Measuring poverty in multidimensional contexts

Abstract

When measuring multidimensional poverty it is reasonable to expect that the trade-offs between variable pairs can differ depending on whether the concerned pairs are complements or substitutes. Yet, currently existing approaches based on deprivation count distributions unrealistically assume that all pairs of variables are related in the same way—an unfortunate circumstance that undermines the possibilities of identifying the poor, aggregating their poverty levels and modeling non-trivial interactions between variables in highly flexible ways. This paper, which aims at modeling non-trivial relational structures across variables both in the identification and aggregation steps, is a first contribution towards addressing these inadequacies. The approach has been axiomatically characterized to flesh out the normative foundations upon which it is based and has a vast potential for application.

Appendix

Proof of Proposition 1

We start with the ‘if’ part of the proof. Assume that $\rho \in {\mathcal {I}}_{d}$. We have to prove that $({\mathcal {L}} \left( P_{\rho }\right) )^{\uparrow }=P_{\rho }$.

(1) We start proving $({\mathcal {L}}\left( P_{\rho }\right) )^{\uparrow }\subset P_{\rho }$. Take ${\varvec{x}}\in ({\mathcal {L}}\left( P_{\rho }\right) )^{\uparrow }$. Then, there exists some ${\varvec{z}}\in {\mathcal {L}}\left( P_{\rho }\right)$ such that ${\varvec{z}}\le {\varvec{x}}$ (if ${\varvec{x}}\in {\mathcal {L}}\left( P_{\rho }\right)$, then ${\varvec{z}}={\varvec{x}}$). Since ${\mathcal {L}}\left( P_{\rho }\right) \subset P_{\rho }$, ${\varvec{z}}\in P_{\rho }$. In addition, since ${\varvec{x}}\in {\varvec{z}}^{\uparrow }$ and $\rho \in {\mathcal {I}}_{d}$, one can conclude that ${\varvec{x}}\in P_{\rho }$.

(2) We now prove $\left( {\mathcal {L}}\left( P_{\rho }\right) \right) ^{\uparrow }\supset P_{\rho }$. Take ${\varvec{x}}\in P_{\rho }$. If it turns out that ${\varvec{x}}\in {\mathcal {L}}\left( P_{\rho }\right)$ then we are done. If ${\varvec{x}}\notin {\mathcal {L}}\left( P_{\rho }\right)$ then there must exist some ${\varvec{y}}\in P_{\rho }\backslash \{{\varvec{x}}\}$ such that ${\varvec{y}}\le {\varvec{x}}$. Now, if ${\varvec{y}}\in {\mathcal {L}}\left( P_{\rho }\right) \subset P_{\rho }$ then ${\varvec{x}}\in {\varvec{y}}^{\uparrow }$. Since $\rho \in {\mathcal {I}}_{d}$, one can conclude that ${\varvec{x}}\in \left( {\mathcal {L}}\left( P_{\rho }\right) \right) ^{\uparrow }$. Otherwise, if ${\varvec{y}}\notin {\mathcal {L}}\left( P_{\rho }\right)$ then we can proceed iteratively until reaching an element belonging to ${\mathcal {L}} \left( P_{\rho }\right)$. That is: since $X^{d}$ is finite ($\left| X^{d}\right| =2^{d}$) there must exist a finite sequence of vector dominations ${\varvec{z}}_{i}\le {\varvec{z}}_{i+1}$ from some element $\varvec{ z}_{1}\in {\mathcal {L}}\left( P_{\rho }\right)$ up to ${\varvec{x}}$ (i.e.: ${\varvec{z}}_{1}\le {\varvec{z}}_{2}\cdots \le {\varvec{z}}_{n}\le {\varvec{x}}$ ), so that ${\varvec{x}}\in {\varvec{z}}_{1}{}^{\uparrow }$. Since $\rho \in {\mathcal {I}}_{d}$, one can conclude that ${\varvec{x}}\in \left( {\mathcal {L}} \left( P_{\rho }\right) \right) ^{\uparrow }$.

This proves the ‘if’ part of the proposition. The ‘only if’ part of the proof goes as follows. Assume $P_{\rho }$ is a subset of $X^{d}$ such that $\left( {\mathcal {L}}\left( P_{\rho }\right) \right) ^{\uparrow }=P_{\rho }$. We have to prove that $\rho \in {\mathcal {I}}_{d}$. Take any ${\varvec{x}}\in P_{\rho }$. Since $\left( {\mathcal {L}}\left( P_{\rho }\right) \right) ^{\uparrow }=P_{\rho }$ we can say that ${\varvec{x}}\in {\varvec{z}}^{\uparrow }$ for some ${\varvec{z}}\in {\mathcal {L}}\left( P_{\rho }\right)$. Consider now any ${\varvec{y}}\in {\varvec{x}}^{\uparrow }$. By the transitivity of $\le$ one has that ${\varvec{y}}\in {\varvec{z}}^{\uparrow }$. Since $\left( {\mathcal {L}}\left( P_{\rho }\right) \right) ^{\uparrow }=P_{\rho }$, we can conclude that ${\varvec{y}}\in P_{\rho }$. $\square$

Proof of Theorem 1

It is easy to verify that when $S_{d}=\mathcal {C }_{d}$, then any $\rho \in S$ satisfies MON, IND, VAN and NTR. Therefore, we will prove that when a group of identification functions $S_{d}\subseteq \Omega _{d}$ satisfies these four axioms then it must be equal to $\mathcal {C }_{d}$.

We will first prove the following auxiliary lemma:

Auxiliary Lemma 1

Whenever IND and VAN hold for some $\rho \in S_{d}\subseteq \Omega _{d}$, then, one has that $\left| {\varvec{x}} \right| =\left| {\varvec{y}}\right| \Rightarrow \rho ({\varvec{x}} )=\rho ({\varvec{y}})$ for all ${\varvec{x}},{\varvec{y}}\in X^{d}$ and all $\rho \in S_{d}$.

