Consistent Conjectural Variations Equilibrium for a Financial Model

Abstract

We consider a general model of financial flows and prices with multiple sectors and instruments. Each sector optimizes the composition of assets and liabilities in its portfolio, whose utility is given by a quadratic function constrained to satisfy the accounting identity that appears in flow-of-funds accounts and the equilibrium conditions that guarantee market clearance. To define the financial equilibrium, we make use of the concept of conjectural variations, in which each sector conjectures the possible dependence of the instruments’ prices upon its portfolio structure. The problem is modeled as a continuous two-stage game. In the first stage, the set of strategies for each sector consists of its possible conjectures about its influence on the prices. The second stage is the financial model’s equilibrium problem where, according to the conjectures selected in the first stage, each sector decides its portfolio composition. Each sector aims to minimize the risks, while at the same time maximizing the value of its assets and minimizing the value of its liabilities.

Appendices

Proof of Proposition 3.1

Proposition 3.1 For any $w \in \mathbb {R}_+^{mn}$, if $(x^*,y^*,r^*)$ is the financial equilibrium, then the vector $(x^*,y^*,\rho ^*)$, where $\rho ^*=(r_1^*-r_n^*, \dots , r_{n-1}^*-r_n^*)$, is the regularized equilibrium and the identity

$$\begin{aligned} -\sum _{j=1}^n r_j^*(x_{ij}^*-y_{ij}^*)=-\sum _{j=1}^{n-1}\rho _j^*(x_{ij}^*-y_{ij}^*) \end{aligned}$$

(10)

holds for all i, $i \in \{1,\dots ,m\}$. Conversely, if $(x^*,y^*,\rho ^*)$ is the regularized equilibrium, then there exists $r^*$ such that $\rho ^*=(r_1^*-r_n^*, \dots , r_{n-1}^*-r_n^*)$, the vector $(x^*,y^*,r^*)$ is the financial equilibrium, and identity (10) holds for all i, $i \in \{1,\dots ,m\}$.

Proof

Substituting the values $r_{ij}$ from (4) into (1), the objective function of sector i, $i \in \{1,\dots ,m\}$, takes the form:

$$\begin{aligned} \begin{pmatrix} x_i\\ y_i \end{pmatrix}^T Q^i \begin{pmatrix} x_i\\ y_i \end{pmatrix} +\sum _{j=1}^{n} w_{ij} {(x_{ij}-y_{ij})}^2-\sum _{j=1}^{n} r_{j} (x_{ij}-y_{ij}). \end{aligned}$$

(32)

Moreover, from the constraints (2), we can see that any feasible portfolio composition $(x_i,y_i)$ for sector i must satisfy that

$$\begin{aligned} x_{in}=s_i-\sum _{j=1}^{n-1}x_{ij},\quad y_{in}=s_i-\sum _{j=1}^{n-1} y_{ij}. \end{aligned}$$

(33)

Partially substituting the values of $x_{in}$ and $y_{in}$ from (33) into (32), we get the following relationship:

$$\begin{aligned} \begin{aligned}&\left( \begin{array}{r} x_i\\ y_i \end{array} \right) ^T Q^i \left( \begin{array}{r} x_i\\ y_i \end{array} \right) +\sum _{j=1}^{n}w_{ij}(x_{ij}-y_{ij})^2-\sum _{j=1}^{n}r_j(x_{ij}-y_{ij})\\&=\left( \begin{array}{r} x_i\\ y_i \end{array} \right) ^T Q^i \left( \begin{array}{r} x_i\\ y_i \end{array} \right) +\sum _{j=1}^{n}w_{ij}(x_{ij}-y_{ij})^2-\sum _{j=1}^{n-1}r_j(x_{ij}-y_{ij})\\&\quad -r_n\left( s_i-\sum _{j=1}^{n-1}x_{ij}-s_i+\sum _{j=1}^{n-1} y_{ij}\right) \\&=\left( \begin{array}{r} x_i\\ y_i \end{array} \right) ^T Q^i \left( \begin{array}{r} x_i\\ y_i \end{array} \right) +\sum _{j=1}^{n}w_{ij}(x_{ij}-y_{ij})^2-\sum _{j=1}^{n-1}(r_j-r_n)(x_{ij}-y_{ij}), \end{aligned} \end{aligned}$$

(34)

and defining $\rho _j=r_j-r_n$, $j\in \{1,\dots ,n-1\}$, yields the identity:

$$\begin{aligned} \begin{aligned}&\left( \begin{array}{r} x_i\\ y_i \end{array} \right) ^T Q^i \left( \begin{array}{r} x_i\\ y_i \end{array} \right) +\sum _{j=1}^{n}w_{ij}(x_{ij}-y_{ij})^2-\sum _{j=1}^{n}r_j(x_{ij}-y_{ij})\\&=\left( \begin{array}{r} x_i\\ y_i \end{array} \right) ^T Q^i \left( \begin{array}{r} x_i\\ y_i \end{array} \right) +\sum _{j=1}^{n}w_{ij}(x_{ij}-y_{ij})^2-\sum _{j=1}^{n-1}\rho _j(x_{ij}-y_{ij}). \end{aligned} \end{aligned}$$

