A stochastic frontier model with structural breaks in efficiency and technology

Abstract

We have developed a stochastic frontier model with appropriate priors to estimate the locations and number of structural breaks for both production efficiency and technology, which experience different regime changes. We assume different units could have unknown common break dates. Although panel data with large cross-sectional size can help identify the break locations, it could render posterior simulation very inefficient. Hence, care must be taken to avoid such problems. We apply our method to study the world production over the period of 1960–2007 and find that the data support structural breaks in technology rather than in efficiency. For most countries under study, the most important source of growth is capital accumulation. The technology adopted by different countries shows signs of convergence. Changes of technology usually happen after economic crises to compensate for negative capital growth. Alternative modelling approach and priors are used for robustness check.

Appendices

Appendix 1: Features of the prior in (4)

It is possible to model the breaks in terms of the locations of change points. Denote the $m$th break point as $\tau _m=\lbrace t:S_{t+1}=m+1,S_t=m\rbrace$ for $m=1,2,\dots ,M-1$. Suppose $D$ only takes integer values. The unrestricted uniform prior suggested by Koop and Potter (2009) can have the following form:

$$\begin{aligned} \begin{aligned}&Pr(\tau _1)=\frac{1}{D}\qquad \text {for } \tau _1=1,2,\dots ,D,\\&Pr(\tau _m|\tau _{m-1})=\frac{1}{D}\qquad \text {for } \tau _m=\tau _{m-1}+1,\tau _{m-1}+2,\dots ,\tau _{m-1}+D. \end{aligned} \end{aligned}$$

(20)

The prior in (20) implies $Pr(\tau _m|\tau _{m-1})$ is uniform over the support, which is as large as $\tau _1$’s.

In what follows, we use $\widetilde{Pr}(\tau _m|\tau _{m-1})$ to denote the probability derived from our prior in (4).

$$\begin{aligned} \begin{aligned}&\widetilde{Pr}(\tau _m=t|\tau _{m-1}=t-j)= Pr(S_{t+1}=m+1,S_t=m|S_{t-j}=m-1,S_{t-j+1}=m),\\&=Pr(S_{t+1}=m+1|S_t=m,S_{t-j}=m-1,S_{t-j+1}=m)Pr(S_{t}\\&=m|S_{t-j}=m-1,S_{t-j+1}=m),\\&=Pr(S_{t+1}=m+1|S_t=m)Pr(S_{t}=m|S_{t-j+1}=m),\\&=\frac{1}{D-(t-j)+m-1}=\frac{1}{D-\tau _{m-1}+m-1}\quad \text {for }j=1,2,\dots ,t-m+1. \end{aligned} \end{aligned}$$

(21)

In the derivation above, we have used the fact that $S_t$ is a Markov process. The last but one equality sign results since

$$\begin{aligned} \begin{aligned} Pr(S_{t}&=m|S_{t-j+1}=m)\\&=Pr(S_{t}=m|S_{t-1}=m)Pr(S_{t-1}=m|S_{t-2}=m)\\&\cdots Pr(S_{t-j+2}=m|S_{t-j+1}=m)\\&=\left( 1-\frac{1}{D-(t-1-m)}\right) \left( 1-\frac{1}{D-(t-2-m)}\right) \\&\cdots \left( 1-\frac{1}{D-(t-j+1-m)}\right) \\&=\frac{D-(t-m)}{D-(t-j)+m-1}. \end{aligned} \end{aligned}$$

(22)

Comparing $\widetilde{Pr}(\tau _m|\tau _{m-1})$ and $Pr(\tau _m|\tau _{m-1})$, we can see that the prior in (4) implies the conditional probability of the break location is uniform over $\tau _m=\tau _{m-1}+1,\dots , D+m-1$, but has smaller support than $\tau _1$ when $\tau _{m-1}>m-1$. Therefore, $\widetilde{Pr}(\tau _m|\tau _{m-1})$ gets bigger when $\tau _{m-1}$ gets closer to $D$, which also implies the points near the end of the sample are more possible break points than the points before as in Fig. 3. Hence, the prior in (4) is in fact similar to the restricted uniform prior in Koop and Potter (2009), though the uniform prior does not necessarily imply $S_t$ follows a Markov process.

