Testing for spatial association of qualitative data using symbolic dynamics

Abstract

Qualitative spatial variables are important in many fields of research. However, unlike the decades-worth of research devoted to the spatial association of quantitative variables, the exploratory analysis of spatial qualitative variables is relatively less developed. The objective of the present paper is to propose a new test (Q) for spatial independence. This is a simple, consistent, and powerful statistic for qualitative spatial independence that we develop using concepts from symbolic dynamics and symbolic entropy. The Q test can be used to detect, given a spatial distribution of events, patterns of spatial association of qualitative variables in a wide variety of settings. In order to enable hypothesis testing, we give a standard asymptotic distribution of an affine transformation of the symbolic entropy under the null hypothesis of independence in the spatial qualitative process. We include numerical experiments to demonstrate the finite sample behaviour of the test, and show its application by means of an empirical example that explores the spatial association of fast food establishments in the Greater Toronto Area in Canada.

Appendix

1.1 Proofs

Proof of Theorem 1

Under the null H ₀, the joint probability density function of the n variables $\left( {Y_{{\sigma_{1} }} ,Y_{{\sigma_{2} }} , \ldots ,Y_{{\sigma_{{k^{m} }} }} } \right)$ is:

$$ P\left( {Y_{{\sigma_{1} }} = a_{1} ,Y_{{\sigma_{2} }} = a_{2} , \ldots ,Y_{{\sigma_{{k^{m} }} }} = a_{{k^{m} }} } \right) = {\frac{{\left( {a_{1} + a_{2} + \cdots + a_{{k^{m} }} } \right)!}}{{a_{1} !a_{2} ! \cdot \cdots \cdot a_{{k^{m} }} !}}}p_{{\sigma_{1} }}^{{a_{1} }} p_{{\sigma_{2} }}^{{a_{2} }} \cdots p_{{\sigma_{{k^{m} }} }}^{{a_{{k^{m} }} }} $$

(19)

where a ₁ + a ₂ + ··· + a _n  = R. Consequently, the joint distribution of the n variables $\left( {Y_{{\sigma_{1} }} ,Y_{{\sigma_{2} }} , \ldots ,Y_{{\sigma_{{k^{m} }} }} } \right)$ is a multinomial distribution.

The likelihood function of the distribution given by Eq. (19) is:

$$ L\left( {p_{{\sigma_{1} }} ,p_{{\sigma_{2} }} , \ldots ,p_{{\sigma_{{k^{m} }} }} } \right) = {\frac{R!}{{n_{{\sigma_{1} }} !n_{{\sigma_{2} }} ! \cdot \cdots \cdot n_{{\sigma_{{k^{m} }} }} !}}}\,p_{{\sigma_{1} }}^{{n_{{\sigma_{1} }} }} p_{{\sigma_{2} }}^{{n_{{\sigma_{2} }} }} \cdots p_{{\sigma_{{k^{m} }} }}^{{n_{{\sigma_{{k^{m} }} }} }} $$

(20)

and since, $\sum\nolimits_{i = 1}^{{k^{m} }} p_{{\sigma_{i} }} = 1$, it follows that

$$ L\left( {p_{{\sigma_{1} }} ,p_{{\sigma_{2} }} , \ldots ,p_{{\sigma_{{k^{m} }} }} } \right) = {\frac{R!}{{n_{{\sigma_{1} }} !n_{{\sigma_{2} }} ! \cdot \cdots \cdot n_{{\sigma_{{k^{m} }} }} !}}}\,p_{{\sigma_{1} }}^{{n_{{\sigma_{1} }} }} p_{{\sigma_{2} }}^{{n_{{\sigma_{2} }} }} \cdots \left( {1 - p_{{\sigma_{1} }} - p_{{\sigma_{2} }} - \cdots - p_{{\sigma_{{k^{m} - 1}} }} } \right)^{{n_{{\sigma_{{k^{m} }} }} }}. $$

(21)

