On Iterated Nash Bargaining Solutions

Lemma A.1

If f satisfies INV, IIA, and SIR, then, for any ( S , d ) ∈ B , f(S, d) ∈ Par(S).

Proof

By B1 and INV, we assume d = 0 without loss of generality. Set S + ≡ S ∩ R + n . Since S₊ ⊆ S, Par(S₊) ⊆ Par(S). By B2, ( S + , 0 ) ∈ B and by SIR, f(S, 0) ∈ S₊. Hence, by IIA, f(S, 0) = f(S₊, 0).

Let T ≡ com(S₊). Then S + ⊆ T ⊆ R + n , Par(T) = Par(S₊), and ( T , 0 ) ∈ B by B3. Suppose v ≡ f(T, 0) ∉ Par(T). Then there exists w ∈ T such that w ≥ v and w ≠ v. Since v ≫ 0 by SIR, we also have w ≫ 0. Set β _i ≡ v _i /w _i > 0 for i = 1, …, n. Then β ≡ (β₁, …, β _n ) ≤ (1, …, 1) and β ≠ (1, …, 1). Let T′ ≡ {(β₁u₁, …, β _n u _n ): u ∈ T}. By B1, ( T ′ , 0 ) ∈ B . By construction T′ ⊆ T and f(T, 0) = v ∈ T′. Consequently, by IIA, f(T′, 0) = v. However, by INV, f(T′, 0) = (β₁v₁, …, β _n v _n ) ≠ v, a contradiction. Therefore, f(T, 0) ∈ Par(T).

Finally, since S₊ ⊆ T and f(T, 0) ∈ Par(T) = Par(S₊) ⊆ S₊, IIA implies f(S₊, 0) = f(T, 0). In summary, f(S, 0) = f(S₊, 0) = f(T, 0) ∈ Par(T) = Par(S₊) ⊆ Par(S). □

Lemma A.2

Suppose f satisfies INV, IIA, and SIR.

1. ≻ _f is a strongly monotone strict linear order; that is, for any u , v , w ∈ R + + n , (i) exactly one of u ≻ _f v, v ≻ _f u, and u = v holds, (ii) u ≻ _f v and v ≻ _f w imply u ≻ _f w, and (iii) u ≥ v and u ≠ v imply u ≻ _f v.

2. For any ( S , 0 ) ∈ B , f(S, 0) is the unique maximizer of ≻ _f on S ∩ R + + n ; that is, f ( S , 0 ) ∈ S ∩ R + + n and

Proof

Throughout this proof f is fixed and we drop the subscript “f” in ≻ _f .

We first establish part (a). Property (i) is obvious and property (iii) follows from Lemma A.1. To show property (ii), suppose u ≻ v and v ≻ w. Then property (i) implies u ≠ w. Again by property (i), it suffices to show that w ≻ u is false. Suppose on the contrary w ≻ u. By Lemma A.1, f(com(u, v, w), 0) ∈ {u, v, w}. If f(com(u, v, w), 0) = u, then IIA implies u ≻ w, contradicting w ≻ u. Similarly, f(com(u, v, w), 0) = v contradicts u ≻ v and f(com(u, v, w), 0) = w contradicts v ≻ w. Therefore, we must have u ≻ w and hence property (ii).

We next establish part (b). Set S + + ≡ S ∩ R + + n and S + ≡ S ∩ R + n . That f(S, 0) ∈ S₊₊ directly follows from SIR. Set u* ≡ f(S, 0). We need to show u* ≻ u, or equivalently, u* = f(com(u, u*), 0), for any u ∈ S₊₊\{u*}. By B2, ( S + , 0 ) ∈ B . By IIA, u* = f(S₊, 0). By B3, ( c o m ( S + ) , 0 ) ∈ B and ( c o m ( u , u * ) , 0 ) ∈ B . By Lemma A.1, we have f(com(S₊), 0) ∈ S₊. Thus, by IIA, we have f(com(S₊), 0) = f(S₊, 0) = u* ∈ com(u, u*). By IIA again, f(com(S₊), 0) = f(com(u, u*), 0). This shows u* = f(com(u, u*), 0). □