Appendix 2: Simulate draws of $\gamma$, $S^\gamma$ and $D^\gamma$

To draw $\gamma$, we numerically invert its cumulative posterior distribution function. The posterior kernel of $\gamma _{iS_t=j}$ ($\gamma _{ij}$) takes the following form:

$$\begin{aligned} \begin{aligned}&p(\gamma _{ij}|S^\gamma ,\mu ,\phi ,a,S^a,\sigma ^2_\epsilon ,Y)\propto \\&exp\Biggl \lbrace -\frac{d_j}{2\sigma ^2_\epsilon } \biggl [\exp (\gamma _{ij})-\frac{1}{d_j}\underset{t\in \lbrace t|S_t=j\rbrace }{\sum }\left( x_{it}'a_{S_t^a}-y_{it}\right) \biggl ]^2 \Biggl \rbrace \exp \left[ -\frac{(\gamma _{ij}-\mu _j)^2}{2\exp (\phi _j)}\right] . \end{aligned} \end{aligned}$$

(23)

To simulate a draw of $S^\gamma$, we follow the algorithm by Chib (1998). Once we have a draw of $S^\gamma$, we could simulate $D^\gamma$, whose posterior kernel is

$$\begin{aligned} p(D^\gamma |S^\gamma ,Y)\propto I(T\le D^\gamma \le D_\mathrm{max})\prod _{t=2}^TPr(S^\gamma _{t}|S^\gamma _{t-1},D^\gamma ). \end{aligned}$$

(24)

We first find the integrating constant of the posterior kernel over $[T,D_\mathrm{max}]$ and then draw $D^\gamma$ by numerically inverting its cumulative distribution function. The same method can be used to draw regime duration for technology.

Appendix 3: Simulate draws of $S^a$ and $a$

Define

$$\begin{aligned} z^a_{it}=y_{it}+exp(\gamma _{iS_t^\gamma }). \end{aligned}$$

(25)

Stack up all the $z^a_{it}$ in regime $j$ of $a$ into a vector (i.e. for $t$ with $S^a_t=j$) to obtain $z^a_{ij}$. Similarly, we concatenate all the $x_{it}$ in regime $j$ to form $\zeta _{ij}$.²² Denote $z^a_j=(z^{a\prime }_{1j},z^{a\prime }_{1j},\dots ,z^{a\prime }_{Nj})'$ and $\varXi _j=(\zeta _{1j}',\zeta _{1j}',\dots ,\zeta _{Nj}')'$. Then we can have the following state space model:

$$\begin{aligned} z^a_j&= \varXi _j a_j+e^a_j,\qquad e^a_j\sim N(0,\sigma ^2_\epsilon I_{d^a_j N\times d^a_j N}),\end{aligned}$$

(26)

$$\begin{aligned} a_{j+1}&= a_j+u^a_{j},\qquad u^a_{j}\sim N(0,\sigma ^2_aI_{k_a\times k_a}), \end{aligned}$$

(27)

where $d^a_j$ is the duration of the $j$th regime, $j=1,\dots ,m_a$ and $m_a$ is the number of in-sample regimes for $a$. We assume the cross-sectional size ($N$) is large and $\varXi _j$ has full column rank. The Kalman filter takes the following form:

$$\begin{aligned} \begin{aligned}&R_j=\varXi _j(V_{j-1}+\sigma ^2_a I)\varXi _j'+\sigma ^2_\epsilon I, J_j=(V_{j-1}+\sigma ^2_aI)\varXi _j',\\&M_j=J_jR_j^{-1},C_j=J_jR_j^{-1}J^{'}_j,V_j=V_{j-1}+\sigma ^2_aI-C_j,\\&m_j=m_{j-1}+M_j(z^a_j-\varXi _j m_{j-1}),j=2,3\dots ,m_a. \end{aligned} \end{aligned}$$

