January 19, 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 tradable instruments and their terminal cash flows contingent on the realized outcome. If there are no arbitrage opportunities available then prices are subject to geometric constraints determined by the cash flows. Arbitrage must always be defined relative to a particular model. These constraints are applied to zero coupon bonds, the binomial model, and the interval model.
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 in prices, much less any consideration of credit or tax issues. We also ignore the fact instruments can only be purchased in integral multiples of their minimum share/lot size. The Appendix posits a model that can incorporate more realistic assumptions.
Quants turn mathematical models into software used for trading. If a model is deployed without ensuring it is arbitrage-free then buy-side clients will exploit mispricings by buying trades that are undervalued and pass on overpriced ones. They don’t need to develop sophisticated models to do this, they just get quotes from several sell-side firms and take the lowest offer. Eventually trading reality catches up and the sell-side firm loses money. Even worse, a “clever” trader might find an internal arbitrage that gives the illusion of making profits until risk management figures out what is going on.
The Fundamental Theorem of Asset Pricing characterizes arbitrage-free models and provides an arbitrages if they are not. As (Ross 1978) showed, this is a purely geometric result having nothing to do with probability. Positive measures having mass one make a showing, but they are not the probability of anything.
Unlike (Black and Scholes 1973), and (Merton 1973), Ross’s model applies to any collection of instruments, not just a bond, stock, and option. It seems to be underappreciated that Ross also showed there is no need for Ito processes, partial differential equations, or a so-called real-world measure that gets immediately thrown out for a risk-neutral measure.
However, Ross made the untenable assumption that continuous time trading is possible and the category error of defining a cash flow as a jump in price. Stock prices jump between market close and market open but there is no associated cash flow. A cash flow is a payment made by the instrument issuer to instrument holders. Stocks pay dividends, bonds pay coupons, futures pay the change in end-of-day quotes (and always have price zero).
The Fundamental Theorem of Asset Pricing shows models of cash flows entail geometric constraints on arbitrage-free prices.
The One-Period Model specifies a finite set of tradable instruments I. The set of possible outcomes \Omega represents 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 cash flows are defined by a vector-valued function {C\colon\Omega\to \boldsymbol{R}^I} where {C(\omega)\in\boldsymbol{R}^I} are the cash flows for each instrument corresponding to outcome {\omega\in\Omega}.
A position \xi\in\boldsymbol{R}^I is the number of shares purchased in each instrument at the beginning of the period. The cost of acquiring the position is {\xi\cdot x = \sum_{i\in I}\xi(i) x(i)} and results in {\xi\cdot C(\omega)} at the end of the period. The realized return of a position \xi\in\boldsymbol{R}^I is {R_\xi = \xi\cdot C/\xi\cdot x} provided \xi\cdot x\not=0.
Exercise. Show R_{t\xi} = R_{\xi} for any non-zero t\in\boldsymbol{R}.
This is actually a deleterious feature of the model. Going long (t > 0) or short (t < 0) typically affects the realized return. It also implies a portfolio strategy can be scaled to arbitrarily large positions. At some point you will run out of instruments to buy or sell.
We assume redundant instruments are removed from the model. If {\xi\cdot C = 0} then one instrument is a linear combination of the others and can be removed. This can be repeated until {\xi\cdot C\not=0} for any \xi\in\boldsymbol{R}^I. If so, the map \xi\in\boldsymbol{R}^I to \xi\cdot C\in\boldsymbol{R}^\Omega 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 possible outcomes.
Classical literature often specifies terminal prices as a function X\colon\Omega\to\boldsymbol{R}^I rather than cash flows C. From a rigorous standpoint, prices do not actually exist at the end of the period since there is no further economic activity available. The classical approach implicitly assumes the initial position is liquidated at the end of the period at prevailing prices yielding a payment of \xi\cdot X. In practice, cash flows are paid proportional to position whether or not any trading occurs.
For example, a zero coupon bond has an initial price/discount and pays a unit cash flow at termination. If you are uncomfortable using cash flows C instead of prices X when the instrument is a stock, C may be interpreted as the firm’s liquidation value or the proceeds from a stock buy back by the company at the end of the period.