(35)

Now, assume that $(x^*,y^*,r^*)$ is the financial equilibrium. Then, $x^*$ and $y^*$ provide an optimal solution for the optimization problem (1)–(4) when $r=r^*$, for all i, $i \in \{1,\dots ,m\}$. Let $\rho ^*=(r_1^*-r_n^*, \dots , r_{n-1}^*-r_n^*)$; from equation (35), we see that the optimization problems (1)–(4) and (6)–(8) are equivalent; thus, $x^*$ and $y^*$ also provide an optimal solution for the optimization problem (6)–(8) when $\rho =\rho ^*$, for all i, $i \in \{1,\dots ,m\}$.

Next, from the equilibrium condition (5) we have that

$$\begin{aligned} \sum _{i=1}^m(x_{ij}^*-y_{ij}^*)\ge 0,\quad \forall j\in \{1,\dots ,n\}. \end{aligned}$$

(36)

Substituting the equations given by (33), when $x=x^*$ and $y=y^*$, in the inequality (36) corresponding to $j=n$, we obtain

$$\begin{aligned} \begin{aligned} 0&\le \sum _{i=1}^m(x_{in}^*-y_{in}^*)=\sum _{i=1}^m\left( s_i-\sum _{j=1}^{n-1}x_{ij}^*-s_i+\sum _{j=1}^{n-1} y_{ij}^*\right) \\&=-\sum _{i=1}^m\sum _{j=1}^{n-1}( x_{ij}^*- y_{ij}^*) =-\sum _{j=1}^{n-1}\sum _{i=1}^m( x_{ij}^*- y_{ij}^*), \end{aligned} \end{aligned}$$

(37)

then, we have that

$$\begin{aligned} \sum _{j=1}^{n-1}\sum _{i=1}^m( x_{ij}^*- y_{ij}^*)\le 0. \end{aligned}$$

(38)

Therefore, from (36) and (38), we can conclude that condition (9) is valid for all j, $j \in \{1, \dots , n-1\}$; hence, the vector $(x^*, y^*, \rho ^*)$ is the regularized equilibrium. Finally, from equation (35), we can conclude that equation (10) holds.

Conversely, suppose that $(x^*, y^*, \rho ^*)$ is the regularized equilibrium. Define the values $r_n^*=-\min \{0,\rho _1^*, \dots ,\rho _{n-1}^*\}$ and $r_j^*=\rho _j^*+r_n^*$, $j\in \{1,\dots ,n-1\}$; then, $r^*=(r_1^*,\dots ,r_n^*)$ is a nonnegative vector of prices satisfying $\rho _j^*=r_j^*-r_n^*$ for all j, $j \in \{1, \dots , n-1\}$. By the definition of regularized equilibrium, the vectors $x^*$ and $y^*$ provide a solution for the optimization problem (6)–(8) when $\rho =\rho ^*$, for all i, $i \in \{1,\dots ,m\}$. Again, from equation (35), the optimization problems (1)–(4) and (6)–(8) are equivalent, then the vectors $x^*$ and $y^*$ are a solution for the optimization problem given by (1)–(4) when $r=r^*$, for all i, $i \in \{1,\dots ,m\}$.

Next, from (9) and (37), we can conclude that

$$\begin{aligned} \sum _{i=1}^m(x_{in}^*-y_{in}^*)=-\sum _{j=1}^{n-1}\sum _{i=1}^m(x_{ij}^*-y_{ij}^*)=0. \end{aligned}$$

(39)

Then, from (9) and (39), the market clearance condition (5) holds. Therefore, the vector $(x^*,y^*, r^*)$ is the financial equilibrium. Finally, from equation (35), we can again conclude that equation (10) holds. $\square$

Proof of Theorem 3.1

Theorem 3.1 For any $w \in \mathbb {R}_+^{mn}$, the vector $(x^*,y^*,\rho ^*) \in \mathbb {R}^{2mn+n-1}$ is the regularized equilibrium if and only if it provides an optimal solution for the following strictly convex quadratic programming problem:

$$\begin{aligned} \min _{x,y}&\displaystyle \sum _{i=1}^{m}\left[ \begin{pmatrix} x_i\\ y_i \end{pmatrix}^T Q^i \begin{pmatrix} x_i\\ y_i \end{pmatrix} +\sum _{j=1}^{n} w_{ij} {(x_{ij}-y_{ij})}^2\right] , \end{aligned}$$

