Sep 12, 2026
The One‑Period Model is the most naive framework for rigorously representing a financial market over a single period of time. It ignores salient features of actual markets by assuming there is no bid/ask spread, instruments can be bought or sold in any amount with no effect on the price, and has nothing whatsoever to say about counterparty risk.
The model defines the initial vector of tradeable instruments prices and their terminal prices contingent on the realized outcome. If there are no arbitrage opportunities available then prices are subject to a geometric constraint determined by the final prices: the initial price vector must belong to the smallest closed cone containing all possible terminal prices. This is equivalent to what (Ross 1976) calls a “(not generally unique) positive linear operator that can be used to value all marketed assets.” We will show such operators correspond to positive, finitely-additive measures on the space of outcomes and call them valuation measures.
There is no need for probability. Everything follows from geometry.
The Model section defines the (unrealistic) one-period model and proves the “easy” direction of the Fundamental Theorem of Asset Pricing. This motivates why cones make a showing. In the Examples section we consider various models and how the FTAP can be usefully applied. The FTAP section proves the “hard” direction and shows how to find an arbitrage if it exists.
Let I be the set of market instruments and x\in\boldsymbol{R}^I be an element of the set exponential1 denoting initial prices. The sample space \Omega is the set of what can happen over the period. The function X\colon\Omega\to\boldsymbol{R}^I determines the final prices X(\omega)\in\boldsymbol{R}^I given \omega\in\Omega occurs.
Arbitrage (in this model) exists if we can find a position \xi\in\boldsymbol{R}^I that makes money when you put it on and never loses money when you unwind it. In math speak, \xi\cdot x < 0 and \xi\cdot X(\omega)\ge0 for all \omega\in\Omega. Note that this definition does not involve probability.
This mathematical definition of arbitrage is not good enough for traders or risk managers. They will compare |\xi\cdot x| with |\xi|\cdot |x| to determine their return on investment. It makes no financial sense to tie up a million bucks to make one buck.
A common definition of arbitrage in mathematical finance literature is to superfluously assume there is a probability measure P on \Omega and define arbitrage to be a portfolio {\xi\in\boldsymbol{R}^I} with {\xi\cdot x = 0} and {\xi\cdot X\ge0} where {\xi\cdot X > 0} on some set {A\subseteq\Omega} with {P(A) > 0}. This mathematical definition of arbitrage is fatuous.
Assuming the trader is still talking to you…
…at this point it is only a matter of how far away the quant will land after getting booted off the trading floor.
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 mispricing by buying trades that are undervalued and pass on overvalued 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 has to take losses when changing to the latest and greatest new model. Even worse, a “clever” sell-side 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 for the one-period model is that there are no arbitrage opportunities if and only if the initial prices belong to the smallest closed cone containing the final prices.
A cone in a vector space is a subset that is closed under multiplication by positive scalars and vector addition.
Exercise. Show the set of all arbitrages is a cone.
Hint: Show if \xi\in\boldsymbol{R}^I is an arbitrage then so is t x for all positive t\in\boldsymbol{R} and if \xi,\eta\in\boldsymbol{R}^I are arbitrages then so is \xi + \eta.
Another reason this mathematical definition of arbitrage does not accurately model reality is that you will eventually run out of instruments to purchase as t > 0 gets large. Every instrument has a finite float. Just ask any large hedge fund.
There is a connection between cones and convex sets.
Exercise. Show cones are convex.
Hint: A set C\subseteq\boldsymbol{R}^I is convex if and only if the line segment connecting two points in the set is contained in the set. For x,y\in C we have {(1 - t)x + ty\in C} for {0 \le t \le 1}.
Exercise. Show if C\subseteq\boldsymbol{R}^I is convex then \boldsymbol{R}^+ C is a cone.
Hint: Where \boldsymbol{R}^+C = \{t x\mid t > 0, x\in C\}.
