Banking competition and stability: evidence from West Africa

Joana Cobbinah; Tan Zhongming; Albert Henry Ntarmah; Joana Cobbinah; Tan Zhongming; Albert Henry Ntarmah

doi:10.3934/NAR.2020015

National Accounting Review

2020, Volume 2, Issue 3: 263-284. doi: 10.3934/NAR.2020015

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Research article

Banking competition and stability: evidence from West Africa

School of Finance and Economics, Jiangsu University, Zhenjiang, Jiangsu, 212013, P. R. China

Received: 09 June 2020 Accepted: 27 July 2020 Published: 03 August 2020
JEL Codes: G21, G28

This study explored the relationship between banking competition and bank stability. We used data of 10 countries in West Africa over the period of 2000–2014. Employing Generalized Method of Moments style Panel Vector Autoregressive estimation model, we found out that banking competition is stability-enhancing. Again, findings from the granger causality test show that the relationship between banking competition and bank stability is bidirectional. We also found out that banking competition explains 50 percent of the variations in bank stability in the region. The results from the impulse response function indicated that the impact of one standard deviation shock in the rise of Boone indicator as a proxy for banking competition on stability (bank Z-score) was zero for the first year and dropped to negative through to the tenth year. Overall, our findings largely support the "competition-stability view".

Keywords:

Citation: Joana Cobbinah, Tan Zhongming, Albert Henry Ntarmah. Banking competition and stability: evidence from West Africa[J]. National Accounting Review, 2020, 2(3): 263-284. doi: 10.3934/NAR.2020015

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Abstract

1. Introduction

Since the 1960s, the rapid development of high-speed rail has made it a very important means of transportation. However, the vibration will be caused because of the contact between the wheels of the train and the train tracks during the operation of the high-speed train. Therefore, the analytical vibration model can be mathematically summarized as a quadratic palindromic eigenvalue problem (QPEP) (see ^[1,2])

$\begin{equation*} (\lambda^2 A_1+\lambda A_0+A_1^\top)x = 0, \end{equation*}$

with $A_i \in {\mathbb{R}}^{n \times n}, \ i = 0, 1$ and $A_0^\top = A_0.$ The eigenvalues $\lambda$ , the corresponding eigenvectors $x$ are relevant to the vibration frequencies and the shapes of the vibration, respectively. Many scholars have put forward many effective methods to solve QPEP ^[3,4,5,6,7]. In addition, under mild assumptions, the quadratic palindromic eigenvalue problem can be converted to the following linear palindromic eigenvalue problem (see ^[8])

$\begin{equation} Ax = \lambda A^\top x, \end{equation}$

(1)

with $A \in {\mathbb{R}}^{n \times n}$ is a given matrix, $\lambda\in {\mathbb{C}}$ and nonzero vectors $x\in {\mathbb{C}}^{n}$ are the wanted eigenvalues and eigenvectors of the vibration model. We can obtain $\frac{1}{\lambda} x^\top A^\top = x^\top A$ by transposing the equation (1). Thus, $\lambda$ and $\frac{1}{\lambda}$ always come in pairs. Many methods have been proposed to solve the palindromic eigenvalue problem such as URV-decomposition based structured method ^[9], QR-like algorithm ^[10], structure-preserving methods ^[11], and palindromic doubling algorithm ^[12].

On the other hand, the modal data obtained by the mathematical model are often evidently different from the relevant experimental ones because of the complexity of the structure and inevitable factors of the actual model. Therefore, the coefficient matrices need to be modified so that the updated model satisfies the dynamic equation and closely matches the experimental data. Al-Ammari ^[13] considered the inverse quadratic palindromic eigenvalue problem. Batzke and Mehl ^[14] studied the inverse eigenvalue problem for T-palindromic matrix polynomials excluding the case that both $+1$ and $-1$ are eigenvalues. Zhao et al. ^[15] updated $\ast$ -palindromic quadratic systems with no spill-over. However, the linear inverse palindromic eigenvalue problem has not been extensively considered in recent years.

In this work, we just consider the linear inverse palindromic eigenvalue problem (IPEP). It can be stated as the following problem:

Problem IPEP. Given a pair of matrices $(\Lambda, X)$ in the form

$\begin{equation*} \Lambda = \text{diag} \{\lambda_{1}, \cdots, \lambda_{p}\} \in {\mathbb{C}}^{p \times p}, \end{equation*}$

and

$\begin{equation*} X = [x_{1}, \cdots, x_{p}] \in {\mathbb{C}}^{n \times p}, \end{equation*}$

where diagonal elements of $\Lambda$ are all distinct, $X$ is of full column rank $p$ , and both $\Lambda$ and $X$ are closed under complex conjugation in the sense that $\lambda_{2i} = \bar{\lambda}_{2i-1} \in {\mathbb{C}}$ , $x_{2i} = \bar{x}_{2i-1} \in {\mathbb{C}}^{n}$ for $i = 1, \cdots, m$ , and $\lambda_{j} \in {\mathbb{R}}$ , $x_{j} \in {\mathbb{R}}^{n}$ for $j = 2m+1, \cdots, p$ , find a real-valued matrix $A$ that satisfy the equation

$\begin{equation} AX = A^\top X\Lambda. \end{equation}$

(2)

Namely, each pair $(\lambda_{t}, x_{t})$ , $t = 1, \cdots, p$ , is an eigenpair of the matrix pencil

$P(\lambda) = A x- \lambda A^\top x.$

It is known that the mathematical model is a "good" representation of the system, we hope to find a model that is closest to the original model. Therefore, we consider the following best approximation problem:

Problem BAP. Given $\tilde{A} \in {\mathbb{R}}^{n \times n}$ , find $\hat{A} \in {\mathcal{S}}_{A}$ such that

$\begin{equation} \|\hat{A}-\tilde{A}\| = \min\limits_{{A} \in {\mathcal{S}_{A}}} \|A-\tilde{A}\|, \end{equation}$

(3)

where $\| \cdot \|$ is the Frobenius norm, and ${\mathcal{S}}_{A}$ is the solution set of Problem IPEP.

In this paper, we will put forward a new direct method to solve Problem IPEP and Problem BAP. By partitioning the matrix $\Lambda$ and using the QR-decomposition, the expression of the general solution of Problem IPEP is derived. Also, we show that the best approximation solution $\hat{A}$ of Problem BAP is unique and derive an explicit formula for it.

2. The solution of Problem IPEP

We first rearrange the matrix $\Lambda$ as

$\begin{equation} \begin{array}{ccc} \Lambda = \left[ \begin{array}{ccc} 1 & 0 & \ \ \ 0 \\ 0 & \Lambda_1 & \ \ \ 0 \\ 0 & 0 & \ \ \ \Lambda_2 \\ \end{array} \right]&\begin{array}{c} t \\ 2s \\ 2(k+2l) \\ \end{array} \\ \begin{array}{ccc} \ \ \ \ \ \ \ \ \ \ \ t & \ 2s & 2(k+2l) \\ \end{array}& \end{array}, \end{equation}$

(4)

where $t+2s+2(k+2l) = p$ , $t = 0 \ or \ 1,$

$\begin{eqnarray*} && \Lambda_1 = \text{diag} \{\lambda_1, \lambda_2, \cdots, \lambda_{2s-1}, \lambda_{2s}\}, \lambda_i \in \mathbb{R}, \ \lambda_{2i-1}^{-1} = \lambda_{2i}, \ 1\leq i\leq s, \\ && \Lambda_2 = \text{diag} \{\delta_1, \cdots, \delta_k, \delta_{k+1}, \delta_{k+2}, \cdots, \delta_{k+2l-1}, \delta_{k+2l}\}, \ \delta_j \in \mathbb{C}^{2 \times 2}, \end{eqnarray*}$

with

$\begin{eqnarray*} &&\delta_j = \left[ \begin{array}{cc} \alpha_j+\beta_{j} i & 0 \\ 0 & \alpha_j-\beta_{j} i \\ \end{array} \right], \ i = \sqrt{-1}, \ 1\leq j\leq k+2l, \\ &&\delta_j^{-1} = \bar{\delta}_j, \ 1 \leq j \leq k, \\ &&\delta_{k+2j-1}^{-1} = \delta_{k+2j}, \ 1 \leq j \leq l, \end{eqnarray*}$

and the adjustment of the column vectors of $X$ corresponds to those of $\Lambda$ .