Lemma A.3

Suppose f satisfies INV, IIA, and SIR. For any u ∈ R + + n define

1. log  H _f (u) = log  H _f (1) + log  u;

2. log  L _f (u) = log  L _f (1) + log  u;

3. log  H _f (1) = −log  L _f (1);

4. log  H _f (1) and log  L _f (1) are convex.

Proof

Throughout this proof f is fixed and we drop the subscript “f” in ≻ _f , H _f (⋅), and L _f (⋅).

Step 1. We claim that, for any λ ∈ R and u 0 , u 1 , v 0 , v 1 ∈ R + + n with u¹ ≻ u⁰ and log  v¹ − log  v⁰ = λ(log  u¹ − log  u⁰), we have v¹ ≻ v⁰ if λ > 0 and v⁰ ≻ v¹ if λ < 0.

To see this, first define u ^t for t ∈ R by

Substep 1.1. We show that the claim of this step holds for λ = 1 and λ = −1. When λ = 1, log  v¹ − log  v⁰ = log  u¹ − log  u⁰, or equivalently v i 1 / u i 1 = v i 0 / u i 0 , for each i ∈ {1, …, n}. Thus, by INV and u¹ ≻ u⁰,

For the case of λ = −1, the claim can be similarly proved with the roles of v⁰ and v¹ interchanged.

Substep 1.2. We show that the claim of this step holds when λ = 1/m for any positive integer m. By the assumptions on u⁰, u¹, v⁰, v¹ and Substep 1.1, it suffices to show u^1/m ≻ u⁰. Suppose on the contrary u⁰ ≻ u^1/m. Since log  u^(i−1)/m − log  u^i/m for i = 1, 2, …, m are all equal to 1 m ( log u 0 − log u 1 ) , it follows from Substep 1.1 that u⁰ ≻ u^1/m ≻ ⋯ ≻ u^(m−1)/m ≻ u¹. The transitivity of ≻ established in Lemma A.2(a) then implies that u⁰ ≻ u¹, which contradicts u¹ ≻ u⁰.

Substep 1.3. We show that the claim of this step holds for all λ > 0. Given that λ > 0, it follows from the assumptions on u⁰, u¹, v⁰, v¹ and Substep 1.1 that it suffices to show u ^λ ≻ u⁰. Consider the “log-convex hull” of {u⁰, u ^λ } defined by T = {u ^t : 0 ≤ t ≤ λ}. From B3, ( c o m ( T ) , 0 ) ∈ B . Suppose on the contrary f(com(T), 0) = u ^t for some t ∈ [0, λ). Then we can pick a large enough positive integer m such that t + 1 m < λ . By construction, u^t+1/m ∈ T and by IIA, f(com(u ^t , u^t+1/m), 0) = u ^t . It follows that u ^t ≻ u^t+1/m. Hence, from Substep 1.1 it follows that u⁰ ≻ u^1/m. This contradicts Substep 1.2. Therefore, f(com(T), 0) ≠ u ^t for any t ∈ [0, λ). But f(com(T), 0) ∈ T from Lemma A.1. Consequently, f(com(T), 0) = u ^λ . Thus, by IIA, f(com(u⁰, u ^λ ), 0) = u ^λ or equivalently, u ^λ ≻ u⁰.

Substep 1.4. Substep 1.1 with λ = −1 and Substep 1.3 together imply that the claim of this step also holds for all λ < 0.

Step 2. Establish parts (a), (b), and (c).

