Non-cooperative Support for the Asymmetric Nash Bargaining Solution

where we use (4) for the second equality, so∑n−1d i j = δ m ki d k j − θj n .k=1We have found thatθj n = δ ∑m ki d k j + (δm ii − 1)d i j, j ∈ N, i ∈ N \ {j, n}. (5)k /∈{i,n}Similarly, for j ≠ n,n∑d j j (1 − δ) = θj j − δ m kn θjkk=1n∑= θ j j − δ m kn (θj k − θj n ) − δθjnk=1= θ j j − θn j − δn∑m kn (θj k − θj n ) + (1 − δ)θj n ,k=1where we use (4) for the second inequality, son−1d j j = dj j − δ ∑m kn d k j + θj n .k=1We have found that∑n−1θj n = δ m kn d k j , j ∈ N \ {n}. (6)k=1We write (5)–(6) in vector–matrix notation asθ n j 1 ⊤ = d j (δM − I) −j−n, j ∈ N.The matrix (M − I) −j−n is invertible by Proposition 4.4, and so is the matrix (δM − I) −j−nfor δ close enough to one. Thus, for every j ∈ N, we can solve the above system for d j asd j = θ n j 1 ⊤ [(δM − I) −j−n] −1 .As δ m goes to one, the sequence θ n j (δ m ) converges to ¯θ j by Proposition 4.3. Thus thesequence d j (δ m ) converges to ¯θ j 1 ⊤ L −1 (j), as desired.Proposition 4.5 expresses each row j of the matrix ¯D as the sum of the rows of the matrixL −1 (j) multiplied by the scalar ¯θ j .We show now that each column of the matrix ¯D is orthogonal to the normal vector ofV at the point ¯θ, which is unique by Assumption A3. This is equivalent to saying thateach column of the matrix ¯D belongs to the tangent space of ∂V at ¯θ. We let span( ¯D)denote the column span of the matrix ¯D.15✷