The Multi-period model clarifies the relationship between prices and cash flows.
The Capital Asset Pricing Model is a one-period model where a probability measure on possible outcomes is specified.
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 C(\omega)\ge0} for all {\omega\in\Omega}: you make money acquiring the initial position and never lose money at the end of the period.
Some authors define arbitrage as a portfolio satisfying {\xi\cdot x = 0} and {\xi\cdot C\ge0} is strictly positive on some set having positive probability. We haven’t specified a probability measure so we can’t use this definition. Moreover, no trader would consider this to be an arbitrage anyway. Even though the position costs nothing other than agita to put on, the above definition has nothing definite to say about how much they will make nor how likely it is they will make it.
Our stronger probability-free definition is still not good enough for traders and risk managers. Even though {\xi\cdot x} is strictly negative they will slap absolute value signs around every number and compute {|\xi|\cdot|x|} as a proxy of how much capital will be tied up putting on the position. No business would approve using a million dollars from their funding account just to make a penny up front even though that technically satisfies our mathematical definition of arbitrage.
The assumption of no arbitrage places constraints on initial prices that are determined by cash flows. The constraints involve a cone.
Recall a cone K is a subset of a vector space closed under positive scalar multiplication and vector addition: if x\in K then tx\in K for t > 0 and if x,y\in K then {x + y\in K}.
Exercise. A cone is convex.
Hint: Recall a subset K\subseteq\boldsymbol{R}^n is convex if and only if every point on the line between two points in K also belongs to K; K, x,y\in K implies {tx + (1-t)y\in K} for {0 < t < 1}.
Exercise. The set of arbitrage positions is a cone.
The smallest cone containing the possible cash flows C is the set of finite linear combinations with positive coefficients {\{\sum_i C(\omega_i) d_i\mid \omega_i\in\Omega, d_i > 0\}}. If x = \sum_i C(\omega_i) d_i is in the cone and \xi\cdot C 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 C then there is no arbitrage.
The contrapositive is also true.
Theorem. Arbitrage exists in a one-period model if x does not belong to the smallest closed cone containing the range of C. 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 K is a closed cone in \boldsymbol{R}^n with x\not\in K then there exists {\xi\in\boldsymbol{R}^n} with {\xi\cdot x < 0} and {\xi\cdot y \ge0} for {y\in K}.
Proof. Let x^* be the point in K closest to x. It exists since K is closed and is 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}.
The lemma proves the FTAP and that \xi = x^* - x implements an arbitrage.
A risk-neutral pricing measure is any positive, finitely additive measure D on \Omega with x = \int_\Omega C\,dD. The FTAP shows no arbitrage implies this set is not empty. Every such measure corresponds to a positive linear functional on the vector space of bounded functions on \Omega. See (Dunford and Schwartz 1958). Risk-neutral pricing measures are not generally unique.
If D is a risk-neutral pricing measure then P = D/D(\Omega) is a positive measure having mass 1 so it satisfies the definition of a probability measure. 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 the expected realized return R = E^P[R_\xi] = 1/D(\Omega) is constant for any portfolio \xi\in\boldsymbol{R}^I with \xi\cdot x\not=0.
Hint: The expectation is with respect to the “probability” measure P = D/D(\Omega).
This exercise is a wake-up call to the fact risk-neutral measures are useless for risk management. The variance of a realized return can be arbitrarily large but a risk-neutral measure cannot detect excess returns to compensate for this risk.
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 C(\omega) = 1} for all {\omega\in\Omega} is a zero coupon bond. The discount of a zero coupon bond is its price {\zeta\cdot x = \int_\Omega \zeta\cdot C(\omega)\,dD(\omega) = D(\Omega)}.
Exercise. If \zeta is a zero coupon bond with only one non-zero component then that component is equal to the discount.
Exercise. If x = \int_\Omega C\,dD show x = E^P[C]D(\Omega).