(28)

Due to the diffuse prior for $a_1$, we initialize the filter by $V_1=\sigma ^2_\epsilon (\varXi _1'\varXi _1)^{-1}$ and $m_1=(\varXi _1'\varXi _1)^{-1}\varXi _1'z^a_1$ (e.g. Koopman 1997). We can then use MH algorithm to draw $\sigma ^2_a$ from the following posterior,

$$\begin{aligned} p(\sigma ^2_a|z^a,S^a,\sigma ^2_\epsilon )&\propto p_{IG}(\sigma ^2_a)\prod _{j=1}^{m_a-1}|R_{j+1}|^{-\frac{1}{2}}\exp \left[ -\frac{1}{2}\sum _{j=1}^{m_a-1}(z^{a}_{j+1}\right. \nonumber \\&\left. -\varXi _{j+1} m_{j})'R_{j+1}^{-1}(z^{a}_{j+1}-\varXi _{j+1} m_{j})\right] , \end{aligned}$$

(29)

To draw $a|z^a$ (smoothing), we use the algorithm by Koopman and Durbin (2002), while we use the algorithm by Gerlach et al. (2000). To draw $S^a$ and let $S^a$ be unconditional on $a$. Define $z^{*a}_{it}=y_{it}+exp(\gamma _{iS_t^\gamma })$, $z^{*a}_{t}=(z^{*a}_{1t},z^{*a}_{2t},\dots ,z^{*a}_{Nt})'$ and $\varXi ^*_{t}=(x_{1t},x_{2t},\dots ,x_{Nt})'$. We can then write down the following state space model:

$$\begin{aligned} \begin{aligned} z^{*a}_{t}&=\varXi ^{*}_ta_t+e^{*a}_t,\qquad e^{*a}_t\sim N(0,\sigma ^2_\epsilon I_N)\\ a_{t+1}&=a_{t}+K_tu_{t},\qquad u_{t}\sim i.i.d.N(0,\sigma ^2_a I). \end{aligned} \end{aligned}$$

(30)

where $K_t=S_{t+1}-S_{t}$ takes the value of either 1 or 0 indicating whether period $t$ is a breakpoint or not. We draw the posterior of $K^a=(K_1,\dots ,K_{T-1})'$ and then transform $K^a$ into $S^a=(S_1,\dots ,S_T)'$. We still use the prior in (4) and represent it as (15). The algorithm by Gerlach et al. (2000) is a Gibbs sampler, which draws $K_t|K_{s\ne t},Y$, and the details are as follows.

Algorithm 6.1

• 1. Given the initial draw of $K^a$, we run through the following backward recursion, where the notations are just confined to this section.