Define $T_p$ as

$\begin{equation} T_p = \text{diag}\left\{ I_{t+2s}, \frac{1}{\sqrt{2}}\left[\begin{array}{cc} 1 & -i \\ 1 & i \\ \end{array}\right], \cdots, \frac{1}{\sqrt{2}}\left[\begin{array}{cc} 1 & -i \\ 1 & i \\ \end{array}\right]\right\} \in \mathbb{C}^{p \times p}, \end{equation}$

(5)

where $i = \sqrt{-1}$ . It is easy to verify that $T_p^\mathrm{H} T_p = I_p$ . Using this matrix of (5), we obtain

$\begin{equation} \tilde{\Lambda} = T_p^\mathrm{H} \Lambda T_p = \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & \Lambda_1 & 0 \\ 0 & 0 & \tilde{\Lambda}_2 \\ \end{array} \right], \end{equation}$

(6)

$\begin{equation} \tilde{X} = XT_p = [x_t, \cdots, x_{t+2s}, \sqrt{2}y_{t+2s+1}, \sqrt{2}z_{t+2s+1}, \cdots, \sqrt{2}y_{p-1}, \sqrt{2}z_{p-1}], \end{equation}$

(7)

where

$\begin{equation*} \tilde{\Lambda}_2 = \text{diag} \left\{ \left[\begin{array}{cc} \alpha_1 & \beta_1 \\ -\beta_1 & \alpha_1 \\ \end{array}\right], \cdots, \left[\begin{array}{cc} \alpha_{k+2l} & \beta_{k+2l} \\ -\beta_{k+2l} & \alpha_{k+2l} \\ \end{array}\right] \right\}\triangleq \text{diag} \{\tilde{\delta}_1, \cdots, \tilde{\delta}_{k+2l}\}, \end{equation*}$

and $\tilde{\Lambda}_2 \in {\bf \mathbb{R}}^{2(k+2l) \times 2(k+2l)}$ , $\tilde{X} \in {\bf \mathbb{R}}^{n \times p}$ . $y_{t+2s+j}$ and $z_{t+2s+j}$ are, respectively, the real part and imaginary part of the complex vector $x_{t+2s+j}$ for $j = 1, 3, \cdots, 2(k+2l)-1$ . Using (6) and (7), the matrix equation (2) is equivalent to

$\begin{equation} A\tilde{X} = A^\top \tilde{X}\tilde{\Lambda}. \end{equation}$

(8)

Since $\text{rank}(X) = \text{rank}(\tilde{X}) = p$ . Now, let the QR-decomposition of $\tilde{X}$ be

$\begin{equation} \tilde{X} = Q\left[ \begin{array}{c} R \\ 0 \\ \end{array} \right], \end{equation}$

(9)

where $Q = [Q_1, Q_2]\in \mathbb{R}^{n \times n}$ is an orthogonal matrix and $R \in \mathbb{R}^{p \times p}$ is nonsingular. Let

$\begin{equation} \begin{array}{cc} Q^\top AQ = \left[ \begin{array}{cc} A_{11}& A_{12} \\ A_{21}& A_{22} \\ \end{array}\right]&\begin{array}{c} p \\ n-p \\ \end{array} \\ \begin{array}{cc} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ p & \ n-p \\ \end{array}& \end{array}. \end{equation}$

(10)

Using (9) and (10), then the equation of (8) is equivalent to

$\begin{eqnarray} && A_{11}R = A_{11}^\top R\tilde{\Lambda}, \end{eqnarray}$

(11)

$\begin{eqnarray} && A_{21}R = A_{12}^\top R\tilde{\Lambda}. \end{eqnarray}$

(12)

Write

$\begin{equation} \begin{array}{ccc} R^\top A_{11}R\triangleq F = \left[ \begin{array}{ccc} f_{11} & F_{12} & \ \ \ F_{13} \\ F_{21} & F_{22} & \ \ \ F_{23} \\ F_{31} & F_{32} & \ \ \ F_{33} \\ \end{array} \right]&\begin{array}{c} t \\ 2s \\ 2(k+2l) \\ \end{array} \\ \begin{array}{ccc} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ t & \ \ \ 2s & \ 2(k+2l) \\ \end{array}& \end{array}, \end{equation}$

(13)

then the equation of (11) is equivalent to

$\begin{eqnarray} && F_{12} = F_{21}^\top \Lambda_1, \ F_{21} = F_{12}^\top, \end{eqnarray}$

(14)

$\begin{eqnarray} && F_{13} = F_{31}^\top \tilde{\Lambda}_2, \ F_{31} = F_{13}^\top, \end{eqnarray}$

(15)

$\begin{eqnarray} && F_{23} = F_{32}^\top \tilde{\Lambda}_2, \ F_{32} = F_{23}^\top \Lambda_1, \end{eqnarray}$

(16)

$\begin{eqnarray} && F_{22} = F_{22}^\top \Lambda_1, \end{eqnarray}$

(17)

$\begin{eqnarray} && F_{33} = F_{33}^\top \tilde{\Lambda}_2. \end{eqnarray}$

(18)

Because the elements of $\Lambda_1, \tilde{\Lambda}_2$ are distinct, we can obtain the following relations by Eqs (14)-(18)

$\begin{eqnarray} &&F_{12} = 0, \ F_{21} = 0, \ F_{13} = 0, \ F_{31} = 0, \ F_{23} = 0, \ F_{32} = 0, \end{eqnarray}$

(19)

$\begin{eqnarray} &&F_{22} = \text{diag} \left\{ \left[\begin{array}{cc} 0 & h_{1} \\ \lambda_1 h_{1} & 0 \\ \end{array}\right], \cdots, \left[\begin{array}{cc} 0 & h_{s} \\ \lambda_{2s-1} h_{s} & 0 \\ \end{array}\right] \right\}, \end{eqnarray}$

(20)

$\begin{eqnarray} &&F_{33} = \text{diag}\left\{G_{1}, \cdots, G_{k}, \left[\begin{array}{cc} 0 & G_{k+1} \\ G_{k+1}^\top\tilde{\delta}_{k+1} & 0 \\ \end{array}\right], \cdots, \left[\begin{array}{cc} 0 & G_{k+l} \\ G_{k+l}^\top\tilde{\delta}_{k+2l-1} & 0 \\ \end{array}\right]\right\}, \end{eqnarray}$

(21)

where

$\begin{eqnarray*} && G_i = a_iB_i, \ G_{k+j} = a_{k+2j-1}D_1+a_{k+2j}D_2, \ G_{k+j}^\top = G_{k+j}, \\ && B_i = \left[\begin{array}{cc} 1 & \frac{1-\alpha_{i}}{\beta_{i}} \\ -\frac{1-\alpha_{i}}{\beta_{i}} & 1 \\ \end{array}\right], \ D_1 = \left[\begin{array}{cc} 1 & 0 \\ 0 & -1 \\ \end{array}\right], \ D_2 = \left[\begin{array}{cc} 0 & 1 \\ 1 & 0 \\ \end{array}\right], \end{eqnarray*}$

and $1\leq i\leq k, 1\leq j\leq l$ . $h_{1}, \cdots, h_{s}, a_{1}, \cdots, a_{k+2l}$ are arbitrary real numbers. It follows from Eq (12) that

$\begin{equation} A_{21} = A_{12}^\top E, \end{equation}$

(22)

where $E = R\tilde{\Lambda}R^{-1}$ .