Let $\rho \in S_{d}$ and let $m>1$ be an integer for which $\left| {\varvec{x}}\right| =\left| {\varvec{y}}\right|$ entails $\rho ( {\varvec{x}})=\rho ({\varvec{y}})$ for all ${\varvec{x}},{\varvec{y}}\in X^{d}$ such that $\left| {\varvec{x}}\right| =\left| {\varvec{y}} \right| <m$. We will now prove the result also holds true whenever $\left| {\varvec{x}}\right| =\left| {\varvec{y}}\right| =m$. Let ${\varvec{x}},{\varvec{y}}\in X^{d}$ be two vectors with $\left| {\varvec{x}} \right| =\left| {\varvec{y}}\right| =m$. There are now two mutually exclusive cases.

Case 1 Assume there exists a variable $j\in \{1,\ldots ,d\}$ such that $x_{j}=y_{j}=1$. Consider the vectors ${\varvec{x}}-{\varvec{e}}_{j}, {\varvec{y}}-{\varvec{e}}_{j}$. Since $\left| {\varvec{x}}-{\varvec{e}} _{j}\right| =\left| {\varvec{y}}-{\varvec{e}}_{j}\right| =m-1<m$, the induction hypothesis holds, so $\rho ({\varvec{x}}-{\varvec{e}}_{j})=\rho ( {\varvec{y}}-{\varvec{e}}_{j})$. Now, by IND one has that $\rho ({\varvec{x}} )=\rho ({\varvec{y}})$, as desired.

Case 2 Assume there does not exist any variable $j\in \{1,\ldots ,d\}$ such that $x_{j}=y_{j}=1$. Consider two different variables $j,l\in \{1,\ldots ,d\}$ such that $x_{j}=0,y_{j}=1$ and $x_{l}=1,y_{l}=0$. Since $\left| {\varvec{x}}-{\varvec{e}}_{l}\right| =\left| {\varvec{y}}- {\varvec{e}}_{j}\right| =m-1<m$, the induction hypothesis holds, so $\rho ( {\varvec{x}}-{\varvec{e}}_{l})=\rho ({\varvec{y}}-{\varvec{e}}_{j})$. Now, by IND one has that

$$\begin{aligned} \rho ({\varvec{x}})=\rho ({\varvec{y}}-{\varvec{e}}_{j}+{\varvec{e}}_{l}). \end{aligned}$$

(A1)

Observe that

$$\begin{aligned} {\varvec{e}}_{j}= & {} {\varvec{y}}-\bigcup _{i\in \delta ({\varvec{y}})\backslash \{j\}} {\varvec{e}}_{i}, \end{aligned}$$

(A2)

$$\begin{aligned} {\varvec{e}}_{l}= & {} {\varvec{y}}-\bigcup _{i\in \delta ({\varvec{y}})\backslash \{j\}} {\varvec{e}}_{i}-{\varvec{e}}_{j}+{\varvec{e}}_{l} \end{aligned}$$

(A3)

where $\delta ({\varvec{y}}):=\{i\in \{1,\ldots ,d\}|y_{i}=1\}\$is the subset of variables in profile ${\varvec{y}}$ where $y_{i}=1$. By VAN, one has that $\rho ({\varvec{e}}_{j})=\rho ({\varvec{e}}_{l})$. Applying now IND several times (once per variable $i\in \delta ({\varvec{y}})\backslash \{j\}$), one can conclude that

$$\begin{aligned} \rho \left( {\varvec{e}}_{j}+\bigcup _{i\in \delta ({\varvec{y}})\backslash \{j\}} {\varvec{e}}_{i}\right) =\rho \left( {\varvec{e}}_{l}+\bigcup _{i\in \delta ({\varvec{y}} )\backslash \{j\}}{\varvec{e}}_{i}\right) \end{aligned}$$

(A4)

By (A2) and (A3), (A4) can be rewritten as

$$\begin{aligned} \rho ({\varvec{y}})=\rho ({\varvec{y}}-{\varvec{e}}_{j}+{\varvec{e}}_{l}). \end{aligned}$$

(A5)

Comparing (A1) with (A5), we can conclude that $\rho ({\varvec{x}} )=\rho ({\varvec{y}})$, as desired. This proves the auxiliary lemma 1. $\square$

Take now any two vectors ${\varvec{x}},{\varvec{y}}\in X^{d}$ such that $\left| {\varvec{x}}\right| \ge \left| {\varvec{y}}\right|$. Define now ${\varvec{w}}\in X^{d}$ in such a way that $\left| {\varvec{w}} \right| =\left| {\varvec{y}}\right|$ and $\delta ({\varvec{w}} )\subseteq \delta ({\varvec{x}})$. By auxiliary lemma 1, one has $\rho ( {\varvec{w}})=\rho ({\varvec{y}})$. On the other hand, by MON $\rho ({\varvec{x}} )\ge \rho ({\varvec{w}})$, so one can conclude that $\rho ({\varvec{x}})\ge \rho ({\varvec{y}})$. This ensures that $\rho$ is a counting measure for all $\rho \in S_{d}$.

Observe that NTR and MON imply that $\rho ({\varvec{0}})=0$ and $\rho ( {\varvec{1}})=1$ (if $\rho ({\varvec{0}})=1$, then $\rho ({\varvec{x}})=1$ for all ${\varvec{x}}\in X^{d}$ and if $\rho ^{b}({\varvec{1}})=0$, then $\rho ^{b}( {\varvec{x}})=0$ for all ${\varvec{x}}\in X^{d}$—in both cases contradicting NTR). Lastly, by MON and NTR there must exist a $k\in \{1,\ldots ,d\}$ such that $\rho ({\varvec{x}})=0$ whenever $\left| {\varvec{x}}\right| <k$ and $\rho ({\varvec{x}})=1$ whenever $\left| {\varvec{x}}\right| \ge k$ . Therefore, $\rho \in {\mathcal {C}}_{d}$, as desired. This proves Theorem 1. $\square$

Statement: If COM holds for a given $S_{d}\subseteq \Omega _{d}$, then IND holds as well for all $\rho \in S_{d}$; however, the opposite is not necessarily true.

