On intertemporal poverty measures: the role of affluence and want

Abstract

This paper proposes classes of intertemporal poverty measures which take into account both the debilitating impact of prolonged spells in poverty and the mitigating effect of periods of affluence on subsequent poverty. The weight assigned to the level of poverty in each time period depends on the length of the preceding spell of poverty or of non-poverty. The proposed classes of intertemporal poverty measures are quite general and allow for a range of possible judgements as to the overall impact on a poor period of preceding spells of poverty or affluence. We discuss the properties of the proposed classes of measures and axiomatically characterize these measures.

Appendix: Proofs

Proof of Proposition 1:

We concentrate on the “only if” part of the proof, as it is immediate to verify that $P_{R}$ satisfies the axioms stated in Proposition 1.

Suppose that $P$ satisfies single period equivalence (Axiom 1), time decomposability (Axiom ), normalization (Axiom 3), constant-relative poverty mitigation (Axiom ) and relative poverty intensification (Axiom ). We need to show that for any time period $T\in \mathbb N$ and any poverty profile $\mathbf p \in [0,1]^{T}$ we have $P( \mathbf p )=P_{R}(\mathbf p )$ to for an exogenously determined $\alpha , \beta \ge 0$ and $\theta \ge 1$. So, take any $T\in \mathbb N$ and any poverty profile $\mathbf p \in [0,1]^{T}$.

Suppose $T=1$. In this case single period equivalence holds and $P(\mathbf p )=P(p)=P_{R}(p)$ for some $p\in (0,1]$. Hence, $P(\mathbf p )=P_{R}( \mathbf p )$ follows.

Assume now that $T>1$. If $\mathbf p =\mathbf 0 ^{T}$, then by normalization we obtain $P(\mathbf p )=P(\mathbf 0 ^{T})=0=P_{R}(\mathbf 0 )$. Thus, $P( \mathbf p )=P_{R}(\mathbf p )$ follows.

Next we proceed by induction on the number of poor periods, when $\mathbf p\ne 0 ^{T}$. Consider first the case where there is exactly one poor period. Recall that we represent a $T$-period poverty profile $\mathbf p= (p_{1},\ldots ,p_{s},\ldots p_{T})$, where $\forall t\ne s,$ $p_{t}=0$ and $p_{s}=1$ as $\mathbf e _{s}^{T}$. Thus $\mathbf p= p\cdot \mathbf e _{s}^{T}$ would stand for a $T$-period poverty profile $\mathbf p= (p_{1},\ldots ,p_{s},\ldots p_{T})$, where $\forall t\ne s,$ $p_{t}=0$ and $p_{s}=p$.

Without loss of generality, we can write $\mathbf p =p\cdot \mathbf e _{t}^{T}$ where $p>0$ and $t\in \left\{ 1,\ldots ,T\right\}$. Thus $P( \mathbf p )=P(p\cdot \mathbf e _{t}^{T})$. Then

$$\begin{aligned} P(p\cdot \mathbf e _{t}^{T})&= \frac{t}{T}P(p\cdot \mathbf e _{t}^{t})+ \frac{T-t}{T}P(\mathbf 0 ^{T-t})\text{,} \text{ by} \text{ time} \text{ decomposability,} \nonumber \\&= \frac{t}{T}P(p\cdot \mathbf e _{t}^{t})\text{,} \text{ by} \text{ normalization,} \nonumber \\&= \frac{t}{T}\frac{P(p\cdot \mathbf e _{1}^{t})}{t^{\beta }}\text{,} \text{ by} \text{ constant-relative} \text{ poverty} \text{ mitigation,} \nonumber \\&= \frac{1}{Tt^{\beta -1}}\cdot \left[\frac{1}{t}P(p)+\frac{t-1}{t}P( \mathbf 0 ^{t-1})\right]\text{,} \text{ by} \text{ time} \text{ decomposability,} \nonumber \\&= \frac{P(p)}{Tt^{\beta }}\text{,} \text{ by} \text{ normalization,} \nonumber \\&= \frac{p^{\theta }}{Tt^{\beta }}, \text{ by} \text{ single} \text{ period} \text{ equivalence.} \end{aligned}$$

