On absolute socioeconomic health inequality comparisons

Abstract

This paper introduces a new graphical tool: the mean deviation concentration curve. Using a unified approach, we derive the associated dominance conditions that identify robust rankings of absolute socioeconomic health inequality for all indices obeying Bleichrodt and van Doorslaer’s (J Health Econ 25:945–957, 2006) principle of income-related health transfer. We also derive dominance conditions that are compatible with other transfer principles available in the literature. To make the identification of all robust orderings implementable using survey data, we discuss statistical inference for these dominance tests. To illustrate the empirical relevance of the proposed approach, we compare joint distributions of income and health-related behavior in the United States.

Appendices

A Sets of indices

Proofs are based on the following mathematical definition of the set of indices.

$\Lambda _A^2$ is the set of all absolute socioeconomic health inequality indices, $I_A(f_{Y,H})$ such that

• (a) $\nu (p)\in \Re$,

• (b) $\nu (p)$ is continuous and differentiable almost everywhere over $\left[ 0,1\right]$,

• (c) $\int _0^1\nu (p)\mathrm {d}{p}=0$,

• (d) $\nu ^{(1)}\left( p\right) > 0, \ \forall p \in [0,1]$.

$\Lambda _R^2$ is the set of all absolute socioeconomic health inequality indices, $I_R(f_{Y,H})$ such that

• (a) $\nu (p)\in \Re$,

• (b) $\nu (p)$ is continuous and differentiable almost everywhere over $\left[ 0,1\right]$,

• (c) $\int _0^1\nu (p)\mathrm {d}{p}=0$,

• (d) $\nu ^{(1)}\left( p\right) > 0, \ \forall p \in [0,1]$.

$\Lambda _{AW}^s$ is the set of all absolute socioeconomic health inequality indices, $I_A(f_{Y,H})\in \Lambda _A^2$ such that

• (a) $\nu (p)\in (-\infty ,1]$,

• (b) $\nu (p)$ is continuous and $s-1$-time differentiable almost everywhere over $\left[ 0,1\right]$,

• (c) $\nu ^{(i)}(1)=0 \ \forall i\in \{1,2,\dots ,s-1\}$,

• (d) $(-1)^{i+1}\nu ^{(i)}(p)\geqslant 0 \ \forall p\in [0,1] \ \forall i\in \{1,2,\dots ,s-1\}$.

$\Lambda _{A\rho }^s$ is the set of all absolute socioeconomic health inequality indices, $I_A(f_{Y,H})\in \Lambda _A^2$ such that

• (a) $\nu (1-p)=\nu (p) \ \forall p\in [0,1]$,

• (b) $\nu (p)$ is continuous and $s-1$-time differentiable almost everywhere over $\left[ 0,1\right]$,

• (c) $\nu ^{(i)}(0.5)=0 \ \forall i\in \{1,2,\dots ,s-1\}$,

• (d) $(-1)^{i+1}\nu ^{(i)}(p)\geqslant 0 \ \forall p\in [0,0.5] \ \forall i\in \{1,2,\dots ,s-1\}$.

B Proofs of dominance theorems

Proof of the result in Eqs. (5) and (6) Integrating by parts Eq. (2) yields

$$\begin{aligned} I_A(f_{Y,H})=\nu (p)GC(p)|_0^1-\int _0^1\nu ^{(1)}(p)GC(p)\mathrm {d}{p} \end{aligned}$$

(16)

Since by definition $GC(0)=0$ and $GC(1)=\mu _h$ for all indices $I_A(f_{Y,H})\in \Lambda ^2_A$, Eq. (16) can be rewritten as

$$\begin{aligned} I_A(f_{Y,H})=\nu (1)\mu _h-\int _0^1\nu ^{(1)}(p)GC(p)\mathrm {d}{p} \end{aligned}$$

(17)

From Eq. (17), we get

$$\begin{aligned} \Delta I_{A12}=\nu (1)(\mu _{h2}-\mu _{h1})+\int _0^1\nu ^{(1)}(p)\left[ GC_1(p)-GC_2(p)\right] \mathrm {d}{p} \end{aligned}$$

(18)

Note that $\nu ^{(1)}(p)$ is non negative. This implies that if $GC_1(p)\ge GC_2(p)$ for all $p\in [0,1]$, then $\int _0^1\nu ^{(1)}(p)\left[ GC_1(p)-GC^s_2(p)\right] \mathrm {d}{p}\ge 0$. If in addition, $\mu _{h2}\ge \mu _{h1}$, then $\Delta I_{A12}\ge 0$. This proves for sufficiency of the condition. $\square$

