A Practical Diagnostic Framework
In a finite market, arbitrage, equivalent martingale measures, and completeness can all be tested with finite-dimensional calculations. The central distinction is between which payoffs the available trading strategies can produce and which probability weights make discounted asset prices fair. The first is a rank question; the second is a system of linear equations with strict positivity constraints.
Consider a finite probability space with states \(\omega_1,\ldots,\omega_m\), each having positive probability under \(\mathbb{P}\). In a one-period model, let \(S_0\in\mathbb{R}^d\) be the vector of current discounted risky-asset prices, and let \(S_1(\omega_j)\in\mathbb{R}^d\) be their discounted prices at the terminal date. The discounted cash account has price \(1\) at both dates. A portfolio holds \(\theta\in\mathbb{R}^d\) risky assets and a cash position. Its zero-cost gain in state \(\omega_j\) is \(\theta\cdot(S_1(\omega_j)-S_0)\).
For a claim with terminal payoff vector \(h=(h_1,\ldots,h_m)\), a replicating portfolio is one whose terminal payoff equals \(h_j\) in every state. The market is complete if every vector in \(\mathbb{R}^m\) can be replicated. Prices and payoffs here are discounted, so no further discount factor is needed.
No-Arbitrage and Risk-Neutral Probabilities
Proof. Write \(x_j=S_1(\omega_j)-S_0\), and let \(C\) be the convex hull of the vectors \(x_1,\ldots,x_m\). If \(q_j>0\), \(\sum_jq_j=1\), and \(\sum_jq_jx_j=0\), then for any \(\theta\) with \(\theta\cdot x_j\geq0\) in every state, \(\sum_jq_j(\theta\cdot x_j)=0\). A strictly positive weighted average of nonnegative numbers can equal zero only if every number is zero. Thus no such \(\theta\) gives an arbitrage. Conversely, the finite-dimensional separating-hyperplane theorem implies that if \(0\) is not in the relative interior of \(C\), there is a linear functional that is nonnegative on every \(x_j\) and strictly positive on at least one, or is strictly positive on all of them. This gives an arbitrage. Therefore absence of arbitrage implies \(0\) is in the relative interior of \(C\). A point in the relative interior of the convex hull of a finite list has a convex representation using strictly positive weights on every listed point: if an initial representation has zero weights, a sufficiently small move from \(0\) away from the average of the omitted points remains in \(C\), and combining that point with the omitted points gives a representation with positive weights throughout. These weights satisfy \(\sum_jq_jx_j=0\); this is exactly the martingale-measure equation. \(\square\)
The separation result used in this proof is the standard finite-dimensional separating-hyperplane theorem. This criterion is useful computationally: solve the martingale equations, then check that a solution has every component strictly positive. A solution with a zero component is not an equivalent measure.
Worked Example: A Complete Two-State Market
A discounted stock costs \(4\) now and is worth \(2\) or \(6\) at the terminal date. Both states have positive \(\mathbb{P}\)-probability. Test for arbitrage, determine completeness, and price a claim paying \((0,2)\) in the two states.
Let \(q\) be the risk-neutral probability of the state where the stock is worth \(2\). The martingale equation is \(4=2q+6(1-q)\). Expanding gives \(4=6-4q\), so \(q=\tfrac12\); the other state also has probability \(\tfrac12\). Both probabilities are strictly positive, so the theorem gives no arbitrage.
The terminal payoff matrix for cash and stock is \(\begin{pmatrix}1&2\\1&6\end{pmatrix}\). Its determinant is \(1\cdot6-1\cdot2=4\), so its rank is \(2\), the number of states. The claim is therefore attainable. Directly, if \(\alpha\) is the cash position and \(\beta\) the stock position, solve \(\alpha+2\beta=0\) and \(\alpha+6\beta=2\). Subtraction gives \(4\beta=2\), hence \(\beta=\tfrac12\) and \(\alpha=-1\). Its current cost is \(-1+\tfrac12\cdot4=1\).
Check: at the terminal date the portfolio pays \(-1+\tfrac12\cdot2=0\) in the first state and \(-1+\tfrac12\cdot6=2\) in the second. The risk-neutral value is also \(\tfrac12\cdot0+\tfrac12\cdot2=1\).
Rank, Completeness, and Claim Prices
Put the attainable terminal payoffs into a matrix \(A\), with states as rows and traded payoff vectors as columns. For a one-period market with cash and \(d\) risky assets, its columns are the vector of ones and the \(d\) terminal asset-payoff vectors. A portfolio produces a vector \(A\gamma\), where \(\gamma\) lists its holdings.