(2)

Since $k_{t}=1$, $t=(1+n_{t})$ and $\forall$ $i\ne t$, $p_{i}=0$ we can write (2) as

$$\begin{aligned} P(\mathbf p )=P(p\cdot \mathbf e _{t}^{T})=\frac{1}{T}\sum \limits _{i=1}^{T} \frac{k_{t}^{\alpha }p_{i}^{\theta }}{(1+n_{t})^{\beta }}=P_{R}(\mathbf p ). \end{aligned}$$

Thus we obtain $P(\mathbf p )=P_{R}(\mathbf p )$.

Suppose now that $P(\hat{\mathbf{p }})=P_{R}(\hat{\mathbf{p }})$ whenever $\hat{\mathbf{p }}$ contains $m$ poor periods, for some $m\in \left\{ 1,\ldots ,T-1\right\}$. Let $\mathbf p \in [0,1]^{T}$ be any poverty profile, such that the number of poor periods is $m+1$. Let $t\in \left\{ 2,\ldots ,T\right\}$ be such that the final poor period is period $t$. Thus $t=\max \{s:2\le s\le T,p_{s}>0\}$. From time decomposability we derive

$$\begin{aligned} P(\mathbf p )&= P(p_{1},\ldots ,p_{t},\ldots ,p_{T}) \\&= \frac{t}{T}P(p_{1},\ldots ,p_{t})+\frac{T-t}{T}P(p_{t+1},\ldots ,p_{T}). \end{aligned}$$

Now $t$ being the final poor period means that by normalization we get

$$\begin{aligned} P(\mathbf p )=\frac{t}{T}P(p_{1},\ldots ,p_{t}). \end{aligned}$$

(3)

Let $s\ne t$ be maximal with $p_{s}>0$. So $s$ is the last poor period prior to $t$. Suppose $s\ne t-1$. Then by time decomposability we obtain

$$\begin{aligned} P(p_{1},\ldots ,p_{s},\ldots ,p_{t})=\frac{s}{t}P(p_{1},\ldots ,p_{s})+\frac{ t-s}{t}P(p_{s+1},\ldots ,p_{t}). \end{aligned}$$

(4)

Further, $P(p_{s+1},\ldots ,p_{t})=P(p_{t}\cdot \mathbf e _{t-s}^{t-s})$ since $p_{i}=0$ for all $i\in \left\{ s+1,\ldots ,t-1\right\}$. Applying single period equivalence, time decomposability, normalization and constant-relative poverty mitigation and noting that $k_{t}=1$ and $n_{t}=t-s-1$, we obtain

$$\begin{aligned} P(p_{s+1},\ldots ,p_{t})=\frac{k_{t}^{\alpha }}{(1+n_{t})^{1+\beta }} p_{t}^{\theta }. \end{aligned}$$

(5)

Now consider the case when $s=t-1$. Then using relative poverty intensification we get

$$\begin{aligned} P(p_{1},\ldots ,p_{s},\ldots ,p_{t})=P(p_{1},\ldots ,p_{s},0)+k_{t}^{\alpha }P(p_{t}\mathbf e _{1}^{t}). \end{aligned}$$

(6)

Using time decomposability, single period equivalence and noting that $n_{t}=0$, (6) can be written as

$$\begin{aligned} P(p_{1},\ldots ,p_{s},\ldots ,p_{t})=\frac{s}{t}P(p_{1},\ldots ,p_{s})+\frac{ k_{t}^{\alpha }}{(1+n_{t})^{\beta }}\frac{p_{t}^{\theta }}{t} \end{aligned}$$

(7)

Now $(p_{1},\ldots ,p_{s})$ contains $m$ poor periods. Thus, by the induction hypothesis, we have

$$\begin{aligned} P(p_{1},\ldots ,p_{s})=P_{R}(p_{1},\ldots ,p_{s})=\frac{1}{s} \sum \limits _{i=1}^{s}w_{i}p_{i}^{\theta } \text{ where} w_{i}=\frac{ k_{i}^{\alpha }}{(1+n_{i})^{\beta }}. \end{aligned}$$