Proof of the statement

To verify this claim let’s start assuming that COM applies for a certain set of identification functions $S_{d}\subseteq \Omega _{d}$. Consider now the deprivation vectors ${\varvec{x}} ,{\varvec{y}},{\varvec{x}}^{\prime },{\varvec{y}}^{\prime }\in X^{d}$ as in the statement of IND. Then, it is trivial to verify that $({\varvec{x}},{\varvec{y}} ^{\prime })$ and $({\varvec{y}},{\varvec{x}}^{\prime })$ are equivalent societies (with $m=2$). Imposing COM, one has that $\rho ({\varvec{x}})\ge \rho ({\varvec{y}})$ implies $\rho ({\varvec{x}}^{\prime })\ge \rho ({\varvec{y}} ^{\prime })$ for all $\rho \in S_{d}$: this is precisely what IND states. On the other hand, one can find infinitely many examples of sets of identification functions satisfying IND but failing to satisfy COM. A very simple example for the case $d=3$ can be $S_{d}=\{\rho _{0}\}$, where

$$\begin{aligned} \rho _{0}({\varvec{x}})\varvec{=}\left\{ \begin{array}{c} 0\text { if }\sum _{i}x_{i}<2 \\ 1\text { if }\sum _{i}x_{i}\ge 2 \end{array} \right\} . \end{aligned}$$

(A6)

It is trivial to check that $S_{d}=\{\rho _{0}\}$ satisfies IND. However, it does not satisfy COM. To verify this, consider the following pair of three-person societies $({\varvec{x}}_{1},{\varvec{x}}_{2},{\varvec{x}}_{3})$ and $({\varvec{y}}_{1},{\varvec{y}}_{2},{\varvec{y}}_{3})$ with ${\varvec{x}}_{1}=(111), {\varvec{x}}_{2}=(101),{\varvec{x}}_{3}=(001),{\varvec{y}}_{1}=(101),{\varvec{y}} _{2}=(011),{\varvec{y}}_{3}=(101)$. Clearly, both societies are equivalent (the number of individuals experiencing deprivation in each variable coincides). However, we have that $\rho _{0}({\varvec{x}}_{1})\le \rho _{0}( {\varvec{y}}_{1})$ and $\rho _{0}({\varvec{x}}_{2})\le \rho _{0}({\varvec{y}} _{2})$ for all $\rho \in S_{d}$ and yet $\rho _{0}({\varvec{x}}_{3})<\rho _{0}({\varvec{y}}_{3})$, thus contradicting COM. $\square$

Proof of Theorem 2

It is easy to verify that when $S_{d}=\mathcal {W }_{d}$, then any $\rho \in S_{d}$ satisfies MON, COM and NTR. Therefore, we will prove that when a group of identification functions $S_{d}\subseteq \Omega _{d}$ satisfies these three axioms then it must be equal to $\mathcal { W}_{d}(k)$ for some $k\in (0,1]$. Since (i) $X^{d}$ is finite, (ii) each $\rho \in S_{d}$ induces a complete ordering in $X^{d}\times X^{d}$, and (iii) COM holds for all $\rho \in S_{d}$, the hypotheses of Theorem 4.1.B in Fishburn (1970) are satisfied. Therefore, for all ${\varvec{x}},{\varvec{y}}\in X^{d}$ and for all $\rho \in S_{d}$ one has that

$$\begin{aligned} \rho ({\varvec{x}})\le \rho ({\varvec{y}})\Leftrightarrow \sum _{j=1}^{d}u_{j}(x_{j})\le \sum _{j=1}^{d}u_{j}(y_{j}) \end{aligned}$$

(A7)

for some real-valued functions $u_{1},\ldots ,u_{d}$ on $\{0,1\}$. One can rewrite the last expression as follows

$$\begin{aligned} \rho ({\varvec{x}})\le \rho ({\varvec{y}})\Leftrightarrow \sum _{j=1}^{d}\left( u_{j}(x_{j})-u_{j}(0)\right) \le \sum _{j=1}^{d}\left( u_{j}(y_{j})-u_{j}(0)\right) \end{aligned}$$

(A8)

In turn, this expression can be rewritten as

$$\begin{aligned} \rho ({\varvec{x}})\le \rho ({\varvec{y}})\Leftrightarrow \sum _{j=1}^{d} {\widetilde{u}}_{j}(x_{j})\le \sum _{j=1}^{d}{\widetilde{u}}_{j}(y_{j}) \end{aligned}$$

(A9)

where ${\widetilde{u}}_{j}(x_{j}):=u_{j}(x_{j})-u_{j}(0)$. Clearly, ${\widetilde{u}}_{j}(0)=0$. If we define $w_{j}:={\widetilde{u}}_{j}(1)$, (A9) can be written as

$$\begin{aligned} \rho ({\varvec{x}})\le \rho ({\varvec{y}})\Leftrightarrow \sum _{j=1}^{d}w_{j}x_{j}\le \sum _{j=1}^{d}w_{j}y_{j} \end{aligned}$$

(A10)

By MON, one must have that $w_{j}\ge 0\forall j$. This ensures that $\rho$ is a counting measure for all $\rho \in S$. By MON and NTR, $\rho (\varvec{ 0})=0$ and $\rho ({\varvec{1}})=1$.¹⁶ Lastly, by MON and NTR there must exist a real number $q\in (0,\sum _{i}w_{i}]$ such that $\rho ({\varvec{x}})=0$ whenever $\sum _{j=1}^{d}w_{j}x_{j}<q$ and $\rho ({\varvec{x}})=1$ whenever $\sum _{j=1}^{d}w_{j}x_{j}\ge q$. Defining $a_{i}:=w_{i}/\sum _{i}w_{i}$ and $k:=q/\sum _{i}w_{i}$ we have found a vector of weights ${\varvec{a}}\in \Delta _{d}$ and a deprivation threshold $k\in (0,1]$ such that $\rho \in \mathcal { W}_{d}$, as desired. This proves Theorem 2. $\square$

