Quantile–DEA classifiers with interval data

Abstract

This research intends to develop the classifiers for dealing with binary classification problems with interval data whose difficulty to be tackled has been well recognized, regardless of the field. The proposed classifiers involve using the ideas and techniques of both quantiles and data envelopment analysis (DEA), and are thus referred to as quantile–DEA classifiers. That is, the classifiers first use the concept of quantiles to generate a desired number of exact-data sets from a training-data set comprising interval data. Then, the classifiers adopt the concept and technique of an intersection-form production possibility set in the DEA framework to construct acceptance domains with each corresponding to an exact-data set and thus a quantile. Here, an intersection-form acceptance domain is actually represented by a linear inequality system, which enables the quantile–DEA classifiers to efficiently discover the groups to which large volumes of data belong. In addition, the quantile feature enables the proposed classifiers not only to help reveal patterns, but also to tell the user the value or significance of these patterns.

Appendices

Appendix 1: Proof of Theorem 1(i)

Theorem 1

Let $L<\bar{\beta}<\hat{\beta},$ and

$$ T_{\bar{\beta}}=\left\{ x\left\vert \sum\limits_{j=1}^{n}x_{j}^{\bar{\beta}} \lambda_{j}\leq x,\,\sum\limits_{j=1}^{n}\lambda_{j}\geq1,\,\lambda_{j}\geq 0,\,j=1,\ldots,n\right. \right\}, $$

and

$$ T_{\hat{\beta}}=\left\{ x\left\vert \sum\limits_{j=1}^{n}x_{j}^{\hat{\beta}} \lambda_{j}\leq x,\,\sum\limits_{j=1}^{n}\lambda_{j}\geq1,\,\lambda_{j}\geq 0,\,j=1,\ldots,n\right. \right\}. $$

Then, $T_{\hat{\beta}}\subset T_{\bar{\beta}}.$

Proof

Since b _j > a _j , j = 1, …, n, if $L<\bar{\beta} <\hat{\beta},$ then

$$ x_{j}^{\bar{\beta}}=a_{j}+\bar{\beta}\left( b_{j}-a_{j}\right) <a_{j} +\hat{\beta}\left( b_{j}-a_{j}\right) =x_{j}^{\hat{\beta}},\quad j=1,\ldots,n. $$

It follows that if $\sum\nolimits_{j=1}^{n}\lambda_{j}\geq1,\,\lambda_{j} \geq0,\,j=1,\ldots,n,$ then

$$ \sum\limits_{j=1}^{n}x_{j}^{\bar{\beta}}\lambda_{j}< \sum\limits_{j=1}^{n}x_{j}^{\hat{\beta}}\lambda_{j}. $$

Thus, if $x\in T_{\hat{\beta}},$ then $x\in T_{\bar{\beta}};$ that is, $T_{\hat{\beta}}\subset T_{\bar{\beta}}.$ □

Appendix 2: Proof of Theorem 2

To prove Theorem 2, we first present the following two lemmas:

Lemma 1

If $L<\bar{\beta}<\hat{\beta},$ and $\tilde{x}\in T_{\hat{\beta}}=\left\{ x\left\vert \sum\nolimits_{j=1}^{n}x_{j}^{\hat{\beta}} \lambda_{j}\leq x,\,\sum\nolimits_{j=1}^{n}\lambda_{j}\geq1,\,\lambda_{j}\geq 0,\,j=1,\ldots,n\right. \right\},$ then the optimal objective function value of the following linear program is less than one; i.e., $\hat{\theta }( \bar{\beta}) <1.$

$$ \begin{array}{ll} & \hat{\theta}( \bar{\beta}) =\min\theta,\\ \left( P_{\bar{\beta}}\right) & \hbox{s.t.}\; \sum\limits_{j=1} ^{n}x_{j}^{\bar{\beta}}\lambda_{j}\leq\theta\tilde{x},\\ & \sum\limits_{j=1}^{n}\lambda_{j}\geq1,\\ & \lambda_{j}\geq0,\quad j=1,\ldots,n. \end{array} $$

Proof

Let $T_{\bar{\beta}}=\left\{ x\left\vert \sum\nolimits _{j=1}^{n}x_{j}^{\bar{\beta}}\lambda_{j}\leq x,\,\sum\nolimits_{j=1}^{n}\lambda_{j} \geq1,\,\lambda_{j}\geq0,\,j=1,\ldots,n\right. \right\}.$ Since $\tilde {x}\in T_{\hat{\beta}},$ there exist $\tilde{\lambda}_{1},\,\tilde{\lambda} _{2},\ldots,\tilde{\lambda}_{n}$ that satisfy

$$ \begin{aligned} &{\sum\limits_{j=1}^{n}x_{j}^{\hat{\beta}}{\tilde{\lambda}_{j}}\leq\tilde{x}},\\ &{{\sum\limits_{j=1}^{n}{\tilde{\lambda}_{j}}{\geq}1}},\\ &{{\tilde{\lambda}_{j}}{\geq} 0,\quad j=1,\ldots,n}. \end{aligned} $$