(8)

Substituting (8) and (5) into (4) (for the $s\ne t-1$ case) or substituting (8) into (7) (for the $s=t-1$ case) yields

$$\begin{aligned} P(p_{1},\ldots ,p_{s},\ldots ,p_{t})=\left[\frac{1}{t}\sum \limits _{i=1}^{s} \frac{k_{i}^{\alpha }p_{i}^{\theta }}{(1+n_{i})^{\beta }}\right] +\frac{ k_{t}^{\alpha }}{(1+n_{t})^{\beta }}\frac{p_{t}^{\theta }}{t}. \end{aligned}$$

(9)

Further, substituting (9) into (3) we obtain

$$\begin{aligned} P(\mathbf p )=\frac{1}{T}\left[ \left( \sum \limits _{i=1}^{s}\frac{ p_{i}^{\theta }}{(1+n_{i})^{\beta }}\right) +\frac{k_{t}^{\alpha }p_{t}^{\theta }}{(1+n_{t})^{\beta }}\right]. \end{aligned}$$

Finally, since $p_{i}=0$ for all $i\in \left\{ s+1,\ldots ,t-1\right\}$ and all $i\in \left\{ t+1,\ldots ,T\right\}$, we have

$$\begin{aligned} P(\mathbf p )=\frac{1}{T}\sum \limits _{i=1}^{T}\frac{k_{i}^{\alpha }p_{i}^{\theta }}{(1+n_{i})^{\beta }}=P_{R}(\mathbf p ). \end{aligned}$$

This concludes the proof for the case of $m+1$ poor periods, and by induction it follows that $P(\mathbf p )\,{=}\,P_{R}(\mathbf p )$ for any poverty profile $\mathbf p$. This completes the proof of Proposition .$\square$

Proof of Proposition 2

We demonstrate that axioms single period equivalence (Axiom 1), time decomposability (Axiom ), normalization (Axiom ), absolute poverty mitigation (Axiom ), constant-relative poverty mitigation (Axiom 4) and relative poverty intensification (Axiom 5) are independent by presenting a separate poverty measure that satisfies all the axioms except one. We do this one at a time for each of the five axioms.

Consider a poverty profile $\mathbf p \in [0,1]^{T}$. Then the following measure violates single period equivalence (Axiom 1) but satisfies the other axioms.

$$\begin{aligned} P_{1}(\mathbf p )=\frac{1}{T}\sum \limits _{i=1}^{T}\frac{k_{i}^{\alpha }2p_{i}}{(1+n_{i})^{\beta }}. \end{aligned}$$

The next measure violates time decomposability (Axiom ) but satisfies the other axioms.

$$\begin{aligned} P_{2}(\mathbf p )=\sum \limits _{i=1}^{T}\frac{k_{i}^{\alpha }p_{i}^{\theta } }{(1+n_{i})^{\beta }}. \end{aligned}$$

A measure which violates normalization (Axiom ) but satisfies the other axioms is given by

$$\begin{aligned} P_{3}(\mathbf p )=\left\{ \begin{array}{ll} 10&\mathrm{if} \mathbf p\in 0 ^{T} \\ \frac{1}{T}\sum \limits _{i=1}^{T}\frac{k_{i}^{\alpha }p_{i}^{\theta . }}{ (1+n_{i})^{\beta }}&\mathrm{otherwise}. \end{array} \right. \end{aligned}$$

A measure which violates constant-relative poverty mitigation (Axiom 4) but satisfies the others is

$$\begin{aligned} P_{4}(\mathbf p )=\frac{1}{T}\sum \limits _{i=1}^{T}\frac{k_{i}^{\alpha }p_{i}^{\theta }}{\ln (1+n_{i})}. \end{aligned}$$

A measure which violates relative poverty intensification (Axiom 5) and satisfies the rest is as follows.

$$\begin{aligned} P_{5}(\mathbf p )=\frac{1}{T}\sum \limits _{i=1}^{T}\frac{p_{i}^{\theta }}{ (1+n_{i})^{\beta }}. \end{aligned}$$

$\square$

Proof of Proposition 3

The proof is similar to that of Proposition and is omitted. $\square$

Proof of Proposition 4

We concentrate on the “only if” part of the proof, as it is immediate to verify that $P_{A}$ satisfies the axioms stated in Proposition 4.

