Segregation, informativeness and Lorenz dominance

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.

1. See Reardon and Firebaugh (2002) for an enumeration and analysis of various multigroup segregation measures. For the two-group case, Massey and Denton (1988) provide a comprehensive survey.

2. For papers that model segregation differently, see Echenique and Fryer (2007) and Ballester and Vorsatz (2013).

3. See also Grant et al. (1998), and Andreoli and Zoli (2013) for related results.

4. Recall that a Markov matrix is a non-negative matrix with each row summing to one. Recall also that a permutation matrix is one that is obtained by permuting the columns of an identity matrix.

5. Given \(\succcurlyeq \), the associated relations \(\succ \) and \(\sim \) are defined as usual: \(X\succ Y\Leftrightarrow X\succcurlyeq Y\) and not \( Y\succcurlyeq X\), and \(X\sim Y\Leftrightarrow X\succcurlyeq Y\) and \( Y\succcurlyeq X\).

6. Massey and Denton (1988) define evenness as the “differential distribution of two social groups among areal units in a city” and James and Taeuber (1985) see segregation as the “differential distribution of students to schools by race regardless of the overall racial proportions in the system concerned.

7. The generalized inverse of a distribution function \(F: {\mathbb {R}} _{+}\rightarrow [0,1]\) is defined as \(F^{-1}:[0,1]\rightarrow {\mathbb {R}} _{+}\) such that \(F^{-1}(p)=\inf _{s}\left\{ s>0:F(s)>p\right\} \).

8. Otherwise, the whole analysis can be done using \(A\cdot P\) instead of \(A\), \(S_{k}\cdot P\) instead of \(S_{k},\,P^{T}\cdot \Pi \) instead of \(\Pi \) and \(P^{T}\cdot \Pi ^{\prime }\cdot P\) instead of \(\Pi ^{\prime }\).

We thank Mikel Bilbao, Sergiu Hart and Robert Hutchens for their valuable comments. Sergiu Hart helped us shorten the proof of Theorem 2 by pointing to Sherman’s (1951) version of the Blackwell–Sherman–Stein theorem. We also thank the Spanish Ministerio de Economía y Competitividad (Project ECO2012-31346) and the Gobierno Vasco (Project IT568-13) for research support.

Authors and Affiliations

1. University of the Basque Country, Bilbao, Spain

   Casilda Lasso de la Vega

2. Ben-Gurion University of the Negev, Beer Sheva, Israel

   Oscar Volij

Corresponding author

Correspondence to Oscar Volij.

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,

where

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

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,

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

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:

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.^{Footnote 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

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

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

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

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 \(q ^{h}v_{*k}/q ^{h}v_{*k+1}\)). \(\square \)

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