Proof of Theorem 1

Integrating by parts equation (??), we get

$$\begin{aligned} I_A(f_{Y,H})=\omega (p)MDC^2(p)|_0^1-\int _0^1\omega ^{(1)}(p)MDC^2(p)\mathrm {d}{p} \end{aligned}$$

(19)

Since by definition $MDC^2(0)=MDC^2(1)=0$, $I_A(f_{Y,H})\in \Lambda ^2_{A}$, the first term on the right hand side of the equation is nil. Also note that $\omega ^{(1)}(p)=\nu ^{(1)}(p)$. This yields to

$$\begin{aligned} I_A(f_{Y,H})=\int _0^1\nu ^{(1)}(p)MDC^2(p)\mathrm {d}{p}. \end{aligned}$$

(20)

Let $\Delta I_{A12}=I_A(f_{Y,H}^2)-I_A(f_{Y,H}^1)$. From equation (20), we get

$$\begin{aligned} \Delta I_{A12}=\int _0^1\nu ^{(1)}(p)\left[ MDC_2^s(p)-MDC_1^s(p)\right] \mathrm {d}{p}. \end{aligned}$$

(21)

Note that $\nu ^{(1)}(p)$ is non negative. This implies that if $MDC_2^2(p)\ge MDC_1^2(p)$ for all $p\in [0,1]$, then $\Delta I_{A12}\ge 0$. This proves for sufficiency of the condition.

Having provided a sufficiency condition let us now prove for the necessity of the condition. Consider now the set of indices $I_A(f_{Y,H})\in \Lambda ^2_{A}$ for which $\nu (p)$ takes the following form:

$$\begin{aligned} \nu (p)=\left\{ \begin{array}{cc} -\theta &{} 0\le p_c\\ \left[ p_c+\varepsilon -p\right] &{} p_c\le p \le p_c+\varepsilon \\ \varepsilon -\theta &{} p \ge p_c+\varepsilon \end{array} \right. , \end{aligned}$$

(22)

where $p_c\in [0,1]$ and $\theta$ is chosen so that $\int _0^1\nu (p)\mathrm {d}{p}=0$. Since $\nu (p)$ is differentiable almost everywhere, it satisfies the conditions in the definition of $\Lambda ^2_{A}$. Differentiating Eq. (22) yields

$$\begin{aligned} \nu ^{(1)}(p)=\left\{ \begin{array}{cc} 0 &{} 0\le p_c\\ 1 &{} p_c\le p \le p_c+\varepsilon \\ 0 &{} p \ge p_c+\varepsilon \end{array} \right. \end{aligned}$$

(23)

Imagine now that $MDC_2^2(p)<MDC_1^2(p)$ on an interval $[p_c,p_c+\varepsilon ]$ for $\varepsilon$ that can be arbitrarily close to 0. For any $\nu (p)$ obeying the relation in Eq. (23), the expression in Eq. (21) is negative. Hence it cannot be that $MDC_2^s(p)<MDC_1^s(p)$ for $p\in [p_c,p_c+\varepsilon ]$. This proves the necessity of the condition. $\square$

Proof of Theorem 2

First note that for $I_A(f_{Y,H})\in \Lambda ^s_{AW}$, equation (2) can be rewritten as

$$\begin{aligned} I_A(f_{Y,H})=\int _0^1(1-\omega (p))h(p)\mathrm {d}{p}=\int _0^1\omega (p)\widetilde{h}(p)\mathrm {d}{p} \end{aligned}$$

(24)

Integrating by parts Eq. (24), we get

$$\begin{aligned} I_A(f_{Y,H})=\omega (p)MDC^2(p)|_0^1-\int _0^1\omega ^{(1)}(p)MDC^2(p)\mathrm {d}{p} \end{aligned}$$

(25)

Since by definition $MDC^2(0)=MDC^2(1)=0$, $I_A(f_{Y,H})\in \Lambda ^2_{A}$, the first term on the right hand side of the equation is nil. This yields to

$$\begin{aligned} I_A(f_{Y,H})=-\int _0^1\omega ^{(1)}(p)MDC^2(p)\mathrm {d}{p} \end{aligned}$$

(26)

Now assume that for $s-1$, we have

$$\begin{aligned} I_A(f_{Y,H})=(-1)^{s-2}\int _0^1\omega ^{(s-2)}(p)MDC^{s-1}(p)\mathrm {d}{p}. \end{aligned}$$