(11)

$$\begin{aligned} {\text {s.t.}}\,\,&\displaystyle \sum _{j=1}^{n}x_{ij}=s_i,\quad \sum _{j=1}^{n}y_{ij}=s_i,\quad i\in \{1,\dots ,m\}, \end{aligned}$$

(12)

$$\begin{aligned}&\displaystyle \sum _{i=1}^{m}(x_{ij}-y_{ij})=0,\quad j\in \{1,\dots ,n-1\}, \end{aligned}$$

(13)

$$\begin{aligned}&x_{ij}\ge 0,\ y_{ij}\ge 0,\quad i\in \{1,\dots ,m\},\ j\in \{1,\dots ,n\}, \end{aligned}$$

(14)

whereas the vector $\rho ^*$ serves as the Lagrange multipliers for the set of constraints (13).

Proof

First, for any two vectors $\alpha =(\alpha _1,\dots ,\alpha _t) \in \mathbb {R}^t$ and $\beta =(\beta _1,\dots ,\beta _t) \in \mathbb {R}^t$ we will say that $\alpha \le \beta$ if and only if $\alpha _k \le \beta _k$ for all $k \in \{1,\dots ,t\}$, and define the element-wise product $\alpha * \beta =(\alpha _1\beta _1,\dots ,\alpha _t\beta _t)$ and the vector $z_i=\begin{pmatrix} x_i\\ y_i \end{pmatrix} \in \mathbb {R}^{2n}$, $i \in \{1,\dots ,m\}$, which, together with the notation defined in Sect. 3, allow us to rewrite the optimization problem (6)–(8) of sector i, $i \in \{1,\dots ,m\}$, in its matrix form as follows:

$$\begin{aligned} \min _{\,\,z_i}&z_i^T H_i z_i+\rho ^T C^T z_i, \end{aligned}$$

(40)

$$\begin{aligned} \text {s.t.}\,\,&B^Tz_i=b_i, \end{aligned}$$

(41)

$$\begin{aligned}&z_i\ge 0_{2n}; \end{aligned}$$

(42)

while the condition (9) can be rewritten as:

$$\begin{aligned} \sum _{i=1}^m C^Tz_i=0_{n-1}. \end{aligned}$$

(43)

For every sector, the quadratic programming problem (40)–(42) is strictly convex, then the symmetric matrices $H_i$, $i \in \{1,\dots ,m\}$, are positive definite. Hence, using the Karush–Kuhn–Tucker (KKT) conditions, this problem can be equivalently replaced by the following system of equalities and inequalities:

$$\begin{aligned}&z_i* (2H_iz_i+C\rho +B\mu _i)=0_{2n}, \end{aligned}$$

(44)

$$\begin{aligned}&B^Tz_i=b_i, \end{aligned}$$

(45)

$$\begin{aligned}&2H_iz_i+C\rho +B\mu _i\ge 0_{2n}, \end{aligned}$$

(46)

$$\begin{aligned}&z_i\ge 0_{2n}; \end{aligned}$$

(47)

where $\mu _i=\begin{pmatrix} \mu _{i1}\\ \mu _{i2} \end{pmatrix} \in \mathbb {R}^2$.

Thus, the vector $(x^*,y^*, \rho ^*)$ is the regularized equilibrium if and only if there exist vectors $\mu _i^*$, $i \in \{1,\dots ,m\}$, such that $x^*$, $y^*$, $\rho ^*$, and $\mu _i^*$, $i \in \{1,\dots ,m\}$, satisfy the system of equalities and inequalities (44)–(47) for all i, $i \in \{1,\dots , m\}$, and the condition (43) holds. Moreover, the system of equalities and inequalities obtained by gathering (44)–(47) for all $i \in \{1,\dots , m\}$, together with (43), is equivalent (using the KKT conditions) to the quadratic programming problem

$$\begin{aligned} \min _{z}&\displaystyle \sum _{i=1}^m (z_i^T H_i z_i), \end{aligned}$$

(48)

$$\begin{aligned} \text {s.t.}&B^Tz_i=b_i,\quad i \in \{1,\dots , m\}, \end{aligned}$$

(49)

$$\begin{aligned}&\displaystyle \sum _{i=1}^m C^Tz_i=0_{n-1}, \end{aligned}$$

(50)

$$\begin{aligned}&z_i\ge 0_{2n}, \end{aligned}$$

(51)

where $\mu _i$, $i \in \{1,\dots , m\}$, are the Lagrange multipliers for the constraints (49) and $\rho$ are the Lagrange multipliers for the constraints (50).

