Segregation, informativeness and Lorenz dominance

Abstract

It is possible to partially order cities according to the informativeness of neighborhoods about their ethnic groups. It is also possible to partially order cities with two ethnic groups according to the Lorenz criterion. We show that a segregation order satisfies four well-established segregation principles if and only if it is consistent with the informativeness criterion. We then use this result to show that for the two-group case, the Lorenz and the informativeness criteria are equivalent.

Appendix

Proof of Lemma 1

Let $A$ be an $n\times m$ Markov matrix and let $B$ be an $n\times (m+1)$ Markov matrix that is obtained from $A$ by splitting one of $A$’s columns into two. Assume that $A$’s $k$th column is the one that is split. Alternatively, $A$ is obtained from $B$ by replacing $B$’s $k$th and $(k+1)$th columns by their sum. Consequently,

$$\begin{aligned} A=B\cdot S_{k} \end{aligned}$$

(8)

where

$$\begin{aligned} S_{k}=\left( \begin{array}{ccc} I_{k-1} &{} \quad 0 &{} \quad 0 \\ 0 &{} \quad 1 &{} \quad 0 \\ 0 &{} \quad 1 &{} \quad 0 \\ 0 &{} \quad 0 &{} \quad I_{m-k} \end{array}\right) . \end{aligned}$$

Let us now assume that there is an $m\times (m+1)$ Markov matrix $\Pi$ such that

$$\begin{aligned} B=A\cdot \Pi . \end{aligned}$$

(9)

We will show that $B$ is necessarily obtained from $A$ by splitting $A$’s $k$ th column proportionally.

Let $\Pi ^{\prime }$ be the matrix that is obtained from $\Pi$ by replacing $\Pi$’s $k$th and $(k+1)$th columns by their sum. That is,

$$\begin{aligned} \Pi ^{\prime }=\Pi \cdot S_{k}. \end{aligned}$$

(10)

Note that $\Pi ^{\prime }$ is a square $m\times m$ Markov matrix. Moreover, by (10), (9) and (8),

$$\begin{aligned} A\cdot \Pi ^{\prime }=A \end{aligned}$$

(11)

which means that each row of $A$ is an invariant distribution of the matrix $\Pi ^{\prime }$.

Since $\Pi ^{\prime }$ is a square Markov matrix, there exists $r\ge 1$ and a permutation matrix $P$ such that $P^{T}\cdot \Pi ^{\prime }\cdot P$ can be written in the following (almost block diagonal) form:

$$\begin{aligned} \left( \begin{array}{ccccc|c} R_{1}^{\prime } &{} \quad &{} \quad &{} \quad &{} \quad &{} \quad 0 \\ &{} \quad R_{2}^{\prime } &{} \quad &{} \quad 0 &{} \quad &{} \quad 0 \\ &{} \quad &{} \quad R_{3}^{\prime } &{}\quad &{} \quad &{} \quad 0 \\ &{} \quad 0 &{} \quad &{} \quad \ddots &{} \quad &{} \quad \vdots \\ &{} \quad &{} \quad &{} \quad &{} R_{r}^{\prime } &{} \quad 0 \\ \hline S_{r+1,1}^{\prime } &{} \quad S_{r+1,2}^{\prime } &{} \quad S_{r+1,3}^{\prime } &{} \quad \cdots &{} \quad S_{r+1,r}^{\prime } &{} \quad Q^{\prime } \end{array}\right) \end{aligned}$$

where for all $j=1,\ldots ,r,\,R_{j}^{\prime }$ are square $\left( m_{j}\times m_{j}\right)$ irreducible Markov matrices and $Q^{\prime }$ is an $\left( n-\sum _{j=1}^{r}m_{j}\right) \times \left( n-\sum _{j=1}^{r}m_{j}\right)$ reducible matrix. We can assume without loss of generality that $P$ is the identity matrix and thus that $\Pi ^{\prime }$ has the above form.⁸

