Almost Lorenz dominance

Abstract

This paper extends Leshno and Levy’s (Manag Sci 48:1074–1085, 2002) approach of “almost stochastic dominance” to inequality orderings. We define and characterize the notion of “almost Lorenz dominance” (ALD) and illustrate it with the US income data. An income distribution is said to “almost” Lorenz dominate another distribution when its Lorenz curve lies almost everywhere but not entirely above the other Lorenz curve. We show that this condition is equivalent to requiring that “almost” all Gini-type inequality measures rank the former distribution to have less inequality than the latter distribution; the condition on the Gini-type inequality measures has a clear interpretation and is easy to apply. We further define an almost composite transfer (ACT) and show that ALD amounts to a sequential application of such transfers. The empirical illustration with the US income data (1967–1986) demonstrates the utility of this generalized notion of inequality ordering.

Appendix

Proof of Proposition 2

To prove the sufficiency, consider a sequence of m ACTs, $\hbox {ACT}_{1}(\varepsilon )$, $\hbox {ACT}_{2}(\varepsilon )$, ..., $\hbox {ACT}_{m}(\varepsilon )$, that transfer G into F. Let $G^{(j)}$ be the distribution after the first j ACTs have been sequentially applied to G with $j=1,2,\ldots ,m$, $G^{(1)}\equiv G$ and $G^{(m)}\equiv F$. By Proposition 1, $G^{(j)}$ $\varepsilon$-ALD $G^{(j-1)}$ for $j=2,\ldots ,m$. That is, for $j=2,\ldots ,m$, $I_{G^{(j)}}(x)\le I_{G^{(j-1)}}(x)$ for all $I(\cdot )\in \mathfrak {G} ^{*}(\varepsilon )$). It follows that $I_{F}(x)\le I_{G}(x)$ for all $I(\cdot )\in \mathfrak {G}^{*}(\varepsilon )$. Again, by Proposition 1, F $\varepsilon$-ALD G.

For the necessity, we prove a more specific result. We first define a specific almost composite transfer: in an ACT($\varepsilon$) if

$$\begin{aligned} \rho =\sigma \left( \frac{l-k}{j-i}\right) \left( \frac{1}{\varepsilon } -1\right) , \end{aligned}$$

(A1)

then it becomes a specific almost composite transfer and we denote it SACT($\varepsilon$). That is, a SACT($\varepsilon$) leads to the equality in Definition 3. We then wish to prove that if

$$\begin{aligned} \int _{S_{L}(F,G)}[L_{G}(p)-L_{F}(p)]dp=\varepsilon \int _{0}^{1}|L_{F}(p)-L_{G}(p)|dp,\quad 0< \varepsilon < \frac{1}{2} \end{aligned}$$

(A2)

then F can be obtained from G by a sequence of SACT($\varepsilon$)s.

We first consider the case of a single crossing between the two Lorenz curves. Suppose that the Lorenz curve of F crosses with that of G once from above. For convenience, denote the area of $L_{F}$ above $L_{G}$ as $\alpha$ and refer to it as the $\alpha$-area; denote the area of $L_{G}$ above $L_{F}$ as $\beta$ and refer to it as the $\beta$-area. By assumption, $\frac{ \beta }{\alpha +\beta }=\varepsilon$ or $\alpha =\beta (\frac{1}{ \varepsilon }-1)$.

By the Fields–Fei Lorenz domination characterization theorem, the $\alpha$-area can be eliminated through a sequence of progressive transfers applied to G and the $\beta$-area can be eliminated through a sequence of regressive transfers applied to G. Note that these transfers are not unique since any single transfer can be split into multiple transfers that shift smaller amount of income between the same two sets of dp individuals (for example, a transfer of $5 from jth quantile to ith quantile can be split into five transfers of $1 from jth quantile to ith quantile). We call a transfer distinct if it transfers the largest possible amount of income between the two sets of individuals (the transfer of $1 in the above example is not a distinct transfer).

Suppose there are M distinct progressive transfers involved in eliminating the $\alpha$-area and there are N distinct regressive transfers involved in eliminating the $\beta$-area. Each of these transfers can be split into multiple smaller transfers (in the sense as explained above). The required SACT($\varepsilon$)s are constructed from these two sets of transfers using the following 3-step procedure.

Step 1::

Compute the changes in the areas caused by these transfers and rank them in increasing order. Denote these areas as $\alpha _{i}$s and $\beta _{j}$s with $0<\alpha _{1}\le \alpha _{2}\le \cdots \le \alpha _{M}$ and $0<\beta _{1}\le \beta _{2}\le \cdots \le \beta _{N}$ with $\alpha =\sum _{i=1}^{M}\alpha _{i}$ and $\beta =\sum _{j=1}^{N}\beta _{j}$. Label the corresponding transfers as $PT_{1},PT_{2},\ldots ,PT_{M}$ and $RT_{1},RT_{2},\ldots ,RT_{N}$, respectively. Note that the sequential application of the transfers is path independent, that is, it does not matter in which order the transfers are applied. Thus, the values of $\alpha _{i}$s and $\beta _{j}$s are not affected by the order of the application.

Step 2::

Consider first to pair $PT_{1}$ with $RT_{1}$. There are three possibilities between $\alpha _{1}$ and $\beta _{1}$: $\alpha _{1}=\beta _{1}(\frac{1}{\varepsilon }-1)$, $\alpha _{1}<\beta _{1}(\frac{1}{ \varepsilon }-1)$, and $\alpha _{1}>\beta _{1}(\frac{1}{\varepsilon }-1)$.