Therefore, $(x^*,y^*,\rho ^*)$ is the financial equilibrium if and only if $x^*$ and $y^*$ are the optimal solution for the strictly convex (given that $H_i$ is positive definite for all i) quadratic programming problem (48)–(51) and $\rho ^*$ serves as the Lagrange multipliers for the set of constraints given by (50). By the matrix notation defined before, the quadratic programming problem (48)–(51) coincides with the quadratic programming problem (11)–(14). $\square$

Proof of Theorem 3.2

Theorem 3.2 Let $(x^*,y^*, \rho ^*)$ be the regularized equilibrium for the vector of influence coefficients $w \in \mathbb {R}_+^{mn}$. If $x^*, y^* >0$, then the regularized equilibrium is unique and continuous with respect to w.

Proof

For any vector $w \in \mathbb {R}_+^{mn}$, there exists, by Theorem 3.1 and Corollary 3.1, the regularized equilibrium $(x^*,y^*,\rho ^*)$ such that the vectors $x^*$ and $y^*$ are unique, so they can be considered as functions of w, $x^*\equiv x^*(w)$ and $y^*\equiv y^*(w)$. Moreover, it is easy to see that the objective function (11) is a continuous function of $w \in \mathbb {R}_+^{mn}$ (since it is a polynomial), and the constraints (12)–(14) do not depend upon w. Then, by the stability theory for the optimal solution of a strictly convex quadratic programming problem (e.g., see [3]), the vectors $x^*(w)$ and $y^*(w)$ are continuous with respect to $w \in \mathbb {R}_+^{mn}$.

Now, recall from the proof of Theorem 3.1 that $(x^*,y^*,\rho ^*)$ is the regularized equilibrium if and only if there exists a vector $\mu ^*=(\mu _1,\dots ,\mu _m) \in \mathbb {R}^{2m}$ such that $(x^*,y^*,\rho ^*,\mu ^*)$ is a solution for the following system of equalities and inequalities:

$$\begin{aligned}&z_i* (2H_iz_i+C\rho +B\mu _i)=0_{2n}, \quad i\in \{1,\dots ,m\}, \end{aligned}$$

(52)

$$\begin{aligned}&2H_iz_i+C\rho +B\mu _i\ge 0_{2n}, \quad i\in \{1,\dots ,m\}, \end{aligned}$$

(53)

$$\begin{aligned}&z_i\ge 0_{2n}, \quad i\in \{1,\dots ,m\}, \end{aligned}$$

(54)

$$\begin{aligned}&B^Tz_i=b_i, \quad i\in \{1,\dots ,m\}, \end{aligned}$$

(55)

$$\begin{aligned}&\sum _{i=1}^m C^Tz_i=0_{n-1}. \end{aligned}$$

(56)

If $x^*,y^*>0$, then we know that any solution of (52)–(56) must satisfy $z_i>0_{2n}$ for all i, $i \in \{1,\dots ,m\}$, so the equations given by (52) can be reduced to:

$$\begin{aligned} 2H_iz_i+C\rho +B\mu _i=0_{2n}, \quad i \in \{1,\dots ,m\}. \end{aligned}$$

(57)

Substituting the values $x^*(w)$ and $y^*(w)$ from the regularized equilibrium into (57), by defining $z_i^*\equiv z_i^*(w)=\begin{pmatrix} x_i^*(w)\\ y_i^*(w) \end{pmatrix}$, we obtain the following system of linear equations with the variables $\rho \in \mathbb {R}^{n-1}$ and $\mu \in \mathbb {R}^{2m}$:

$$\begin{aligned} C\rho +B\mu _i=-2H_iz_i^*, \quad i \in \{1,\dots ,m\}. \end{aligned}$$

(58)

The system of equations (58) has $2m+n-1$ variables and 2mn equations; moreover, by Theorem 3.1 and Corollary 3.1, system (58) has a solution $(\rho ^*,\mu ^*)$. Using the matrix notation defined before, we can rewrite system (58) as follows:

$$\begin{aligned}&-\rho _j+\mu _{i1}=-2h_{ij}z_i^*, \quad i \in \{1,\dots ,m\},\ j \in \{1,\dots ,n-1\}, \end{aligned}$$

(59)

$$\begin{aligned}&\mu _{i1}=-2h_{in}z_i^*, \quad i \in \{1,\dots ,m\}, \end{aligned}$$

(60)

$$\begin{aligned}&\rho _j+\mu _{i2} =-2h_{i(n+j)}z_i^*, \quad i \in \{1,\dots ,m\},\ j \in \{1,\dots ,n-1\}, \end{aligned}$$

(61)

$$\begin{aligned}&\mu _{i2}=-2h_{i(2n)}z_i^*,\quad i \in \{1,\dots ,m\}, \end{aligned}$$

(62)

where $h_{ik}$, $k \in \{1,\dots ,2m\}$, is the kth row of the matrix $H_i$, $i \in \{1,\dots ,m\}$.

