Technological inefficiency indexes: a binary taxonomy and a generic theorem

Abstract

Over the years following Debreu’s (1951) seminal formulation of a “coefficient of resource utilization”, a large number of indexes of technological inefficiency have been specified and a spate of papers has examined the properties satisfied by these indexes. This paper approaches the subject more synthetically, presenting generic results on classes of indexes and their properties. In particular, we consider a broad class of indexes containing almost all known indexes and a partition of this class into two subsets, slacks-based indexes and path-based indexes. Slacks-based indexes are expressed in terms of additive or multiplicative slacks for all inputs and outputs, and particular indexes are generated by specifying the form of aggregation over the coordinate-wise slacks. Path-based indexes are expressed in terms of a common contraction/expansion factor, and particular indexes are generated by specifying the form of the path to the frontier of the technology. Owing to an impossibility result in one of our earlier papers, we know that the set of all inefficiency indexes can be partitioned into three subsets: those that satisfy continuity (in quantities and technologies) and violate indication (equal to some specified value if and only if the quantity vector is efficient), those that satisfy indication and violate continuity, and those that satisfy neither. We prove two generic theorems establishing the equivalence of these two partitions: all slacks-based indexes satisfy indication and hence violate continuity, and all path-based indexes satisfy continuity and hence violate indication. We also discuss the few indexes that do not belong to either of these two sets. Our hope is that these results will help guide decisions about specification of the form of efficiency indexes used in empirical analysis.

Appendix: Proof of the Theorem

1.1 Proof of (a) Slacks-based indexes violate continuity and satisfy indication

It suffices to prove that I ^s satisfies (I), since the R–S Impossibility Result then implies that this class of indexes violates (C).

Suppose that $\left\langle {x,y} \right\rangle$∈T, with $\left\langle {x,y} \right\rangle \gg 0^{[n + m]}$, is not efficient so that there exists a production vector $\left\langle {x{\prime},y{\prime}} \right\rangle \in T$ satisfying $\left\langle {x{\prime},y{\prime}} \right\rangle \gg 0^{[n + m]}$ and $\left\langle {x{\prime}, - y{\prime}} \right\rangle < \left\langle {x, - y} \right\rangle$. If $\left\langle {s{\prime},t{\prime}} \right\rangle$ is the solution to

$$\mathop {{\max}}\limits_{\langle s,t\rangle \in {\bf{R}}_ + ^{n + m}} \left\{ {\psi \left. {\left( {x{\prime},y{\prime},s,t} \right)} \right)\left| {\left\langle {x{\prime} - s,y{\prime} + t} \right\rangle \in T} \right.} \right\},$$

then $\left\langle {x{\prime} - s{\prime},y{\prime} + t{\prime}} \right\rangle$ is a boundary point of T. As $\left\langle {x{\prime} - s{\prime},y{\prime} + t{\prime}} \right\rangle \le \left\langle {x{\prime}, - y{\prime}} \right\rangle < \left\langle {x, - y} \right\rangle$, there exists an $\left\langle {s^o,t^o} \right\rangle \in {\bf{R}}_ + ^{n + m}$ such that $\left\langle {s^o,t^o} \right\rangle >\left\langle {s{\prime},t{\prime}} \right\rangle$ and $\left\langle {x - s^o,y + y^o} \right\rangle = \left\langle {x{\prime} - s{\prime},y{\prime} + t{\prime}} \right\rangle \in T$. As ψ is increasing in $\left\langle {s,t} \right\rangle$, it must be that $I^s(x,y,T) >I^s\left( {x{\prime},y{\prime},T} \right) \ge 0$.

Next suppose that I ^s(x,y,T) > 0. Then $\left\langle {s,t} \right\rangle$ > 0^[n+m], so that there exists a point $\left\langle {x{\prime},y{\prime}} \right\rangle \in T$ satisfying $\left\langle {x{\prime},y{\prime}} \right\rangle \gg 0^{[n + m]}$ and $\left\langle {x{\prime}, - y{\prime}} \right\rangle < \left\langle {x, - y} \right\rangle$. Therefore, (x, y) is inefficient.

Consequently, I ^s satisfies indication (I) and therefore violates (C). ♢

1.2 Proof of (b) Path-Based Indexes Satisfy Continuity and Violate Indication

It suffices to prove that I ^p satisfies (C), since the R–S Impossibility Result then implies that this class of indexes violates (I).

