A semiparametric stochastic input distance frontier model with application to the Indonesian banking industry

Abstract

This paper proposes a semiparametric smooth-varying coefficient input distance frontier model with multiple outputs and multiple inputs, panel data, and determinants of technical inefficiency for the Indonesian banking industry during the period 2000 to 2015. The technology parameters are unknown functions of a set of environmental factors that shift the input distance frontier non-neutrally. The computationally simple constraint weighted bootstrapping method is employed to impose the regularity constraints on the distance function. As a by-product, total factor productivity (TFP) growth is estimated and decomposed into technical change, scale component, and efficiency change. The distance elasticities, marginal effects of the environmental factors on the distance elasticities, temporal behavior of technical efficiency, and also TFP growth and its components are investigated.

Appendices

Appendix A

This appendix explains how we select the bandwidths to estimate the smooth coefficients and their derivatives in (11). Following Li and Racine (2010), we employ the most commonly used least-squares cross-validation (LSCV) method, which is a fully automatic data-driven approach, to select the bandwidth vector h; that is,

$$C{V}_{ll}(h)=\mathop{\min }\limits_{h}\ \mathop{\sum }\limits_{i = 1}^{N}\mathop{\sum }\limits_{t = 1}^{T}{[{{\mathcal{Y}}}_{it}-{W}_{it}^{\prime}{\hat{\rho }}_{-it}({Z}_{it})]}^{2}M({Z}_{it}),$$

(33)

where CV_ll(h) determines the cross-validation bandwidth vector h for local-linear estimator, ${W}_{it}^{\prime}{\hat{\rho }}_{-it}({Z}_{it})$ is the leave-one-out local-linear kernel conditional mean, and 0 ≤ M(⋅) ≤ 1 is a weight function that serves to avoid difficulties caused by dividing by zero. The same bandwidth vector is used to estimate the constrained smooth coefficients.

Appendix B

This appendix details the TFP growth decomposition given in Section 2.3.

First, take the time derivatives of both sides of (4) (ignoring the noise term), and we would have:

$$-{\dot{X}}_{1}=\frac{\partial \alpha (Z)}{\partial t}+\mathop{\sum }\limits_{k = 1}^{K}\frac{\partial {\gamma }_{k}(Z)}{\partial t}{\mathrm{ln}}\,{Y}_{k}+\mathop{\sum }\limits_{k = 1}^{K}{\gamma }_{k}(Z){\dot Y}_{k}+\mathop{\sum }\limits_{j = 2}^{J}\frac{\partial {\beta }_{j}(Z)}{\partial t}{\mathrm{ln}}\,{\widetilde{X}}_{j}+\mathop{\sum }\limits_{j = 2}^{J}{\beta }_{j}(Z){\dot{\widetilde{X}}}_{j}-\frac{u}{t}.$$

(34)

The TC component measures the non-neutral shift of the IDF over time; therefore,

$$-{\dot{X}}_{1}=TC+\mathop{\sum }\limits_{k = 1}^{K}{\gamma }_{k}(Z){\dot{Y}}_{k}+\mathop{\sum }\limits_{j = 2}^{J}{\beta }_{j}(Z){\dot{\widetilde{X}}}_{j}+EC,$$

(35)

where $TC=\frac{\partial \alpha (Z)}{\partial t}+\mathop{\sum }\nolimits_{k = 1}^{K}\frac{\partial {\gamma }_{k}(Z)}{\partial t}{\mathrm{ln}}\,{Y}_{k}+\mathop{\sum }\nolimits_{j = 2}^{J}\frac{\partial {\beta }_{j}(Z)}{\partial t}{\mathrm{ln}}\,{\widetilde{X}}_{j}$, and $EC=-\frac{u}{t}$ measures the efficiency change.

Add $\dot{TFP}$ to both sides of (35) and rearrange to obtain:

$$\begin{array}{ll}\dot{TFP}&=TC+\dot{TFP}+{\dot{X}}_{1}+\mathop{\sum }\limits_{k = 1}^{K}{\gamma }_{k}(Z){\dot{Y}}_{k}+\mathop{\sum }\limits_{j = 2}^{J}{\beta }_{j}(Z){\dot{\widetilde{X}}}_{j}+EC\\ &=TC+\mathop{\sum }\limits_{k = 1}^{K}{R}_{k}{\dot{Y}}_{k}-\mathop{\sum }\limits_{j = 1}^{J}{S}_{j}{\dot{X}}_{j}+{\dot{X}}_{1}+\mathop{\sum }\limits_{k = 1}^{K}{\gamma }_{k}(Z){\dot{Y}}_{k}+\mathop{\sum }\limits_{j = 2}^{J}{\beta }_{j}(Z){\dot{\widetilde{X}}}_{j}+EC\\ &=TC+SC+{\dot{X}}_{1}-\mathop{\sum }\limits_{j = 1}^{J}{S}_{j}{\dot{X}}_{j}+\mathop{\sum }\limits_{j = 2}^{J}{\beta }_{j}(Z){\dot{\widetilde{X}}}_{j}+EC,\end{array}$$