(27)

Integrating by parts Eq. (27) yields

$$\begin{aligned} I_A(f_{Y,H})=(-1)^{s-2}\left\{ \omega ^{(s-2)}(p)MDC^s(p)|_0^1-\int _0^1\omega ^{(s-1)}MDC^s(p)\mathrm {d}{p} \right\} . \end{aligned}$$

(28)

Since by definition $MDC^s(0)=0$ and $\omega ^{(s-2)}(1)=0$ (because $\nu ^{(s-2)}(1)=0$) for all indices $I_A(f_{Y,H})\in \Lambda ^s_{AW}$, the first term in the braces on the right hand side of the equation is nil. This yield

$$\begin{aligned} I_A(f_{Y,H})=(-1)^{s-1}\int _0^1\omega ^{(s-1)}(p)MDC^s(p)\mathrm {d}{p}. \end{aligned}$$

(29)

Given that Eqs. (27) and (29) both conform to the relation depicted in Eq. (26), it follows that Eq. (29) holds for all $s\in \{2,3,\dots \}$. Let $\Delta I_{A12}=I_A(f_{Y,H}^2)-I_A(f_{Y,H}^1)$. From Eq. (29), we get

$$\begin{aligned} \Delta I_{A12}=(-1)^{s-1}\int _0^1\omega ^{(s-1)}(p)\left[ MDC_2^s(p)-MDC_1^s(p)\right] \mathrm {d}{p}. \end{aligned}$$

(30)

Note that $(-1)^{s-1}\omega ^{(s-1)}(p)$ is non negative. This implies that if $MDC_2^s(p)\ge MDC_1^s(p)$ for all $p\in [0,1]$, then $\Delta I_{A12}\ge 0$. This proves for sufficiency of the condition.

Having provided a sufficiency condition let us now prove for the necessity of the condition. Consider now the set of indices $I_A(f_{Y,H})\in \Lambda ^s_{AW}$ for which $\omega ^{(s-2)}(p)$ takes the following form:

$$\begin{aligned} \omega ^{(s-2)}(p)=\left\{ \begin{array}{cc} (-1)^{s-2}\varepsilon &{} 0\le p_c\\ (-1)^{s-2}\left[ p_c+\varepsilon -p\right] &{} p_c\le p \le p_c+\varepsilon \\ 0 &{} p \ge p_c+\varepsilon \end{array} \right. \end{aligned}$$

(31)

where $p_c\in [0,1]$. Since $\omega (p)$ is differentiable almost everywhere, it satisfies the conditions in the definition of $\Lambda ^s_{AW}$. Differentiating Eq. (31) yields

$$\begin{aligned} \omega ^{(s-1)}(p)=\left\{ \begin{array}{cc} 0 &{} 0\le p_c\\ (-1)^{s-1} &{} p_c\le p \le p_c+\varepsilon \\ 0 &{} p \ge p_c+\varepsilon \end{array} \right. \end{aligned}$$

(32)

Imagine now that $MDC^s(p)<MDC_1^s(p)$ on an interval $[p_c,p_c+\varepsilon ]$ for $\varepsilon$ that can be arbitrarily close to 0. For any $\omega (p)$ obeying the relation in Eq. (32), the expression in Eq. (30) is negative. Hence it cannot be that $MDC_2^s(p)<MDC_1^s(p)$ for $p\in [p_c,p_c+\varepsilon ]$. This proves the necessity of the condition. $\square$

Proof of Theorem 3

First note that for $I_A(f_{Y,H})\in \Lambda ^s_{A\rho }$, Eq. (2) can be rewritten as

$$\begin{aligned} I_A(f_{Y,H})=-\int _0^{0.5}\nu (p)r(p)\mathrm {d}{p} \end{aligned}$$

(33)

Integrating by parts Eq. (33), we get

$$\begin{aligned} I_A(f_{Y,H})=-\nu (p)GR^2(p)|_0^{0.5}+\int _0^{0.5}\nu ^{(1)}(p)GR^2(p)\mathrm {d}{p}. \end{aligned}$$

(34)

Since by definition $GR^2(0)=0$ and $\nu (0.5)=0$ for all indices $I_A(f_{Y,H})\in \Lambda ^s_{A\rho }$, the first term on the right hand side of the equation is nil. This yields to

$$\begin{aligned} I_A(f_{Y,H})=\int _0^{0.5}\nu ^{(1)}(p)GR^2(p)\mathrm {d}{p}. \end{aligned}$$

