Superhedging as a Statewise Requirement
In an incomplete market, a contingent claim need not be exactly replicable. A seller who promises the claim can instead choose a trading strategy whose terminal wealth is at least the claim in every state. The least initial capital that permits this is the superhedging price. Its dual description is an upper pricing bound: it is the largest expected claim value among martingale measures, with care needed about whether the maximizing measure is equivalent to the physical measure.
We work in a one-period, finite-state market. Let the states be \(\omega_1,\ldots,\omega_m\), each with positive physical probability under \(\mathbb{P}\). Take the bank account as numeraire, so all asset prices and payoffs below are discounted; equivalently, the bank account has value one at both dates. There are \(d\) risky assets, with initial discounted price vector \(s_0\in\mathbb{R}^d\) and terminal discounted price vector \(S_1(\omega_i)\in\mathbb{R}^d\). A fixed vector \(\theta\in\mathbb{R}^d\) of shares held from time zero to time one gives terminal gain \(\theta\cdot(S_1-s_0)\). No stochastic integral is used in this one-period model.
Write \(D_i=S_1(\omega_i)-s_0\), and form the \(m\times d\) matrix \(A\) whose \(i\)-th row is \(D_i^\mathsf{T}\). Thus the gain vector across states is \(A\theta\). A martingale measure in this model is a probability vector \(q=(q_1,\ldots,q_m)\) satisfying \(q_i\ge0\), \(\sum_iq_i=1\), and \(\sum_iq_iD_i=0\). It is equivalent to \(\mathbb{P}\) when every \(q_i>0\), since every state has positive physical probability. The finite-state no-arbitrage theorem supplies an equivalent martingale measure under no arbitrage. For the dual result below, we assume directly that at least one such strictly positive vector exists.
The Dual Representation
Proof. Let \(L=\{A\theta:\theta\in\mathbb{R}^d\}\), the linear subspace of attainable gain vectors, and let \(K=L+\mathbb{R}^m_+\), where \(\mathbb{R}^m_+\) is the set of vectors with nonnegative coordinates. The inequalities \(x\mathbf{1}+A\theta\ge H\) hold for some \(\theta\) exactly when \(x\mathbf{1}-H\in K\): indeed, the difference \(x\mathbf{1}+A\theta-H\) is nonnegative, and \(x\mathbf{1}-H\) is that difference minus \(A\theta\), which belongs to \(L+\mathbb{R}^m_+\).
The set \(K\) is a closed convex cone. To check closedness, project orthogonally onto \(L^\perp\), with projection \(P\). Then \(K=\{y:Py\in P(\mathbb{R}^m_+)\}\). The cone \(P(\mathbb{R}^m_+)\) is generated by the finitely many vectors \(Pe_i\), where \(e_i\) are the coordinate vectors, and every finitely generated cone in finite-dimensional space is closed. One way to verify this last fact is to reduce each nonnegative combination to one using linearly independent generators: a linear dependence lets one adjust the coefficients until at least one becomes zero, without making any coefficient negative. For a convergent sequence of combinations, a subsequence uses the same independent subset, since there are only finitely many subsets. The coefficients for that subset converge, by linear independence, to nonnegative coefficients representing the limit. Thus the cone is closed, and so is its inverse image under \(P\).
If \(q\ge0\) and \(A^\mathsf{T}q=0\), then \(q\cdot k\ge0\) for every \(k\in K\). For any superhedge, this gives \(x-q\cdot H=q\cdot(x\mathbf{1}-H)\ge0\). When \(\sum_iq_i=1\), we have \(x\ge q\cdot H\). Taking the infimum over superhedging \(x\) proves that the dual supremum is at most \(p^*(H)\).
For the reverse inequality, let \(x<p^*(H)\), so \(z=x\mathbf{1}-H\notin K\). Since \(K\) is a nonempty closed convex cone in finite-dimensional Euclidean space, there is a closest point \(y\in K\) to \(z\). Set \(q=y-z\). The minimizing property gives \(q\cdot(k-y)\ge0\) for each \(k\in K\). Taking \(k=0\) and \(k=2y\) gives \(q\cdot y=0\); consequently \(q\cdot k\ge0\) for every \(k\in K\), and \(q\cdot z=-\lVert y-z\rVert^2<0\). Because \(K\) contains every nonnegative coordinate vector and contains \(L\) and \(-L\), these inequalities imply \(q\ge0\) and \(A^\mathsf{T}q=0\). The vector \(q\) is nonzero, so \(c=\sum_iq_i>0\). The normalized vector \(q/c\) is a martingale measure, and \(q\cdot(x\mathbf{1}-H)<0\) gives \((q/c)\cdot H>x\). This holds for every \(x<p^*(H)\); hence the dual supremum is at least \(p^*(H)\). The primal price is finite: \(x=\max_iH_i,\theta=0\) is feasible, while the strictly positive martingale measure \(q^0\) gives the lower bound \(x\ge q^0\cdot H\) for every superhedge.
Finally, if \(q\) is any nonnegative martingale measure, then \(q^\varepsilon=(1-\varepsilon)q+\varepsilon q^0\), for \(0<\varepsilon\le1\), is a strictly positive martingale measure. As \(\varepsilon\) decreases to zero, \(q^\varepsilon\cdot H\) tends to \(q\cdot H\). Thus the supremum over equivalent martingale measures is at least the value for each nonnegative martingale measure, and cannot exceed the supremum over all of them. This proves the equality of the two suprema, including when a boundary measure is needed to attain the value. \(\square\)
Three Payoffs in an Incomplete Market
Worked Example: A Claim with a Boundary Maximizer
Consider three states and one risky asset with initial discounted price \(s_0=2\) and terminal prices \(S_1=(0,2,4)\). Thus \(D=(-2,0,2)\). Find the martingale measures, then compute the superhedging price of \(H=(0,0,4)\) by both methods.