This formula can be read “Prices are expected discounted cash flows.” It is a mathematically rigorous one-period example of the method used by (Graham and Dodd 1934) in Security Analysis for valuing equities.
We now apply the FTAP to particular models.
A common misconception is that the price of a zero coupon bond must not be greater than its notional since this would imply negative interest rates. Negative rates actually occured in Europe between 2014 and 2020 but did not give rise to arbitrage opportunities. As we have seen above, the only constraint is the price of the zero coupon bond must be positive.
A very simple and unrealistic one-period model consists of a bond with price 1 at the beginning of the period that has a cash flow 2 at the end and a stock with price 1 that has a cash flow of either 1 or 3. The model is x = (1, 1) and {C(\omega) = (2, \omega)} where {\omega\in\{1,3\} = \Omega}. This is arbitrage-free if and only if we can find {d_1,d_3 \ge0} with {x = C(1)d_1 + C(3)d_3}. The bond component implies {1 = 2d_1 + 2d_3} and the stock component implies {1 = 1d_1 + 3d_3} so {d_1 = d_3 = 1/4}.
This shows the risk-neutral measure is D(\{1\}) = D(\{3\}) = 1/4. The risk-neutral probability measure P = D/D(\Omega) is P(\{1\}) = P(\{3\}) = 1/2.
Exercise. Show E^P[C]D(\Omega) = \int_\Omega C\,dD = (1, 1) is the initial bond and stock price.
If we add a call option with strike 2 and price v then the model becomes {x = (1, 1, v)}, {C(\omega) = (2, \omega, \max\{\omega - 2,0\})} where v is the option value. Since the bond and stock components determine {d_1 = d_3 = 1/4} the option component is {v = \max\{1 - 2, 0\}(1/4) + \max\{3 - 2, 0\}(1/4) = 1/4}.
A similar argument shows any European option paying \nu(\omega) at expiration has value {v = (\nu(1) + \nu(3))/4}. Every option payoff is linear in a binomial model.
Note this argument does not depend on probability. If the real-world probability of the stock staying at 1 is 0.1 and the stock tripling to 3 is 0.9 then the discounted expected payoff is (0(0.1) + 1(0.9))/2 = 0.45. As John Illuzi at Banc of America securities pointed out when I showed him this, “Do you mean I can buy the option for 0.25 and get 0.45 on average? I’d take that trade all day long!” He also identified the risk-blind nature of risk-neutral probability. “But not if I got shot in the head if the option ever finished out-of-the-money.”
We have already seen every risk-neutral measure has the same expected realized return. This example shows even if the measure is unique it implies infinite risk aversion.
The 1-2-3 model is a special case of the binomial model having instruments a bond and stock where {x = (1, s)}, {C(\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. This is arbitrage-free if and only we can find {d_L,d_H \ge 0} with {x = C(L)d_L + C(H)d_H}. Considering the bond and stock components we have {d_L = (H - sR)/R(H - L)} and {d_H = (sR - L)/R(H - L)}. The model is arbitrage-free if and only if these are both non-negative so {L/R \le s \le H/R}.
Exercise. Find an arbitrage if s < L/R or s > H/R.
Hint: If s < L/R then buy the stock and short the bond. If s > H/R then sell the stock and buy the bond.
If we have a call option with payoff \max\{\omega - K,0\}, where L < K < H, its arbitrage-free value is v = 0 d_L + (H - K) d_H = (H - K)(s - L/R)/(H - L)
Exercise. Show this agrees with the 1-2-3 Model for a call option with strike 2.
A somewhat more realistic model than the binomial model is to allow the final stock price to take any value between a low and a high, {\omega\in\Omega = [L, H]}. The arbitrage-free conditions are still the same.
Exercise. The smallest cone containing C(L) and C(H) is the same as the smallest cone containing C(\omega), L\le\omega\le H.
Hint: Show if L\le\omega\le H then C(\omega) = (1 - t)C(L) + tC(H) for some t\in[0,1].
Since cones are convex, this shows the smallest cone containing C(L) and C(H) also contains C(\omega) for \omega\in(L,H).