Theorem 1. Suppose that $\Lambda = \mathit{\text{diag}} \{\lambda_{1}, \cdots, \lambda_{p}\} \in {\mathbb{C}}^{p \times p}$ , $X = [x_{1}, \cdots, x_{p}] \in {\mathbb{C}}^{n \times p}$ , where diagonal elements of $\Lambda$ are all distinct, $X$ is of full column rank $p$ , and both $\Lambda$ and $X$ are closed under complex conjugation in the sense that $\lambda_{2i} = \bar{\lambda}_{2i-1} \in {\mathbb{C}}$ , $x_{2i} = \bar{x}_{2i-1} \in {\mathbb{C}}^{n}$ for $i = 1, \cdots, m$ , and $\lambda_{j} \in {\mathbb{R}}$ , $x_{j} \in {\mathbb{R}}^{n}$ for $j = 2m+1, \cdots, p$ . Rearrange the matrix $\Lambda$ as $(4)$ , and adjust the column vectors of $X$ with corresponding to those of $\Lambda$ . Let $\Lambda, X$ transform into $\tilde{\Lambda}, \tilde{X}$ by $(6)-(7)$ and QR-decomposition of the matrix $\tilde{X}$ be given by $(9)$ . Then the general solution of $(2)$ can be expressed as

$\begin{equation} {\mathcal{S}}_{A} = \left\{A \left| A = Q\left[ \begin{array}{cc} R^{-\top}\left[ \begin{array}{ccc} f_{11} & 0 & 0 \\ 0 & F_{22} & 0 \\ 0 & 0 & F_{33} \\ \end{array} \right]R^{-1} & A_{12} \\ A_{12}^\top E & A_{22} \\ \end{array} \right]Q^\top \right. \right\}, \end{equation}$

(23)

where $E = R\tilde{\Lambda}R^{-1},$ $f_{11}$ is arbitrary real number, $A_{12} \in {\mathbb{R}}^{p \times (n-p)}, A_{22}\in {\mathbb{R}}^{(n-p) \times (n-p)}$ are arbitrary real-valued matrices and $F_{22}, F_{33}$ are given by $(20)-(21)$ .

3. The solution of Problem BAP

In order to solve Problem BAP, we need the following lemma.

Lemma 1. ^[16] Let $A, B$ be two real matrices, and $X$ be an unknown variable matrix. Then

$\begin{eqnarray*} && \frac{\partial tr(BX)}{\partial X} = B^\top, \ \frac{\partial tr(X^\top B^\top)}{\partial X} = B^\top, \ \frac{\partial tr(AXBX)}{\partial X} = (BXA+AXB)^\top, \\ && \frac{\partial tr(AX^\top BX^\top)}{\partial X} = BX^\top A+AX^\top B, \ \frac{\partial tr(AXBX^\top)}{\partial X} = AXB+A^\top XB^\top. \end{eqnarray*}$

By Theorem $1$ , we can obtain the explicit representation of the solution set ${\mathcal{S}}_{A}$ . It is easy to verify that $\mathcal{S}_A$ is a closed convex subset of ${\bf \mathbb{R}}^{n \times n}\times {\bf \mathbb{R}}^{n \times n}.$ By the best approximation theorem (see Ref. ^[17]), we know that there exists a unique solution of Problem BAP. In the following we will seek the unique solution $\hat{A}$ in $\mathcal{S}_A.$ For the given matrix $\tilde{A} \in {\bf \mathbb{R}}^{n \times n},$ write

$\begin{equation} \begin{array}{cc} Q^\top \tilde{A}Q = \left[ \begin{array}{cc} \tilde{A}_{11}& \tilde{A}_{12} \\ \tilde{A}_{21}& \tilde{A}_{22} \\ \end{array}\right]&\begin{array}{c} p \\ n-p \\ \end{array} \\ \begin{array}{cc} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ p & \ n-p \\ \end{array}& \end{array}, \end{equation}$

(24)

then

$\begin{eqnarray*} \|A-\tilde{A}\|^2 & = & \left\|\left[ \begin{array}{cc} R^{- \top}\left[ \begin{array}{ccc} f_{11} & 0 & 0 \\ 0 & F_{22} & 0 \\ 0 & 0 & F_{33} \\ \end{array} \right]R^{-1}-\tilde{A}_{11} & A_{12}-\tilde{A}_{12} \\ A_{12}^\top E-\tilde{A}_{21} & A_{22}-\tilde{A}_{22} \\ \end{array} \right]\right\|^2\\ & = &\left\| R^{- \top}\left[ \begin{array}{ccc} f_{11} & 0 & 0 \\ 0 & F_{22} & 0 \\ 0 & 0 & F_{33} \\ \end{array} \right]R^{-1}-\tilde{A}_{11}\right\|^2\\ &+& \| A_{12}-\tilde{A}_{12}\|^2+\|A_{12}^\top E-\tilde{A}_{21} \|^2+\|A_{22}-\tilde{A}_{22}\|^2. \end{eqnarray*}$

Therefore, $\|A-\tilde{A}\| = \min$ if and only if

$\begin{eqnarray} && \left\|R^{-\top}\left[ \begin{array}{ccc} f_{11} & 0 & 0 \\ 0 & F_{22} & 0 \\ 0 & 0 & F_{33} \\ \end{array} \right]R^{-1}-\tilde{A}_{11}\right\|^2 = \min, \end{eqnarray}$

(25)

$\begin{eqnarray} && \| A_{12}-\tilde{A}_{12}\|^2+\|A_{12}^\top E-\tilde{A}_{21} \|^2 = \min, \end{eqnarray}$

(26)

$\begin{eqnarray} &&A_{22} = \tilde{A}_{22}. \end{eqnarray}$

(27)

Let

$\begin{equation} \begin{array}{c} R^{-1} = \left[ \begin{array}{c} R_1 \\ R_2 \\ R_3 \\ \end{array} \right] \end{array}, \end{equation}$

(28)

then the relation of (25) is equivalent to

$\begin{equation} \|R_1^\top f_{11}R_1+ R_2^\top F_{22}R_2 + R_3^\top F_{33} R_3- \tilde{A}_{11} \|^2 = \min. \end{equation}$

(29)

Write

$\begin{equation} R_1 = \left[r_{1, t}\right], \ R_2 = \left[ \begin{array}{c} r_{2, 1} \\ \vdots \\ r_{2, 2s}\\ \end{array} \right], \ R_3 = \left[ \begin{array}{c} r_{3, 1} \\ \vdots \\ r_{3, k+2l} \\ \end{array} \right], \end{equation}$

(30)

where $r_{1, t} \in {\mathbb{R}}^{t \times p}, r_{2, i} \in {\mathbb{R}}^{1 \times p}, r_{3, j} \in {\mathbb{R}}^{2 \times p}, \ i = 1, \cdots, 2s, \ j = 1, \cdots, k+2l$ .

Let

$\begin{equation} \left\{ \begin{array}{rcl} && J_t = r_{1, t}^\top r_{1, t}, \\ && J_{t+i} = \lambda_{2i-1}r_{2, 2i}^\top r_{2, 2i-1}+r_{2, 2i-1}^\top r_{2, 2i} \ (1 \leq i \leq s), \\ && J_{r+i} = r_{3, i}^\top B_i r_{3, i} \ (1 \leq i \leq k), \\ && J_{r+k+2i-1} = r_{3, k+2i}^\top D_1 \tilde{\delta}_{k+2i-1}r_{3, k+2i-1}+r_{3, k+2i-1}^\top D_1 r_{3, k+2i} \ (1 \leq i \leq l), \\ && J_{r+k+2i} = r_{3, k+2i}^\top D_2 \tilde{\delta}_{k+2i-1}r_{3, k+2i-1}+r_{3, k+2i-1}^\top D_2 r_{3, k+2i} \ (1 \leq i \leq l), \end{array} \right. \end{equation}$