  $$\begin{aligned} \begin{aligned} N_{t+1}^{-1}&=\left( \sigma ^2_aK_{t}\varXi ^{*}_{t+1} \varXi ^{*\prime }_{t+1}+\sigma ^2_\epsilon I_N\right) ^{-1}\\&=\frac{1}{\sigma ^2_\epsilon }\left[ I_N- \varXi ^{*}_{t+1}\left( \frac{\sigma ^2_\epsilon }{\sigma ^2_aK_{t}}I +\varXi ^{*\prime }_{t+1}\varXi ^{*}_{t+1}\right) ^{-1} \varXi ^{*\prime }_{t+1}\right] ,\\ B_{t+1}&=\sigma ^2_aK_{t}\varXi ^{*\prime }_{t+1}N^{-1}_{t+1}, A_{t+1}=I_{k_a}-B_{t+1}\varXi ^{*}_{t+1},\\ Z_{t+1}&=\sigma ^a_a K_{t}A_{t+1}A_{t+1}'+\sigma ^2_\epsilon B_{t+1}B_{t+1}',\\ D_{t+1}&=Z_{t+1}-Z_{t+1}(\varOmega _{t+1}^{-1}+Z_{t+1})^{-1}Z_{t+1}\\&=Z_{t+1}^{\frac{1}{2}}\left( I+Z_{t+1}^{\frac{1}{2}}\varOmega _{t+1}Z_{t+1}^ {\frac{1}{2}}\right) ^{-1}Z_{t+1}^{\frac{1}{2}},\\ \varOmega _t&=A_{t+1}'(\varOmega _{t+1}-\varOmega _{t+1}D_{t+1} \varOmega _{t+1})A_{t+1} +\varXi ^{*\prime }_{t+1}N_{t+1}^{-1}\varXi ^{*}_{t+1},\\ \mu _{t}&=A_{t+1}'(I-\varOmega _{t+1}D_{t+1}) (\mu _{t+1}-\varOmega _{t+1}B_{t+1}z_{t+1}^{*a})\\&\quad \ +\varXi ^{*\prime }_{t+1}N_{t+1}^{-1}z_{t+1}^{*a},t=T-1,\dots ,1, \end{aligned} \end{aligned}$$

  (31)

  where $\varOmega _T=0_{k_a\times k_a}$, $(\varOmega _{T}^{-1}+Z_{T})^{-1}=0$ and $\mu _T=0_{k_a\times 1}$.

• 2. For $K_t=0,1$,

  • (a) run the Kalman filter in (28) by replacing the subscript $j$ by $t$ ($t=2,\dots ,T$), $\sigma ^2_a$ by $K_{t-1}\sigma ^2_{a}$, $\varXi _j$ by $\varXi ^*_t$ and $z^a_j$ by $z^{*a}_t$.

  • (b) Calculate the posterior conditionals:

    $$\begin{aligned}&\!\!Pr(K_t|z^{*a},K_{s\ne t})\propto p(z_{t+1}^{*a}|z_{1:t}^{*a},K_{1:t})p(z_{(t+2):T}^ {*a}|z_{1:(t+1)}^{*a},K^a)Pr(K_t|K_{s\ne t})\nonumber \\&\!\!p(z_{t+1}^{*a}|z_{1:t}^{*a},K_{1:t})\propto |R_{t+1}|^{-\frac{1}{2}}\exp \left[ -\frac{1}{2}(z^{*a}_{t+1}-\varXi ^{*}_{t+1} m_{t}) 'R_{t+1}^{-1}(z^{*a}_{t+1}\right. \nonumber \\&\!\!\left. -\varXi ^{*}_{t+1} m_{t})\right] \nonumber \\&\!\!p(z_{(t+2):T}^{*a}|z_{1:(t+1)}^{*a},K^a)\propto |\varOmega _{t+1}V_{t+1}+I|^{-\frac{1}{2}} \exp \Biggl \lbrace -\frac{1}{2}\biggl [m_{t+1}'\varOmega _{t+1} m_{t+1}\nonumber \\&\!\!-2\mu _{t+1}' m_{t+1}-(\mu _{t+1}-\varOmega _{t+1} m_{t+1})'\nonumber \\&\!\!\quad \left( V_{t+1}-V_{t+1}(\varOmega _{t+1}^{-1} +V_{t+1})^{-1}V_{t+1}\right) \nonumber \\&\!\!\quad (\mu _{t+1} -\varOmega _{t+1}m_{t+1})\biggr ]\Biggr \rbrace \end{aligned}$$

    (32)

    where $Pr(K_t|K_{s\ne t})$ is obtained from the prior and $(\varOmega _{T}^{-1}+V_{T})^{-1}=0$.

  • (c) Draw $K_t|z^{*a},K_{s\ne t}$ based on the results from (32) and use the values of $m_{t+1}$ and $V_{t+1}$ accordingly for the next iteration.