Furthermore, since $\sum\nolimits_{j=1}^{n}\tilde{\lambda}_{j}\geq1,\, ( \tilde{\lambda}_{1},\,\tilde{\lambda}_{2},\ldots,\tilde{\lambda} _{n}) \neq0.$ Moreover, since $a_{j}<b_{j},\, 0<x_{j}^{\bar{\beta}}<x_{j}^{\hat{\beta}},$ j = 1, …, n. In summary,

$$ \sum\limits_{j=1}^{n}x_{j}^{\bar{\beta}}\tilde{\lambda}_{j}<\sum\limits_{j=1}^{n}x_{j} ^{\hat{\beta}}\tilde{\lambda}_{j}\leq\tilde{x}. $$

It follows that there exist solutions to the following system of inequalities:

$$ \begin{aligned} &{\sum\limits_{j=1}^{n}x_{j}^{\bar{\beta}}\lambda_{j}<\tilde{x}},\\ &{\sum\limits_{j=1}^{n}\lambda_{j}\geq1},\\ &{\lambda_{j}\geq0,\quad j=1,\ldots,n}. \end{aligned} $$

As a result, $\hat{\theta}( \bar{\beta}) <1$ (i.e., the optimal objective function value of $( P_{\bar{\beta}})$ is less than one). □

Lemma 2

If $L<\bar{\beta},\, \hat{x}>0$ and $\hat{x}\notin T_{\beta},$ then the optimal objective function value of the following linear program is greater than one; i.e., $\hat{\theta}( \beta) >1.$

$$ \begin{array}{lll} && \hat{\theta}( \beta) =\min\theta,\\ \left( P_{\beta}\right) &\hbox{s.t.} &\sum\limits_{j=1}^{n} x_{j}^{\beta}\lambda_{j}\leq\theta\hat{x},\\ && \sum\limits_{j=1}^{n}\lambda_{j}\geq 1,\\ && \lambda_{j}\geq0,\quad j=1,\ldots,n. \end{array} $$

Proof

Let $\hat{\lambda}_{1},\,\hat{\lambda}_{2} ,\ldots,\hat{\lambda}_{n}$ denote the optimal solution to $( P_{\beta })$ and $\hat{\theta}( \beta) =\hat{\theta}.$ If $\hat{\theta}( \beta) =\hat{\theta}\leq1,$ then

$$ \begin{aligned} &{\sum\limits_{j=1}^{n}x_{j}^{\beta}\hat{\lambda}_{j}\leq\hat{\theta}\hat{x}\leq \hat{x}},\\ &{\sum\limits_{j=1}^{n}\hat{\lambda}_{j}\geq1},\\ &{\hat{\lambda}_{j}\geq0,\quad j=1,\ldots,n}. \end{aligned} $$

That is, $\hat{x}\in T_{\beta},$ which is a contradiction. □

In what follows, we give the proof to Theorem 2, first to (i) and then to (ii).

Theorem 2

Let $\hat{x}\in\hat{T}\cap {\rm Int}\,\left\{ x|\sum\nolimits_{j=1}^{n}x_{j}^{L}\lambda_{j}\leq x,\,\sum\nolimits_{j=1}^{n}\lambda_{j} \geq1,\,\lambda_{j}\geq0,\,j=1,\ldots,n\right\},$ and $\hat{\theta}( \beta)$ be the quantile function of DMU-$\hat{x}.$ Then,

• (i) $\hat{\theta}( \beta)$ is a continuous function defined over (L, +∞).

• (ii) $\hat{\theta}( \beta)$ is a strictly monotonically decreasing function over (L, +∞).

Proof

• (i) Consider the following linear program $( P_{\beta}):$

  $$ \begin{array}{lll} && \hat{\theta}( \beta) =\min\theta,\\ \left( P_{\beta}\right) & \hbox{s.t.}& \sum\limits_{j=1}^{n} x_{j}^{\beta}\lambda_{j}\leq\theta\hat{x},\\ && \sum\limits_{j=1}^{n}\lambda_{j}\geq1,\\ && \lambda_{j}\geq0,\quad j=1,\ldots,n. \end{array} $$

  Equivalently,

  $$ \begin{array}{lll} && \hat{\theta}( \beta) =\min\theta,\\ \left( P_{\beta}\right) & \hbox{s.t.}& \sum\limits_{j=1}^{n}\left[ a_{j}+\beta\left( b_{j}-a_{j}\right) \right] \lambda_{j}\leq\theta\hat{x},\\ && \sum\limits_{j=1}^{n}\lambda_{j}\geq1,\\ && \lambda_{j}\geq0,\quad j=1,\ldots,n. \end{array} $$

  According to the stability of linear programming (Ying et al. 1975), the optimal objective function value of (P _β ), $\hat{\theta}(\beta),$ is a continuous function defined over (L, +∞).

