A Non-Monetary Multidimensional Poverty Analysis of Tunisia Using Generalized Sen-Shorrocks-Thon Measures

Abstract

A monetary approach cannot represent the complex and multidimensional phenomena of poverty, as it only takes economic aspects into account. Many attempts have been made to propose multidimensional approaches to poverty using basic needs or capabilities approaches. For our study we adopted a non-monetary approach, using multivariate correspondence analysis to construct a composite welfare indicator as an aggregation index of the various well-being attributes. In order to measure poverty within a composite indicator distribution we first developed new classes of poverty measures. Indeed, we developed classes of ethical generalized Sen-Shorrocks-Thon (SST) poverty measures. These are a generalization of the Shorrocks (Econometrica 63:1225–1230, 1995) poverty measure which itself is a modified version of the Sen (Econometrica 44:219–231, 1976) poverty measure. We analyzed the non-monetary poverty trend in Tunisia between 1994 and 2006 using the composite welfare indicator and the generalized SST poverty measures. In order to achieve this, we constructed confidence intervals and tested hypotheses based on the bootstrap method. We found that poverty decreased between 1994 and 2006, but that there were inequalities within the poorer population. We also found that poverty was essentially rural poverty and that it was concentrated in the North West, Center West, and South East of Tunisia.

Appendices

Appendix A: Proofs

1.1 Proof of Proposition 3

Proof 1:

$\phi \left( m \right) = \sum\nolimits_{j = 1}^{m} {\pi_{j} } ,m \in I.$

$$ \begin{aligned} \varphi \left( m \right) &= \sum\nolimits_{i = 1}^{m} {w_{i} } \\ &= w_{1} + w_{2} + {\ldots} + w_{m} \\ &= w_{m - m + 1} + w_{m - (m - 1) + 1} + {\ldots} + w_{m - 1 + 1} \\ &= \sum\nolimits_{j = 1}^{m} {w_{m - j + 1} } \\ &= \pi_{m} + \pi_{m - 1} + {\ldots} + \pi_{1} \\ &= \sum\nolimits_{j = 1}^{m} {\pi_{j} } . \\ \end{aligned} $$

Proof 2:

$\pi_{j} = \varphi \left( i \right) - \varphi \left( {i - 1} \right)$.

For i = m, j = 1; prove that $\pi_{1} = \varphi \left( m \right) - \varphi \left( {m - 1} \right).$.

By proof 1, we have $\varphi \left( m \right) = \sum\nolimits_{j = 1}^{m} {\pi_{j} } = \sum\nolimits_{j = 1}^{m} {w_{m - j + 1} } = \sum\nolimits_{i = 1}^{m} {w_{i} } .$

We have j = m−i + 1, if $i = 1 \to j = m,$if $i = m \to j = 1$and if $i = m - 1 \to j = 2$.

Then, by change of variable in the precedent equation, we obtain

$$ \varphi \left( {m - 1} \right) = \sum\nolimits_{j = 2}^{m} {\pi_{j} } = \sum\nolimits_{j = 2}^{m} {w_{m - j + 1} } = \sum\nolimits_{i = 1}^{m - 1} {w_{i} } . $$

Verification

$$ \begin{aligned} \varphi \left( {m - 1} \right) &= \sum\nolimits_{i = 1}^{m - 1} {w_{i} } \\ &= w_{1} + w_{2} + {\ldots} + w_{m - 1} \\ &= w_{m - m + 1} + w_{{m - \left( {m - 1} \right) + 1}} + {\ldots} + w_{m - 2 + 1} \\ &= \sum\nolimits_{j = 2}^{m} {w_{m - j + 1} } \\ &= \pi_{m} + \pi_{m - 1} + {\ldots} + \pi_{2} \\ &= \sum\nolimits_{j = 2}^{m} {\pi_{j} } . \\ \end{aligned} $$

.

Then,

$$ \begin{aligned} \pi_{1} &= \left( {\pi_{1} + \pi_{2} + {\ldots} + \pi_{m} } \right) - \left( {\pi_{2} + \pi_{3} + {\ldots} + \pi_{m} } \right) \\ &= \sum\nolimits_{j = 1}^{m} {\pi_{j} } - \sum\nolimits_{j = 2}^{m} {\pi_{j} } \\ &= \varphi \left( m \right) - \varphi \left( {m - 1} \right). \\ \end{aligned} $$

.

Gradually, we obtain $\pi_{j} = \varphi \left( i \right) - \varphi \left( {i - 1} \right).$

Proof 3:

$\pi_{j} = \varphi \left( {m - j + 1} \right) - \varphi \left( {m - j} \right).$

We have, j = m−i + 1, then by change of variable, we obtain:

$\pi_j = \phi{i}-\phi{i-1} = \phi{m-j+1}-\phi{m-j}.$

1.2 Proof of Proposition 4

Similar to the proof of Proposition 1, we just have to replace i by n−j + 1.

$$ \begin{aligned} \pi_{j} &= \phi \left( i \right) - \phi \left( {i - 1} \right) \\ &= \phi \left( {m - j + 1} \right) - \phi \left( {m - j} \right) \\ \end{aligned} $$

1.3 Proof of Proposition 5

We have $\pi_{j} = w_{n - j + 1}$.