The probability and martingale equations are \(q_1+q_2+q_3=1\) and \(-2q_1+2q_3=0\). Hence \(q_1=q_3=a\) and \(q_2=1-2a\). Nonnegative probabilities require \(0\le a\le\tfrac12\); equivalent measures have \(0<a<\tfrac12\). The expected payoff is \(q\cdot H=4a\), whose maximum over the closed interval is \(2\), attained at \(q=(\tfrac12,0,\tfrac12)\).
For the direct hedge, the three statewise inequalities are \(x-2\theta\ge0\), \(x\ge0\), and \(x+2\theta\ge4\). Adding the first and third gives \(2x\ge4\), so \(x\ge2\). Choosing \(x=2,\theta=1\) gives terminal wealth \((0,2,4)\), which dominates \((0,0,4)\). Therefore \(p^*(H)=2\).
Check: the terminal wealth just computed exceeds the claim by \((0,2,0)\). The dual value is \(4a\le4(\tfrac12)=2\), with equality at the boundary measure. For equivalent measures \(a<\tfrac12\), so their expected payoffs are strictly less than \(2\), but approach \(2\) as \(a\) tends to \(\tfrac12\).
Worked Example: A Payoff Concentrated in the Middle State
In the same market, take \(H=(0,2,0)\). Compute its superhedging price and explain why the equivalent-measure supremum is not a maximum.
The martingale-measure parameterization above gives \(q\cdot H=2(1-2a)\). For \(0\le a\le\tfrac12\), this is at most \(2\), with equality at \(a=0\). For equivalent measures, \(a>0\), so the value is strictly below \(2\), but it tends to \(2\) as \(a\) tends to zero.
The statewise constraints are \(x-2\theta\ge0\), \(x\ge2\), and \(x+2\theta\ge0\). The middle-state constraint alone gives \(x\ge2\). Taking \(x=2,\theta=0\) gives terminal wealth \((2,2,2)\), which dominates \((0,2,0)\). Hence \(p^*(H)=2\).
Check: cash of \(2\) gives excess payoffs \((2,0,2)\), all nonnegative. The boundary martingale measure \((0,1,0)\) has expected payoff \(2\), agreeing with the hedge cost; every equivalent martingale measure has expected payoff \(2(1-2a)<2\).
Worked Example: A Replicable Payoff
Still in the same market, let \(H=(1,2,3)\). Show directly that it is replicable and that all martingale measures give the same value.
Choose \(x=2\) and \(\theta=\tfrac12\). The terminal wealth in the three states is \(2+\tfrac12(-2)=1\), \(2+\tfrac12(0)=2\), and \(2+\tfrac12(2)=3\). It equals the claim in every state, so the claim is replicated at cost \(2\). The cost is least: for every martingale measure, the weak-duality inequality in the theorem gives \(x\ge q\cdot H\).
Using \(q=(a,1-2a,a)\), the expectation is \(q\cdot H=a+2(1-2a)+3a=2\). This value is constant over the full interval \(0\le a\le\tfrac12\), so the dual supremum is \(2\), matching the replicating cost.
Check: the stock payoff vector is \((0,2,4)\), and \(1+\tfrac12(0,2,4)=(1,2,3)\). Also, substituting the endpoint values \(a=0\) and \(a=\tfrac12\) into \(a+2(1-2a)+3a\) gives \(2\) in both cases.
Interpreting the Dual and Avoiding a Sign Error
The dual formula says that every superhedging price bounds the expected payoff under every martingale measure. The reverse direction is the substantive part: in a finite-state market, if a price lies below the least superhedging capital, cone separation produces a martingale measure that assigns the claim an expectation above that price. In an incomplete market there may be many such measures, so the upper bound can differ from a lower pricing bound. A claim that is replicable is a special case: every martingale measure gives it the same value.
A common mistake is to take the infimum of martingale expectations when computing a superhedging price. For \(H=(0,0,4)\) in the first example, the expectations are \(4a\), with infimum \(0\) at \(a=0\) and supremum \(2\) at \(a=\tfrac12\). The value \(0\) is not the cost of a statewise dominating hedge: the state-three payoff is \(4\), so cash alone would need to be \(4\), and the least cost using the asset is \(2\). The correct dual operation is the supremum, which returns \(2\).
It is also important not to silently replace “martingale measure” by “equivalent martingale measure” when discussing attainment. The theorem proves equality of the two suprema under the strictly positive-measure assumption, not that the equivalent-measure supremum has a maximizing element. In the first two examples, the maximizing measures assign zero probability to at least one state; equivalent measures approach the same values but do not attain them.
Check Your Understanding
Use the definitions and the finite-state duality theorem to answer the following.
- For the three-state market above, derive the full range of \(a\) for nonnegative martingale measures and for equivalent martingale measures.
- For \(H=(0,0,4)\), which statewise inequality forces \(x\ge2\) after it is added to another inequality?
- For \(H=(0,2,0)\), why does the equivalent-measure supremum equal \(2\) even though no equivalent measure attains \(2\)?
- Verify directly that the hedge \((x,\theta)=(2,\tfrac12)\) replicates \(H=(1,2,3)\) in all three states.
- Explain why the dual representation uses a supremum rather than an infimum for a superhedging price.