From system (59)–(62), we can see that the $n-1$ equations given by (59) when $i=1$ and the 2m equations given by (60) and (62), are linearly independent. Then, the system of linear equations (59)–(62) has $2m+n-1$ linearly independent equations, thus the solution $(\rho ^*,\mu ^*)$ is unique. Moreover, we can easily see that the solution for these $2m+n-1$ linearly independent equations is

$$\begin{aligned}&\rho _j=2h_{1j}z_1^*-2h_{1n}z_1^*, \quad j \in \{1,\dots ,n-1\}, \end{aligned}$$

(63)

$$\begin{aligned}&\mu _{i1}=-2h_{in}z_i^*, \quad i \in \{1,\dots ,m\}, \end{aligned}$$

(64)

$$\begin{aligned}&\mu _{i2}=-2h_{i(2n)}z_i^*, \quad i \in \{1,\dots ,m\}. \end{aligned}$$

(65)

By definition $H_i=Q^i+\varOmega ^i$ where $Q^i$ is a constant matrix and the entries of $\varOmega ^i$ are linear (by (15) and (18)) with respect to w, then the rows $h_{ik}$ of $H_i$ are continuous with respect to w. In addition, $x^*(w)$ and $y^*(w)$ are also continuous with respect to w; hence, the vectors $z_i^*$ are continuous with respect to w, too.

Thus, from (63) to (65), we can easily see that $\rho _j$, $j \in \{1,\dots ,n-1\}$, and $\mu _{i1}$, $\mu _{i2}$, $i \in \{1,\dots ,m\}$, are continuous with respect to w. Therefore, the regularized equilibrium $(x^*,y^*,\rho ^*)\equiv (x^*(w),y^*(w),\rho ^*(w))$ is unique and continuous with respect to $w \in \mathbb {R}_+^{mn}$. $\square$

Proof of Theorem 3.2

Proposition 3.2 Let $(x^*,y^*,\rho ^*)$ be the regularized equilibrium for the vector of influence coefficients $w \in \mathbb {R}_+^{mn}$. Then, $x^*, y^*>0$ if and only if

$$\begin{aligned} \mathcal {H}^{-1} \begin{pmatrix} \mathcal {C}&\mathcal {B} \end{pmatrix} {\left( \begin{pmatrix} \mathcal {C}^T\\ \mathcal {B}^T \end{pmatrix} \mathcal {H}^{-1} \begin{pmatrix} \mathcal {C}&\mathcal {B} \end{pmatrix} \right) }^{-1} \begin{pmatrix} 0_{n-1}\\ b \end{pmatrix}>0. \end{aligned}$$

(21)

Proof

Let $(x^*,y^*,\rho ^*)$ be the regularized equilibrium.

Necessity. Suppose that $x^*,y^*>0$. As mentioned in the previous proof, the solution for (52)–(56) must satisfy $z_i>0$ for all i, $i\in \{1,\dots ,m\}$, then the system of equalities and inequalities (52)–(56) is reduced to the following system of linear equations:

$$\begin{aligned}&2H_iz_i+C\rho +B\mu _i=0_{2n}, \quad i \in \{1,\dots ,m\}, \end{aligned}$$

(66)

$$\begin{aligned}&B^Tz_i=b_i, \quad i \in \{1,\dots ,m\}, \end{aligned}$$

(67)

$$\begin{aligned}&\sum _{i=1}^m C^Tz_i=0_{n-1}. \end{aligned}$$

(68)

Now, we define the vector

$$\begin{aligned} z=\begin{pmatrix} z_1\\ \vdots \\ z_m \end{pmatrix}. \end{aligned}$$

(69)

Using the same matrix notation defined before, together with (69), we can rewrite system (66)–(68) as follows:

$$\begin{aligned} \begin{pmatrix} 2\mathcal {H} &{} \mathcal {C} &{} \mathcal {B}\\ \mathcal {C}^T &{} O_{(n-1)\times (n-1)} &{} O_{(n-1)\times 2m}\\ \mathcal {B}^T &{} O_{2m\times (n-1)} &{} O_{2m\times 2m} \end{pmatrix} \begin{pmatrix} z\\ \rho \\ \mu \end{pmatrix}= \begin{pmatrix} 0_{2mn}\\ 0_{n-1}\\ b \end{pmatrix}. \end{aligned}$$

(70)