(31)

with $r = t+s, q = t+s+k+2l.$ Then the relation of (29) is equivalent to

$\begin{eqnarray*} && g(f_{11}, h_1, \cdots, h_s, a_1, \cdots, a_{k+2l}) = \\ &&\|f_{11}J_t+h_1J_{t+1}+\cdots+h_sJ_{r}+a_1J_{r+1}+\cdots+a_{k+2l}J_{q}-\tilde{A}_{11}\|^2 = \text{min}, \end{eqnarray*}$

that is,

$\begin{eqnarray*} && g(f_{11}, h_1, \cdots, h_s, a_1, \cdots, a_{k+2l}) \\ && = \text{tr} [(f_{11}J_t+h_1J_{t+1}+\cdots+h_sJ_{r}+a_1J_{r+1}+\cdots+a_{k+2l}J_{q}-\tilde{A}_{11})^\top \\ && (f_{11}J_t+h_1J_{t+1}+\cdots+h_sJ_{r}+a_1J_{r+1}+\cdots+a_{k+2l}J_{q}-\tilde{A}_{11})]\\ && = f_{11}^2c_{t, t}+2f_{11}h_1c_{t, t+1}+\cdots+2f_{11}h_sc_{t, r}+2f_{11}a_1c_{t, r+1}+\cdots+2f_{11}a_{k+2l}c_{t, q}-2f_{11}e_t\\ && +h_1^2c_{t+1, t+1}+\cdots+2h_1h_sc_{t+1, r}+2h_1a_1c_{t+1, r+1}+\cdots+2h_1a_{k+2l}c_{t+1, q}-2h_1e_{t+1} \\ && +\cdots \\ && +h_s^2c_{r, r}+2h_sa_1c_{r, r+1}+\cdots+2h_sa_{k+2l}c_{r, q}-2h_se_{r}\\ && +a_1^2c_{r+1, r+1}+\cdots+2a_1a_{k+2l}c_{r+1, q}-2a_1e_{r+1}\\ && +\cdots \\ && +a_{k+2l}^2c_{q, q}-2a_{k+2l}e_{q}+\text{tr} (\tilde{A}_{11}^\top \tilde{A}_{11}), \end{eqnarray*}$

where $c_{i, j} = \text{tr} (J_i^\top J_j), e_i = \text{tr} (J_i^\top\tilde{A}_{11}) (i, j = t, \cdots, t+s+k+2l)$ and $c_{i, j} = c_{j, i}$ .

Consequently,

$\begin{eqnarray*} \frac{\partial g(f_{11}, h_1, \cdots, h_s, a_1, \cdots, a_{k+2l})}{\partial f_{11}}&& = 2f_{11}c_{t, t}+2h_1c_{t, t+1}+\cdots+2h_sc_{t, r}+2a_1c_{t, r+1}\\ &&+\cdots+2a_{k+2l}c_{t, q}-2e_t, \\ \frac{\partial g(f_{11}, h_1, \cdots, h_s, a_1, \cdots, a_{k+2l})}{\partial h_{1}}&& = 2f_{11}c_{t+1, t}+2h_1c_{t+1, t+1}+\cdots+2h_sc_{t+1, r}+2a_1c_{t+1, r+1}\\ &&+\cdots+2a_{k+2l}c_{t+1, q}-2e_{t+1}, \\ &&\cdots\\ \frac{\partial g(f_{11}, h_1, \cdots, h_s, a_1, \cdots, a_{k+2l})}{\partial h_{s}}&& = 2f_{11}c_{r, t}+2h_1c_{r, t+1}+\cdots+2h_sc_{r, r}+2a_1c_{r, r+1}\\ &&+\cdots+2a_{k+2l}c_{r, q}-2e_{r}, \\ \frac{\partial g(f_{11}, h_1, \cdots, h_s, a_1, \cdots, a_{k+2l})}{\partial a_{1}}&& = 2f_{11}c_{r+1, t}+2h_1c_{r+1, t+1}+\cdots+2h_sc_{r+1, r}+2a_1c_{r+1, r+1}\\ &&+\cdots+2a_{k+2l}c_{r+1, q}-2e_{r+1}, \\ &&\cdots\\ \frac{\partial g(f_{11}, h_1, \cdots, h_s, a_1, \cdots, a_{k+2l})}{\partial a_{k+2l}}&& = 2f_{11}c_{q, t}+2h_1c_{q, t+1}+\cdots+2h_sc_{q, r}+2a_1c_{q, r+1}\\ &&+\cdots+2a_{k+2l}c_{q, q}-2e_{q}. \end{eqnarray*}$

Clearly, $g(f_{11}, h_1, \cdots, h_s, a_1, \cdots, a_{k+2l}) = \text{min}$ if and only if

$\frac{\partial g(f_{11}, h_1, \cdots, h_s, a_1, \cdots, a_{k+2l})}{\partial f_{11}} = 0, \cdots, \frac{\partial g(f_{11}, h_1, \cdots, h_s, a_1, \cdots, a_{k+2l})}{\partial a_{k+2l}} = 0.$

Therefore,

$\begin{equation} \begin{split} &f_{11}c_{t, t}+h_1c_{t, t+1}+\cdots+h_sc_{t, r}+a_1c_{t, r+1}+\cdots+a_{k+2l}c_{t, q} = e_t, \\ &f_{11}c_{t+1, t}+h_1c_{t+1, t+1}+\cdots+h_sc_{t+1, r}+a_1c_{t+1, r+1}+\cdots+a_{k+2l}c_{t+1, q} = e_{t+1}, \\ &\cdots\\ &f_{11}c_{r, t}+h_1c_{r, t+1}+\cdots+h_sc_{r, r}+a_1c_{r, r+1}+\cdots+a_{k+2l}c_{r, q} = e_{r}, \\ &f_{11}c_{r+1, t}+h_1c_{r+1, t+1}+\cdots+h_sc_{r+1, r}+a_1c_{r+1, r+1}+\cdots+a_{k+2l}c_{r+1, q} = e_{r+1}, \\ &\cdots\\ &f_{11}c_{q, t}+h_1c_{q, t+1}+\cdots+h_sc_{q, r}+a_1c_{q, r+1}+\cdots+a_{k+2l}c_{q, q} = e_{q}. \end{split} \end{equation}$

(32)

If let

$C = \left[ \begin{array}{ccccccc} c_{t, t}&c_{t, t+1}&\cdots&c_{t, r}&c_{t, r+1}&\cdots&c_{t, q}\\ c_{t+1, t}&c_{t+1, t+1}&\cdots&c_{t+1, r}&c_{t+1, r+1}&\cdots&c_{t+1, q}\\ \vdots&\vdots& &\vdots&\vdots& &\vdots\\ c_{r, t}&c_{r, t+1}&\cdots&c_{r, r}&c_{r, r+1}&\cdots&c_{r, q}\\ c_{r+1, t}&c_{r+1, t+1}&\cdots&c_{r+1, r}&c_{r+1, r+1}&\cdots&c_{r+1, q}\\ \vdots&\vdots& &\vdots&\vdots& &\vdots\\ c_{q, t}&c_{q, t+1}&\cdots&c_{q, r}&c_{q, r+1}&\cdots&c_{q, q}\\ \end{array} \right], \ h = \left[ \begin{array}{c} f_{11}\\ h_1\\ \vdots\\ h_s\\ a_1\\ \vdots\\ a_{k+2l}\\ \end{array} \right], \ e = \left[ \begin{array}{c} e_t\\ e_{t+1}\\ \vdots\\ e_{r}\\ e_{r+1}\\ \vdots\\ e_{q}\\ \end{array} \right],$

where $C$ is symmetric matrix. Then the equation (32) is equivalent to

$\begin{equation} C h = e, \end{equation}$

(33)

and the solution of the equation (33) is

$\begin{equation} h = C^{-1} e. \end{equation}$

(34)

Substituting (34) into (20)-(21), we can obtain $f_{11}, F_{22}$ and $F_{33}$ explicitly. Similarly, the equation of (26) is equivalent to

$\begin{eqnarray*} &&g(A_{12}) = \text{tr} (A_{12}^\top A_{12})+\text{tr} (\tilde{A}_{12}^\top \tilde{A}_{12})-2\text{tr} (A_{12}^\top \tilde{A}_{12})\\ &&+\text{tr} (E^\top A_{12}A_{12}^\top E)+\text{tr} (\tilde{A}_{21}^\top \tilde{A}_{21})-2\text{tr} (E^\top A_{12}\tilde{A}_{21}). \end{eqnarray*}$

Applying Lemma 1, we obtain

$\frac{\partial g(A_{12})}{\partial A_{12}} = 2A_{12}-2\tilde{A}_{12}+2EE^\top A_{12}-2E\tilde{A}_{21}^\top,$

setting $\frac{\partial g(A_{12})}{\partial A_{12}} = 0$ , we obtain

$\begin{equation} A_{12} = (I_p+EE^\top)^{-1}(\tilde{A}_{12}+E\tilde{A}_{21}^\top), \end{equation}$

(35)

Theorem 2. Given $\tilde{A} \in {\mathbb{R}}^{n \times n}$ , then the Problem BAP has a unique solution and the unique solution of Problem BAP is

$\begin{equation} \hat{A} = Q\left[ \begin{array}{cc} R^{-\top}\left[ \begin{array}{ccc} f_{11} & 0 & 0 \\ 0 & F_{22} & 0 \\ 0 & 0 & F_{33} \\ \end{array} \right]R^{-1} & A_{12} \\ A_{12}^\top E & \tilde{A}_{22} \\ \end{array} \right]Q^\top, \end{equation}$

(36)

where $E = R\tilde{\Lambda} R^{-1}$ , $F_{22}, F_{33}, A_{12}, \tilde{A}_{22}$ are given by $(20), (21), (35), (24)$ and $f_{11}, h_{1}, \cdots, h_s, a_1, \cdots, a_{k+2l}$ are given by $(34)$ .