Since $R_{j}^{\prime }$, for $j=1,\ldots ,r$, is an irreducible Markov matrix, it has unique invariant distribution $q ^{j}=(q _{1}^{j},\ldots ,q _{m_{j}}^{j})$, i.e., $q ^{j}$ is the unique probability vector $q$ that satisfies $q =q R_{j}^{\prime }$. Furthermore, any invariant distribution of $\Pi ^{\prime }$ can be written as

$$\begin{aligned} \left( \alpha _{1}q ^{1},\alpha _{2}q ^{2},\ldots ,\alpha _{r}q ^{r},\underbrace{ 0,\ldots ,0}_{n-\sum _{j=1}^{r}m_{j}}\right) \end{aligned}$$

for some $\alpha _{1},\ldots \alpha _{r}\ge 0$ and $\sum _{j=1}^{r}\alpha _{j}=1$ (see, for instance, Lucas and Stokey 1989 (Theorem 11.1, pages 326–330)). Therefore, since each row of $A$ is an invariant distribution of $\Pi ^{\prime }$, it can be written as

$$\begin{aligned} A=\left( \begin{array}{ccccccc} \alpha _{11}q ^{1} &{} \quad \alpha _{12}q ^{2} &{} \quad \cdots &{} \quad \alpha _{1r}q^{r} &{} \quad 0 &{} \quad \cdots &{} \quad 0 \\ \vdots &{} \quad \vdots &{} \quad &{} \quad \vdots &{} \quad \vdots &{}\quad &{} \quad \vdots \\ \alpha _{n1}q ^{1} &{} \quad \alpha _{n2}q ^{2} &{} \quad \cdots &{} \quad \alpha _{nr}q ^{r} &{} \quad 0 &{} \quad \cdots &{} \quad 0 \end{array}\right) , \end{aligned}$$

(12)

where for each $i=1,\ldots ,n$ and $j=1,\ldots ,r,\,\alpha _{ij}\ge 0$ and $\sum _{j=1}^{r}\alpha _{ij}=1$. If $B$ was obtained from $A$ by splitting column $k$ in a disproportional way, it ought to be the case that this column is one that has at least one positive entry.

Assume that column $k$ corresponds to the $h$th block of $\Pi ^{\prime }$. Therefore we can write

$$\begin{aligned} R_{h}^{\prime }=\left( R_{h_{1}}^{\prime },v_{*k}^{\prime },R_{h_{2}}^{\prime }\right) \end{aligned}$$

where $v_{*k}^{\prime }=(v_{1k}^{\prime },\ldots ,v_{m_{h}k}^{\prime })^{T}$ is the column of block $R_{h}^{\prime }$ that corresponds to the $k$th column of $\Pi ^{\prime }$. Since $\Pi$ is obtained from $\Pi ^{\prime }$ by splitting the $k$th column into two, $\Pi$ can be written as

$$\begin{aligned} \Pi =\left( \begin{array}{cccccccc|c} R_{1}^{\prime } &{} \quad 0 &{} \quad &{} \quad 0 &{} \quad 0 &{} \quad &{} \quad \cdots &{} \quad 0 &{} \quad 0 \\ 0 &{} \quad \ddots &{} \quad &{} \quad 0 &{} \quad 0 &{} \quad &{} \quad \cdots &{} \quad \vdots &{} \quad \vdots \\ &{} \quad &{} \quad R_{h_{1}}^{\prime } &{} \quad v_{*k} &{} \quad v_{*k+1} &{}\quad R_{h_{2}}^{\prime } &{}\quad &{} \quad 0 &{} \quad 0 \\ \cdots &{} \quad \cdots &{} \quad &{} \quad 0 &{} \quad 0 &{} \quad &{} \quad \ddots &{} \quad 0 &{} \quad 0 \\ 0 &{} \quad \cdots &{} \quad &{} \quad 0 &{} \quad 0 &{} \quad &{} \quad 0 &{} \quad R_{r}^{\prime } &{} \quad 0 \\ \hline S_{r+1,1}^{\prime } &{} \quad \cdots &{} \quad S_{r+1,h_{1}}^{\prime } &{} \quad s_{*k} &{} \quad s_{*k+1} &{} \quad S_{r+1,h_{2}}^{\prime } &{} \quad \cdots &{} \quad S_{r+1,r}^{\prime } &{} \quad Q^{\prime } \end{array}\right) \qquad \end{aligned}$$

(13)

where $v_{*k}$ and $v_{*k+1}$ are column vectors such that $v_{*k}+v_{*k+1}=v_{*k}^{\prime }$. Consequently, since $B=A\cdot \Pi$, using (12) and (13) we obtain that $B$’s $k$th column is $\left( \alpha _{1h}q ^{h}v_{*k},\ldots ,\alpha _{nh}q ^{h}v_{*k}\right) ^{T}$ and, $B$’s $(k+1)$th column is $\left( \alpha _{1h}q ^{h}v_{*k+1},\ldots ,\alpha _{nh}q ^{h}\right.$ $\left. v_{*k+1}\right) ^{T}$, which are proportional to each other (the proportion is $q ^{h}v_{*k}/q ^{h}v_{*k+1}$). $\square$