The system of equations (70) is linear and (by Theorem 3.2) has a unique solution; then, its coefficient matrix is invertible and its solution is given by

$$\begin{aligned} \begin{pmatrix} z^*\\ \rho ^*\\ \mu ^* \end{pmatrix}= {\begin{pmatrix} 2\mathcal {H} &{} \mathcal {C} &{} \mathcal {B}\\ \mathcal {C}^T &{} O_{(n-1)\times (n-1)} &{} O_{(n-1)\times 2m}\\ \mathcal {B}^T &{} O_{2m\times (n-1)} &{} O_{2m\times 2m} \end{pmatrix}}^{-1} \begin{pmatrix} 0_{2mn}\\ 0_{n-1}\\ b \end{pmatrix}. \end{aligned}$$

(71)

Next, we define

$$\begin{aligned} \varLambda =\begin{pmatrix} \mathcal {C}&\mathcal {B} \end{pmatrix},\quad \mathcal {O}=\begin{pmatrix} O_{(n-1)\times (n-1)} &{} O_{(n-1)\times 2m}\\ O_{2m\times (n-1)} &{} O_{2m\times 2m} \end{pmatrix}, \end{aligned}$$

(72)

and rewrite

$$\begin{aligned} \begin{pmatrix} 2\mathcal {H} &{} \mathcal {C} &{} \mathcal {B}\\ \mathcal {C}^T &{} O_{(n-1)\times (n-1)} &{} O_{(n-1)\times 2m}\\ \mathcal {B}^T &{} O_{2m\times (n-1)} &{} O_{2m\times 2m} \end{pmatrix}=\begin{pmatrix} 2\mathcal {H} &{} \varLambda \\ \varLambda ^T&{}\mathcal {O} \end{pmatrix}. \end{aligned}$$

(73)

In addition, the matrix $\mathcal {H}$ is invertible (since it is a diagonal block matrix whose diagonal elements are the invertible matrices $H_i$), then we make use of the Schur complement

$$\begin{aligned} \begin{pmatrix} 2\mathcal {H} &{} \varLambda \\ \varLambda ^T&{}\mathcal {O} \end{pmatrix}/(2\mathcal {H}) =-\frac{1}{2}\varLambda ^T\mathcal {H}^{-1}\varLambda \end{aligned}$$

(74)

to calculate the inverse matrix

$$\begin{aligned} {\begin{pmatrix} 2\mathcal {H} &{} \varLambda \\ \varLambda ^T&{}\mathcal {O} \end{pmatrix}}^{-1}=\begin{pmatrix} \dfrac{1}{2}\mathcal {H}^{-1}-\dfrac{1}{2}\mathcal {H}^{-1} \varLambda {\left( \varLambda ^T\mathcal {H}^{-1} \varLambda \right) }^{-1} \varLambda ^T\mathcal {H}^{-1} &{} \mathcal {H}^{-1} \varLambda {\left( \varLambda ^T\mathcal {H}^{-1} \varLambda \right) }^{-1}\\ {\left( \varLambda ^T\mathcal {H}^{-1} \varLambda \right) }^{-1} \varLambda ^T \mathcal {H}^{-1} &{} -2{\left( \varLambda ^T\mathcal {H}^{-1} \varLambda \right) }^{-1} \end{pmatrix},\nonumber \\ \end{aligned}$$

(75)

thus,

$$\begin{aligned} \begin{pmatrix} z^*\\ \rho ^*\\ \mu ^* \end{pmatrix}&= \begin{pmatrix} \dfrac{1}{2}\mathcal {H}^{-1}-\dfrac{1}{2}\mathcal {H}^{-1} \varLambda {\left( \varLambda ^T\mathcal {H}^{-1} \varLambda \right) }^{-1} \varLambda ^T\mathcal {H}^{-1} &{} \mathcal {H}^{-1} \varLambda {\left( \varLambda ^T\mathcal {H}^{-1} \varLambda \right) }^{-1}\\ {\left( \varLambda ^T\mathcal {H}^{-1} \varLambda \right) }^{-1} \varLambda ^T \mathcal {H}^{-1} &{} -2{\left( \varLambda ^T\mathcal {H}^{-1} \varLambda \right) }^{-1} \end{pmatrix} \begin{pmatrix} 0_{2mn}\\ 0_{n-1}\\ b \end{pmatrix} \nonumber \\&=\begin{pmatrix} \mathcal {H}^{-1}\varLambda {\left( \varLambda ^T\mathcal {H}^{-1} \varLambda \right) }^{-1} \begin{pmatrix} 0_{n-1}\\ b \end{pmatrix}\\ -2{\left( \varLambda ^T\mathcal {H}^{-1} \varLambda \right) }^{-1} \begin{pmatrix} 0_{n-1}\\ b \end{pmatrix} \end{pmatrix}. \end{aligned}$$

(76)