• (ii) Let $L<\bar{\beta}<\hat{\beta},$ and consider the following problem $( P_{\hat{\beta}}):$

  $$ \begin{array}{lll} && \hat{\theta}( \hat{\beta}) =\min\theta,\\ \left( P_{\hat{\beta}}\right) & \hbox{s.t.}& \sum\limits_{j=1} ^{n}x_{j}^{\hat{\beta}}\lambda_{j}\leq\theta\hat{x},\\ && \sum\limits_{j=1}^{n}\lambda_{j}\geq1,\\ && \lambda_{j}\geq0,\quad j=1,\ldots,n. \end{array} $$

  It is clear that $\hat{\theta}( \hat{\beta}) \hat{x}\in T_{\hat{\beta}}.$ Consider also the following problem $( \tilde{P}_{\bar{\beta}}):$

  $$ \begin{array}{lll} && \hat{\theta}( \bar{\beta}) =\min\theta,\\ \left( \tilde{P}_{\bar{\beta}}\right) &\hbox{s.t.}& \sum\limits_{j=1}^{n}x_{j}^{\bar{\beta}}\lambda_{j}\leq\theta( \hat{\theta}( \hat{\beta}) \hat{x}), \\ && \sum\limits_{j=1}^{n}\lambda_{j}\geq1,\\ && \lambda_{j}\geq0,\quad j=1,\ldots,n. \end{array} $$

Let $\tilde{\theta},\,\tilde{\lambda}_{1},\,\tilde{\lambda} _{2},\ldots,\tilde{\lambda}_{n}$ denote the optimal solution to $( \tilde{P}_{\bar{\beta}}) .$ It is easy to check that $\tilde{\theta} >0.$ Furthermore, since $\hat{\theta}( \hat{\beta}) \hat{x}\in T_{\hat{\beta}}$ and $L<\bar{\beta}<\hat{\beta},$ from Lemma 1, $\tilde {\theta}<1.$ Moreover, since $\hat{\theta}( \bar{\beta})$ is the optimal objective function value of $( \tilde{P}_{\bar{\beta}}) ,\, \hat{\theta}( \bar{\beta}) \leq\tilde{\theta}\hat{\theta }( \hat{\beta}) <\hat{\theta}( \hat{\beta}).$□

Appendix 3: Existence of β*

The following Theorem 3 shows the existence of β*.

Theorem 3

Let $\hat{x}\in\hat{T}\cap {\rm Int}\,\left\{ x|\sum\nolimits_{j=1}^{n}x_{j}^{L}\lambda_{j}\leq x,\,\sum\nolimits_{j=1}^{n}\lambda_{j} \geq1,\,\lambda_{j}\geq0,\,j=1,\ldots,n\right\},$ and $\hat{\theta}( \beta)$ be the quantile function of DMU-$\hat{x}.$ Then, there exists β* ∈ (L, +∞) such that the optimal objective function value of the following problem (P _β ) is equal to one; i.e., $\hat{\theta}(\beta^{\ast})=1.$

$$ \begin{array}{lll} && \hat{\theta}( \beta) =\min\theta,\\ \left( P_{\beta}\right) & \hbox{s.t.}& \sum\limits_{j=1}^{n}x_{j}^{\beta}\lambda_{j}\leq\theta\hat{x},\\ && \sum\limits_{j=1}^{n}\lambda_{j}\geq1,\\ && \lambda_{j}\geq0,\quad j=1,\ldots,n. \end{array} $$

Proof

• (i) If $\hat{x}$ is located on the frontier of T ₁, then $\hat{\theta}( 1) =1,$ i.e., β* = 1.

• (ii) If $\hat{x}$ is not located on the frontier of T ₁, and $\hat{x}\in{\rm Int}\,T_{1},$ then there exist $\lambda_{j}^{0}\geq 0,\,j=1,\ldots,n,\,\sum\nolimits_{j=1}^{n}\lambda_{j}^{0}\geq1$ such that

  $$ \sum\limits_{j=1}^{n}\left[ a_{j}+1\times\left( b_{j}-a_{j}\right) \right] \lambda_{j}^{0}=\sum\limits_{j=1}^{n}b_{j}\lambda_{j}^{0}<\hat{x}, $$

  (1)

  and $\hat{\theta}( 1) <1.$ Let

  $$ \hat{\beta}>\max\left\{ \underset{1\leq i\leq m;1\leq j\leq n}{\max}\left\{ \left( \hat{x}_{ij}-a_{ij}\right) /\left( b_{ij}-a_{ij}\right) \right\} ,\,L\right\}. $$