Then the logarithm of this likelihood function remains as

$$ \begin{aligned} \ln \left[ {L\left( {p_{{\sigma_{1} }} ,p_{{\sigma_{2} }} , \ldots ,p_{{\sigma_{{k^{m} }} }} } \right)} \right] = & \ln \left( {{\frac{R!}{{n_{{\sigma_{1} }} !n_{{\sigma_{2} }} ! \cdot \cdots \cdot n_{{\sigma_{{k^{m} }} }} !}}}} \right) + \sum\limits_{i = i}^{{k^{m} - 1}} n_{{\sigma_{i} }} \ln \left( {p_{{\sigma_{i} }} } \right) \\ & + n_{{\sigma_{{k^{m} }} }} \ln \left( {1 - p_{{\sigma_{1} }} - p_{{\sigma_{2} }} - \cdots - p_{{\sigma_{{k^{m} - 1}} }} } \right). \\ \end{aligned}$$

(22)

In order to obtain the maximum likelihood estimators $\hat{p}_{{\sigma_{i} }}$ of $p_{{\sigma_{i} }}$ for all i = 1, 2,…, n, we solve the following equation

$$ {\frac{\partial }{{\partial p_{{\sigma_{i} }} }}}\ln \left[ {L\left( {p_{{\sigma_{1} }} ,p_{{\sigma_{2} }} , \ldots ,p_{{\sigma_{n} }} } \right)} \right] = 0 $$

(23)

to get that:

$$ \hat{p}_{{\sigma_{i} }} = {\frac{{n_{{\sigma_{i} }} }}{R}}. $$

(24)

Then the likelihood ratio statistic is (see for example Lehman 1986):

$$ \begin{aligned} \lambda \left( Y \right) & = {\frac{{{\frac{R!}{{n_{{\sigma_{1} }} !n_{{\sigma_{2} }} ! \cdot \ldots \cdot n_{{\sigma_{{k^{m} }} }} !}}}\,p_{{\sigma_{1} }}^{{(0)n_{{\sigma_{1} }} }} p_{{\sigma_{2} }}^{{(0)n_{{\sigma_{2} }} }} \cdots p_{{\sigma_{{k^{m} }} }}^{{(0)n_{{\sigma_{{k^{m} }} }} }} }}{{{\frac{R}{{n_{{\sigma_{1} }} !n_{{\sigma_{2} }} ! \cdot \ldots \cdot n_{{\sigma_{{k^{m} }} }} !}}}\mathop {\widehat{p}}\nolimits_{{\sigma_{1} }}^{{n_{{\sigma_{1} }} }} \mathop {\widehat{p}}\nolimits_{{\sigma_{2} }}^{{n_{{\sigma_{2} }} }} \cdots \mathop {\widehat{p}}\nolimits_{{\sigma_{{k^{m} }} }}^{{n_{{\sigma_{{k^{m} }} }} }} }}} = {\frac{{\prod\nolimits_{i = 1}^{{k^{m} }} p_{{\sigma_{i} }}^{{(0)n_{{\sigma_{i} }} }} }}{{\prod\nolimits_{i = 1}^{{k^{m} }} \mathop {\left( {{\frac{{n_{{\sigma_{i} }} }}{R}}} \right)}\nolimits^{{n_{{\sigma_{i} }} }} }}} \\ &= R^{{\sum\nolimits_{i = 1}^{{k^{m} }} n_{{\sigma_{i} }} }} \prod\limits_{i = 1}^{{k^{m} }} \mathop {\left( {{\frac{{p_{{\sigma_{i} }}^{(0)} }}{{n_{{\sigma_{i} }} }}}} \right)}\nolimits^{{n_{{\sigma_{i} }} }} = R^{R} \prod\limits_{i = 1}^{{k^{m} }} \mathop {\left( {{\frac{{p_{{\sigma_{i} }}^{(0)} }}{{n_{{\sigma_{i} }} }}}} \right)}\nolimits^{{n_{{\sigma_{i} }} }} \\ \end{aligned} $$

(25)

where $p_{{\sigma_{i} }}^{(0)}$ denotes the probability of the symbol σ _i under the null hypothesis.