The result in Step 1 with λ = 1 can be written as: if log  v¹ − log  v⁰ = log  u¹ − log  u⁰, then v¹ ≻ v⁰ and u¹ ≻ u⁰ are equivalent; v⁰ ≻ v¹ and u⁰ ≻ u¹ are equivalent. Since log  1 = 0, parts (a) and (b) follow. The result in Step 1 with λ = −1 can be written as: if log v 1 − log v 0 = − log u 1 − log u 0 , then v¹ ≻ v⁰ and u⁰ ≻ u¹ are equivalent; v⁰ ≻ v¹ and u¹ ≻ u⁰ are equivalent. Part (c) follows.

Step 3. Establish part (d).

We only prove log  L(u) is convex; the proof for log  H(u) is analogous. Suppose that log  v⁰ and log  v¹ are in log  L(u) or equivalently, u ≻ v⁰ and u ≻ v¹. Consider log  v ^λ = λ log  v¹ + (1 − λ) log  v⁰ for λ ∈ (0, 1). We need to show that u ≻ v ^λ . If v⁰ = v¹, then u ≻ v⁰ = v ^λ . If v⁰ ≻ v¹, then Step 1 implies u ≻ v⁰ ≻ v ^λ . If v¹ ≻ v⁰, then Step 1 implies u ≻ v¹ ≻ v ^λ . This shows u ≻ v ^λ . □

Lemma A.4

Suppose f satisfies INV, IIA, and SIR. Define H _f (⋅) and L _f (⋅) as in Lemma A.3. There exist pairwise orthogonal non-zero vectors α 1 , … , α n ∈ R n with α¹ ≥ 0 such that x ∈ log  H _f (1) (resp. x ∈ log  L _f (1)) if and only if there is some j ∈ {1, …, n} such that α ^j ⋅ x > 0 (resp. α ^j ⋅ x < 0) and α ^k ⋅ x = 0 for k = 1, …, j − 1. Moreover, every such α ⁱ (i = 1, …, n) is unique up to positive scalar multiplication.

Proof

For notational convenience, let H ̂ ≡ log H f ( 1 ) and L ̂ ≡ log L f ( 1 ) . By Lemma A.2(a), H ̂ ∩ L ̂ = ∅ , H ̂ ∪ L ̂ = R n \ { 0 } , and R + + n ⊆ H ̂ . By Lemma A.3(c), H ̂ = − L ̂ . By Lemma A.3(d), H ̂ , L ̂ are convex.

Note that H ̂ and L ̂ are nonempty. Denote their common boundary by E¹. Since H ̂ and L ̂ are convex, E¹ is a hyperplane with some non-zero normal vector α 1 ∈ R n . Since H ̂ = − L ̂ , we have 0 ∈ E¹. This concludes that E¹ is an (n − 1)-dimensional vector subspace of R n . Choose the direction of the normal vector α¹ so that { x : α 1 ⋅ x > 0 } ⊆ H ̂ and { x : α 1 ⋅ x < 0 } ⊆ L ̂ . Since R + + n ⊆ H ̂ , we have α¹ ≥ 0.

Since E¹ is a nontrivial vector space, H ̂ ∪ L ̂ = R n \ { 0 } , H ̂ = − L ̂ , and H ̂ , L ̂ , E 1 are all convex, it follows that H ̂ ∩ E 1 and L ̂ ∩ E 1 are nonempty, convex, and they share the common boundary E² ⊆ E¹ with some non-zero normal vector α² ∈ E¹. Applying the previous reasoning, E² is an (n − 2)-dimensional vector subspace of R n . Choose the direction of the normal vector α² so that { x : α 2 ⋅ x > 0 } ∩ E 1 ⊆ H ̂ and { x : α 2 ⋅ x < 0 } ∩ E 1 ⊆ L ̂ . Since α² ∈ E¹ and α¹ ⊥ E¹, we have α² ⊥ α¹.