Consider a sequence $\left\{ {x^\nu ,y^\nu ,T^\nu } \right\}_{\nu = 1}^\infty$ converging (in the topology of closed convergence) to $\left\langle {x^o,y^o,T^o} \right\rangle$. We need to show that I ^p(x ^ν, y ^ν, T ^ν) converges to I(x ^o, y ^o, T ^o).

Before addressing directly the issue of continuity of I ^p, we formally establish two (intuitively obvious) facts.

• (i) The intersection of the curve $\left\{ {\Omega (x,y,\lambda )\left| {\lambda \in \Lambda } \right.} \right\}$ and the frontier ∂(T) is a singleton for each $\left\langle {x,y,T} \right\rangle$∈ Ξ (and hence for each element of the sequence $\left\{ {x^\nu ,y^\nu ,T^\nu } \right\}_{\nu = 1}^\infty$ and its limit $\left\langle {x^o,y^o,T^o} \right\rangle$ ).

Suppose not: for some $\left\langle {x,y,T} \right\rangle$ ∈ Ξ, $\left\{ {\Omega (x,y,\lambda )\left| {\lambda \in \Lambda } \right.} \right\} \cap \partial (T)$ contains two points, $\left\langle {x{\prime},y{\prime}} \right\rangle$ and $\left\langle {\hat x,\hat y} \right\rangle$. Since $\Omega _i^{in}$ is decreasing in λ for all i and $\Omega _j^{out}$ is increasing in λ for all j, either $\left\langle {x{\prime}, - y{\prime}} \right\rangle \gg \left\langle {\hat x, - \hat y} \right\rangle$ or $\left\langle {\hat x, - \hat y} \right\rangle \gg \left\langle {x{\prime}, - y{\prime}} \right\rangle$. As these are frontier points, each inequality violates the free disposability assumption (FD), which proves the result.

• (ii) Since Ω is continuous, the sequence of paths $\left\{ {\Omega \left( {x^\nu ,y^\nu ,\lambda } \right)\left| {\lambda \in \Lambda } \right.} \right\}_{\nu = 1}^\infty$ converges to the path $\left\{ {\Omega \left( {x^o,y^o,\lambda } \right)\left| {\lambda \in \Lambda } \right.} \right\}$ (in the topology of closed convergence).

Consider a sequence, $\left\{ {\Omega \left( {x^\nu ,y^\nu ,\lambda ^\nu } \right)} \right\}_{\nu = 1}^\infty$, converging to Ω(x ^o, y ^o, λ′), where (by continuity of Ω) $\lambda {\prime} = {\mathrm{lim}}_{\nu \to \infty }\,\lambda ^\nu$. As Λ is a closed set, λ′ ∈ Λ, establishing that $\Omega \left( {x^o,y^o,\lambda {\prime}} \right) \in \left\{ {\Omega \left( {x^o,y^o,\lambda } \right)\left| {\lambda \in \Lambda } \right.} \right\}$.

Now consider a point Ω(x ^o, y ^o, λ′) with λ′ ∈ Λ and construct the sequence $\left\{ {\Omega \left( {x^\nu ,y^\nu ,\lambda {\prime}} \right)} \right\}_{\nu = 1}^\infty$. Clearly, $\left\{ {\Omega \left( {x^\nu ,y^\nu ,\lambda {\prime}} \right)} \right\} \in \left\{ {\Omega \left( {x^\nu ,y^\nu ,\lambda } \right)\left| {\lambda \in \Lambda } \right.} \right\}$ for all ν and ${\mathrm{lim}}_{\nu \to \infty }\Omega \left( {x^\nu ,y^\nu ,\lambda {\prime}} \right) = \Omega \left( {x^o,y^o,\lambda {\prime}} \right)$, establishing convergence.

• (iii) Continuity of I ^p.

Assume I ^p is not continuous. Then there exists a sequence $\left\{ {\left\langle {x^\nu ,y^\nu ,T^\nu } \right\rangle } \right\}_{\nu = 1}^\infty$ converging to $\left\langle {x^o,y^o,T^o} \right\rangle$ for which the sequence $\left\{ {I^p\left( {x^\nu ,y^\nu ,T^\nu } \right)} \right\}_{\nu = 1}^\infty$ does not converge to I ^p(x ^o, y ^o, T ^o).