(35)

Now assume that for $s-1$, we have

$$\begin{aligned} I_A(f_{Y,H})=(-1)^{s-1}\int _0^{0.5}\nu ^{(s-2)}(p)GR^{s-1}(p)\mathrm {d}{p}. \end{aligned}$$

(36)

Integrating by parts Eq. (36) yields

$$\begin{aligned} I_A(f_{Y,H})=(-1)^{s-1}\left\{ \nu ^{(s-2)}(p)GR^s(p)|_0^{0.5}-\int _0^{0.5}\nu ^{(s-1)}GR^s(p)\mathrm {d}{p} \right\} . \end{aligned}$$

(37)

Since by definition $GR^s(0)=0$ and $\nu ^{(s-2)}(0.5)=0$ for all indices $I_A(f_{Y,H})\in \Lambda ^s_{A\rho }$, the first term in the braces on the right hand side of the equation is nil. This yield

$$\begin{aligned} I_A(f_{Y,H})=(-1)^s\int _0^{0.5}\nu ^{(s-1)}(p)GR^s(p)\mathrm {d}{p}. \end{aligned}$$

(38)

Given that Eqs. (35) and (38) both conform to the relation depicted in Eq. (36), it follows that Eq. (38) holds for all $s\in \{2,3,\dots \}$. Let $\Delta I_{A12}=I_A(f_{Y,H}^2)-I_A(f_{Y,H}^1)$. From Eq. (38), we get

$$\begin{aligned} \Delta I_{A12}=(-1)^s\int _0^{0.5}\nu ^{(s-1)}(p)\left[ GR_2^s(p)-GR_1^s(p)\right] \mathrm {d}{p}. \end{aligned}$$

(39)

Note that $(-1)^s\nu ^{(s-1)}(p)$ is non negative. This implies that if $GR_2^s(p)\ge GR_1^s(p)$ for all $p\in [0,0.5]$, then $\Delta I_{A12}\ge 0$. This proves for sufficiency of the condition.

Having provided a sufficiency condition let us now prove for the necessity of the condition. Consider now the set of indices $I_A(f_{Y,H})\in \Lambda ^s_{A\rho }$ for which $\nu ^{(s-2)}(p)$ takes the following form:

$$\begin{aligned} \nu ^{(s-2)}(p)=\left\{ \begin{array}{cc} (-1)^{s-1}\varepsilon &{} 0\le p_c\\ (-1)^{s-1}\left[ p_c+\varepsilon -p\right] &{} p_c\le p \le p_c+\varepsilon \\ 0 &{} p \ge p_c+\varepsilon \end{array} \right. \end{aligned}$$

(40)

where $p_c\in [0,0.5]$. Since $\nu (p)$ is differentiable almost everywhere, it satisfies the conditions in the definition of $\Lambda ^s_{A\rho }$. Differentiating Eq. (40) yields

$$\begin{aligned} \nu ^{(s-1)}(p)=\left\{ \begin{array}{cc} 0 &{} 0\le p_c\\ (-1)^s &{} p_c\le p \le p_c+\varepsilon \\ 0 &{} p \ge p_c+\varepsilon \end{array} \right. \end{aligned}$$

(41)

Imagine now that $GR_2^s(p)<GR_1^s(p)$ on an interval $[p_c,p_c+\varepsilon ]$ for $\varepsilon$ that can be arbitrarily close to 0. For any $\nu (p)$ obeying the relation in Eq. (40), the expression in Eq. (39) is negative. Hence it cannot be that $GR_2^s(p)<GR_1^s(p)$ for $p\in [p_c,p_c+\varepsilon ]$. This proves the necessity of the condition. $\square$

C Estimator for $MDC^s(p)$

Computation of integrals containing indicator variables involving inverse of $\widehat{F}^1_Y$. Even though $\widehat{F}_Y$ is a step function, the following standard result holds: $y_i \leqslant \widehat{F}_Y^{- 1} (p)$ if and only if $\widehat{F}_Y (y_i) \leqslant p$. In what follows, We will check the formula for the estimator by induction.