Repeating the preceding process and reasoning, a collection of subspaces {E¹, …, E ⁿ } and a collection {α¹, …, α ⁿ } of corresponding normal vectors can be generated, such that E¹ ⊃ E² ⊃ ⋯ ⊃ E ⁿ , each E ^j is an (n − j)-dimensional vector subspace of R n so that E ⁿ = {0} and each normal vector α ^j of E ^j satisfies { x : α j ⋅ x > 0 } ∩ E j − 1 ⊆ H ̂ and { x : α j ⋅ x < 0 } ∩ E j − 1 ⊆ L ̂ (we let E 0 ≡ R n ). The process stops at E ⁿ = {0} because H ̂ ∩ E n and L ̂ ∩ E n are empty. Hence, the separation between H ̂ and L ̂ is complete, in that given any point x ∈ R n \ { 0 } , we can determine whether x ∈ H ̂ or x ∈ L ̂ using {α¹, …, α ⁿ }.

It is easy to verify that H ̂ and L ̂ are exactly described as in the lemma. Furthermore, the direction of each α ^j is uniquely determined. It remains to show that α¹, …, α ⁿ are pairwise orthogonal. To this end, pick any j, k ∈ {1, …, n} with j < k. By our construction, α ^k ∈ E^k−1 ⊆ E ^j and α ^j  ⊥ E ^j . It follows that α ^j  ⊥ α ^k . □

Proof of Proposition 2

Suppose f satisfies INV, IIA, and SIR. Let α 1 , … , α n ∈ R n be vectors with the properties claimed in Lemma A.4. Let W be the n × n matrix with ith row being α ⁱ . Since α¹, …, α ⁿ are pairwise orthogonal and non-zero, they are linearly independent.

By Lemma A.3(b), log  v ∈ log  L _f (u) if and only if for some j ∈ {1, …, n}, α ^j ⋅ (log  v − log  u) < 0 and α ^k ⋅ (log  v − log  u) = 0 for k < j. Equivalently, u ≻ _f v if and only if there is some j ∈ {1, …, n} such that g ^j (v, 0) < g ^j (u, 0) and g ^k (v, 0) = g ^k (u, 0) for k < j, where g j ( u , d ) ≡ ∏ i = 1 n ( u i − d i ) α i j for u ≥ d and j = 1, …, n. Now, Lemma A.2(b) implies that f is the W-bargaining solution.

Notice that α¹ is the unique maximizer of g¹(⋅, 0) on S°. Hence, α¹ = f(S°, 0). By SIR, α¹ ≫ 0. Therefore, W is an admissible weight matrix. From Lemma A.4 we know the α¹, …, α ⁿ above can be uniquely normalized to be in Δ ⁿ . □

Proof of Lemma 1

“If” part. The fact that any row operation of the first kind does not change the represented solution is rather obvious. Now let us consider a row operation of the second kind. Consider replacing the jth row α ^j by α ^j + θα ^k , where θ is a scalar (not necessarily positive) and α ^k is a preceding row (i.e. 1 ≤ k < j). From (2), all elements in Σ^j−1(S, d) must have a common value of ∏ i = 1 n u i − d i α i k . This common value, denoted as π, must be positive. Therefore, the new maximization problem in the jth round can be written as

which is clearly equivalent to the original maximization problem in the jth round.

“Only if” part. First, clearly the row operations in this lemma are reversible; that is, if W ̃ can be obtained from W through an application of those row operations, then W can also be obtained from W ̃ through an application of those row operations. Also note that our row operations of the second kind are the operations needed in Gram–Schmidt orthogonalization process. Therefore, a finite sequence of our row operations allows us to transform W through Gram–Schmidt orthogonalization process into some n × n matrix whose rows are pairwise orthogonal. This new matrix has full rank because W has. Therefore, each row of the new matrix must be non-zero and hence we can apply our row operations of the first kind to normalize each row to be in Δ ⁿ . Thus, we know that a finite sequence of our row operations can transform W into some n × n matrix A whose rows are pairwise orthogonal and in Δ ⁿ . Similarly, W ̃ can also be transformed into some n × n matrix A ̃ with the same properties through a finite sequence of our row operations. From the “if” part we proved above, we know that f is the A-bargaining solution and f ̃ is the A ̃ -bargaining solution. Now, if it is the case that f = f ̃ , then from the uniqueness part of Proposition 2(b) we know that A = A ̃ . Therefore, we can from W obtain A and then from A obtain W ̃ through a finite sequence of our row operations. □