Then $w_{1} \le w_{2} \le {\ldots} \le w_{n} \Leftrightarrow \pi_{1} \ge \pi_{2} \ge {\ldots} \ge \pi_{n}$.

So the function $\varphi$ with $\pi_{1} \ge \pi_{2} \ge {\ldots} \ge \pi_{n}$.

satisfies the equation $\varphi (kn) = \varphi (k)\varphi (n),$

if and only if $\varphi (n) = n^{\nu }$.

1.4 Proof of lemma 1

I/$\sum\nolimits_{i = 1}^{n} {\left[ {i^{\nu } - \left( {i - 1} \right)^{\nu } } \right]} = \sum\nolimits_{j = 1}^{n} {\left[ {\left( {n - j + 1} \right)^{\nu } - \left( {n - j} \right)^{\nu } } \right]} = n^{\nu }.$

Let, $k = i - 1 \Rightarrow i = 1 \to k = 0$ and $i = n \to k = n - 1$.

Proof 1:

$\sum\nolimits_{i = 1}^{n} {\left[ {i^{\nu } - \left( {i - 1} \right)^{\nu } } \right]} = n^{\nu } .$

$$ \begin{aligned} \sum\nolimits_{i = 1}^{n} {\left[ {i^{\nu } - \left( {i - 1} \right)^{\nu } } \right]} &= \sum\nolimits_{i = 1}^{n} {i^{\nu } - \sum\nolimits_{i = 1}^{n} {\left( {i - 1} \right)^{\nu } } } \\ &= \sum\nolimits_{i = 1}^{n} {i^{\nu } - \sum\nolimits_{k = 0}^{n - 1} {k^{\nu } } } \\ &= \sum\nolimits_{i = 1}^{n} {i^{\nu } - \left[ {\sum\nolimits_{k = 1}^{n} {k^{\nu } + 0^{\nu } - n^{\nu } } } \right]} \\ &= \sum\nolimits_{i = 1}^{n} {i^{\nu } - } \sum\nolimits_{i = 1}^{n} {i^{\nu } + n^{\nu } } \\ &= n^{\nu } . \\ \end{aligned} $$

Proof 2:

$\sum\nolimits_{j = 1}^{n} {\left[ {\left( {n - j + 1} \right)^{\nu } - \left( {n - j} \right)^{\nu } } \right]} = n^{\nu } .$

We have j = n−i + 1, then $\sum\nolimits_{j = 1}^{n} {\left[ {\left( {n - j + 1} \right)^{\nu } - \left( {n - j} \right)^{\nu } } \right]} = \sum\nolimits_{i = 1}^{n} {\left[ {i^{\nu } - \left( {i - 1} \right)^{\nu } } \right]} .$

By proof 1, we have $\sum\nolimits_{i = 1}^{n} {\left[ {i^{\nu } - \left( {i - 1} \right)^{\nu } } \right]} = n^{\nu }.$

Then $\sum\nolimits_{j = 1}^{n} {\left[ {\left( {n - j + 1} \right)^{\nu } - \left( {n - j} \right)^{\nu } } \right]} = n^{\nu }.$

II/$\sum\nolimits_{i = 1}^{n} {\left[ {i^{\nu } - \left( {i - 1} \right)^{\nu } } \right]} \tilde{x}_{i}^{*} = \sum\nolimits_{j = 1}^{n} {\left[ {\left( {n - j + 1} \right)^{\nu } - \left( {n - j} \right)^{\nu } } \right]\hat{x}_{j}^{*} }.$

We have j = n−i + 1, then $i^{\nu } - \left( {i - 1} \right)^{\nu } = \left( {n - j + 1} \right)^{\nu } - \left( {n - j} \right)^{\nu } .$

And we have $\sum\nolimits_{i = 1}^{n} {\tilde{x}_{i}^{*} } = \sum\nolimits_{j = 1}^{n} {\hat{x}_{j}^{*} } .$ $\Rightarrow$ $\sum\nolimits_{i = 1}^{n} {\left[ {i^{\nu } - \left( {i - 1} \right)^{\nu } } \right]} \tilde{x}_{i}^{*} = \sum\nolimits_{j = 1}^{n} {\left[ {i^{\nu } - \left( {i - 1} \right)^{\nu } } \right]} \hat{x}_{j}^{*} .$ $\Rightarrow$ $\sum\nolimits_{i = 1}^{n} {\left[ {i^{\nu } - \left( {i - 1} \right)^{\nu } } \right]} \tilde{x}_{i}^{*} = \sum\nolimits_{j = 1}^{n} {\left[ {\left( {n - j + 1} \right)^{\nu } - \left( {n - j} \right)^{\nu } } \right]} \hat{x}_{j}^{*} .$