If we introduce a call option with strike K\in (L,H) then the smallest cone containing the range of C coincides the the smallest cone containing C(L), C(K), and C(H).
Exercise. Show the smallest cone containing C(L), C(K), and C(H) is the same as the smallest cone containing C(\omega), L\le\omega\le H.
Hint: Show C(\omega) is on the line segment from C(L) to C(K) if L \le \omega \le K and is on the line segment from C(K) to C(H) if K \le \omega \le H.
Unlike the binomial model, the value of the option in the interval model is not determined by the bond and the stock. The interval model is x = (1, s, v) and C(\omega) = (R, \omega, (\omega - K)), where \omega\in [L,H]. There is no arbitrage if and only if there exist non-negative d_L, d_K, d_H with x = C(L)d_L + C(K)d_K + C(H)d_H.
The components of the three instruments are \begin{aligned} 1 &= R d_L + R d_K + R d_H \\ s &= L d_L + K d_K + H d_H \\ v &= (H - K) d_H \\ \end{aligned}
Substituting d_H = v/(H - K) and solving for d_L and d_K yields \begin{aligned} d_L &= \frac{v + K/R - s}{K - L} \\ d_K &= \frac{(s - L/R)(H - K) - v(H - L)}{(K - L)(H - K)} \\ \end{aligned} The no arbitrage constraints are v\ge 0, v \ge s - K/R, and (s - L/R)(H - K)/(H - L) \ge v. Note the binomial model option value is an upper bound in the interval model. Fixing R and s we have the allowable arbitrage-free initial option value must satisfy s - K/R \le v \le (s - L/R)(H - K)/(H - L).
If we have n call options with values v_i and strikes K_i where L < K_1 < \cdots K_n < H then {x = (1, s, v_1,\ldots,v_n)} and {C(\omega) = (R, \omega, (\omega - K_1)^+,\ldots,(\omega - K_n)^+)}, \omega\in[L,H]. There is no arbitrage if and only if there exist non-negative d_L, d_1, d_2,\ldots,d_H with x = C(L)d_L + \sum_{i=1}^n C(K_i)d_i + C(H)d_H The components are \begin{aligned} 1 &= R d_L + R d_1 + R d_2 + \cdots + R d_H \\ s &= L d_L + K_1 d_1 + K_2 d_2 + \cdots + H d_H \\ v_1 &= (K_2 - K_1) d_2 + \cdots + (H - K_1) d_H \\ v_2 &= (K_3 - K_2) d_3 + \cdots + (H - K_2) d_H \\ &\cdots \\ v_{n-1} &= (K_n - K_{n-1}) d_n + (H - K_{n-1}) d_H \\ v_n &= (H - K_n) d_H \\ \end{aligned} In matrix form \begin{bmatrix} 1 \\ s \\ v_1 \\ \vdots \\ v_{n-1} \\ v_n \end{bmatrix} = \begin{bmatrix} R & R & R & \cdots & R & R \\ L & K_1 & K_2 & \cdots & K_n & H \\ 0 & 0 & K_2 - K_1 & \cdots & K_n - K_1 & H - K_1 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & K_n - K_{n-1} & H - K_{n-1} \\ 0 & 0 & 0 & \cdots & 0 & H - K_n \\ \end{bmatrix} \begin{bmatrix} d_L \\ d_1 \\ d_2 \\ \vdots \\ d_n \\ d_H \end{bmatrix} We can write this as a block matrix on the first row and first column [A\,B; C\,D] where A = [R] B = [R \cdots R], and C = [L\,0 \cdots 0]^T. If we let K_0 = 0 and K_{n+1} = H then D is upper triangular with non-zero entries {D_{i,j} = K_j - K_{i-1}} for 1\le i \le j\le n + 1. The symmetry can be used to show the inverse is upper bidiagonal with main diagonal {D_{i,i} = 1/(K_{i+1} - K_i)}, 0\le i\le n and upper diagonal {D_{i,i+1} = -1/(K_{i + 1} - K_i)}.