Note any finite linear combination {x = \sum_{\omega_j\in\Omega} X(\omega_j) D_j}, D_j > 0 is in any cone containing the range of X, {X(\Omega) = \{X(\omega)\mid\omega\in\Omega\}}.
Exercise. Show for any such x that \xi\cdot X(\omega)\ge0 for all \omega\in\Omega implies \xi\cdot x\ge0.
This shows there is no arbitrage for such x.
Since x_n\to x and \xi\cdot x_n\ge0 imply \xi\cdot x\ge0 we have the “easy” direction of the FTAP.
Theorem. (FTAP “easy”) If x belongs to the smallest closed cone containing the range of X then there are no arbitrage opportunities.
The converse is also true but requires more mathematical machinery. The elementary proof given below also shows how to find an arbitrage when one exists.
Before proving the one-period FTAP let’s consider some examples of one-period models.
The simplest possible one-period model has one instrument with initial price x = (1) and final price X(\omega) = (R) for all \omega. Since X is constant we can let \Omega be any one element set. This is a riskless zero coupon bond/cash deposit having realized return R.
A common misconception is the lack of arbitrage implies the realized return R\ge 1 since R < 1 implies negative interest rates. Interest rates have been negative in many financial markets. For example, from 2016 to 2024 the Bank of Japan had a negative interest rate policy. This was only available to large corporations depositing in the central bank, not the general public, but we are ignoring counterparty considerations.
We can use elementary algebra to prove the no-arbitrage condition is R > 0. If R\le0 and \xi=-1 then \xi\cdot x = \xi x = -1 and \xi R \ge0 so arbitrage exists. If R > 0 and \xi\cdot x = \xi < 0 then \xi R < 0 so there is no arbitrage.
In terms of cones there is no arbitrage if and only if x belongs to the smallest closed cone containing the range of X. If R > 0 the cone is [0,\infty). Since 1\in[0,\infty) there is no arbitrage. If R\le 0 the cone is (-\infty, 0]. Since 1\not\in(-\infty, 0] there is arbitrage.
Consider a model with a bond and stock where their initial prices are both 1, the bond always goes to 2, and the stock can go to either 1 or 3. The model is x = (1,1), X(\omega) = (2, \omega) where \omega\in\{1,3\} = \Omega. Note how the sample space models in the most direct fashion “what can happen” over the interval. If x = X(1)D_1 + X(3)D_3 for some D_1,D_3\in\boldsymbol{R} then D_1 = D_3 = 1/4 so the model is arbitrage free.
What are the arbitrage-free models given initial stock price s instead of 1? In this case D_1 = (3 - 2s)/4 and D_3 = (-1 + 2s)/4 so D_1,D_3\ge0 imply s\le 3/2 and s\ge 1/2. This is intuitively obvious. If s > 3/2 then the stock is rich so we should short the stock and buy the bond to finance that. If s < 1/2 then the stock is cheap so we should buy that and invest in the bond.
Exercise. If s > 3/2 then \xi = (2,-3) is an arbitrage.
Exercise. If s < 1/2 then \xi = (-2,1) is an arbitrage.
Do you notice a pattern here?
Stop reading here.
Option with strike 2. R > 0, 1/2 \le s \le 3/2.
Logical entropy 1 - sum p_i^2 vs Shannon entropy -sum p_i log p_i.
90-100-110
{L,H} to [L,H]. Same no-arbitrage condition but non-unique measure.
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 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.
The assumption of no arbitrage places constraints on initial prices determined by possible final prices. The constraints involve a cone.
Theorem. Arbitrage exists in a 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 establish the theorem by a purely geometric lemma residing between Farkas’ Lemma and the Hahn-Banach theorem. The Hahn-Banach theorem states this holds for locally convex topological vector spaces when K has an interior point (cite?). Every finite dimensional vector space is a locally convex topological vector space(cite?). In this case we do not require K have an interior point.
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 valuation 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.