(36)

where $SC=\mathop{\sum }\nolimits_{k = 1}^{K}{R}_{k}{\dot{Y}}_{k}+\mathop{\sum }\nolimits_{k = 1}^{K}{\gamma }_{k}(Z){\dot{Y}}_{k}=\mathop{\sum }\nolimits_{k = 1}^{K}({R}_{k}+{\gamma }_{k}(Z)){\dot{Y}}_{k}$. To further decompose the SC into one sub-component related to RTS and the other related to market power:

$$\begin{array}{ll}SC&=\mathop{\sum }\limits_{k = 1}^{K}{R}_{k}{\dot{Y}}_{k}+\mathop{\sum }\limits_{k = 1}^{K}{\gamma }_{k}(Z){\dot{Y}}_{k}+RTS\cdot \mathop{\sum }\limits_{k = 1}^{K}{\gamma }_{k}(Z){\dot{Y}}_{k}-RTS\cdot \mathop{\sum }\limits_{k = 1}^{K}{\gamma }_{k}(Z){\dot{Y}}_{k}\\ &=(1-RTS)\mathop{\sum }\limits_{k = 1}^{K}{\gamma }_{k}(Z){\dot{Y}}_{k}+\mathop{\sum }\limits_{k = 1}^{K}{R}_{k}{\dot{Y}}_{k}-\mathop{\sum }\limits_{k = 1}^{K}\frac{{\gamma }_{k}(Z)}{\mathop{\sum }\nolimits_{k = 1}^{K}{\gamma }_{k}(Z)}{\dot{Y}}_{k}\\ &=(1-RTS)\mathop{\sum }\limits_{k = 1}^{K}{\gamma }_{k}(Z){\dot{Y}}_{k}+\mathop{\sum }\limits_{k = 1}^{K}\left({R}_{k}-\frac{{\gamma }_{k}(Z)}{\mathop{\sum }\nolimits_{k = 1}^{K}{\gamma }_{k}(Z)}\right){\dot{Y}}_{k},\end{array}$$

(37)

given that $RTS=-1/\mathop{\sum }\nolimits_{k = 1}^{K}{\gamma }_{k}(Z)$. Finally, it can be shown that:

$$\begin{array}{ll}AC&={\dot{X}}_{1}-\mathop{\sum }\limits_{j = 1}^{J}{S}_{j}{\dot{X}}_{j}+\mathop{\sum }\limits_{j = 2}^{J}{\beta }_{j}(Z){\dot{\widetilde{X}}}_{j}\\ &=\mathop{\sum }\limits_{j = 1}^{J}{S}_{j}{\dot{X}}_{1}-\mathop{\sum }\limits_{j = 1}^{J}{S}_{j}{\dot{X}}_{j}+\mathop{\sum }\limits_{j = 2}^{J}{\beta }_{j}(Z){\dot{\widetilde{X}}}_{j}\\ &=\mathop{\sum }\limits_{j = 1}^{J}({\dot{X}}_{1}-{\dot{X}}_{j}){S}_{j}+\mathop{\sum }\limits_{j = 2}^{J}{\beta }_{j}(Z){\dot{\widetilde{X}}}_{j},\end{array}$$

(38)

using the fact that $\mathop{\sum }\nolimits_{j = 1}^{J}{S}_{j}=1$. Furthermore,

$$\begin{array}{ll}\mathop{\sum }\limits_{j = 1}^{J}({\dot{X}}_{1}-{\dot{X}}_{j}){S}_{j}&=-\mathop{\sum }\limits_{j = 1}^{J}({\dot{X}}_{j}-{\dot{X}}_{1}){S}_{j}=-\mathop{\sum }\limits_{j = 1}^{J}{\dot{\widetilde{X}}}_{j}{S}_{j}\\ &=-\left({\dot{\widetilde{X}}}_{1}{S}_{1}+\mathop{\sum }\limits_{j = 2}^{J}{\dot{\widetilde{X}}}_{j}{S}_{j}\right)=-\mathop{\sum }\limits_{j = 2}^{J}{\dot{\widetilde{X}}}_{j}{S}_{j},\end{array}$$

(39)

using the fact that ${\dot{\widetilde{X}}}_{1}=0$. Therefore, plug (39) into (38) to obtain $AC=\mathop{\sum }\nolimits_{j = 2}^{J}({\beta }_{j}(Z)-{S}_{j}){\dot{\widetilde{X}}}_{j}$. QED.