If E = (D - CA^{-1}B)^{-1} is the Schur complement then \begin{bmatrix} A & B \\ C & D \\ \end{bmatrix}^{-1} = \begin{bmatrix} A^{-1} + A^{-1}BECA^{-1} & -A^{-1}BE \\ -ECA^{-1} & E \\ \end{bmatrix}
The classical one-period model neglects, among other things, the fact there are many market participants. It assumes there is only a price taker that decides how much to buy or sell. There must also be price makers exhibiting what they offer to sell or buy. We only consider accurate accounting of transactions.
The price of an instrument must be an integral multiple of its minimal trading increment, or tick size. Let \epsilon(i)\in\boldsymbol{R} be the tick size of instrument i\in I. Initial prices x\in\boldsymbol{Z}^I correspond to the actual prices x(i)\epsilon(i).
The amount purchased in each instrument 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. Shares \xi\in\boldsymbol{Z}^I correspond to actual amounts \xi(i)\delta(i).
The atoms of finance are holdings: an amount, instrument, and owner. Let {\eta = (a,i,o)} denote amount a of instrument i is held by owner o.
A position is a collection of holdings. Positions are modified by exchanges. An exchange at time t is a triple {(t,\eta,\eta')} where \eta and \eta' are holdings. If the taker o holds \eta and the maker o' holds \eta' at time t then after the exchange settles the taker holds (a',i',o) and the maker holds (a,i,o'). The price of the exchange is X = a/a'.
After an exchange, the price is a well-defined number recorded in the books and records of each counterparty.
Makers quote prices for potential exchanges. Takers decide whether or not to execute the exchange. For example, if a maker quotes Ford stock for 12 USD per share then the taker can exchange \$24 for 2 shares of Ford stock. This corresponds to the exchange where {\eta = (24, \$, o)} and {\eta' = (2, F, o')}. The price is {X = a/a' = 24/2} and we write 12 USD/F to indicate a price of \$12 per share of Ford stock. Replacing the virgule ‘/’ by ‘= 1’ yields the mnemonic 12 USD = 1 F. After the exchange settles the taker holds {\eta = (2, F, o)} and the maker holds {\eta' = (24, \$, o')} in their respective positions.
Prior to an exchange, the price is not a well-defined number. Makers quote a bid price and an ask price to takers. When a taker buys one share they pay the maker’s ask price and when the taker sells one share they only get the maker bid price. The difference between the bid and the ask is the bid-ask spread and is usually positive.2 This is the maker vigorish to get paid for providing liquidity to takers in the market.
The quotes are only valid for a small number of shares. As the number of shares becomes larger maker increase the bid-ask spread. On an exchange the quote only holds out to level 1 depth before switching to level 2. Sellers might not know why so many shares are being purchased, but know demand will move the price up and they will increase their ask. At some point the number of shares to purchase will bump into the total number of shares issued and price becomes meaningless. When a taker sells shares they face even more restrictions and are also charged a borrow cost over the period of time they are short. Prices can also depend on the particular taker and maker of the exchange due to credit issues or regulations, among other things.
A more realistic model for prices in a one-period model is to replace x\colon I\to\boldsymbol{R} by x\colon I\times A\to\boldsymbol{R} where I is the set of instruments and A is the set of possible amounts. The cost of setting up the position \xi\in\boldsymbol{R}^I is {\sum_{i\in I} \xi(i) x(i, \xi(i))}. If a limit order book is available then a very good approximation to the execution price is the function of cumulative limit order amounts versus order levels.
Exercise. Show if the order book is (l_i,a_i) and b(l) = \sum_i a_i 1(l > l_i) then the….
Not every transaction occurs on an exchange. An over-the-counter trade involves only the taker and maker as counterparties. While exchange traded instruments often settle within milliseconds, for these sort of trades, settlement time is typically measured in days and often involve risk mitigation measures such as posting collateral instead of margin accounts.
A broker is an intermediary between a taker and a maker and charge a transaction fee based on the instrument and amount transacted. A dealer is a broker that may hold transactions from taker or maker over a period of time. The amount they make is subject to market movements and settlement times are on the order of hours.