Oct 11, 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 to say about counterparty risk.
The model defines the initial vector of tradable 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 risk-neutral valuation measures. Positive measures having mass 1 make a showing, but it is a mistake to assume they are the actual probability of anything.
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 show up. 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. We conclude with some remarks about the Capital Asset Pricing Model.
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. For {x\in\boldsymbol{R}^I} we define the Euclidean norm by {\|x\|_2^2 = \sum_i x(i)^2} and for {X\colon\Omega\to\boldsymbol{R}^I} the supremum norm is {\|X\|_\infty = \sup_{\omega\in\Omega} \|X(\omega)\|_2}. We assume the final prices are bounded, as they are in the real world.
Arbitrage (in this model) exists if we can find a position \xi\in\boldsymbol{R}^I that makes money when put on and never loses money when unwound: \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.
Q(uant): Hey, I've got an arbitrage for you!
T(rader): Bring it on! How much do I make on the front end?
Q: You make 0.
T: Oh. Well how much do I make on the back end?
Q: A positive amount.
T: How "positive" are we talking here?
Q: All I can tell you is that it is greater than 0.Assuming the trader is still talking to you…
T: What is the probability of making this positive amount?
Q: A strictly positive probability.
T: More than 1%?
Q: All I can tell you is that it is greater than 0.…at this point it is only a matter of how far away the quant will land after getting booted off the trading floor.
It is high time for academics to stop spouting mathematically correct, but totally useless statements. Not only have I seen this happen in person on trading floors, I am also guilty of this!2
Quants turn mathematical models into software used for trading. If a model is deployed without ensuring it is arbitrage-free then savvy buy-side clients will exploit mispricing and lift 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. Or their 233-year old bank goes out of business. (Board of Banking Supervision 1995) Board of Banking Supervision Report
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 buy or sell as t gets large. Every instrument has a finite float. This is a real problem for large hedge funds.
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.
A zero coupon pays one unit at the end of the period. Let its initial price/discount be x = D and X(\omega) = 1 for all \omega\in\Omega. A common misconception is the absence of arbitrage implies D\le 1 since D > 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.
The smallest closed cone containing the range of X is [0,\infty) so there is no arbitrage if and only if D\ge0.
Exercise. If D < 0 find an 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 directly “what can happen” over the interval.
Exercise. Show x = X(1)/4 + X(3)/4.
Hint. Solve {x = X(1)D_1 + X(3)D_3} for D_1 and D_3.
This shows x belongs to the cone determined by the range of X so there is no arbitrage.
Suppose the initial stock price is s instead of 1. What are the arbitrage-free values for s? Solving {x = X(1)D_1 + X(3)D_3} gives {D_1 = (3 - 2s)/4} and {D_3 = (-1 + 2s)/4} where we write {D_\omega} for Ross’s linear pricing measure {D(\{\omega\})}, {\omega\in\{1,3\}}. If x is in the smallest closed cone containing the range of X then {D_1,D_3\ge0} so {1/2\le s\le 3/2}.
If s < 1/2 then the stock is cheap so we should buy that and invest in the bond.
Exercise. If s < 1/2 then \xi = (-2,1) is an arbitrage.
If s > 3/2 then the stock is rich so we should short it and buy the bond to finance that.
Exercise. If s > 3/2 then \xi = (2,-3) is an arbitrage.
Do you notice a pattern here? A line perpendicular to a line with slope m has slope {m^\perp = -1/m}. A line perpendicular to a line with slope 1/2 has slope -2/1. A line perpendicular to a line with slope 3/2 has slope 2/-3.
We will see later how this generalizes to finding arbitrage in any one-period model.
Now let’s add a call option with strike 2 to the model. We have {x = (1, 1, v)}, where v is the value of the option, and {X(\omega) = (2, \omega, \max\{\omega - 2, 0\})} for {\omega\in\{1,3\}}.
As we’ve already seen, the bond and the stock determine the linear pricing measure D_1 = 1/4 and D_3 = 1/4. This implies the arbitrage-free option price is {v = \max\{1 - 2, 0\}/4 + \max\{3 - 1, 0\}/4 = .25}.
Notice this result does not involve probability, only geometry.
Exercise: What is the no-arbitrage value of the call option if the initial stock price is s?
Also notice this result does not make sense to traders if they think the price goes to 3 with probability .9 and to 1 with probability .1. For a quarter they can get a dollar 90% of the time. Discounted by 1/2 that works out to 45 cents. The mathematical theory gives the same call option price if it goes to 3 with probability .99 and to 1 with probability .01. The one-period model is oblivious to the possibility of multiple trading opportunities.
As John Illuzi pointed out to me back when we worked at Banc of America Securities after the Glass-Steagall legislation was dismantled in the 90’s and Bank of America still had to spell the equity group they acquired from Montgomery Securities with a ‘c’, this only makes sense if someone puts a gun to your head if you ever lose money on the call option over a period.
Risk-neutral pricing is risk blind.
The binomial model generalizes the 1-2-3 model by assuming the bond has realized return R and the stock can go to either a low or high value {\Omega = \{L, H\}}. It specifies {x = (1, s)} and {X(\omega) = (R, \omega)}.
The model is arbitrage-free if and only if x = X(L) D_L + X(H) D_H for some non-negative D_L and D_H. The general solution is D_L = (H - Rs)/R(H - L) and D_H = (-L + Rs)/R(H - L) and the no arbitrage condition holds if and only if L/R \le s \le H/R.
Exercise. Find an arbitrage if s < L/R or s > H/R.
Hint. Consider -R/L and R/-H.
A somewhat more realistic model would allow the stock price to be any value between the low and the high, \Omega = [L,H]. There seems to be no common name for this so we will call it the interval model.
Exercise. Show the smallest closed cone containing X([L,H]) is the same as the smallest closed cone containing X(L) and X(H).
Hint: X(\omega) = (1 - t) X(L) + t X(H) for some t\in[0,1] when L\le\omega\le H.
This shows the binomial no-arbitrage condition L/R\le s\le H/R is the same as the more general interval model.
If we add a call option with strike k the model is {x = (1, s, v)} and {X(\omega) = (R, \omega, \max\{\omega - k, 0\})}, {\omega\in\{L,H\}}. The bond and stock determine the linear pricing operator and the arbitrage-free price of the option is
The Fundamental Theorem of Asset Pricing characterizes arbitrage-free models and provides an arbitrages if they are not.
If we were being intellectually honest about the one-period model we would have to acknowledge the final prices must be zero since there is no further economic activity. The final prices are really cash flows that accrue in proportion to the position held, just like for stock dividends and bond coupons. The assumption we liquidate the position at the end of the period obscures this. The Multi-period model clarifies the relationship between prices and cash flows.
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.
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.