It is similar to draw $\phi$ and $\mu$, though we need to use the algorithm by Carter and Kohn (1994) to simulate draws from partially Gaussian state space model.

Appendix 4: A case when Chib’s algorithm is inconsistent

Consider the following model, which is a simplified version of (1) without the exogenous regressors and the inefficiency terms:

$$\begin{aligned} y_{it}=a_{S_t}+\epsilon _{it},\qquad \epsilon _{it}\sim i.i.d.N(0,\sigma ^2_\epsilon ). \end{aligned}$$

(33)

Suppose $T=2$ and set the initial draw of $(S_1,S_2)$ to be $(1,2)$. We can then draw $a_1$ and $a_2$ from the Gibbs sampler below,

$$\begin{aligned} \begin{aligned}&a_1|y_1,a_2,\sim N\left( \frac{\sigma ^2_a\frac{\sum _{i=1}^Ny_{i1}}{N} +a_2\frac{\sigma ^2_\epsilon }{N}}{\sigma ^2_a +\frac{\sigma ^2_\epsilon }{N}},\frac{\sigma ^2_\epsilon \sigma ^2_a}{N\sigma ^2_a +\sigma ^2_\epsilon }\right) ,\\&a_2|y_2,a_1,\sim N\left( \frac{\sigma ^2_a\frac{\sum _{i=1}^Ny_{i2}}{N} +a_1\frac{\sigma ^2_\epsilon }{N}}{\sigma ^2_a +\frac{\sigma ^2_\epsilon }{N}},\frac{\sigma ^2_\epsilon \sigma ^2_a}{N\sigma ^2_a+\sigma ^2_\epsilon }\right) . \end{aligned} \end{aligned}$$

(34)

Now if we draw $S_1$ and $S_2$ conditional on the draws of $a_1$ and $a_2$ using Chib’s algorithm, we can obtain

$$\begin{aligned} \begin{aligned} \frac{Pr(S_2=2|a_2,y_2)}{Pr(S_2=1|a_1,y_2)}&=\frac{p(y_2|a_2) \left[ Pr(S_{2}=2|S_{1}=1)Pr(S_{1}=1)\right] }{p(y_2|a_1)\left[ Pr(S_{2}=1|S_{1}=1)Pr(S_{1}=1)\right] },\\&=\exp \left[ \frac{\sum _{i=1}^N(y_{i2}-a_1)^2-\sum _{i=1}^N (y_{i2}-a_2)^2}{2\sigma ^2_\epsilon }\right] , \end{aligned} \end{aligned}$$

(35)

where $Pr(S_{1}=1)=1$ and $Pr(S_{2}=2|S_{1}=1)=Pr(S_{2}=1|S_{1}=1)=\frac{1}{2}$ if we set $D=M=T=2$ in (4). If $N$ goes to infinity, the posterior draws of $a_1$ and $a_2$ in (34) will have the properties of $a_1-\bar{y}_1\overset{p}{\rightarrow }0$ and $a_2-\bar{y}_2\overset{p}{\rightarrow }0$ where $\bar{y}_1=\frac{\sum _{i=1}^Ny_{i1}}{N}$ and $\bar{y}_2=\frac{\sum _{i=1}^Ny_{i2}}{N}$. We can therefore replace $a_1$ and $a_2$ by $\bar{y}_1$ and $\bar{y}_2$, respectively, for asymptotic analysis. Denote $a_1^*$ and $a_2^*$ as the true values of $a_1$ and $a_2$, respectively. We can have $\bar{y}_1=a_1^*+\frac{\sum _{i=1}^N\epsilon _{i1}}{N}=a_1^*+\bar{\epsilon }_1$ and $\bar{y}_2=a_2^*+\bar{\epsilon }_2$. Substituting these terms into the right-hand side of (35), we can obtain