On the other hand, Q(m) = −2 ln(λ(Y)) asymptotically follows a Chi-squared distribution with k ^m − 1 degrees of freedom (see Lehman 1986). Hence:

$$ Q(m) = - 2\ln \left( {\lambda \left( Y \right)} \right) = - 2\left[ {R\ln \left( R \right) + \sum\limits_{i = 1}^{{k^{m} }} n_{{\sigma_{i} }} \ln \left( {{\frac{{p_{{\sigma_{i} }}^{(0)} }}{{n_{{\sigma_{i} }} }}}} \right)} \right] \sim \chi_{{k^{m} - 1}}^{2}. $$

(26)

Denote by α _ij the number of times that class a _j appears in symbol σ _i and by q _j  = P(X = a _j ). Then under the null we have that $p_{{\sigma_{i} }}^{(0)} = \prod\nolimits_{j = 1}^{k} {q_{j}^{{\alpha_{ij} }} }$ and hence, it follows that

$$ \begin{aligned} Q(m) & = - 2R\left[ {\ln (R) + \sum\limits_{i = 1}^{{k^{m} }} {\frac{{n_{{\sigma_{i} }} }}{R}}\ln \left( {{\frac{{\prod\nolimits_{j = 1}^{k} {q_{j}^{{\alpha_{ij} }} } }}{{n_{{\sigma_{i} }} }}}} \right)} \right] \\ & = - 2R\left[ {\ln (R) + \sum\limits_{i = 1}^{{k^{m} }} {\frac{{n_{{\sigma_{i} }} }}{R}}\left( {\ln \left( {\prod\limits_{j = 1}^{k} {q_{j}^{{\alpha_{ij} }} } } \right) - \ln \left( {n_{{\sigma_{i} }} } \right)} \right)} \right] \\ & = - 2R\left[ {\ln (R) + \sum\limits_{i = 1}^{{k^{m} }} {\frac{{n_{{\sigma_{i} }} }}{R}}\left( {\ln \left( {\prod\limits_{j = 1}^{k} {q_{j}^{{\alpha_{ij} }} } } \right) - \ln \left( {{\frac{{n_{{\sigma_{i} }} }}{R}}} \right) - \ln \left( R \right)} \right)} \right] \\ & = - 2R\left[ {\ln (R) + \sum\limits_{i = 1}^{{k^{m} }} {\frac{{n_{{\sigma_{i} }} }}{R}}\left( {\sum\limits_{j = 1}^{k} {\alpha_{ij} } \ln \left( {q_{j} } \right) - \ln \left( {{\frac{{n_{{\sigma_{i} }} }}{R}}} \right) - \ln \left( R \right)} \right)} \right]. \\ \end{aligned}$$

(27)

Now, taking into account that $h(m) = - \sum\nolimits_{i = 1}^{{k^{m} }} {p_{{\sigma_{i} }} { \ln }\left( {p_{{\sigma_{i} }} } \right)} = - \sum\nolimits_{i = 1}^{{k^{m} }} {{\frac{{n_{{\sigma_{i} }} }}{R}}{ \ln }\left( {{\frac{{n_{{\sigma_{i} }} }}{R}}} \right)}$, we have that

$$ Q\left( m \right) = - 2R\left[ {\sum\limits_{i = 1}^{{k^{m} }} {\frac{{n_{{\sigma_{i} }} }}{R}}\sum\limits_{j = 1}^{k} {\alpha_{ij} } { \ln }\left( {q_{j} } \right) + h\left( m \right)} \right]. $$

(28)

Notice that if the spatial process is independent identically distributed then $q_{j} = {\frac{1}{k}}$ and therefore $Q(m) = 2R\left( {{ \ln }(k^{m} ) - h(m)} \right)$ which finishes the proof of the theorem.

Proof of Theorem 2

First, notice that the estimator of h(m), $\widehat{h}\left( m \right) = - \sum\nolimits_{{\sigma \in S_{m} }} {\hat{p}_{\sigma } { \ln }\left( {\hat{p}_{\sigma } } \right)}$, where $\hat{p}_{\sigma } = n_{\sigma } /R$, is consistent because, $p\mathop { \lim }\nolimits_{R \to \infty } \hat{p}_{\sigma } = p_{\sigma }$, and hence:

$$ p\mathop { \lim }\limits_{R \to \infty } \widehat{h}\left( m \right) = h\left( m \right). $$

(29)