4. A numerical example

Based on Theorems $1$ and $2$ , we can describe an algorithm for solving Problem BAP as follows.

Algorithm 1.

1) Input matrices $\Lambda$ , $X$ and $\tilde{A}$ ;

2) Rearrange $\Lambda$ as (4), and adjust the column vectors of $X$ with corresponding to those of $\Lambda$ ;

3) Form the unitary transformation matrix $T_p$ by (5);

4) Compute real-valued matrices $\tilde{\Lambda}, \tilde{X}$ by (6) and (7);

5) Compute the QR-decomposition of $\tilde{X}$ by (9);

6) $F_{12} = 0, F_{21} = 0, F_{13} = 0, F_{31} = 0, F_{23} = 0, F_{32} = 0$ by (19) and $E = R\tilde{\Lambda} R^{-1}$ ;

7) Compute $\tilde{A}_{ij} = Q_i^\top \tilde{A}Q_j, i, j = 1, 2$ ;

8) Compute $R^{-1}$ by (28) to form $R_1, R_2, R_3$ ;

9) Divide matrices $R_1, R_2, R_3$ by (30) to form $r_{1, t}, r_{2, i}, r_{3, j},$ $i = 1, \cdots, 2s, j = 1, \cdots, k+2l$ ;

10) Compute $J_i,$ $i = t, \cdots, t+s+k+2l,$ by (31);

11) Compute $c_{i, j} = \mbox{tr} (J_i^\top J_j), e_i = \mbox{tr} (J_i^\top\tilde{A}_{11}),$ $i, j = t, \cdots, t+s+k+2l$ ;

12) Compute $f_{11}, h_1, \cdots, h_s, a_1, \cdots, a_{k+2l}$ by (34);

13) Compute $F_{22}, F_{33}$ by (20), (21) and $A_{22} = \tilde{A}_{22}$ ;

14) Compute $A_{12}$ by (35) and $A_{21}$ by (22);

15) Compute the matrix $\hat{A}$ by (36).

Example 1. Consider a $11$ -DOF system, where

$\tilde{A} = \left[ {\begin{array}{rrrrrrrrrrr} 96.1898 & 18.1847 & 51.3250 & 49.0864 & 13.1973 & 64.9115 & 62.5619 & 81.7628 & 58.7045 & 31.1102 & 26.2212 \\ 0.4634 & 26.3803 & 40.1808 & 48.9253 & 94.2051 & 73.1722 & 78.0227 & 79.4831 & 20.7742 & 92.3380 & 60.2843 \\ 77.4910 & 14.5539 & 7.5967 & 33.7719 & 95.6135 & 64.7746 & 8.1126 & 64.4318 & 30.1246 & 43.0207 & 71.1216 \\ 81.7303 & 13.6069 & 23.9916 & 90.0054 & 57.5209 & 45.0924 & 92.9386 & 37.8609 & 47.0923 & 18.4816 & 22.1747 \\ 86.8695 & 86.9292 & 12.3319 & 36.9247 & 5.9780 & 54.7009 & 77.5713 & 81.1580 & 23.0488 & 90.4881 & 11.7418 \\ 8.4436 & 57.9705 & 18.3908 & 11.1203 & 23.4780 & 29.6321 & 48.6792 & 53.2826 & 84.4309 & 97.9748 & 29.6676 \\ 39.9783 & 54.9860 & 23.9953 & 78.0252 & 35.3159 & 74.4693 & 43.5859 & 35.0727 & 19.4764 & 43.8870 & 31.8778 \\ 25.9870 & 14.4955 & 41.7267 & 38.9739 & 82.1194 & 18.8955 & 44.6784 & 93.9002 & 22.5922 & 11.1119 & 42.4167 \\ 80.0068 & 85.3031 & 4.9654 & 24.1691 & 1.5403 & 68.6775 & 30.6349 & 87.5943 & 17.0708 & 25.8065 & 50.7858 \\ 43.1414 & 62.2055 & 90.2716 & 40.3912 & 4.3024 & 18.3511 & 50.8509 & 55.0156 & 22.7664 & 40.8720 & 8.5516 \\ 91.0648 & 35.0952 & 94.4787 & 9.6455 & 16.8990 & 36.8485 & 51.0772 & 62.2475 & 43.5699 & 59.4896 & 26.2482 \\ \end{array}} \right],$

the measured eigenvalue and eigenvector matrices $\Lambda$ and $X$ are given by

$\begin{eqnarray*} &&\Lambda = \mbox{diag} \{1.0000, \ -1.8969, \ -0.5272, \ -0.1131+0.9936i, -0.1131-0.9936i, \\ &&1.9228+2.7256i, \ 1.9228-2.7256i, \ 0.1728-0.2450i, \ 0.1728+0.2450i\}, \end{eqnarray*}$

and

$\begin{eqnarray*} &&X = \left[ {\begin{array}{rrrrr} -0.0132& -1.0000& 0.1753& 0.0840 + 0.4722i& 0.0840 - 0.4722i \\ -0.0955& 0.3937& 0.1196& -0.3302 - 0.1892i& -0.3302 + 0.1892i \\ -0.1992& 0.5220& -0.0401& 0.3930 - 0.2908i& 0.3930 + 0.2908i \\ 0.0740& 0.0287& 0.6295& -0.3587 - 0.3507i& -0.3587 + 0.3507i \\ 0.4425& -0.3609& -0.5745& 0.4544 - 0.3119i& 0.4544 + 0.3119i \\ 0.4544& -0.3192& -0.2461& -0.3002 - 0.1267i& -0.3002 + 0.1267i \\ 0.2597& 0.3363& 0.9046& -0.2398 - 0.0134i& -0.2398 + 0.0134i \\ 0.1140& 0.0966& 0.0871& 0.1508 + 0.0275i& 0.1508 - 0.0275i \\ -0.0914& -0.0356& -0.2387& -0.1890 - 0.0492i& -0.1890 + 0.0492i \\ 0.2431& 0.5428& -1.0000& 0.6652 + 0.3348i& 0.6652 - 0.3348i \\ 1.0000& -0.2458& 0.2430& -0.2434 + 0.6061i& -0.2434 - 0.6061i \\ \end{array}} \right. \\ &&\left. {\begin{array}{rrrr} 0.6669 + 0.2418i& 0.6669 - 0.2418i& 0.2556 - 0.1080i& 0.2556 + 0.1080i \\ -0.1172 - 0.0674i& -0.1172 + 0.0674i& -0.5506 - 0.1209i& -0.5506 + 0.1209i \\ 0.5597 - 0.2765i& 0.5597 + 0.2765i& -0.3308 + 0.1936i& -0.3308 - 0.1936i \\ -0.7217 - 0.0566i& -0.7217 + 0.0566i& -0.7306 - 0.2136i& -0.7306 + 0.2136i \\ 0.0909 + 0.0713i& 0.0909 - 0.0713i& 0.5577 + 0.1291i& 0.5577 - 0.1291i \\ 0.1867 + 0.0254i& 0.1867 - 0.0254i& 0.2866 + 0.1427i& 0.2866 - 0.1427i \\ -0.5311 - 0.1165i& -0.5311 + 0.1165i& -0.3873 - 0.1096i& -0.3873 + 0.1096i \\ 0.2624 + 0.0114i& 0.2624 - 0.0114i& -0.6438 + 0.2188i& -0.6438 - 0.2188i \\ -0.0619 - 0.1504i& -0.0619 + 0.1504i& 0.2787 - 0.2166i& 0.2787 + 0.2166i \\ 0.3294 - 0.1718i& 0.3294 + 0.1718i& 0.9333 + 0.0667i& 0.9333 - 0.0667i \\ -0.4812 + 0.5188i& -0.4812 - 0.5188i& 0.6483 - 0.1950i& 0.6483 + 0.1950i \\ \end{array}} \right]. \end{eqnarray*}$

