Goodness--of--fit tests for stochastic frontier models based on the characteristic function

Abstract

We consider goodness–of–fit tests for the distribution of the composed error in Stochastic Frontier Models. The proposed test statistic utilizes the characteristic function of the composed error term, and is formulated as a weighted integral of properly standardized data. The new test statistic is shown to be consistent and computationally convenient. Simulation results are presented whereby resampling versions of the new tests are compared to classical goodness–of–fit methods.

Appendices

Appendix A

Starting from Eq. (6) we obtain

$$\begin{array}{lll}{D}_{n}^{2}(t)&=&{S}_{n}^{2}(t)+{t}^{2}{C}_{n}^{2}(t)+2t{S}_{n}(t){C}_{n}(t)\\ &=&{\left(\frac{1}{n}\mathop{\sum }\limits_{j = 1}^{n}\sin ({\widehat{\widetilde{\varepsilon }}}_{j})\right)}^{2}+{t}^{2}{\left(\frac{1}{n}\mathop{\sum }\limits_{j = 1}^{n}\cos ({\widehat{\widetilde{\varepsilon }}}_{j})\right)}^{2}\\ &&+\,2t\left(\frac{1}{n}\mathop{\sum }\limits_{j=1}^{n}\sin ({\widehat{\widetilde{\varepsilon }}}_{j})\right)\left(\frac{1}{n}\mathop{\sum }\limits_{j=1}^{n}\cos ({\widehat{\widetilde{\varepsilon }}}_{j})\right)\\ &=&\frac{1}{{n}^{2}}\mathop{\sum }\limits_{j,k=1}^{n}\sin ({\widehat{\widetilde{\varepsilon }}}_{j})\sin ({\widehat{\widetilde{\varepsilon }}}_{k})+\frac{{t}^{2}}{{n}^{2}}\mathop{\sum }\limits_{j,k=1}^{n}\cos ({\widehat{\widetilde{\varepsilon }}}_{j})\cos ({\widehat{\widetilde{\varepsilon }}}_{k})\\ &&+\,\frac{2t}{{n}^{2}}\mathop{\sum }\limits_{j,k=1}^{n}\sin ({\widehat{\widetilde{\varepsilon }}}_{j})\cos ({\widehat{\widetilde{\varepsilon }}}_{k}),\end{array}$$

where we write ∑_j,k for the double sum ∑_j∑_k. Also recall the trigonometric identities

$$\begin{array}{l}\sin {z}_{1}\sin {z}_{2}=\frac{1}{2}[\cos ({z}_{1}-{z}_{2})-\cos ({z}_{1}+{z}_{2})]\\ \cos {z}_{1}\cos {z}_{2}=\frac{1}{2}[\cos ({z}_{1}-{z}_{2})+\cos ({z}_{1}+{z}_{2})]\\ \sin {z}_{1}\cos {z}_{2}=\frac{1}{2}[\sin ({z}_{1}-{z}_{2})+\sin ({z}_{1}+{z}_{2})]\end{array}$$

Now plug the above expression for ${D}_{n}^{2}(t)$ into the test statistic (7) and substitute the above product formulae, and integrate term-by-term the resulting expression. Then after some grouping we obtain (8) by making use of the integrals

$$\int\nolimits_{-\infty }^{\infty }\cos (tz){e}^{-\lambda | t| }dt=\frac{2\lambda }{{z}^{2}+{\lambda }^{2}},$$

$$\int\nolimits_{-\infty }^{\infty }{t}^{2}\cos (tz){e}^{-\lambda | t| }dt=\frac{4\lambda ({\lambda }^{2}-3{z}^{2})}{{\left({z}^{2}+{\lambda }^{2}\right)}^{3}},$$

$$\int\nolimits_{-\infty }^{\infty }t\sin (tz){e}^{-\lambda | t| }dt=\frac{4z\lambda }{{\left({z}^{2}+{\lambda }^{2}\right)}^{2}}.$$