$$\begin{aligned} \begin{aligned}&\frac{Pr(S_2=2|a_2,y_2)}{Pr(S_2=1|a_1,y_2)} -\exp \left[ \frac{N(\bar{y}_{1}-\bar{y}_2)^2}{2\sigma ^2_\epsilon }\right] \overset{p}{\rightarrow }0,\\ \text {or }&\frac{Pr(S_2=2|a_2,y_2)}{Pr(S_2=1|a_1,y_2)}-\exp \left\{ \frac{N\left( a_1^*-a_2^*+\bar{\epsilon }_{2}-\bar{\epsilon }_1\right) ^2}{2\sigma ^2_\epsilon }\right\} \overset{p}{\rightarrow }0. \end{aligned} \end{aligned}$$

(36)

Note that if $a_1^*-a_2^*\ne 0$, i.e. there are indeed two regimes, $\frac{Pr(S_2=2|a_2,y_2)}{Pr(S_2=1|a_1,y_{2_{}})}$ will go to infinity with $N$ and the Chib’s algorithm will give consistent inference. But if there is only one regime, $\frac{Pr(S_2=2|a_2,y_2)}{Pr(S_2=1|a_1,y_2)}$ will be asymptotically equivalent to $\exp \left[ \frac{N (\bar{\epsilon }_{2}-\bar{\epsilon }_1)^2}{2\sigma ^2_\epsilon }\right] =O_p(1)$. Note that due to (33) $\frac{N(\bar{\epsilon }_{2}-\bar{\epsilon }_1)^2}{2\sigma ^2_\epsilon }$ follows a chi-squared distribution with 1 degree of freedom.²³ Hence, $\frac{Pr(S_2=2|a_2,y_2)}{Pr(S_2=1|a_1,y_2)}$ is greater than 1 and $S_2=2$ will be preferred. The inference result is hence inconsistent.

Using the algorithm by Gerlach et al. (2000) described in the previous section based on (28) and (32) to analyse the same example, we can obtain

$$\begin{aligned} \frac{Pr(K_1=1|y,\sigma ^2_\epsilon ,\sigma ^2_a)}{Pr(K_1=0|y,\sigma ^2_\epsilon ,\sigma ^2_a)}&= \sqrt{\frac{\sigma ^2_\epsilon }{\sigma ^2_\epsilon +N\sigma ^2_a} \frac{2N}{\frac{N\sigma ^2_\epsilon }{\sigma ^2_\epsilon +N\sigma ^2_a}+N}}\nonumber \\&\exp \left[ \frac{1}{2\sigma ^2_\epsilon }\left( \sum _{i=1}^Ny_{i2} -N\bar{y}_1\right) ^2\left( \frac{1}{\frac{N\sigma ^2_\epsilon }{\sigma ^2_\epsilon +N\sigma ^2_a}+N} -\frac{1}{2N}\right) \right] ,\nonumber \\&= \sqrt{\frac{2\sigma ^2_\epsilon }{2\sigma ^2_\epsilon +N\sigma ^2_a}} \exp \left[ \frac{N(a_2^*-a_1^*+\bar{\epsilon }_2-\bar{\epsilon }_1)^2 \sigma ^2_a}{4\sigma ^2_\epsilon (\frac{2\sigma ^2_\epsilon }{N}+\sigma ^2_a)} \right] . \end{aligned}$$

(37)

It is not hard to observe that if there exists a change point, i.e. $a_2^*-a_1^*\ne 0$, $\frac{Pr(K_1=1|y,\sigma ^2_\epsilon ,\sigma ^2_a)}{Pr(K_1=0|y,\sigma ^2_\epsilon ,\sigma ^2_a)}$ will tend to infinity; if there is no change point, $\frac{Pr(K_1=1|y,\sigma ^2_\epsilon ,\sigma ^2_a)}{Pr(K_1=0|y,\sigma ^2_\epsilon ,\sigma ^2_a)}$ will tend to 0. Therefore, the inference of the break point is consistent.