Using Algorithm 1, we obtain the unique solution of Problem BAP as follows:

$\begin{eqnarray*} \hat{A} = \left[ {\begin{array}{rrrrrrrrrrr} 34.2563 & 41.7824 & 33.3573 & 33.6298 & 23.8064 & 42.0770 & 50.0641 & 37.5705 & 31.0908 & 48.6169 & 19.0972\\ 18.8561 & 35.2252 & 35.9592 & 44.3502 & 31.9918 & 55.2920 & 55.3052 & 54.3793 & 31.3909 & 60.8345 & 16.9540\\ 29.6359 & 7.6805 & 19.1249 & 17.7183 & 16.7082 & 40.0636 & 18.2916 & 49.9437 & 37.6913 & 15.6027 & 4.9603\\ 58.8782 & 51.4906 & 47.8974 & 35.6985 & 45.6889 & 56.0434 & 53.0908 & 56.5402 & 55.5120 & 38.3447 & 35.8894\\ 33.4087 & 46.9635 & 9.7767 & 41.4215 & 51.4466 & 52.1058 & 65.6724 & 60.1293 & 5.8061 & 62.0139 & 16.5231\\ 31.6580 & 51.2359 & 24.7978 & 65.5567 & 61.7840 & 62.5494 & 58.9363 & 74.7099 & 52.2105 & 55.8532 & 44.3925\\ 19.2961 & 51.2333 & 22.4280 & 56.9340 & 42.6348 & 45.8453 & 56.3729 & 61.5555 & 31.6836 & 67.9525 & 40.2012\\ 41.2796 & 71.3821 & 34.4140 & 33.2817 & 77.4393 & 60.8944 & 32.1411 & 108.5056 & 49.6078 & 19.8351 & 85.7434\\ 64.0890 & 57.6524 & 19.1280 & 25.0394 & 39.0524 & 66.7740 & 20.9023 & 48.8512 & 14.4695 & 18.9284 & 24.8348\\ 37.2550 & 32.3254 & 38.3534 & 59.7358 & 33.5902 & 54.0265 & 50.7770 & 70.2011 & 65.4159 & 58.0720 & 40.0652\\ 28.1301 & 14.7638 & 8.9507 & 20.0963 & 25.5907 & 59.6940 & 30.8558 & 66.8781 & 30.4807 & 23.6107 & 12.9984\\ \end{array}} \right], \end{eqnarray*}$

and

$\|\hat{A}X-\hat{A}^\top X\Lambda\| = 8.2431\times 10^{-13}.$

Therefore, the new model $\hat{A}X = \hat{A}^\top X\Lambda$ reproduces the prescribed eigenvalues (the diagonal elements of the matrix $\Lambda$ ) and eigenvectors (the column vectors of the matrix $X$ ).

Example 2. (Example 4.1 of ^[12]) Given $\alpha = \cos(\theta)$ , $\beta = \sin(\theta)$ with $\theta = 0.62$ and $\lambda_1 = 0.2, \lambda_2 = 0.3, \lambda_3 = 0.4$ . Let

$\begin{equation*} J_0 = \left[ {\begin{array}{rr} 0_2 & \Gamma\\ I_2 & I_2\\ \end{array}} \right], \ J_s = \left[ {\begin{array}{cc} 0_3 & \mbox{diag} \{\lambda_1, \lambda_2, \lambda_3 \}\\ I_3 & 0_3\\ \end{array}} \right], \end{equation*}$

where $\Gamma = \left[{\begin{array}{cc} \alpha & -\beta\\ \beta & \alpha\\ \end{array}} \right].$ We construct

$\begin{equation*} \tilde{A} = \left[ {\begin{array}{rr} J_0 & 0\\ 0 & J_s\\ \end{array}} \right], \end{equation*}$

the measured eigenvalue and eigenvector matrices $\Lambda$ and $X$ are given by

$\begin{eqnarray*} \Lambda = \mbox{diag} \{5, 0.2, 0.8139+0.5810i, 0.8139-0.5810i\}, \end{eqnarray*}$

and

$\begin{eqnarray*} X = \left[ {\begin{array}{rrrr} -0.4155& 0.6875& -0.2157 - 0.4824i& -0.2157 + 0.4824i\\ -0.4224& -0.3148& -0.3752 + 0.1610i& -0.3752 - 0.1610i\\ -0.0703& -0.6302& -0.5950 - 0.4050i& -0.5950 + 0.4050i\\ -1.0000& -0.4667& 0.2293 - 0.1045i& 0.2293 + 0.1045i\\ 0.2650& 0.3051& -0.2253 + 0.7115i& -0.2253 - 0.7115i\\ 0.9030& -0.2327& 0.4862 - 0.3311i& 0.4862 + 0.3311i\\ -0.6742& 0.3132& 0.5521 - 0.0430i& 0.5521 + 0.0430i\\ 0.6358& 0.1172& -0.0623 - 0.0341i& -0.0623 + 0.0341i\\ -0.4119& -0.2768& 0.1575 + 0.4333i& 0.1575 - 0.4333i\\ -0.2062& 1.0000& -0.1779 - 0.0784i& -0.1779 + 0.0784i\\ \end{array}} \right]. \end{eqnarray*}$

Using Algorithm 1, we obtain the unique solution of Problem BAP as follows:

$\begin{equation*} \hat{A} = \left[ {\begin{array}{rrrrrrrrrr} -0.1169& -0.2366& 0.6172& -0.7195& -0.0836& 0.2884& 0.0092& -0.0490& -0.0202& 0.0171\\ -0.0114& -0.0957& 0.1462& 0.6194& 0.3738& -0.1637& 0.1291& -0.0071& 0.0972& 0.1247\\ 0.7607& -0.0497& 0.5803& -0.0346& 0.0979& 0.2959& 0.0937& -0.1060& 0.1323& -0.0339\\ -0.0109& 0.6740& -0.3013& 0.7340& 0.1942& -0.0872& 0.0054& 0.0051& 0.0297& 0.0814\\ 0.1783& 0.2283& 0.2643& 0.0387& 0.0986& -0.3125& -0.0292& 0.2926& -0.0717& -0.0546\\ 0.0953& 0.1027& 0.0360& 0.2668& -0.2418& 0.1206& 0.1406& -0.0551& 0.3071& 0.2097\\ -0.0106& -0.2319& 0.1946& -0.0298& -0.1935& 0.0158& -0.0886& 0.0216& -0.0560& 0.2484\\ 0.1044& 0.1285& 0.1902& 0.2277& 0.6961& 0.1657& 0.0728& -0.0262& -0.0831& -0.0001\\ 0.0906& 0.0021& 0.0764& -0.1264& 0.2144& 0.6703& -0.0850& 0.0764& -0.0104& -0.0149\\ -0.1245& 0.0813& 0.1952& -0.0784& 0.0760& -0.0875& 0.7978& -0.0093& 0.0206& -0.1182\\ \end{array}} \right], \end{equation*}$

and

$\|\hat{A}X-\hat{A}^\top X\Lambda\| = 1.7538\times 10^{-8}.$

5. Concluding remarks

In this paper, we have developed a direct method to solve the linear inverse palindromic eigenvalue problem by partitioning the matrix $\Lambda$ and using the QR-decomposition. The explicit best approximation solution is given. The numerical examples show that the proposed method is straightforward and easy to implement.