Suppose that $P$ satisfies single period equivalence (Axiom 1), time decomposability (Axiom ), normalization (Axiom ), relative poverty intensification (Axiom ), absolute poverty mitigation (Axiom 7), and monotonic poverty mitigation (Axiom 8). We need to show that for any time period $T\in \mathbb N$ and any poverty profile $\mathbf p \in [0,1]^{T}$ we have $P( \mathbf p )=P_{A}(\mathbf p )$ for a monotonically increasing function $f: \mathbb R _{+}\rightarrow \mathbb R _{+}$ such that $f(0)=0$ and $f(n_{t}+1)\ge f(n_{t})$.

Take any $T\in \mathbb N$ and any poverty profile $\mathbf p \in [0,1]^{T}$. Suppose $T=1$. Note that $k_{1}=1$ and $n_{1}=0$. By absolute poverty mitigation, we have

$$\begin{aligned} P(p\mathbf e _{1}^{1})=\max \left( P(p\mathbf e _{1}^{1})-h(1),0\right) . \end{aligned}$$

(10)

By construction let $f(n_{t})=th(t)$, $t=n_{t}+1$. Given $h(t)\in \mathbb R _{+}$ and $t\in \left\{ 1,\ldots ,T\right\} , f: \mathbb Z _{+}\rightarrow \mathbb R _{+}$. When $t=1$, it implies $f(0)=h(1)=0$. Due to single period equivalence and $k_{1}=1$ we can write Eq. (10) as

$$\begin{aligned} P(\mathbf p )=\max \left( k_{1}^{\alpha }p^{\theta },0\right) =P_{A}(\mathbf p ). \end{aligned}$$

Assume now that $T>1$. If $\mathbf p =\mathbf 0 ^{T}$, then by normalization we obtain $P(\mathbf p )=P(\mathbf 0 ^{T})=0=P_{A}(\mathbf p )$ . Thus, $P(\mathbf p )=P_{A}(\mathbf p )$.

Next, we proceed by induction on the number of poor periods when $\mathbf p\ne 0 ^{T}$. Consider the case where there is exactly one poor period. Without loss of generality, we can write $\mathbf p =p\cdot \mathbf e _{t}^{T}$ where $p\in (0,1]$ and $t\in \left\{ 1,\ldots ,T\right\}$. Thus $P(\mathbf p )=P(p\cdot \mathbf e _{t}^{T})$. If $t=1$ then, applying single period equivalence, time decomposability and normalization, and noting that $k_{1}=1$ and $f(0)=0$, yields

$$\begin{aligned} P(p\cdot \mathbf e _{1}^{T})&= \frac{1}{T}p^{\theta }, \\&= \frac{1}{T}\max \left( k_{1}^{\alpha }p^{\theta }-f(0),0\right). \end{aligned}$$

For $t>1$ applying time decomposability and normalization, we obtain

$$\begin{aligned} P(p\cdot \mathbf e _{t}^{T})&= \frac{t}{T}P(p\cdot \mathbf e _{t}^{t}). \nonumber \\&= \frac{t}{T}\max \left( P(p\cdot \mathbf e _{1}^{t})-h(t),0\right) ,\text{ by} \text{ absolute} \text{ poverty} \text{ mitigation,} \nonumber \\&= \frac{1}{T}\max \left( t(P(p\cdot \mathbf e _{1}^{t})-h(t)),0\right) \nonumber \\&= \frac{1}{T}\max \left( t\left(\frac{1}{t}P(p)-h(t)\right),0\right) , \text{ by} \text{ time} \text{ decomposability,} \nonumber \\&= \frac{1}{T}\max \left( P(p)-f(n_{t}),0\right) , \text{ where} f(n_{t})=th(t) \text{ by} \text{ construction.} \nonumber \\ \end{aligned}$$