Dividing by $n^{\nu }$, we obtain $\Upxi_{S - SST}^{n} \left( {\tilde{x}^{*} } \right) = \hat{\Upxi }_{S - SST}^{n} \left( {\hat{x}^{*} } \right).$

1.5 Proof of Proposition 6

Proof 1:

$S - SST^{R} \left( {\underline{x}^{*} ;z} \right) = \frac{{\left[ {z - \hat{\Upxi }_{{G^{\nu } }} \left( {\underline{{\hat{x}}}^{*} } \right)} \right]}}{z}.$

By Eq. (6.41), we have

$$ \begin{aligned} S - SST^{R} \left( {\underline{x}^{*} ;z} \right) &= S - S\hat{S}T^{R} \left( {\underline{x}^{*} ;z} \right) \\ &= 1 - \frac{{\sum\nolimits_{j = 1}^{n} {\left[ {\left( {n - j + 1} \right)^{\nu } - \left( {n - j} \right)^{\nu } } \right]} \hat{x}_{j}^{*} }}{{n^{\nu } z}} \\ &= SST^{\nu } \left( {\underline{x}^{*} ;z} \right) \\ &= 1 - \frac{{\hat{\Upxi }_{S - SST}^{n} \left( {\underline{{\hat{x}}}^{*} } \right)}}{z} \\ &= \frac{{z - \hat{\Upxi }_{S - SST}^{n} \left( {\underline{{\hat{x}}}^{*} } \right)}}{z} \\ &= \frac{{z - \hat{\Upxi }_{{G^{\nu } }} \left( {\underline{{\hat{x}}}^{*} } \right)}}{z}. \\ \end{aligned} $$

.

Proof 2:

$S - SST^{R} \left( {\underline{x}^{*} ;z} \right) = \hat{\Upxi }_{{G^{\nu } }} \left( {\underline{y}^{*} } \right).$

We have $\sum\nolimits_{j = 1}^{n} {\left[ {\left( {n - j + 1} \right)^{\nu } - \left( {n - j} \right)^{\nu } } \right]} = n^{\nu } .$.

Hence,

$$ \begin{aligned} S - SST^{R} \left( {\underline{x}^{*} ;z} \right) & = S - S\hat{S}T^{R} \left( {\underline{x}^{*} ;z} \right) \\ &= 1 - \frac{{\sum\nolimits_{j = 1}^{n} {\left[ {\left( {n - j + 1} \right)^{\nu } - \left( {n - j} \right)^{\nu } } \right]} \hat{x}_{j}^{*} }}{{n^{\nu } z}} \\ &= \frac{{n^{\nu } z - \sum\nolimits_{j = 1}^{n} {\left[ {\left( {n - j + 1} \right)^{\nu } - \left( {n - j} \right)^{\nu } } \right]} \hat{x}_{j}^{*} }}{{n^{\nu } z}} \\ &= \frac{{\sum\nolimits_{j = 1}^{n} {\left[ {\left( {n - j + 1} \right)^{\nu } - \left( {n - j} \right)^{\nu } } \right]} }}{{n^{\nu } }}\left( {\frac{{z - \hat{x}_{j}^{*} }}{z}} \right) \\ &= \frac{{\sum\nolimits_{j = 1}^{n} {\left[ {\left( {n - j + 1} \right)^{\nu } - \left( {n - j} \right)^{\nu } } \right]} }}{{n^{\nu } }}y_{j}^{*} \\ &= \hat{\Upxi }_{{G^{\nu } }} \left( {\underline{y}^{*} } \right). \\ \end{aligned} $$

where $y_{j}^{*} = \frac{{z - \hat{x}_{j}^{*} }}{z}$ so $\hat{x}_{j}^{*} \uparrow \Rightarrow y_{j}^{*} \downarrow$.

1.6 Proof of Lemma 2

Similar to the proof of Lemma 1, we just have to replace $i^{\nu } - (i - 1)^{\nu }$ by $w_{i}$ and $\left( {n - j + 1} \right)^{\nu } - \left( {n - j} \right)^{\nu }$ by $\pi_{j}$.

1.7 Proof of Proposition 7

Similar to the proof of Proposition 6, we just have to replace $i^{\nu } - \left( {i - 1} \right)^{\nu }$by $w_{i}$ and $\left( {n - j + 1} \right)^{\nu } - \left( {n - j} \right)^{\nu }$ by $\pi_{j}$.

Appendix B: Multivariate Correspondence Analysis

Table B.1–B.4

Appendix C: Analysis of Poverty in Tunisia

Table C.1–C.8