Conflict of interest

The authors declare no conflict of interest.

References

[1]	Abedifar P, Hasan I, Tarazi A (2016) Finance-growth nexus and dual-banking systems: Relative importance of Islamic banks. J Econ Behav Organ 132: 198-215. doi: 10.1016/j.jebo.2016.03.005
[2]	Abrigo MR, Love I (2015) Estimation of Panel Vector Autoregression in Stata: A Package of Programs. Available from: http://paneldataconference2015.ceu.hu/Program/Michael-Abrigo.pdf.
[3]	Ahmed MM, Mallick S (2017) Does Regulatory Forbearance Matter for Bank Stability? Evidence from Creditors' Perspective. J Financ Stab 28: 163-180.
[4]	Agoraki MEK, Delis MD, Pasiouras F (2011) Regulations, competition and bank risk-taking in transition countries. J Financ Stab 7: 38-48. doi: 10.1016/j.jfs.2009.08.002
[5]	Alam N, Hamid BA, Tan DT (2019) Does competition make banks riskier in dual banking system? Borsa Istanbul Rev 19: S34-S43. doi: 10.1016/j.bir.2018.09.002
[6]	Ahi K, Laidroo L (2019) Banking market competition in Europe-financial stability or fragility enhancing? Quant Financ Econ 2: 257-285.
[7]	Albaity M, Mallek RS, Noman AHM (2019) Competition and bank stability in the MENA region: The moderating effect of Islamic versus conventional banks. Emerg Mark Rev 38: 310-325. doi: 10.1016/j.ememar.2019.01.003
[8]	Allen F, Carletti E, Marquez R (2011) Credit Market Competition and Capital Regulation. Rev Financ Stud 24: 983-1018. doi: 10.1093/rfs/hhp089
[9]	Allen F, Gale D (2000) Comparing Financial Systems, MIT press.
[10]	Arellano M, Bover O (1995) Another look at the instrumental variable estimation of error-components models. J Econometrics 68: 29-51. doi: 10.1016/0304-4076(94)01642-D
[11]	Ariss RT (2010) On the implications of market power in banking: Evidence from developing countries. J Bank Financ 34: 765-775. doi: 10.1016/j.jbankfin.2009.09.004
[12]	Beccalli E, Anolli M, Borello G (2015) Are European banks too big? Evidence on economies of scale. J Bank Financ 58: 232-246.
[13]	Beltratti A, Stulz RM (2012) The credit crisis around the globe: Why did some banks perform better? J Financ Econ 105: 1-17. doi: 10.1016/j.jfineco.2011.12.005
[14]	Berger AN, Klapper LF, Turk-Ariss R (2009) Bank competition and financial stability. J Financ Serv Res 35: 99-118. doi: 10.1007/s10693-008-0050-7
[15]	Beck T, De Jonghe O, Schepens G (2013) Bank competition and stability: cross-country heterogeneity. J Financ Intermed 22: 218-244. doi: 10.1016/j.jfi.2012.07.001
[16]	Bikker JA, Shaffer S, Spierdijk L (2012) Assessing competition with the Panzar-Rosse model: The role of scale, costs, and equilibrium. Rev Econ Stat 94: 1025-1044. doi: 10.1162/REST_a_00210
[17]	Boone J, Griffith R, Harrison R (2005) Measuring competition. AIM Research Working Paper Series, Tilburg University.
[18]	Boone J (2008) A new way to measure competition. Econ J 118: 1245-1261. doi: 10.1111/j.1468-0297.2008.02168.x
[19]	Boyd JH, De Nicolo G (2005) The theory of bank risk taking and competition revisited. J Financ 60: 1329-1343. doi: 10.1111/j.1540-6261.2005.00763.x
[20]	Brunnermeier MK, Crockett A, Goodhart CA, et al. (2009) The fundamental principles of financial regulation, (Vol. 11), ICMB, Internation, Center for Monetary and Banking Studies, Geneva, Switzerland.
[21]	Bushman RM, Hendricks BE, Williams CD (2016) Bank competition: measurement, decision-making, and risk-taking. J Accounting Res 54: 777-826. doi: 10.1111/1475-679X.12117
[22]	Canova F, Ciccarelli M (2013) Panel Vector Autoregression Models. ECB Working Paper Series 1507.
[23]	Carletti E (2008) Competition and regulation in banking. Handb Financ Intermed bank 126: 449-482.
[24]	Caminal R, Matutes C (2002) Market power and banking failures. Int J Ind Organ 20: 1341-1361. doi: 10.1016/S0167-7187(01)00092-3
[25]	Carbo-Valverde S, Humphrey D, Maudos J, et al. (2009) Cross-country comparisons of competition and pricing power in European banking. J Int Money Financ 28: 115-134. doi: 10.1016/j.jimonfin.2008.06.005
[26]	Choi I (2001) Unit root tests for panel data. J Int Money Financ 20: 249-272. doi: 10.1016/S0261-5606(00)00048-6
[27]	Choi I (2002) Combination unit root tests for cross‐sectionally correlated panels, In Essays in Honor of Peter C. B. Phillips, Corbae D, Durlauf SN, Hansen B (eds.), Econometric Theory and Practice: Frontiers of Analysis and Applied Research, Cambridge University Press: Cambridge, UK, 311-333.
[28]	Čihák M, Hesse H (2010) Islamic banks and financial stability: an empirical analysis. J Financ Serv Res 38: 95-113 doi: 10.1007/s10693-010-0089-0
[29]	Claessens S (2009) Competition in the financial sector: Overview of competition policies. World Bank Res Obs 24: 83-118. doi: 10.1093/wbro/lkp004
[30]	Claessens S, Laeven L (2004) What drives bank competition? Some international evidence. J Money Credit Bank 36: 563-583. doi: 10.1353/mcb.2004.0044
[31]	Clark E, Mare DS, Radić N (2018) Cooperative banks: What do we know about competition and risk preferences? J Int Financ Mark Inst Money 52: 90-101. doi: 10.1016/j.intfin.2017.09.008
[32]	Clerides S, Delis MD, Kokas S (2015) A new data set on competition in national banking markets. Financ Mark Inst Instrum 24: 267-311. doi: 10.1111/fmii.12030
[33]	Diallo B (2015) Bank competition and crises revisited: New results. Econ Lett 129: 81-86. doi: 10.1016/j.econlet.2015.02.015
[34]	Demirguc-Kunt A, Detragiache E (1997) The Determinants of Banking Crisis: Evidence from developed and developing countries. Int Monetary Fund.
[35]	Dwumfour RA (2017) Explaining banking stability in Sub-Saharan Africa. Res Int Bus Financ 41: 260-279. doi: 10.1016/j.ribaf.2017.04.027
[36]	Yaldiz E, Bazzana F (2010) The effect of market power on bank risk taking in Turkey. Financ Theory Pract 34: 297-314.
[37]	Elzinga KG, Mills DE (2011) The Lerner index of monopoly power: Origins and uses. Am Econ Rev 101: 558-564. doi: 10.1257/aer.101.3.558
[38]	Fosu S (2013) Banking competition in Africa: Sub-regional comparative studies. Working paper no.13/12. Leicester: University of Leicester.
[39]	Fiordelisi F, Mare DS (2014) Competition and financial stability in European cooperative banks. J Int Money Financ 45: 1-16. doi: 10.1016/j.jimonfin.2014.02.008
[40]	Fungáčová Z, Weill L (2013) Does competition influence bank failures? Evidence from Russia. Econ Transit 21: 301-322.
[41]	Goetz MR (2017) Competition and bank stability. J Financ Intermed 35: 57-69.
[42]	Hellmann TF, Murdock KC, Stiglitz JE (2000) Liberalization, moral hazard in banking, and prudential regulation: are capital requirements enough? Am Econ Rev 90: 147-165. doi: 10.1257/aer.90.1.147
[43]	Hermes N, Lensink R (2004) Foreign bank presence, domestic bank performance and financial development. J Emerg Mark Financ 3: 207-229. doi: 10.1177/097265270400300206
[44]	Hughes JP, Mester LJ (2013) Who said large banks don't experience scale economies? Evidence from a risk-return-driven cost function. J Financ Intermed 22: 559-585.
[45]	Im KS, Pesaran MH (2003) On the panel unit root tests using nonlinear instrumental variables. Cambridge Working Papers in Economics NO. 0347, University of Cambridge.