First compute

$$\begin{aligned} I_3(p)= & {} \int _0^p \mathbbm {1} (y_i \leqslant \widehat{F}_Y^{- 1} (u)) \mathrm {d}{u} \end{aligned}$$

(42)

$$\begin{aligned}= & {} \int _0^p \mathbbm {1} (\widehat{F}_Y (y_i) \leqslant u) \mathrm {d}{u}\end{aligned}$$

(43)

$$\begin{aligned}= & {} (p - \widehat{F}_Y (y_i)) \mathbbm {1} (\widehat{F}_Y (y_i) \leqslant p) \end{aligned}$$

(44)

$$\begin{aligned}= & {} \frac{(p - \widehat{F}_Y (y_i))^{3-2}}{(3-2)!} \mathbbm {1} (\widehat{F}_Y (y_i) \leqslant p). \end{aligned}$$

(45)

Then recursively compute for $s > 3$

$$\begin{aligned} I_{s} (p)= & {} \int _0^p I_{s-1} (u) \mathrm {d}{u}\end{aligned}$$

(46)

$$\begin{aligned}= & {} \int _0^p \frac{(u - \widehat{F}_Y (y_i))^{s-3}}{(s-3)!} \mathbbm {1} (\widehat{F}_Y (y_i) \leqslant u) \mathrm {d}{u}\end{aligned}$$

(47)

$$\begin{aligned}= & {} \frac{(p - \widehat{F}_Y (y_i))^{s-2}}{(s-2)!} \mathbbm {1} (\widehat{F}_Y (y_i) \leqslant p). \end{aligned}$$

(48)

$\square$

Estimator for $MDC^s(p)$. The fact that MDC(p) can be written as

$$\begin{aligned} MDC(p)=\int _0^\infty \int _0^\infty (\mu _h-h)\mathbbm {1}((y \le F_Y^{-1}(p))f_{H,Y}(h,y)\mathrm {d}{h}\mathrm {d}{y}, \end{aligned}$$

(49)

results into the simple estimator for MDC(p) from a sample $(h_i,y_i)$ for $i=1,\dots ,N$ to be given by:

$$\begin{aligned} \widehat{MDC}(p)=\frac{1}{N}\sum _{i=1}^N\left[ \overline{h} - h_i\right] \mathbbm {1}(y_i\leqslant \widehat{F}^{-1}_Y(p)), \end{aligned}$$

(50)

where $\overline{h}=N^{-1}\sum _{i=1}^N h_i$. From equations (48) and (49), we can also get the simple estimator for $MDC^s(p)$

$$\begin{aligned} \widehat{MDC}^s(p)=\frac{1}{N}\sum _{i=1}^N\left[ \overline{h}-h_i\right] \frac{\left[ p-\widehat{F}_Y^{-1}(p)\right] ^{s-2}}{\left( s-2\right) !}\mathbbm {1}(y_i\leqslant \widehat{F}^{-1}_Y(p)). \end{aligned}$$

(51)

$\square$

D Bootstrap algorithm

The bootstrap algorithm is constructed as follows. Assume that we have an i.i.d. sample of size $n_0$ from the random variable corresponding to first theoretical curve $L_0$ and and and i.i.d. sample of size $n_1$ from the random variable corresponding to the second theoretical curve $L_1$. Denote those samples by $\mathcal {S}_0$ and $\mathcal {S}_1$ respectively. Let $\widehat{L}_0$ and $\widehat{L}_1$ be the nonparametric estimators of $L_0$ and $L_1$ respectively, constructed from those two samples and let $\widehat{L}_{01} (p) = \widehat{L}_0 (p) - \widehat{L}_1 (p)$. Let

$$\begin{aligned} \widehat{\tau } = \sqrt{\frac{n_0 n_1}{n_0 + n_1}} \sup _p \widehat{L}_{01} (p) \end{aligned}$$

• 1. Repeat for $b = 1, \ldots , B$

  • (a) Draw a sample of size $n_0$ from $\mathcal {S}_0$. Compute the nonparametric estimator $\widehat{L}_{0 b}$.

  • (b) Draw a sample of size $n_1$ from $\mathcal {S}_1$. Compute the nonparametric estimator $\widehat{L}_{1 b}$.

  • (c) Compute $\widehat{L}_{01 b} (p) = \widehat{L}_{0 b} (p) - \widehat{L}_{1 b} (p)$.

  • (d) Compute $\widehat{\tau }_b = \sqrt{\frac{n_0 n_1}{n_0 + n_1}} \sup _p \left( \widehat{L}_{01 b} (p)-\widehat{L}_{01} (p)\right) .$

• 2. Using the sample $\widehat{\tau }_1, \ldots , \widehat{\tau }_B$, compute the bootstrap p-value

  $$\begin{aligned} \frac{1}{B} \sum _{b = 1}^B \mathbbm {1} (\widehat{\tau }_b > \widehat{\tau }). \end{aligned}$$