Proof of Proposition 2

It is straightforward to prove that if $S_{d}\in \{{\mathcal {I}}_{d},{\mathcal {W}}_{d},{\mathcal {C}}_{d}\}$ then $S_{d}^{\psi }$ satisfies NTR and MON. We will only show that $S_{d}^{\psi }$ does not satisfy COM. For that purpose we need to prove an auxiliary lemma (see below). According to (11) ${\mathcal {C}}_{d}^{\psi }$ is the set of identification functions for d variables where the unweighted counting approach is used within- and between-dimensions (given $\psi \in \Psi _{G}$ ). To allow proper labelling, this set will be rewritten as ${\mathcal {C}} _{d}^{\psi }(k_{1},\ldots ,k_{G};k^{b}),$ where $k_{1},\ldots ,k_{G}$ denote the domain specific deprivation thresholds and $k^{b}$ the between domain deprivation threshold. Within this set, define

$$\begin{aligned} \widehat{{\mathcal {C}}}_{d}^{\psi }&:=\left\{ \rho \in {\mathcal {C}}_{d}^{\psi }(k_{1},\ldots ,k_{G};k^{b})\mid k^{b}<1\text { and }k_{g_{j}}\right. \nonumber \\&\ge \left. 2/d_{g_{j}} \text { for at least two }g_{j}\in \{1,\ldots ,G\}\right\} , \end{aligned}$$

(A11)

$$\begin{aligned} \widetilde{{\mathcal {C}}}_{d}^{\psi }&:=\left\{ \rho \in {\mathcal {C}}_{d}^{\psi }(k_{1},\ldots ,k_{G};k^{b})\mid k^{b}=1\text { and }k_{g_{j}}\right. \nonumber \\&<\left. 1\text { for at least two }g_{j}\in \{1,\ldots ,G\}\right\} . \end{aligned}$$

(A12)

The set $\widehat{{\mathcal {C}}}_{d}^{\psi }$ contains identification functions where poor individuals do not have to experience deprivation in all dimensions simultaneously and in some of them they must be deprived in at least two variables. The set $\widetilde{{\mathcal {C}}}_{d}^{\psi }$ contains identification functions where poor individuals have to experience deprivation in all dimensions simultaneously but where deprivation needs not to be universal within at least two of these dimensions. The sets $\widehat{ {\mathcal {C}}}_{d}^{\psi }$ and $\widetilde{{\mathcal {C}}}_{d}^{\psi }\$are generalizations of Examples 1 and 2 to the multiple dimension context.

Auxiliary Lemma 2

$\widehat{{\mathcal {C}}}_{d}^{\psi }\cap \mathcal {W }_{d}=\emptyset$ and $\widetilde{{\mathcal {C}}}_{d}^{\psi }\cap {\mathcal {W}} _{d}=\emptyset .$

Proof of Auxiliary Lemma 2

In both cases we follow the same strategy: if $\rho \in \widehat{{\mathcal {C}}}_{d}^{\psi }$ or $\rho \in \widetilde{{\mathcal {C}}}_{d}^{\psi }$ we start assuming that there is a weighting scheme ${\varvec{a}}\in \Delta _{d}$ and a deprivation threshold k such that $\rho \in {\mathcal {W}}_{d}$ to arrive at a contradiction. Given the partition of D in G dimensions $(D_{1},\ldots ,D_{G})\in \Psi _{G}$, we will denote the elements of the weighting vector ${\varvec{a}}$ as $a_{gv}$, where $g\in \{1,\ldots ,G\}$ indexes the member of the partition $D_{g}$ to which the weight belongs and $v\in \{1,\ldots ,d_{g}\}$ indexes the members within domain $D_{g}$. We can assume without loss of generality that within each domain $D_{g}$ the weights are sorted in a non-ascending order, i.e.: $a_{gv}\ge a_{gv+1}$ for all $g\in \{1,\ldots ,G\}$ and all $v\in \{1,\ldots ,d_{g}-1\}$.

We start with $\widehat{{\mathcal {C}}}_{d}^{\psi }$. Without loss of generality, we can assume that the two dimensions $g_{1},g_{2}\in \{1,\ldots ,G\}$ with $k_{g_{1}}\ge 2/d_{g_{1}}$ and $k_{g_{2}}\ge 2/d_{g_{2}}\$are $g_{1}=1$ and $g_{2}=2$. Since $\rho \in \widehat{{\mathcal {C}}}_{d}^{\psi }$, there exist $\rho ^{b}\in {\mathcal {C}}_{G}(k^{b})$ and $\rho _{g}^{w}\in {\mathcal {C}}_{d_{g}}(k_{g})$ such that $\rho ({\varvec{x}})=\rho ^{b}(\rho _{1}^{w}({\varvec{x}}_{1}),\ldots ,\rho _{G}^{w}({\varvec{x}}_{G}))$. Let ${\mathcal {L}}^{b}\subset X^{G}$ be the set of least deprived profiles in $\left( \rho ^{b}\right) ^{-1}(1)$ (i.e. if ${\varvec{x}}\in {\mathcal {L}} ^{b}$ and ${\varvec{y}}\in X^{G}$ is such that ${\varvec{y}}<{\varvec{x}}$, then ${\varvec{y}}\in \left( \rho ^{b}\right) ^{-1}(0)$). Consider ${\varvec{u}}, {\varvec{u}}^{\prime }\in {\mathcal {L}}^{b}$. Without loss of generality we will write them as ${\varvec{u}}=(1\ 0$$u_{3}\ldots u_{G})$, ${\varvec{u}}^{\prime }=(0\ 1$$u_{3}\ldots u_{G}).$ By definition, the following inequalities must hold:

$$\begin{aligned}&a_{11}+a_{12}+\sum _{v=3}^{v=m_{1}}a_{1v}+\sum _{v=3}^{v=m_{2}}a_{2v}+ \sum _{g\in \delta (u_{3}\ldots u_{G})}\sum _{v=1}^{v=m_{g}}a_{gv}\ge k \end{aligned}$$

(A13)

$$\begin{aligned}&a_{21}+a_{22}+\sum _{v=3}^{v=m_{1}}a_{1v}+\sum _{v=3}^{v=m_{2}}a_{2v}+ \sum _{g\in \delta (u_{3}\ldots u_{G})}\sum _{v=1}^{v=m_{g}}a_{gv}\ge k \end{aligned}$$