(11)

To show that $f(n_{t})$ is monotonic, consider another profile $P(p\cdot \mathbf e _{t-1}^{T})$. Applying (11) we get

$$\begin{aligned} P\left(p\cdot \mathbf e _{t-1}^{T}\right)=\frac{1}{T}\max \left(P(p)-f(n_{t-1}),0\right) . \end{aligned}$$

(12)

By monotonic poverty mitigation it must be the case that $P(p\cdot \mathbf e _{t-1}^{T})\ge P(p\cdot \mathbf e _{t}^{T})$. Comparing (11) and (12) and noting that $n_{t}=n_{t-1}+1$, we can show $f(n_{t})\ge f(n_{t-1})$. Applying single period equivalence in (11) and noting that $k_{t}=1$, we can show

$$\begin{aligned} P\left(p\cdot \mathbf e _{t}^{T}\right)=\frac{1}{T}\max \left( k_{t}^{\alpha }p^{\theta }-f(n_{t}),0\right). \end{aligned}$$

(13)

Since $p_{i}=0$ for all $i\ne t$, we can write (13) as

$$\begin{aligned} P(\mathbf p )=P\left(p\cdot \mathbf e _{t}^{T}\right)=\frac{1}{T}\underset{p_{i}\ne 0}{ \sum \limits _{i=1}^{T}}\max \left( k_{i}^{\alpha }p_{i}^{\theta }-f(n_{i}),0\right) =P_{A}(\mathbf p ). \end{aligned}$$

(14)

Suppose now that $P(\hat{\mathbf{p }})=P_{A}(\hat{\mathbf{p }})$ whenever $\hat{\mathbf{p }}$ contains $m$ poor periods, for some $m\in \left\{ 1,\ldots ,T-1\right\}$. Let $\mathbf p \in [0,1]^{T}$ be any poverty profile, such that the number of poor periods is $m+1$. Let $t\in \left\{ 2,\ldots ,T\right\}$ be such that the final poor period is period $t$. Thus $t=\max \{s:2\le s\le T,p_{s}>0\}$. From time decomposability we derive

$$\begin{aligned} P(\mathbf p )&= P(p_{1},\ldots ,p_{t},\ldots ,p_{T}), \\&= \frac{t}{T}P(p_{1},\ldots ,p_{t})+\frac{T-t}{T}P(p_{t+1},\ldots ,p_{T}). \end{aligned}$$

Now $t$ being the final poor period means that by normalization we get

$$\begin{aligned} P(\mathbf p )=\frac{t}{T}P(p_{1},\ldots ,p_{t}). \end{aligned}$$

(15)

Let $s\ne t$ be maximal with $p_{s}>0$. So $s$ is the last poor period prior to $t$. Suppose $s\ne t-1$. Then by time decomposability we obtain

$$\begin{aligned} P(p_{1},\ldots ,p_{s},\ldots ,p_{t})=\frac{s}{t}P(p_{1},\ldots ,p_{s})+\frac{ t-s}{t}P(p_{s+1},\ldots ,p_{t}). \end{aligned}$$

(16)

Further, $P(p_{s+1},\ldots ,p_{t})=P(p_{t}\cdot \mathbf e _{t-s}^{t-s})$ since $p_{i}=0$ for all $i\in \left\{ s+1,\ldots ,t-1\right\}$. Using (14) and noting that $k_{t}=1$ and $n_{t}=t-s-1$, we obtain

$$\begin{aligned} P(p_{s+1},\ldots ,p_{t})=\frac{1}{t-s}\underset{p_{t}\ne 0}{ \sum \limits _{i=s+1}^{t}}\max \left( k_{i}^{\alpha }p_{i}^{\theta }-f(n_{i}),0\right) . \end{aligned}$$