  Then,

  $$ a_{j}+\hat{\beta}\left( b_{j}-a_{j}\right) >\hat{x},\quad j=1,\ldots,n. $$

  Therefore, for any $\lambda_{j}\geq0,\,j=1,\ldots,n,\,\sum\nolimits_{j=1}^{n} \lambda_{j}\geq1,$ we have

  $$ \sum\limits_{j=1}^{n}x_{j}^{\hat{\beta}}\lambda_{j}>\hat{x}. $$

  (2)

  From (2), $\hat{x}\notin T_{\hat{\beta}},$ and from Lemma 2, $\hat{\theta}( \hat{\beta}) >1.$ As a result, since $\hat{\theta }( 1) <1,\,\hat{\theta}( \hat{\beta}) >1,\,\hat{\beta}\in(L,\,+\infty),$ from Theorem 2(i), $\hat{\theta}( \beta)$ is a continuous function defined over (L, +∞). It follows that there exists $\beta^{\ast}\in(L,\,+\infty)$ such that $\hat{\theta}( \beta^{\ast}) =1.$

• (iii) If $\hat{x}\notin T_{1},$ from Lemma 2, $\hat{\theta}( 1) >1.$ In addition, since

  $$ \hat{x}\in\text{Int}\,\left\{x |\sum\limits_{j=1}^{n}x_{j}^{L}\lambda_{j}\leq x,\,\sum\limits_{j=1}^{n}\lambda_{j}\geq1,\,\lambda_{j}\geq0,\,j=1,\ldots,n\right\}, $$

  there exist $\lambda_{j}^{0}\geq0,\,j=1,\ldots,n,\,\sum\nolimits_{j=1}^{n} \lambda_{j}^{0}\geq1$ such that

  $$ \sum\limits_{j=1}^{n}\left[a_{j}+L\times\left( b_{j}-a_{j}\right) \right]\lambda_{j}^{0}=\sum\limits_{j=1}^{n}x_{j}^{L}\lambda_{j}^{0}<\hat{x}. $$

  Therefore, there exists $\hat{\beta}$ that satisfies $\hat{\beta}>L$ such that

  $$ \sum\limits_{j=1}^{n}x_{j}^{\hat{\beta}}\lambda_{j}^{0}<\hat{x}. $$

  That is, $\hat{x}\in {\rm Int}\,T_{\hat{\beta}},$ and thus $\hat{\theta}( \hat{\beta}) <1.$ Consequently, since $\hat{\theta}( 1) >1,\,\hat{\theta}( \hat{\beta}) <1,\,\hat{\beta}\in(L,\,+\infty),$ from Theorem 2(i), $\hat{\theta}( \beta)$ is a continuous function defined over (L, +∞). It follows that there exists β* ∈ (L, +∞) such that $\hat{\theta }( \beta^{\ast}) =1.$ □

Appendix 4: Uniqueness of β*

The following Theorem 4 shows the uniqueness of β*.

Theorem 4

Let $b_{j}>a_{j},\,j=1,\ldots,n,\, L<\bar{\beta} <\hat{\beta},$ and $\hat{x}\in\hat{T}.$ Then

• (i) There is no intersection between the frontiers of $T_{\hat{\beta }}$ and $T_{\bar{\beta}}.$

• (ii) The quantile of DMU-$\hat{x},$ i.e., β*, is uniquely determined.

Proof

The proof to (i) is achieved by contradiction. If there exists $x^{0}\in\Re_{+}^{m},$ and x ⁰ is located on the frontiers of both $T_{\hat{\beta}}$ and $T_{\bar{\beta}},$ then, from Theorem 2, $1=\hat{\theta }( \bar{\beta}) <\hat{\theta}( \hat{\beta}) =1,$ which is a contradiction. That is, there is no intersection between the frontiers of $T_{\hat{\beta}}$ and $T_{\bar{\beta}}.$

The proof to (ii) is also achieved by contradiction. Assume that there exist two quantiles of DMU-$\hat{x},$ i.e., $\beta_{1}^{\ast}$ and $\beta_{2}^{\ast}.$ Without loss of generality, assume that $L<\beta_{1} ^{\ast}<\beta_{2}^{\ast}.$ Since both $\beta_{1}^{\ast}$ and $\beta_{2}^{\ast }$ are the quantiles of DMU-$\hat{x},\,\hat{\theta}( \beta_{1}^{\ast }) =\hat{\theta}( \beta_{2}^{\ast}) =1.$ However, from Theorem 2, $\hat{\theta}( \beta_{1}^{\ast}) <\hat{\theta}( \beta_{2}^{\ast}).$ That is, there is a contradiction. It follows that β* is uniquely determined. □