlying instrument. The author implemented this for the option trading book at Banc of America Securities. We found the only arbitrage opportunities were negative price butterfly spreads at adjacent traded strikes, but the bid/ask spread turned the price positive.
We can make the one-period model more realistic. The price of an instrument must be an integral multiple of its minimal trading increment, or tick size. Let \epsilon(i) be the tick size of instrument i\in I. Initial prices x\in\boldsymbol{Z}^I correspond to actual prices x(i)\epsilon(i). Likewise for final prices X\colon\Omega\to\boldsymbol{Z}^I.
Price depends on the amount being bought or sold. The amount must be an integral multiple of its minimum share size, or lot size. Let \delta(i) be the lot size of instrument i\in I. Amounts \xi\in\boldsymbol{Z}^I correspond to amounts \xi(i)\delta(i). The bid and ask are a function of each instrument and the amount being purchased. Initial price is a function x\colon\boldsymbol{Z}\times I\to\boldsymbol{R} where x(\xi,i) is the price of \xi lot sizes of i\in I. For example x(\xi,i) = x_i + \operatorname{sgn}(\xi)\eta where x_i is the mid price models a fixed bid/ask spread. Recall the signum \operatorname{sgn}(\xi) = 1 if \xi > 0, \operatorname{sgn}(\xi) = -1 if \xi < 0, and \operatorname{sgn}(0) = 0.
A more realistic model is x(\xi,i) = x_i + \operatorname{sgn}(\xi)|\xi|\eta where the bid/ask spread is proportional to how much is being transacted. For exchange traded instruments for which order book levels are available we can model the price quite accurately.
Recall the set exponential B^A = \{f\colon A\to B\} is the set of all functions from the set A to the set B. If x\colon I\to\boldsymbol{R} then x(i) is the price of instrument i\in I. The dot product of x,y\in\boldsymbol{R}^I is {x\cdot y = \sum_{i\in I}x_i y_i} if I is finite. The Euclidean norm of {x\in\boldsymbol{R}^I} is defined by {\|x\| = \sqrt{x\cdot x}}.↩︎