Case 1. :

If $\alpha _{1}=\beta _{1}(\frac{1}{\varepsilon }-1)$, then $PT_{1}$ and $RT_{1}$ together constitute a SACT($\varepsilon$).

Case 2. :

If $\alpha _{1}<\beta _{1}(\frac{1}{\varepsilon }-1)$, then the areal change generated by $PT_{1}$ is smaller than what is required to pair with $RT_{1}$ in forming a SACT. We can split $RT_{1}$ into two smaller transfers $RT_{1}^{\prime }$ and $RT_{1}^{\prime \prime }$ such that $\alpha _{1}=\beta _{1}^{\prime }(\frac{1}{\varepsilon }-1)$ where $\beta _{1}^{\prime }$ is the areal reduction by $RT_{1}^{\prime }$. Note that since the areal reduction is a continuous function of the amount of income transfer, such a split can always be performed. Now pair $PT_{1}$ with $RT_{1}^{\prime }$ to form the desired SACT.

Case 3. :

If $\alpha _{1}>\beta _{1}(\frac{1}{\varepsilon }-1)$, then the areal change generated by $PT_{1}$ is larger than what is needed to pair with $RT_{1}$ in forming a SACT. We can split $PT_{1}$ into two smaller transfers $PT_{1}^{\prime }$ and $PT_{1}^{\prime \prime }$ such that $\alpha _{1}^{\prime }=\beta _{1}(\frac{1}{\varepsilon }-1)$ where $\alpha _{1}^{\prime }$ is the areal increase by $PT_{1}^{\prime }$. Then pair $PT_{1}^{\prime }$ with $RT_{1}$ to form the desired SACT.

Step 3::

For Case 1 above, move on to examine $PT_{2}$ and $RT_{2}$; for Case 2, move on to examine $PT_{2}$ and $RT_{1}^{\prime \prime }$; for Case 3, move on to examine $PT_{1}^{\prime \prime }$ and $RT_{2}$. For each of them, there are again three possible cases to entertain.

Steps 2 and 3 are repeatedly applied to all $PT_{i}$ and $RT_{j}$ until they are perfectly pair matched into a sequence of SACTs. This is feasible and necessary since $\alpha =\sum _{i=1}^{M}\alpha _{i}$ and $\beta =\sum _{j=1}^{N}\beta _{j}$ and any splitting of $\alpha _{i}$ and $\beta _{j}$ keeps the total areas unchanged.³ By construction, the sequence of SACTs is finite.

The conclusion of this single crossing case is: if the two areas (the $\alpha$ type and the $\beta$ type) are in the ratio of $\frac{1}{\varepsilon }-1$, then a sequence of SACT($\varepsilon$) can always be constructed to characterize the $\varepsilon$-ALD. We can use this result to prove the general case of multiple crossings.

Suppose now the Lorenz curve of F crosses with that of G K times (think of a situation as in Fig. 1). With a slight abuse of notation, we denote the areas of $L_{F}$ above $L_{G}$ also as the $\alpha$-areas and the areas of $L_{F}$ below $L_{G}$ as the $\beta$-areas, then there are $K_{1}=\left[ \frac{K+1}{2}\right]$ $\alpha$-areas and $K_{2}=K-K_{1}+1$ $\beta$-areas; $K_{1}$ and $K_{2}$ are either equal or differ by 1. Rank the $\alpha$-areas and $\beta$-areas in increasing order and denote them $\alpha _{1},\alpha _{2},\ldots ,\alpha _{K_{1}}$ and $\beta _{1},\beta _{2},\ldots ,\beta _{K_{2}}$, respectively. By assumption, $\alpha =\sum _{i=1}^{M}\alpha _{i}=\left( \sum _{j=1}^{N}\beta _{j}\right) \left( \frac{1}{\varepsilon }-1\right) =\beta \left( \frac{1}{ \varepsilon }-1\right)$. Utilizing the result from the single-crossing case, we want to match pair $\alpha _{i}$s and $\beta _{j}$s so that the ratio between each pair is exactly $\frac{1}{\varepsilon }-1$.

The 3-step procedure outlined above can also be modified to be applied to this general case. First, consider $\alpha _{1}$ and $\beta _{1}$, the two smallest areas in each category. If $\alpha _{1}=\beta _{1}\left( \frac{1}{ \varepsilon }-1\right)$, then the procedure employed in the special case can be directly applied to generate a sequence of SACTs. If $\alpha _{1}<\beta _{1}\left( \frac{1}{\varepsilon }-1\right)$, then split the area $\beta _{1}$ into $\beta _{1}^{\prime }$ and $\beta _{1}^{\prime \prime }$ (creating smaller transfers) such that $\alpha _{1}=\beta _{1}^{\prime }\left( \frac{1}{\varepsilon }-1\right)$. The transfers within $\alpha _{1}$ and $\beta _{1}^{\prime }$ can then be combined into a sequence of SACTs. The remaining area $\beta _{1}^{\prime \prime }$ is then moved forward to compare with $\alpha _{2}$. If $\alpha _{1}>\beta _{1}\left( \frac{1}{ \varepsilon }-1\right)$, then it is the $\alpha _{1}$-area needs to be split into $\alpha _{1}^{\prime }$ and $\alpha _{1}^{\prime \prime }$ such that $\alpha _{1}^{\prime }=\beta _{1}\left( \frac{1}{\varepsilon }-1\right)$. The transfers within $\alpha _{1}^{\prime }$ and $\beta _{1}$ can then be combined to form a sequence of SACTs. The remaining area $\alpha _{1}^{\prime \prime }$ is then moved forward to be compared with $\beta _{2}$. This procedure is repeated until all areas are exhausted and the required sequence of SACTs are generated. $\square$