(17)

Substituting (17) in (16) we get

$$\begin{aligned} P(p_{1},\ldots ,p_{s},\ldots ,p_{t})=\frac{s}{t}P(p_{1},\ldots ,p_{s})+\frac{ 1}{t}\underset{p_{t}\ne 0}{\sum \limits _{i=s+1}^{t}}\max \left( k_{i}^{\alpha }p_{i}^{\theta }-f(n_{i}),0\right) . \end{aligned}$$

(18)

Now consider the case when $s=t-1$. Using relative poverty intensification we get

$$\begin{aligned} P(p_{1},\ldots ,p_{s},p_{t})=P(p_{1},\ldots ,p_{s},0)+k_{t}^{\alpha }P(p_{t}\cdot \mathbf e _{1}^{t}). \end{aligned}$$

(19)

Applying time decomposability, single period equivalence and using (11 ) we can write (19) as

$$\begin{aligned} P(p_{1},\ldots ,p_{s},p_{t})&= \frac{s}{t}P(p_{1},\ldots ,p_{s})+\frac{ k_{t}^{\alpha }}{t}\max \left(p_{t}^{\theta }-f(n_{1}),0\right), \nonumber \\&= \frac{s}{t}P(p_{1},\ldots ,p_{s})+\frac{1}{t}\max \left(k_{t}^{\alpha }p_{t}^{\theta }-f(n_{1}),0\right). \end{aligned}$$

(20)

Note that in (20) $f(n_{1})=0$, since $n_{1}=0$ by definition. Now $(p_{1},\ldots ,p_{s})$ contains $m$ poor periods. By the induction hypothesis, we have

$$\begin{aligned} P(p_{1},\ldots ,p_{s})=P_{R}(p_{1},\ldots ,p_{s})=\frac{1}{s}\underset{ p_{i}\ne 0}{\sum \limits _{i=1}^{s}}\max \left( k_{i}^{\alpha }p_{i}^{\theta }-f(n_{i}),0\right) . \end{aligned}$$

(21)

Substituting (21) into (18) (for the $s\ne t-1$ case) or substituting (21) into (20) (for the $s=t-1$ case) yields

$$\begin{aligned} P(p_{1},\ldots ,p_{s},\ldots ,p_{t})&= \frac{1}{t}\underset{p_{i}\ne 0}{ \sum \limits _{i=1}^{s}}\max \left( k_{i}^{\alpha }p_{i}^{\theta }-f(n_{i}),0\right) \nonumber \\ \quad&\quad +&\frac{1}{t}\underset{p_{i}\ne 0}{\sum \limits _{i=s+1}^{t}}\max \left(k_{i}^{\alpha }p_{i}^{\theta }-f(n_{i}),0\right). \end{aligned}$$

(22)

Further, substituting (22) into (15) we obtain

$$\begin{aligned} P(\mathbf p )=\frac{1}{T}\left[ \underset{p_{i}\ne 0}{\sum \limits _{i=1}^{s} }\max \left( k_{i}^{\alpha }p_{i}^{\theta }-f(n_{i}),0\right) +\underset{ p_{i}\ne 0}{\sum \limits _{i=s+1}^{t}}\max \left(k_{i}^{\alpha }p_{i}^{\theta }-f(n_{i}),0\right)\right] . \end{aligned}$$

Finally, since $p_{i}=0$ for all $i\in \left\{ t+1,\ldots ,T\right\}$, we have

$$\begin{aligned} P(\mathbf p )=\frac{1}{T}\underset{p_{t}\ne 0}{\sum \limits _{t=1}^{T}}\max \left( k_{t}^{\alpha }p_{t}^{\theta }-f(n_{t}),0\right) =P_{A}(\mathbf p ). \end{aligned}$$

This concludes the proof for the case of $m+1$ poor periods, and by induction it follows that $P(\mathbf p )=P_{A}(\mathbf p )$ for any poverty profile $\mathbf p$. It therefore follows that $P=P_{A}$, which completes the proof of Proposition .$\square$