Let λ ^ν = I ^p(x ^ν, y ^ν, T ^ν) for ν = 1…,∞ and λ ^o = I ^p(x ^o, y ^o, T ^o) so that the sequence $\left\{ {\lambda ^\nu } \right\}_{\nu _k = 1}^\infty$ does not converge to λ ^o. Therefore, there exists a subsequence $\left\{ {\lambda ^{\nu _k}} \right\}_{\nu _k = 1}^\infty$ such that

$${\mathrm{either}}\quad \lambda ^{\nu _k} >\lambda ^o\quad {\mathrm{or}}\quad \lambda ^{\nu _k} < \lambda ^o$$

for all elements of the subsequence.

By the definition of path-based indexes (2), Ω(x, y, I ^p(x, y, T)) is the intersection of the paths $\left\{ {\Omega (x,y,T)\left| {\lambda \in \Lambda } \right.} \right\}$ and ∂(T). Moreover, (i) above implies that this intersection consists of a single point. Finally, condition (ii) above implies that

$$\left\{ {\Omega \left( {x^\nu ,y^\nu ,\lambda ^\nu } \right)} \right\}_{\nu = 1}^\infty \to \Omega \left( {x^o,y^o,\lambda ^\nu } \right)$$

and

$$\left\{ {\Omega \left( {x^\nu ,y^\nu ,\lambda ^o} \right)} \right\}_{\nu = 1}^\infty \to \Omega \left( {x^o,y^o,\lambda ^o} \right).$$

Since Ω is continuous and $\lambda ^\nu \not \to \lambda ^o$, the sequence $\left\{ {\Omega \left( {x^o,y^o,\lambda ^\nu } \right)} \right\}_{\nu = 1}^\infty$ does not converge to Ω(x ^o, y ^o , λ ^o).

To ease the notation, denote Ω(x ^ν, y ^ν, I ^p(x ^ν, y ^ν, T ^ν)) by $\left\langle {\tilde x^\nu ,\tilde y^\nu } \right\rangle$ for all ν and Ω(x ^o, y ^o, I ^p(x ^o, y ^o, T ^o)) by $\left\langle {\tilde x^o,\tilde y^o} \right\rangle$.

The condition (c) on Ω implies that

$${\mathrm{either}}\quad \left\langle {\tilde x^{\nu _k}, - \tilde y^{\nu _k}} \right\rangle \gg \left\langle {\tilde x^o, - \tilde y^o} \right\rangle \quad {\mathrm{or}}\quad \left\langle {\tilde x^{\nu _k}, - \tilde y^{\nu _k}} \right\rangle \ll \left\langle {\tilde x^o, - \tilde y^o} \right\rangle $$

for all elements in the subsequence $\left\{ {\nu _k} \right\}_{k = 1}^\infty$.

As T ^ν → T ^o and $\left\{ {\Omega \left( {x^\nu ,y^\nu ,\lambda } \right)\left| {\lambda \in \Lambda } \right.} \right\}_{\nu = 1}^\infty$ converges to $\left\{ {\Omega \left( {x^o,y^o,\lambda } \right)\left| {\lambda \in \Lambda } \right.} \right\}$, there exists a sequence $\left\{ {\bar x^\nu ,\bar y^\nu } \right\}_{\nu = 1}^\infty \subset \partial \left( {T^\nu } \right)$ converging to $\left\langle {\tilde x^o,\tilde y^o} \right\rangle$. Hence, there exists a subsequence of {ν _k } denoted $\left\{ {\nu _{k_j}} \right\}$ such that

$${\kern 1pt} {\mathrm{either}}\quad \left\langle {\tilde x^{\nu _{k_j}}, - \tilde y^{\nu _{k_j}}} \right\rangle \gg \left\langle {\bar x^{\nu _{k_j}}, - \bar y^{\nu _{k_j}}} \right\rangle \quad {\mathrm{or}}\quad \left\langle {\tilde x^{\nu _{k_j}}, - \tilde y^{\nu _{k_j}}} \right\rangle \ll \left\langle {\bar x^{\nu _{k_j}}, - \bar y^{\nu _{k_j}}} \right\rangle $$

for all elements of the subsequence.

As $\left\langle {\tilde x^\nu ,\tilde y^\nu } \right\rangle$ and $\left\langle {\bar x^\nu ,\bar y^\nu } \right\rangle$ are contained in the frontier of T ^ν, either of these strict inequalities violates free disposability (FD).

This contradiction implies that I ^p(x ^ν, y ^ν, T ^ν) converges to I ^p(x ^o, y ^o, T ^o) so that I ^p satisfies continuity (C), in which case it must violate indication (I). ♢