Proof of Proposition 3

Existence. Let W be an admissible weight matrix (i.e. is n × n, of full rank, and with first row in R + + n ). Let M k denote the set of k × k full rank matrices with its first row belonging to Δ + k and the first non-zero entry in every column being positive. First, W is equivalent to some matrix in M n because the first row of W is in R + + n and hence can be multiplied by a positive scalar to become in Δ + + n . In the following, we shall show that, for any positive integer k, if B ∈ M k + 1 , then B is equivalent to some matrix D ∈ M k + 1 such that, for some i* ∈ {1, …, k + 1}, (i) the i*th column of D has all entries being zero except the first entry (which must be positive), and (ii) the submatrix of D after deleting the first row and i*th column of D belongs to M k . (B and D must also have the same first row since they are equivalent and both are in M k + 1 .) Once the last claim is proved, it is easy to see that we can iteratively transform W into equivalent matrices such that an equivalent triangularly standard weight matrix is ultimately obtained.

Now, let k be a positive integer and B ∈ M k + 1 with its (j, i)th entry denoted as β i j . Recursively define

Note that the above sets are nonempty because K¹ is nonempty (since β 1 ∈ Δ + k + 1 ). Note also that K^k+1 must be a singleton (otherwise B must not have full rank). Now, let i* be the unique element of K^k+1. Then we use the first row of B to eliminate the i*th entry in all other rows. That is, we replace each jth row β ^j (j = 2, …, k + 1) by β j − β i * j / β i * 1 β 1 . After the above row operations, the second row must be non-negative and non-zero. Then we multiply the second row by a positive constant to make it belonging to Δ + k . Now, the resulting matrix is our desired matrix D.

Uniqueness. It suffices to show that, if A and B are n × n full rank matrices which have all rows in Δ + n and are upper triangular up to permutations of columns, then A and B are equivalent only when they are equal. In the rest of this proof, we show the last claim by a proof by contradiction.

Take any A, B that have the above properties. Suppose on the contrary that A and B are equivalent but A ≠ B. Let α ^j denote the jth row of A, and β ^j the jth row of B. Since A ≠ B, there is a unique row index k ∈ {1, …, n} such that

Since all rows in A and B are in Δ + n , we have

Note that any upper triangular matrix has full rank if and only if all of its diagonal entries are non-zero. Therefore, any n × n full rank matrix that is upper triangular up to permutations of columns has only one permuting of columns that can make it upper triangular. To save one notation, we without loss of generality assume that this permuting for A is the identity permuting (i.e. A itself is upper triangular). Let σ: {1, …, n} → {1, …, n} denote the permuting of columns that makes B upper triangular. Thus,

Since A and B are of full rank, we have

Since A and B are equivalent, we know from our definition of matrix equivalence that there exist (unique) constants θ¹, …, θ ^k such that

and

Now, if it is the case that θ ^j = 0 whenever j < k, then (B.6) becomes θ ^k α ^k = β ^k , which together with (B.3) implies

so that (B.6) becomes α ^k = β ^k , contradicting (B.1). Therefore, k ≥ 2 and at least one of θ¹, …, θ^k−1 is non-zero. Then there is a unique index k′ ∈ {1, …, k − 1} such that

Using (B.9) to simplify (B.6), we have

Considering the k′th entry of (B.10) and using (B.4) to simplify, we have

Since we have α k ′ k ′ > 0 from (B.5) and β k ′ k ≥ 0 from (B.3), it follows from (B.8) and (B.11) that