From (72) and (76), we find that

$$\begin{aligned} z^*&=\mathcal {H}^{-1}\varLambda {\left( \varLambda ^T\mathcal {H}^{-1} \varLambda \right) }^{-1} \begin{pmatrix} 0_{n-1}\\ b \end{pmatrix}= \mathcal {H}^{-1}\begin{pmatrix} \mathcal {C}&\mathcal {B} \end{pmatrix} {\left( \begin{pmatrix} \mathcal {C}^T\\ \mathcal {B}^T \end{pmatrix}\mathcal {H}^{-1} \begin{pmatrix} \mathcal {C}&\mathcal {B} \end{pmatrix}\right) }^{-1} \begin{pmatrix} 0_{n-1}\\ b \end{pmatrix}, \end{aligned}$$

(77)

$$\begin{aligned}&\begin{pmatrix} \rho ^*\\ \mu ^* \end{pmatrix}= -2{\left( \varLambda ^T\mathcal {H}^{-1} \varLambda \right) }^{-1} \begin{pmatrix} 0_{n-1}\\ b \end{pmatrix}= -2{\left( \begin{pmatrix} \mathcal {C}^T\\ \mathcal {B}^T \end{pmatrix}\mathcal {H}^{-1} \begin{pmatrix} \mathcal {C}&\mathcal {B} \end{pmatrix}\right) }^{-1} \begin{pmatrix} 0_{n-1}\\ b \end{pmatrix}, \end{aligned}$$

(78)

and given that this solution must satisfy $z^*>0$ (since $x^*,y^*>0$), we conclude that

$$\begin{aligned} z^*= \mathcal {H}^{-1}\begin{pmatrix} \mathcal {C}&\mathcal {B} \end{pmatrix} {\left( \begin{pmatrix} \mathcal {C}^T\\ \mathcal {B}^T \end{pmatrix}\mathcal {H}^{-1} \begin{pmatrix} \mathcal {C}&\mathcal {B} \end{pmatrix}\right) }^{-1} \begin{pmatrix} 0_{n-1}\\ b \end{pmatrix}>0. \end{aligned}$$

(79)

Sufficiency. Again, we consider the matrix

$$\begin{aligned} \begin{pmatrix} 2\mathcal {H} &{} \varLambda \\ \varLambda ^T&{}\mathcal {O} \end{pmatrix}= \begin{pmatrix} 2\mathcal {H} &{} \mathcal {C} &{} \mathcal {B}\\ \mathcal {C}^T &{} O_{(n-1)\times (n-1)} &{} O_{(n-1)\times 2m}\\ \mathcal {B}^T &{} O_{2m\times (n-1)} &{} O_{2m\times 2m} \end{pmatrix}. \end{aligned}$$

(80)

As mentioned above, the matrix $\mathcal {H}$ is invertible; thus, making use of the Schur complement (74), we can calculate the determinant

$$\begin{aligned} \begin{aligned} \det \begin{pmatrix} 2\mathcal {H} &{} \varLambda \\ \varLambda ^T&{}\mathcal {O} \end{pmatrix}&=\det (2\mathcal {H}) \det (\mathcal {O}-\varLambda ^T(2\mathcal {H})^{-1}\varLambda )\\&={(-1)}^{n-1}2^{(2m-1)(n-1)}\det (\mathcal {H})\det (\varLambda ^T{\mathcal {H}}^{-1}\varLambda ). \end{aligned} \end{aligned}$$

(81)

Next, we can easily see that the matrices

$$\begin{aligned} C=\begin{pmatrix} -I_{n-1}\\ 0_{n-1}^T\\ I_{n-1}\\ 0_{n-1}^T \end{pmatrix},\quad B=\begin{pmatrix} 1_n&{} 0_n\\ 0_n&{}1_n \end{pmatrix}, \end{aligned}$$

(82)

have full column rank, then the matrices

$$\begin{aligned} \mathcal {C}=\begin{pmatrix} C\\ \vdots \\ C \end{pmatrix},\quad \mathcal {B}=\begin{pmatrix} B &{} &{}\\ &{} \ddots &{}\\ &{} &{} B \end{pmatrix}, \end{aligned}$$

(83)

will also have full column rank; moreover, the columns of C are linearly independent from the columns of B and vice versa, then the block matrix $\begin{pmatrix} \mathcal {C}&\mathcal {B} \end{pmatrix}$ has full column rank too, which guarantees that the matrix $\begin{pmatrix} \mathcal {C}^T\\ \mathcal {B}^T \end{pmatrix}\mathcal {H}^{-1} \begin{pmatrix} \mathcal {C}&\mathcal {B} \end{pmatrix}$ is invertible.