(A14)

where $m_{g}:=\lfloor k_{g}d_{g}\rfloor$ and $\delta (u_{3}\ldots u_{G}):=\{g|u_{g}=1\}\$is the subset of elements in vector $(u_{3}\ldots u_{G})$ where $u_{g}=1$. Consider now a third vector ${\varvec{u}}^{\prime \prime }=(0\ 0$$u_{3}\ldots u_{G}).$ Since ${\varvec{u}},{\varvec{u}}^{\prime }\in {\mathcal {L}}^{b}$, one has that ${\varvec{u}}^{\prime \prime }\in \left( \rho ^{b}\right) ^{-1}(0)$. This implies that the following inequalities must hold.

$$\begin{aligned}&a_{11}+\sum _{v=3}^{v=m_{1}}a_{1v}+a_{21}+\sum _{v=3}^{v=m_{2}}a_{2v}+ \sum _{g\in \delta (u_{3}\ldots u_{G})}\sum _{v=1}^{v=m_{g}}a_{gv}<k \end{aligned}$$

(A15)

$$\begin{aligned}&a_{11}+\sum _{v=3}^{v=m_{1}}a_{1v}+a_{22}+\sum _{v=3}^{v=m_{2}}a_{2v}+ \sum _{g\in \delta (u_{3}\ldots u_{G})}\sum _{v=1}^{v=m_{g}}a_{gv}<k \end{aligned}$$

(A16)

$$\begin{aligned}&a_{12}+\sum _{v=3}^{v=m_{1}}a_{1v}+a_{21}+\sum _{v=3}^{v=m_{2}}a_{2v}+ \sum _{g\in \delta (u_{3}\ldots u_{G})}\sum _{v=1}^{v=m_{g}}a_{gv}<k \end{aligned}$$

(A17)

$$\begin{aligned}&a_{12}+\sum _{v=3}^{v=m_{1}}a_{1v}+a_{22}+\sum _{v=3}^{v=m_{2}}a_{2v}+ \sum _{g\in \delta (u_{3}\ldots u_{G})}\sum _{v=1}^{v=m_{g}}a_{gv}<k \end{aligned}$$

(A18)

If one defines

$$\begin{aligned} k^{\prime }:=k-\left( \sum _{v=3}^{v=m_{1}}a_{1v}+\sum _{v=3}^{v=m_{2}}a_{2v}+\sum _{g\in \delta (u_{3}\ldots u_{G})}\sum _{v=1}^{v=m_{g}}a_{gv}\right) \end{aligned}$$

(A19)

the inequalities (A13)–(A18) can be rewritten as

$$\begin{aligned} \left\{ \begin{array}{c} a_{11}+a_{12}\ge k^{\prime }, \\ a_{21}+a_{22}\ge k^{\prime }, \end{array} \begin{array}{c} a_{11}+a_{21}<k^{\prime }, \\ a_{11}+a_{22}<k^{\prime }, \end{array} \begin{array}{c} a_{12}+a_{21}<k^{\prime }, \\ a_{12}+a_{22}<k^{\prime }. \end{array} \right\} \end{aligned}$$

(A20)

It is trivial to show that the inequalities system shown in (A20) does not have feasible solutions. In the first inequality of the system, either $a_{11}$ or $a_{12}$ must be greater or equal than $k^{\prime }/2$. The same goes for $a_{21},a_{22}$ in the second inequality of the system: at least one of them must be greater or equal than $k^{\prime }/2$. Picking the largest elements between $a_{11},a_{12}$ and $a_{21},a_{22}$ and adding them up results in a number that is greater or equal than $k^{\prime }$, therefore contradicting at least one of the four last inequalities of the system. We have reached the contradiction we were looking for.

Let us now consider case $\widetilde{{\mathcal {C}}}_{d}^{\psi }$. Without loss of generality, we can assume that the two dimensions $g_{1},g_{2}\in \{1,\ldots ,G\}$ with $k_{g_{1}}<1$ and $k_{g_{2}}<1\$are $g_{1}=1$ and $g_{2}=2$. Since $\rho \in \widetilde{{\mathcal {C}}}_{d}^{\psi }$, there exist $\rho ^{b}\in {\mathcal {C}}_{G}(1)$ and $\rho _{g}^{w}\in {\mathcal {C}} _{d_{g}}(k_{g})$ such that $\rho ({\varvec{x}})=\rho ^{b}(\rho _{1}^{w}( {\varvec{x}}_{1}),\ldots ,\rho _{G}^{w}({\varvec{x}}_{G}))$. By definition, the following inequalities must hold:

$$\begin{aligned}&a_{11}+\sum _{v=3}^{v=m_{1}+1}a_{1v}+a_{21}+\sum _{v=3}^{v=m_{2}+1}a_{2v}+ \sum _{g=3}^{g=G}\sum _{v=1}^{v=m_{g}}a_{gv}\ge k \end{aligned}$$

(A21)

$$\begin{aligned}&a_{11}+\sum _{v=3}^{v=m_{1}+1}a_{1v}+a_{22}+\sum _{v=3}^{v=m_{2}+1}a_{2v}+ \sum _{g=3}^{g=G}\sum _{v=1}^{v=m_{g}}a_{gv}\ge k \end{aligned}$$

(A22)

$$\begin{aligned}&a_{12}+\sum _{v=3}^{v=m_{1}+1}a_{1v}+a_{21}+\sum _{v=3}^{v=m_{2}+1}a_{2v}+ \sum _{g=3}^{g=G}\sum _{v=1}^{v=m_{g}}a_{gv}\ge k \end{aligned}$$

(A23)

$$\begin{aligned}&a_{12}+\sum _{v=3}^{v=m_{1}+1}a_{1v}+a_{22}+\sum _{v=3}^{v=m_{2}+1}a_{2v}+ \sum _{g=3}^{g=G}\sum _{v=1}^{v=m_{g}}a_{gv}\ge k \end{aligned}$$