Equation (16) may be proved by following analogous steps, but we also need the extra integrals

$$\int\nolimits_{-\infty }^{\infty }{t}^{4}\cos (tz){e}^{-\lambda | t| }dt=\frac{48\lambda (5{z}^{4}-10{z}^{2}{\lambda }^{2}+{\lambda }^{4})}{{\left({z}^{2}+{\lambda }^{2}\right)}^{5}},$$

$$\int\nolimits_{-\infty }^{\infty }{t}^{3}\sin (tz){e}^{-\lambda | t| }dt=\frac{48z\lambda ({\lambda }^{2}-{z}^{2})}{{\left({z}^{2}+{\lambda }^{2}\right)}^{4}}.$$

Appendix B

Starting from Eq. (6) and using $\sin (z)=z-({z}^{3}/3!)+\ldots$ and $\cos (z)=1-({z}^{2}/2!)+\ldots$, we obtain (in increasing powers of t)

$$\begin{array}{lll}{D}_{n}(t)&=&t\left(\frac{1}{n}\mathop{\sum }\limits_{j=1}^{n}{\widehat{\widetilde{\varepsilon }}}_{j}+1\right)-\frac{{t}^{3}}{2!}\left(\frac{1}{3}\frac{1}{n}\mathop{\sum }\limits_{j=1}^{n}{\widehat{\widetilde{\varepsilon }}}_{j}^{3}+\frac{1}{n}\mathop{\sum }\limits_{j=1}^{n}{\widehat{\widetilde{\varepsilon }}}_{j}^{2}\right)\\ &&+\,\ldots \,,\end{array}$$

and by squaring

$$\begin{array}{lll}{D}_{n}^{2}(t)&=&{t}^{2}{\left(\frac{1}{n}\mathop{\sum }\limits_{j = 1}^{n}{\widehat{\widetilde{\varepsilon }}}_{j}+1\right)}^{2}\\ &&-\,{t}^{4}\left(\frac{1}{3}\frac{1}{n}\mathop{\sum }\limits_{j=1}^{n}{\widehat{\widetilde{\varepsilon }}}_{j}^{3}+\frac{1}{n}\mathop{\sum }\limits_{j=1}^{n}{\widehat{\widetilde{\varepsilon }}}_{j}^{2}\right)\left(\frac{1}{n}\mathop{\sum }\limits_{j=1}^{n}{\widehat{\widetilde{\varepsilon }}}_{j}+1\right)\\ &&+\,\ldots \,.\end{array}$$

Plugging the above expression in Eq. (7) and integrating term-by-term leads to

$$\begin{array}{lll}{T}_{n,w}&=&n\,\left[\frac{4}{{\lambda }^{3}}{\left(\frac{1}{n}\mathop{\sum }\limits_{j = 1}^{n}{\widehat{\widetilde{\varepsilon }}}_{j}+1\right)}^{2}\right.\\ &&-\,\frac{48}{{\lambda }^{5}}\left(\frac{1}{n}\mathop{\sum }\limits_{j=1}^{n}{\widehat{\widetilde{\varepsilon }}}_{j}+1\right)\left(\frac{1}{3}\frac{1}{n}\mathop{\sum }\limits_{j=1}^{n}{\widehat{\widetilde{\varepsilon }}}_{j}^{3}+\frac{1}{n}\mathop{\sum }\limits_{j=1}^{n}{\widehat{\widetilde{\varepsilon }}}_{j}^{2}\right)\\ &&+\left.\,\ldots \right],\end{array}$$

and by taking the limit as λ → ∞, we readily obtain (9), where we made use of the integral

$$\int\nolimits_{-\infty }^{\infty }| t{| }^{m}{e}^{-\lambda | t| }dt=\frac{2m!}{{\lambda }^{m+1}},\,m=1,2,\ldots \,.$$