[46]	Im KS, Pesaran MH, Shin Y (2003) Testing for unit roots in heterogeneous panels. J Econometrics 115: 53-74. doi: 10.1016/S0304-4076(03)00092-7
[47]	Jiménez G, Lopez JA, Saurina J (2013) How does competition affect bank risk-taking? J Financ Stab 9: 185-195. doi: 10.1016/j.jfs.2013.02.004
[48]	Kabir MN, Worthington AC (2017) The competition-stability/fragility' nexus: a comparative analysis of Islamic and conventional banks. Int Rev Financ Anal 50: 111-128. doi: 10.1016/j.irfa.2017.02.006
[49]	Kasman A, Carvallo O (2014) Financial stability, competition and efficiency in Latin American and Caribbean banking. J Appl Econ 17: 301-324. doi: 10.1016/S1514-0326(14)60014-3
[50]	Kasman S, Kasman A (2015) Bank competition, concentration and financial stability in the Turkish banking industry. Econ Syst 39: 502-517. doi: 10.1016/j.ecosys.2014.12.003
[51]	Keeley MC (1990) Deposit insurance, risk and market power in banking. Am Econ Rev 80: 1183-1200.
[52]	Kick T, Prieto E (2015) Bank risk and competition: Evidence from regional banking markets. Rev Financ 19: 1185-1222. doi: 10.1093/rof/rfu019
[53]	Klomp J, De Haan J (2014) Bank regulation, the quality of institutions, and banking risk in emerging and developing countries: an empirical analysis. Emerg Mark Financ Trade 50: 19-40. doi: 10.1080/1540496X.2014.1013874
[54]	Lapteacru I (2017) Market power and risk of Central and Eastern European banks: Does more powerful mean safer? Econ Model 63: 46-59. doi: 10.1016/j.econmod.2017.01.022
[55]	Leroy A, Lucotte Y (2017) Is there a competition-stability trade-off in European banking? J Int Financ Mark Inst Money 46: 199-215. doi: 10.1016/j.intfin.2016.08.009
[56]	Love I, Zicchino L (2006) Financial development and dynamic investment behavior: Evidence from Panel VAR. Q Rev Econ Financ 46: 190-210. doi: 10.1016/j.qref.2005.11.007
[57]	Louati S, Boujelbene Y (2015) Banks' stability-efficiency within dual banking system: a stochastic frontier analysis. Int J Islamic Middle East Financ Manage 8: 472-490. doi: 10.1108/IMEFM-12-2014-0121
[58]	Liu H, Molyneux P, Wilson JO (2013) Competition and stability in European banking: A regional analysis. Manch School 81: 176-201. doi: 10.1111/j.1467-9957.2011.02285.x
[59]	Liu H, Wilson JO (2013) Competition and risk in Japanese banking. Eur J Financ 19: 1-18.
[60]	Maddala GS, Wu S (1999) A comparative study of unit root tests with panel data and a new simple test. Oxf Bull Econ Stat 61: 631-652. doi: 10.1111/1468-0084.0610s1631
[61]	Martinez-Miera D, Repullo R (2010) Does competition reduce the risk of bank failure? Rev Financ Stud 23: 3638-3664. doi: 10.1093/rfs/hhq057
[62]	Matutes C, Vives X (2000) Imperfect competition, risk taking, and regulation in banking. Eur Econ Rev 44: 1-34. doi: 10.1016/S0014-2921(98)00057-9
[63]	Moon HR, Perron B (2004) Asymptotic local power of pooled t‐ratio tests for unit roots in panels with fixed effects. Working paper, University of Montreal.
[64]	Moyo B (2018) An analysis of competition, efficiency and soundness in the South African banking sector. South Afri J Econ Manage Sci 21: 1-14.
[65]	Noman AHM, Gee CS, Isa CR (2017) Does competition improve financial stability of the banking sector in ASEAN countries? An empirical analysis. PLoS ONE 12: 1-27.
[66]	Nyantakyi E, Sy M (2015) The banking system in Africa: Main facts and challenges. Afr Dev Bank Group 6: 1-16.
[67]	OECD (2010) Competition, concentration and stability in the banking sector. Organisation for Economic Co-operation and Development, DAF/COMP (2010)9. Available from: http://www.oecd.org/competition/sectors/46040053.pdf.
[68]	Osuagwu ES, Nwokoma N (2017) Empirical assessment of the competitive conduct of Nigerian banks in a post-consolidation era. Res Int Bus Financ 41: 412-422. doi: 10.1016/j.ribaf.2017.04.051
[69]	Ozili PK (2018) Banking Stability Determinants in Africa. Int J Managerial Financ 14: 462-483. doi: 10.1108/IJMF-01-2018-0007
[70]	Pawlowska M (2016) Does the size and market structure of the banking sector have an effect on the financial stability of the European Union? J Econ Asymmetries14: 112-127.
[71]	Pesaran MH (2007) A simple panel unit root test in the presence of cross‐section dependence. J Appl Econometrics 22: 265-312. doi: 10.1002/jae.951
[72]	Risfandy T, Tarazi A, Trinugroho I (2018) Competition in dual markets: Implications for banking system stability. Available at SSRN 3158510.
[73]	Samantas I (2013) Bank competition and financial (in) stability in Europe: A sensitivity analysis. Available at SSRN 2358999.
[74]	Sanya S, Gaertner M (2012) Assessing bank competition within the East African Communit (No. 12-32). IMF Working Paper.
[75]	Saif-Alyousfi AY, Saha A, Md-Rus R (2018) The impact of bank competition and concentration on bank risk-taking behavior and stability: Evidence from GCC countries. North Am J Econ Financ 51: 100867.
[76]	Schaeck K, Cihák M (2014) Competition, efficiency, and stability in banking. Financ Manage 43: 215-241. doi: 10.1111/fima.12010
[77]	Simpasa AM (2011) Competitive conditions in the Tanzanian commercial banking industry. Afr Dev Rev 23: 88-98. doi: 10.1111/j.1467-8268.2010.00274.x
[78]	Simbanegavi W, Greenberg JB, Gwatidzo T (2015) Testing for competition in the South African Banking Sector. J Afr Econ 24: 303-324. doi: 10.1093/jae/eju022
[79]	Sims CA (1980) Macroeconomics and reality. Econometrica J Econometric Society, 1-48.
[80]	Shijaku G (2017) Bank Stability and Competition: Evidence from Albanian Banking Market. Eurasian J Bus Econ 19: 127-154.
[81]	Smith BD (1984) Private information, deposit interest rates, and the 'stability' of the banking system. J Monetary Econ 14: 293-317. doi: 10.1016/0304-3932(84)90045-X
[82]	Soedarmono W, Machrouh F, Tarazi A (2013) Bank competition, crisis and risk taking: Evidence from emerging markets in Asia. J Int Financ Mark Inst Money 23: 196-221. doi: 10.1016/j.intfin.2012.09.009
[83]	Tabak BM, Fazio DM, Cajueiro DO (2012) The relationship between banking market competition and risk-taking: Do size and capitalization matter? J Bank Financ 36: 3366-3381. doi: 10.1016/j.jbankfin.2012.07.022
[84]	Ariss RT (2010) On the implications of market power in banking: Evidence from developing countries. J Bank Financ 34: 765-775. doi: 10.1016/j.jbankfin.2009.09.004
[85]	Wheelock DC, Wilson PW (2012) Do large banks have lower costs? New estimates of returns to scale for U.S. banks. J Money Credit Bank 44: 171-199.
[86]	World Bank (2019) World Development Indicators, World Bank, Washington D.C. Available from: http://data.worldbank.org/data-catalog/world-development-indicators/wdi-2019.
[87]	World Bank (2019) Economic research data. Retrieved from FRED, Federal Reserve Bank of St. Louis. Available from: https://fred.stlouisfed.org/series.

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Banking competition and stability: evidence from West Africa

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Banking competition and stability: evidence from West Africa

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