Proof. A payoff vector \(h\in\mathbb{R}^m\) is attainable exactly when \(h=A\gamma\) for some portfolio vector \(\gamma\), which means exactly that \(h\) belongs to the column space of \(A\). Every payoff is attainable if and only if this column space is \(\mathbb{R}^m\), which is equivalent to \(\operatorname{rank}(A)=m\). For the second claim, an equivalent martingale measure probability vector \(q\) solves \(A^\mathsf{T}q=p\), where \(p\) lists the current prices of the traded assets, including cash. If \(A\) has rank \(m\), any two solutions have a difference orthogonal to every column of \(A\), so their difference is zero. If the rank is less than \(m\), there is a nonzero vector \(z\) orthogonal to every column. The cash column implies \(\sum_jz_j=0\). Starting from a strictly positive solution \(q\), choose \(\varepsilon>0\) small enough that \(q_j+\varepsilon z_j>0\) for every \(j\). Then \(q+\varepsilon z\) is a distinct probability vector solving the same pricing equations. Hence the equivalent martingale measure is not unique. \(\square\)
For any proposed price \(p\) of a new claim \(h\), the enlarged market is arbitrage-free exactly when some existing equivalent martingale measure also satisfies \(p=\sum_jq_jh_j\). Indeed, applying the no-arbitrage criterion to the enlarged list of assets adds precisely this one pricing equation. Thus the set of arbitrage-free prices is the set of expectations \(\{\sum_jq_jh_j:\mathbb{Q}\text{ is an equivalent martingale measure}\}\). In a complete market this set is a singleton. In an incomplete market it can contain an interval; whether its endpoints are included depends on whether positive-probability martingale measures attain them.
Worked Example: An Incomplete Market and Its Price Range
A stock costs \(3\) and has terminal values \(1,3,5\) in three states, each of positive probability under \(\mathbb{P}\). Find the martingale measures and classify the arbitrage-free prices of a claim with payoff \(h=(0,1,0)\).
Let the state probabilities under \(\mathbb{Q}\) be \(q_1,q_2,q_3\). They must satisfy \(q_1+q_2+q_3=1\) and \(q_1+3q_2+5q_3=3\). Subtracting three times the first equation from the second gives \(-2q_1+2q_3=0\), so \(q_1=q_3=a\) and \(q_2=1-2a\). Strict positivity requires \(0<a<\tfrac12\). Such measures exist, so the market has no arbitrage.
The payoff matrix has rows \((1,1),(1,3),(1,5)\). Its rank is \(2<3\), so the market is incomplete. Under the martingale measure indexed by \(a\), the claim price is \(q_2=1-2a\). As \(a\) ranges strictly between \(0\) and \(\tfrac12\), this price ranges strictly between \(0\) and \(1\). For instance, at \(a=\tfrac13\), the probabilities are \((\tfrac13,\tfrac13,\tfrac13)\), and the claim price is \(\tfrac13\).
The claim is not replicable: if \(\alpha+\beta(1,3,5)=(0,1,0)\), the first and third coordinates imply \(\alpha+\beta=0\) and \(\alpha+5\beta=0\). Subtracting gives \(4\beta=0\), so \(\beta=0\) and \(\alpha=0\), contradicting the middle coordinate. At price \(0\), buying the claim is a zero-cost nonnegative payoff, positive in the middle state. At price \(1\), buying one unit of cash and shorting the claim costs zero and pays \((1,0,1)\). Thus neither endpoint is arbitrage-free. Prices below \(0\) or above \(1\) also create arbitrage: for \(p<0\), combine a long claim with cash position \(-p\); for \(p>1\), combine a short claim with cash position \(p\). The respective payoffs are \(h-p\mathbf{1}>0\) and \(p\mathbf{1}-h>0\) in every state.
A Two-Period Market: Test Locally, Replicate Dynamically
In a multi-period market, testing only the initial and terminal stock prices can miss the strategies available after observing intermediate information. Consider a two-period tree with discounted stock price \(4\) initially, \(6\) or \(2\) after the first period, and then \(9\) or \(3\) after the upper node and \(3\) or \(1\) after the lower node. Take the four terminal paths \(UU,UD,DU,DD\) to have positive probability under \(\mathbb{P}\), for example probability \(\tfrac14\) each.