(A24)

where $m_{g}:=\lfloor k_{g}d_{g}\rfloor$. Consider now the following G-dimensional binary vectors: ${\varvec{v}}=(1\ 0$$1\ldots 1)$, ${\varvec{v}} ^{\prime }=(0\ 1$$1\ldots 1).$ Since ${\varvec{v}},{\varvec{v}}^{\prime }\in \left( \rho ^{b}\right) ^{-1}(0)$, the following inequalities must hold:

$$\begin{aligned}&a_{11}+a_{12}+\sum _{v=3}^{v=m_{1}+1}a_{1v}+\sum _{v=3}^{v=m_{2}+1}a_{2v}+ \sum _{g=3}^{g=G}\sum _{v=1}^{v=m_{g}}a_{gv}<k \end{aligned}$$

(A25)

$$\begin{aligned}&a_{21}+a_{22}+\sum _{v=3}^{v=m_{1}+1}a_{1v}+\sum _{v=3}^{v=m_{2}+1}a_{2v}+ \sum _{g=3}^{g=G}\sum _{v=1}^{v=m_{g}}a_{gv}<k \end{aligned}$$

(A26)

Defining

$$\begin{aligned} k^{\prime }:=k-\left( \sum _{v=3}^{v=m_{1}+1}a_{1v}+\sum _{v=3}^{v=m_{2}+1}a_{2v}+ \sum _{g=3}^{g=G}\sum _{v=1}^{v=m_{g}}a_{gv}\right) \end{aligned}$$

(A27)

the inequalities (A21)–(A26) can be rewritten as

$$\begin{aligned} \left\{ \begin{array}{c} a_{11}+a_{21}\ge k^{\prime }, \\ a_{11}+a_{22}\ge k^{\prime }, \end{array} \begin{array}{c} a_{12}+a_{21}\ge k^{\prime }, \\ a_{12}+a_{22}\ge k^{\prime }, \end{array} \begin{array}{c} a_{11}+a_{12}<k^{\prime }, \\ a_{21}+a_{22}<k^{\prime }. \end{array} \right\} \end{aligned}$$

(A28)

Again, it is trivial to prove that the inequalities system shown in (A28) does not have feasible solutions. In the second to last inequality of the system, either $a_{11}$ or $a_{12}$ must be smaller than $k^{\prime }/2$ . The same goes for $a_{21},a_{22}$ in the last inequality of the system: at least one of them must be smaller than $k^{\prime }/2$. Picking the smallest elements between $a_{11},a_{12}$ and $a_{21},a_{22}$ and adding them up results in a number that is smaller than $k^{\prime }$, therefore contradicting at least one of the four first inequalities of the system. We have reached the contradiction we were looking for. This proves auxiliary Lemma 2. $\square$

The proof of Proposition 1 is now almost immediate. According to Auxiliary Lemma 2, ${\mathcal {C}}_{d}^{\psi }\backslash {\mathcal {W}}_{d}\ne \emptyset$ (essentially, it is only when the union or intersection approaches are used both within and across dimensions –i.e. either $k_{g}=1/d_{g}\forall g,k^{b}=1/G$ or $k_{g}=1\forall g,k^{b}=1$—that the counting approach ${\mathcal {W}}_{d}$ is able to generate some identification functions included in ${\mathcal {C}}_{d}^{\psi }$). Since COM uniquely characterizes ${\mathcal {W}} _{d}$ (see Theorem 2), ${\mathcal {C}}_{d}^{\psi }$ cannot satisfy that axiom as well. Finally, since ${\mathcal {I}}_{d}^{\psi }\supset {\mathcal {W}} _{d}^{\psi }\supset {\mathcal {C}}_{d}^{\psi }$ and ${\mathcal {C}}_{d}^{\psi }$ does not satisfy COM, neither ${\mathcal {W}}_{d}^{\psi }$ nor ${\mathcal {I}} _{d}^{\psi }$ can satisfy COM. This proves Proposition 2. $\square$

Proof of Theorem 3

It is clear that the multivariate poverty index shown in Eq. (12) satisfies SDC, MOA, NRM, FOC, SEP, HMG and CON. We are going to prove the opposite implication. Since ${\varvec{f}} =\{f_{d}\}_{d\in {\mathbb {N}} }$ satisfies SDC, it can be written as

$$\begin{aligned} f_{d}(\Gamma )=\frac{1}{n}\sum _{i=1}^{n}p(\varvec{\gamma }_{i}) \end{aligned}$$

(A29)

for some function $p:[0,1]^{d}\rightarrow {\mathbb {R}}$. Clearly, for any $\varvec{\gamma }\in [0,1]^{d},p(\varvec{\gamma } )=f_{d}([\varvec{\gamma }])$ for some deprivation matrix $[\varvec{\gamma } ]\in {\mathcal {G}}_{n\times d}^{S}$ where all rows are equal to $\varvec{ \gamma }$. Therefore, since $f_{d}$ satisfies SDC, CON, HMG, SEP and MOA, p will satisfy them too. It can be shown that MOA implies minimal increasingness and strict essentiality (see Blackorby and Donaldson 1982: 251). Moreover, the domain of p is $[0,1]^{d}$, which is connected and topologically separable. In an analogous way to Blackorby and Donaldson (1982: 252), based on Gorman (1968: 369) and Blackorby, Primont and Russel (1978: 127) it can be shown that p is additively separable and can be written as

$$\begin{aligned} p(\varvec{\gamma })=p^{*}\left( \sum _{j=1}^{d}p_{j}(\gamma _{j})\right) \end{aligned}$$

(A30)

where $p^{*}$ and $p_{j},j\in \{1,\ldots ,d\}$ are continuous real-valued functions and $p^{*}$ is increasing. By HMG, one has that

$$\begin{aligned} p(\lambda \varvec{\gamma })=\lambda p(\varvec{\gamma }) \end{aligned}$$

(A31)

for any $\lambda \in (0,1]$. Using Eq. (A30), for each $l\in \{1,\ldots ,d\}$ we can define the functions

$$\begin{aligned} h_{l}(\gamma _{l}):=p(0,\ldots ,0,\gamma _{l},0,\ldots ,0)=p^{*}\left( p_{l}(\gamma _{l})+\sum _{j\ne l}^{{}}p_{j}(0)\right) . \end{aligned}$$