Recall that:

$$ Q\left( m \right) = 2R\left[ { - \sum\limits_{i = 1}^{{k^{m} }} \,{\frac{{n_{{\sigma_{i} }} }}{R}}\sum\limits_{j = 1}^{k} {\alpha_{ij} } \ln \left( {q_{j} } \right) + \sum\limits_{i = 1}^{{k^{m} }} {\frac{{n_{{\sigma_{i} }} }}{R}}\ln \left( {{\frac{{n_{{\sigma_{i} }} }}{R}}} \right)} \right]. $$

(30)

Now, let us call:

$$ H\left( m \right) = - \sum\limits_{i = 1}^{{k^{m} }} \,{\frac{{n_{{\sigma_{i} }} }}{R}}\sum\limits_{j = 1}^{k} {\alpha_{ij} } \ln \left( {q_{j} } \right) + \sum\limits_{i = 1}^{{k^{m} }} {\frac{{n_{{\sigma_{i} }} }}{R}}\ln \left( {{\frac{{n_{{\sigma_{i} }} }}{R}}} \right) = \sum\limits_{i = 1}^{{k^{m} }} {\frac{{n_{{\sigma_{i} }} }}{R}}\ln \left( {{\frac{{{\frac{{n_{{\sigma_{i} }} }}{R}}}}{{\prod\nolimits_{j = 1}^{k} {q_{j}^{{\alpha_{ij} }} } }}}} \right). $$

(31)

Also, since ln(x) ≤ x − 1 for all x with equality if and only if x = 1, and under the alternative hypothesis of spatial dependence of order ≤m we have that:

$$ {\frac{{{\frac{{n_{{\sigma_{i} }} }}{R}}}}{{\prod\nolimits_{j = 1}^{k} {q_{j}^{{\alpha_{ij} }} } }}} \ne 1 $$

(32)

it follows that:

$$ H\left( m \right) = - \sum\limits_{i = 1}^{{k^{m} }} {\frac{{n_{{\sigma_{i} }} }}{R}}\ln \left( {{\frac{{\prod\nolimits_{j = 1}^{k} {q_{j}^{{\alpha_{ij} }} } }}{{{\frac{{n_{{\sigma_{i} }} }}{R}}}}}} \right) > - \sum\limits_{i = 1}^{{k^{m} }} {\frac{{n_{{\sigma_{i} }} }}{R}}\left( {{\frac{{\prod\nolimits_{j = 1}^{k} {q_{j}^{{\alpha_{ij} }} } }}{{{\frac{{n_{{\sigma_{i} }} }}{R}}}}} - 1} \right) = - \sum\limits_{i = 1}^{{k^{m} }} \prod\limits_{j = 1}^{k} {q_{j}^{{\alpha_{ij} }} } + 1 \ge 0. $$

(33)

Since, also $p\mathop { \lim }\nolimits_{R \to \infty } \hat{q}_{j} = q_{j}$, then by Eq. (29) we have

$$ p\mathop { \lim }\nolimits_{R \to \infty } \widehat{H}\left( m \right) = H\left( m \right). $$

(34)

Let 0 < C < ∞ with $C \in \mathbb{R}$ and take R large enough such that

$$ {\frac{C}{2R}} < H( m ). $$

(35)

Then, under the spatial dependence of order less than or equal to m it follows that H(m) ≠ 0 and, thus,

$$ \begin{aligned} Pr\left[ {\widehat{Q}\left( m \right) > C} \right] & = Pr\left[ {2R\widehat{H}\left( m \right) > C} \right] \\ & = Pr\left[ {2R\left( {\widehat{H}\left( m \right) - H\left( m \right)} \right) > C - 2RH\left( m \right)} \right] \\ & = Pr\left[ {2R\left( {H\left( m \right) - \widehat{H}\left( m \right)} \right) < 2RH\left( m \right) - C} \right] \\ & = Pr\left[ {H\left( m \right) - \widehat{H}\left( m \right) <H\left( m \right) - {\frac{C}{2R}}} \right]. \\ \end{aligned} $$

(36)

Therefore, by Eqs. (34), (35) and (36) we have that:

$$ \mathop { \lim }\limits_{R \to \infty } Pr\left( {\widehat{Q}(m) > C} \right) = 1 $$

(37)

as desired.