Worked Example: A Two-Period Complete Tree
At the initial node, the risk-neutral probability of an up move is determined by \(4=6q+2(1-q)\), giving \(q=\tfrac12\). At the upper node, \(6=9q_U+3(1-q_U)\) gives \(q_U=\tfrac12\). At the lower node, \(2=3q_D+1(1-q_D)\) gives \(q_D=\tfrac12\). Each conditional probability and its complement is positive. The resulting terminal probabilities under \(\mathbb{Q}\) are \(\mathbb{Q}(UU)=\mathbb{Q}(UD)=\mathbb{Q}(DU)=\mathbb{Q}(DD)=\tfrac14\). At each node the conditional expected next stock price equals its current price, so the discounted stock satisfies the martingale condition under this measure.
Consider a claim paying \(4,0,2,0\) on \(UU,UD,DU,DD\), respectively. At the upper node, the value needed to match the two terminal payoffs is \(\tfrac12\cdot4+\tfrac12\cdot0=2\); at the lower node it is \(\tfrac12\cdot2+\tfrac12\cdot0=1\). At the upper node, stock and cash holdings \((\beta,\alpha)=(\tfrac23,-2)\) produce \(4\) when the stock is \(9\), since \(-2+\tfrac23\cdot9=4\), and \(0\) when it is \(3\), since \(-2+\tfrac23\cdot3=0\). Their value at that node is \(-2+\tfrac23\cdot6=2\). At the lower node, holdings \((1,-1)\) produce \(2\) at stock price \(3\) and \(0\) at stock price \(1\); their value at the node is \(-1+1\cdot2=1\).
At time zero, holdings \((\beta,\alpha)=(\tfrac14,\tfrac12)\) have value \(\tfrac14\cdot4+\tfrac12=\tfrac32\). They are worth \(\tfrac14\cdot6+\tfrac12=2\) at the upper node and \(\tfrac14\cdot2+\tfrac12=1\) at the lower node, matching the required continuation values. Rebalancing to the node-specific holdings just computed therefore replicates the claim. Directly, the risk-neutral expectation is \(\tfrac14\cdot4+\tfrac14\cdot0+\tfrac14\cdot2+\tfrac14\cdot0=\tfrac32\), the replication cost.
Dynamic completeness can also be checked by rank. The following four attainable terminal payoff vectors are the cash payoff, the stock payoff from holding one share from time zero, and the zero-cost gains from holding one share only after reaching the upper or lower node, respectively: \((1,1,1,1)\), \((9,3,3,1)\), \((3,-3,0,0)\), and \((0,0,1,-1)\). To check their independence, suppose a linear combination with coefficients \(a,b,c,d\) is zero. Its four coordinates give \(a+9b+3c=0\), \(a+3b-3c=0\), \(a+3b+d=0\), and \(a+b-d=0\). Subtracting the second equation from the first gives \(c=-b\); substitution into the first gives \(a=-6b\). The third then gives \(d=3b\), and the fourth gives \(-8b=0\). Hence \(a=b=c=d=0\), so the rank is \(4\), the number of terminal states.
Common Pitfall: Static Rank Is Not Dynamic Rank
In the two-period example, using only the terminal cash and stock columns gives a matrix with rows \((1,9),(1,3),(1,3),(1,1)\), whose rank is at most \(2\). Concluding that the market is incomplete from this matrix would be wrong: it omits strategies that change holdings after the first-period price is observed. The correct dynamic payoff matrix must include attainable payoffs from trading at intermediate dates, such as the upper-node and lower-node strategies above. Its rank is \(4\), so every terminal payoff can be replicated.
A second diagnostic error is accepting nonnegative state probabilities with a zero component as proof of no arbitrage. Equivalence requires strictly positive probabilities in every state. The martingale equations must be solved together with this positivity check; the distinction matters for both arbitrage detection and whether a candidate claim price is admissible.
Check Your Understanding
For each question, keep strict positivity of equivalent state probabilities in view.
- In a two-state market, a stock costs \(5\) and pays \(3\) or \(7\). Find the equivalent martingale probabilities and test completeness by rank.
- For terminal stock values \(1,3,5\) and initial price \(3\), find the range of prices for a claim paying \((1,0,0)\).
- Explain why a solution of the martingale equations with one zero probability does not establish absence of arbitrage.
- In a four-state, two-period tree, explain which strategies must be included when constructing a payoff matrix for a dynamic completeness test.
- Show that if two equivalent martingale measures give different expectations for a claim, that claim cannot be replicated.