Thus, both $\mathcal {H}$ and $\varLambda ^T{\mathcal {H}}^{-1}\varLambda =\begin{pmatrix} \mathcal {C}^T\\ \mathcal {B}^T \end{pmatrix}\mathcal {H}^{-1} \begin{pmatrix} \mathcal {C}&\mathcal {B} \end{pmatrix}$ are invertible matrices, implying that

$$\begin{aligned} \det \begin{pmatrix} 2\mathcal {H} &{} \varLambda \\ \varLambda ^T&{}\mathcal {O} \end{pmatrix}={(-1)}^{n-1}2^{(2m-1)(n-1)}\det (\mathcal {H})\det (\varLambda ^T{\mathcal {H}}^{-1}\varLambda )\ne 0. \end{aligned}$$

(84)

Therefore, matrix (80) is also invertible and the system of linear equations (70) has the unique solution

$$\begin{aligned} \begin{pmatrix} z^{**}\\ \rho ^{**}\\ \mu ^{**} \end{pmatrix}= {\begin{pmatrix} 2\mathcal {H} &{} \mathcal {C} &{} \mathcal {B}\\ \mathcal {C}^T &{} O_{(n-1)\times (n-1)} &{} O_{(n-1)\times 2m}\\ \mathcal {B}^T &{} O_{2m\times (n-1)} &{} O_{2m\times 2m} \end{pmatrix}}^{-1} \begin{pmatrix} 0_{2mn}\\ 0_{n-1}\\ b \end{pmatrix}, \end{aligned}$$

(85)

hence, we can find (as we did in the first part of this proof) that

$$\begin{aligned} z^{**}&= \mathcal {H}^{-1}\begin{pmatrix} \mathcal {C}&\mathcal {B} \end{pmatrix} {\left( \begin{pmatrix} \mathcal {C}^T\\ \mathcal {B}^T \end{pmatrix}\mathcal {H}^{-1} \begin{pmatrix} \mathcal {C}&\mathcal {B} \end{pmatrix}\right) }^{-1} \begin{pmatrix} 0_{n-1}\\ b \end{pmatrix}, \end{aligned}$$

(86)

$$\begin{aligned}&\begin{pmatrix} \rho ^{**}\\ \mu ^{**} \end{pmatrix}= -2{\left( \begin{pmatrix} \mathcal {C}^T\\ \mathcal {B}^T \end{pmatrix}\mathcal {H}^{-1} \begin{pmatrix} \mathcal {C}&\mathcal {B} \end{pmatrix}\right) }^{-1} \begin{pmatrix} 0_{n-1}\\ b \end{pmatrix}. \end{aligned}$$

(87)

Now, suppose that $\mathcal {H}^{-1}\begin{pmatrix} \mathcal {C}&\mathcal {B} \end{pmatrix} {\left( \begin{pmatrix} \mathcal {C}^T\\ \mathcal {B}^T \end{pmatrix}\mathcal {H}^{-1} \begin{pmatrix} \mathcal {C}&\mathcal {B} \end{pmatrix}\right) }^{-1} \begin{pmatrix} 0_{n-1}\\ b \end{pmatrix}>0$, then by (86) the solution $(z^{**},\rho ^{**},\mu ^{**})$ of system (70) satisfies that $z^{**}>0$. Recall that system (70) is the matrix form of system (66)–(68); thus,

$$\begin{aligned}&2H_iz_i^{**}+C\rho ^{**}+B\mu _i^{**}=0_{2n}, \quad i \in \{1,\dots ,m\}, \end{aligned}$$

(88)

$$\begin{aligned}&B^Tz_i^{**}=b_i, \quad i \in \{1,\dots ,m\}, \end{aligned}$$

(89)

$$\begin{aligned}&\sum _{i=1}^m C^Tz_i^{**}=0_{n-1}. \end{aligned}$$

(90)

In addition, since $z^{**}>0$ the equalities given by (88) are equivalent to the following group of equalities and inequalities:

$$\begin{aligned}&z_i^{**}*(2H_iz_i^{**}+C\rho ^{**}+B\mu _i^{**})=0_{2n}, \quad i \in \{1,\dots ,m\}, \end{aligned}$$

(91)

$$\begin{aligned}&2H_iz_i^{**}+C\rho ^{**}+B\mu _i^{**}\ge 0_{2n}, \quad i \in \{1,\dots ,m\}, \end{aligned}$$

(92)

$$\begin{aligned}&z_i^{**}\ge 0_{2n}, \quad i \in \{1,\dots ,m\}. \end{aligned}$$

(93)

In other words, $(z^{**},\rho ^{**},\mu ^{**})$ is a solution for the KKT conditions given by (52)–(56), i.e., $(z^{**},\rho ^{**},\mu ^{**})$ is the regularized equilibrium.

Finally, since $z^{**}>0$ we have that $x^{**},y^{**}>0$. Therefore, by Corollary 3.1, we have that $x^*=x^{**}>0$ and $y^*=y^{**}>0$.