(A32)

Since p is linearly homogeneous on $\varvec{\gamma }$, so are the functions $h_{l}(\gamma _{l}).$ Therefore

$$\begin{aligned} h_{l}(\lambda \gamma _{l})=\lambda h_{l}(\gamma _{l}) \end{aligned}$$

(A33)

for any $\lambda \in (0,1]$ and for all $l\in \{1,\ldots ,d\}$. As a consequence, there exist constants $c_{l}$ such that

$$\begin{aligned} h_{l}(\gamma _{l})=\gamma _{l}h_{l}(1)=c_{l}\gamma _{l} \end{aligned}$$

(A34)

Plugging Eqs. (A32) and (A34) we have that

$$\begin{aligned} c_{l}\gamma _{l}=p^{*}\left( p_{l}(\gamma _{l})+\sum _{j\ne l}^{{}}p_{j}(0)\right) . \end{aligned}$$

(A35)

Hence

$$\begin{aligned} p_{l}(\gamma _{l})=p^{*-1}\left( c_{l}\gamma _{l}\right) -\sum _{j\ne l}^{{}}p_{j}(0) \end{aligned}$$

(A36)

Substituting Eq. (A36) in Eq. (A30), one has that

$$\begin{aligned} p(\varvec{\gamma })&=p^{*}\left( \sum _{j=1}^{d}\left[ p^{*-1}\left( c_{j}\gamma _{j}\right) -\sum _{l\ne j}^{{}}p_{l}(0)\right] \right) \nonumber \\&=p^{*}\left( \sum _{j=1}^{d}p^{*-1}\left( c_{j}\gamma _{j}\right) +\varsigma \right) \end{aligned}$$

(A37)

for some constants $c_{j},\varsigma$ and a continuous increasing function $p^{*}$. Equation (A37) is essentially the same as Eq. (34) in Blackorby and Donaldson (1982: 260). Therefore, following those authors–who in turn draw from Eichhorn (1978: 32–34)—it can be proven that $p^{*-1}=:f$ must satisfy the following functional equation

$$\begin{aligned} f(\lambda u)=\alpha (\lambda )f(u)+b(\lambda ) \end{aligned}$$

(A38)

Without the domain restrictions on $\lambda$ and u, the solutions to Eq. (A38) are well-known (Aczél et al. 1986). It is straightforward to show that the solution for Eq. (A38) on the present restricted domain is

$$\begin{aligned} f(u)=\left\{ \begin{array}{c} au^{\theta }+b \\ c\ln (u)+d \end{array} \right\} \end{aligned}$$

(A39)

for some parameters $a,b,c,d,\theta$ (with $\theta \ne 0$). Since continuity of f at 0 precludes the logarithmic solution, the general solution of Eq. (A37) can be written as

$$\begin{aligned} p(\varvec{\gamma })=p^{*}\left( \sum _{j=1}^{d}a\left( c_{j}\gamma _{j}\right) ^{\theta }+db+\varsigma \right) =\left( \sum _{j=1}^{d}\left( c_{j}\gamma _{j}\right) ^{\theta }+\frac{b(d-1)+\varsigma }{a}\right) ^{1/\theta }. \end{aligned}$$

(A40)

Since p is linearly homogeneous on $\varvec{\gamma }$, one must have that $\left( b(d-1)+\varsigma \right) /a=0$. Therefore, Eq. (A30) can be rewritten as

$$\begin{aligned} p(\varvec{\gamma })=\psi \left( \left( \sum _{j=1}^{d}\left( c_{j}\gamma _{j}\right) ^{\theta }\right) ^{1/\theta }\right) \end{aligned}$$

(A41)

for some continuous increasing function $\psi$. Since p satisfies HMG one has

$$\begin{aligned} \lambda \psi \left( \left( \sum _{j=1}^{d}\left( c_{j}\gamma _{j}\right) ^{\theta }\right) ^{1/\theta }\right)= & {} \lambda p(\varvec{\gamma })=p(\lambda \varvec{\gamma })=\psi \left( \left( \sum _{j=1}^{d}\left( c_{j}\lambda \gamma _{j}\right) ^{\theta }\right) ^{1/\theta }\right) \nonumber \\= & {} \psi \left( \lambda \left( \sum _{j=1}^{d}\left( c_{j}\gamma _{j}\right) ^{\theta }\right) ^{1/\theta }\right) . \end{aligned}$$

(A42)

Equation (A42) implies that

$$\begin{aligned} \lambda \psi (x)=\psi (\lambda x) \end{aligned}$$

(A43)

for all $x\ge 0,\lambda \in (0,1]$. As a consequence, there exists a constant q such that

$$\begin{aligned} \psi (x)=x\psi (1)=qx \end{aligned}$$

(A44)

Imposing NRM and MOA and rewriting accordingly, one obtains the desired functional form. $\square$

Proof of Theorem 4

Given its similarity with Theorem 3, we will simply present a brief sketch of the proof (the complete proof is available upon request). The sufficiency part of the theorem is clear, so we focus on the reverse implication. Axioms SDC, FOC, MOA and NRM imply that $f_{d}(\Gamma )$ can be written as

$$\begin{aligned} \frac{1}{n}\sum \limits _{i\in Q(P_{\rho })}\Psi (\gamma _{i1},\ldots ,\gamma _{id}) \end{aligned}$$

(A45)

for some increasing function $\Psi :[0,1]^{d}\rightarrow [0,1]$ with $\Psi ({\varvec{1}})=1$ and $\Psi ({\varvec{0}})=0$. Imposing CON, HMG, MOA and WDS, it turns out that $\Psi (\gamma _{i1},\ldots ,\gamma _{id})=\Phi (P_{1}( \varvec{\gamma }_{1}),\ldots ,P_{G}(\varvec{\gamma }_{G}))$, where

$$\begin{aligned} P_{g}(\varvec{\gamma }_{g})=\left[ \sum _{v=1}^{d_{g}}w_{gv}(\gamma _{gv})^{\theta _{g}}\right] ^{1/\theta _{g}} \end{aligned}$$