We are now ready to derive a contradiction as follows:

In the above chain, the first and fourth lines are from (B.4); the second and third lines are from (B.10) and (B.2) respectively; the second last line is from (B.5) and (B.12); the last line is from (B.7) and (B.3). □

Proof of Theorem 2

By Theorem 1(b) or Proposition 2, f is the W-bargaining solution for some admissible weight matrix W. Let α ⁱ (i = 1, …, n) denote the ith row of W. Then α 1 , … , α n ∈ R n are linearly independent, α¹ ≫ 0, and (1) is satisfied. We shall show that for j = 1, …, n,

1. S ^j is nonempty,

2. β j ∈ Δ + + n ,

3. there are constants θ¹, …, θ ^j such that θ¹α¹ + ⋯ + θ ^j α ^j = β ^j and θ ^j > 0,

4. β¹, …, β ^j are linearly independent.

Once (i)–(iv) are established, we would know that W ̂ is an admissible weight matrix that is equivalent to W; hence from Lemma 1 f is the W ̂ -bargaining solution and the proof would be completed.

We prove (i)–(iv) by induction. Observe that claims (i)–(iv) hold for j = 1. Take any k ∈ {2, …, n} and suppose (i)–(iv) hold for j = 1, …, k − 1. Then,

The log-transformation of S ^k can be written as ln S ^k = V₁ ∩ V₂ where

Since (iv) holds for j = k − 1 (i.e. β¹, …, β^k−1 are linearly independent), V₂ is an (n − k + 1)-dimensional flat surface in R n and hence, it is nonempty. On the boundary of V₁, there are many points above V₂ and many below V₂. Indeed, for any u ∈ Δ + + n close enough to (1/n, …, 1/n), it follows from β 1 , … , β k − 1 ∈ Δ + + n and ϕ ∈ (0, 1/n) that ln u lies above V₂; for any u ∈ Δ + + n that has at least one coordinate close enough to 0, ln u lies below V₂. This shows that ln S ^k is nonempty. Consequently, S ^k is nonempty and thus (i) holds for j = k.

Since (iii) holds for j = 1, …, k − 1, from the proof of the “if” part of Lemma 1 it follows that the sets of maximizers in the first k − 1 rounds of the iterative maximization in Definition 2 do not change if we replace α¹, …, α^k−1 with β¹, …, β^k−1. Therefore, Σ^k−1(com(S ^k ), 0) = S ^k . The kth round of the iterative maximization can be written as

Considering u ̂ ≡ ln ⁡ u , (B.13) is equivalent to

Problem (B.14) is a convex optimization problem. Moreover, the constraints satisfy the Slater condition because (ln ϕ, …, ln ϕ) satisfies the affine equality constraints (since (ii) holds for j = 1, …, k − 1) and strictly satisfies the inequality constraint (since ϕ < 1/n).

By definition, β ^k must be a solution of problem (B.13) and hence, ln β ^k must be a solution of problem (B.14). Thus, the following Kuhn–Tucker conditions for problem (B.14) hold:

where λ and μ¹, …, μ^k−1 are the Langrangian multipliers. If λ = 0, then (B.15) implies that α ^k is a linear combination of β¹, …, β^k−1 and hence, it is also a linear combination of α¹, …, α^k−1 because (iii) holds for j = 1, …, k − 1. This contradicts the linear independence of α¹, …, α ⁿ . Therefore, it must be the case that λ > 0. From (B.16), ∑ i = 1 n β i k = 1 , which implies that (ii) holds for j = k. Moreover, from (B.15),

This establishes (iii) for j = k.

Finally, the validity of claim (iii) for j = 1, …, k together with the linear independence of α¹, …, α ^k implies that β¹, …, β ^k are linearly independent. It follows that claim (iv) holds for j = k. □