(A46)

for some $\theta _{g}>0,w_{gv}>0$. Lastly, CON, HMG, MOA and BDS imply that

$$\begin{aligned} \Phi (P_{1},\ldots ,P_{G})=\left( \sum _{g=1}^{g=G}a_{g\cdot }P_{g}^{\theta }\right) ^{1/\theta } \end{aligned}$$

(A47)

for some $\theta>0,a_{g\cdot }>0$, so we obtain the desired functional form. $\square$

Proof of Proposition 3

Let

$$\begin{aligned} \pi _{\varvec{\theta }}:=\left( \sum _{g=1}^{g=G}a_{g\cdot }\left[ \sum _{v=1}^{d_{g}}w_{gv}(\gamma _{gv})^{\theta _{g}}\right] ^{\theta /\theta _{g}}\right) ^{1/\theta } \end{aligned}$$

(A48)

be the individual level poverty function corresponding to (16). Therefore, one has that

$$\begin{aligned} \frac{\partial \pi _{\varvec{\theta }}}{\partial \gamma _{gv}}=\left( \sum _{g=1}^{g=G}a_{g\cdot }\left[ \sum _{v=1}^{d_{g}}w_{gv}(\gamma _{gv})^{\theta _{g}}\right] ^{\theta /\theta _{g}}\right) ^{\frac{1}{\theta } -1}\left[ \sum _{v=1}^{d_{g}}w_{gv}(\gamma _{gv})^{\theta _{g}}\right] ^{ \frac{\theta }{\theta _{g}}-1}w_{gv}(\gamma _{gv})^{\theta _{g}-1} \nonumber \\ \end{aligned}$$

(A49)

After several algebraic manipulations it is easy to show that

$$\begin{aligned} \frac{\partial ^{2}\pi _{\varvec{\theta }}}{\partial \gamma _{gv}\partial \gamma _{gu}}\equiv & {} \left[ (1-\theta )\left( \sum _{v=1}^{d_{g}}w_{gv}(\gamma _{gv})^{\theta _{g}}\right) ^{\theta /\theta _{g}}\right. \nonumber \\&\left. +(\theta -\theta _{g})\left( \sum _{g=1}^{g=G}\left[ \sum _{v=1}^{d_{g}}w_{gv}(\gamma _{gv})^{\theta _{g}}\right] ^{\theta /\theta _{g}}\right) \right] \end{aligned}$$

(A50)

The last expression can be rearranged and written as follows:

$$\begin{aligned} \frac{\partial ^{2}\pi _{\varvec{\theta }}}{\partial \gamma _{gv}\partial \gamma _{gu}}\equiv & {} \left[ (1-\theta _{g})\left( \sum _{v=1}^{d_{g}}w_{gv}(\gamma _{gv})^{\theta _{g}}\right) ^{\theta /\theta _{g}}\right. \nonumber \\&\left. +(\theta -\theta _{g})\left( \sum _{h=1,h\ne g}^{h=G}\left[ \sum _{v=1}^{d_{g}}w_{hv}(\gamma _{hv})^{\theta _{h}}\right] ^{\theta /\theta _{h}}\right) \right] \end{aligned}$$

(A51)

Therefore, one can basically say that

$$\begin{aligned} \frac{\partial ^{2}\pi _{\varvec{\theta }}}{\partial \gamma _{gv}\partial \gamma _{gu}}\equiv A(1-\theta _{g})+B(\theta -\theta _{g}) \end{aligned}$$

(A52)

for some real constants $A,B>0$. Hence, whenever $\theta _{g}<\min \{1,\theta \}$, $\left( \partial ^{2}\pi _{\varvec{\theta }}\right) /\left( \partial \gamma _{gv}\partial \gamma _{gu}\right) >0$, so the attributes u, v belonging to the same dimension are complements. On the other hand, whenever $\theta _{g}>\max \{1,\theta \}$, $\left( \partial ^{2}\pi _{ \varvec{\theta }}\right) /\left( \partial \gamma _{gv}\partial \gamma _{gu}\right) <0$, so the attributes u, v belonging to the same dimension are substitutes. This proves part (i). For part (ii), we need to compute $\left( \partial ^{2}\pi _{\varvec{\theta }}\right) /\left( \partial \gamma _{gv}\partial \gamma _{hu}\right)$. After algebraic manipulations it can be shown that

$$\begin{aligned}&\frac{\partial ^{2}\pi _{\varvec{\theta }}}{\partial \gamma _{gv}\partial \gamma _{hu}}\nonumber \\&\quad \equiv \left( 1-\theta \right) \left( \sum _{g=1}^{g=G}a_{g\cdot }\left[ \sum _{v=1}^{d_{g}}w_{gv}(\gamma _{gv})^{\theta _{g}}\right] ^{\theta /\theta _{g}}\right) ^{\frac{1}{\theta } -2}\left[ \sum _{u=1}^{d_{h}}w_{hu}(\gamma _{hu})^{\theta _{g}}\right] ^{ \frac{\theta }{\theta _{h}}}w_{hu}(\gamma _{hu})^{\theta _{h}-1} \nonumber \\ \end{aligned}$$

(A53)

From the previous equation we can say that

$$\begin{aligned} \frac{\partial ^{2}\pi _{\varvec{\theta }}}{\partial \gamma _{gv}\partial \gamma _{hu}}\equiv C(1-\theta ) \end{aligned}$$

(A54)

for some real constant $C>0$. Therefore, whenever $\theta <1$, $\left( \partial ^{2}\pi _{\varvec{\theta }}\right) /\left( \partial \gamma _{gv}\partial \gamma _{hu}\right) >0$, so the attributes u, v belonging to different dimensions are complements. Analogously, when $\theta >1$, $\left( \partial ^{2}\pi _{\varvec{\theta }}\right) /\left( \partial \gamma _{gv}\partial \gamma _{hu}\right) <0$, so the attributes u, v belonging to different dimensions are substitutes. This proves part (ii). $\square$