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Moran's $I$ statistic is defined to be the quantity $$ \frac{N}{\sum_{i,j} w_{ij}}\frac{\sum_{i,j}w_{ij}(X_i-\bar{X})(X_j-\bar{X})}{\sum_i(X_i-\bar{X})^2} $$

where $w_{ij}$ is some matrix of spatial weights, and $X_1,\dots X_N$ are $N$ readings of a random variable $X$, with sample mean $\bar{X}$. From what I've read, this quantity should always lie between $-1$ and $1$, with $1$ indicating data perfectly correlated over space, and $-1$ indicating data perfectly dispersed.

When I was playing around with some of my own data, I found that Moran's $I$ was turning out greater than $1$. After a little thought, I was able to come up with a simple example.

Let $N=4$, and let $(w_{ij})$ be the matrix $$ \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ \end{pmatrix} $$

Suppose that our readings are: \begin{align} X_1=&3\\ X_2=-&1\\ X_3=-&1\\ X_4=-&1 \end{align}

Then Moran's $I$ comes out to be exactly $3$. What is going wrong here?

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    $\begingroup$ See this answer, stats.stackexchange.com/a/66822/1036. Moran's I is not guaranteed to be between -1 and 1, and your W would not be considered a valid weights matrix typically. If you supply a row-standardized weights matrix (and presumably have zero entries on the diagonal) Moran's I should be between -1 and 1. $\endgroup$
    – Andy W
    Jul 8, 2015 at 12:33

2 Answers 2

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Comparison of $I$ with correlation coefficients is good, but it has its limits. This answer uncovers what those limits are. It derives a tight upper bound for $|I|$ in terms of the weights matrix $W$ and shows that in ordinary applications, where $W$ is symmetric and row-normalized, this bound is $1$ (or less). For the extraordinary weights matrix in the question (which has some zero rows, making it impossible to row-normalize), it shows that $X=\pm(3,-1,-1,-1)$ achieves the most extreme value of $|I|$ possible, which is $3$.


Analysis: Simplifying the Formula for $I$

The repeated appearances of $x_i - \bar x$ in the formula, along with the sum of their squares in the denominator, invite us to re-express $I$ in terms of the standardized variates

$$z_i = \frac{(x_i - \bar x)}{\sqrt{\sum_{i=1}^n (x_i - \bar x)^2}};\quad \bar x = \frac{1}{n}\sum_{i=1}^n x_i,$$

which can be assembled into the column vector $z = (z_1, \ldots, z_n)^\prime$. Because the variables have been standardized, $z$ is a unit vector

$$|z|^2 = z_1^2 + \cdots + z_n^2 = 1.$$

(This differs slightly from the usual statistical concept of "standardization," which recenters and rescales $x$ to have length $\sqrt{n}$.)

Further simplicity is achieved by renormalizing the weights $\omega_{ij}$ so they sum to unity:

$$\sum_{i,j} \omega_{ij} = 1.$$

When $S_0 = \sum_{i,j}\omega_{ij} \ne 0$, this can always be done by dividing every weight by $S_0$. (When the sum of the weights is zero, the formula for $I$ is undefined in the first place.) The renormalized weights form a matrix $W = (\omega_{ij} / S_0) = (w_{ij})$.

In terms of $z$ and $W$,

$$\eqalign{ I &= \sum_{i,j} \frac{n \omega_{ij}}{\sum_{i,j} \omega_{ij}} \left(\frac{(x_i-\bar x)}{\sqrt{\sum_i(x_i-\bar x)^2}}\right)\left(\frac{(x_j-\bar x)}{\sqrt{\sum_i(x_i-\bar x)^2}}\right) \\ &= \sum_{i,j} n w_{ij}z_i z_j = z^\prime (n W z). }$$

This is identical to the formula for the Pearson correlation

$$\rho(z, w) =z^\prime w$$

between any two standardized variables $z$ and $w$.

Comparison to Correlation Coefficients

The difference between $I$ and a correlation is that $w = nWz$ might not be standardized: $|nWz| = \sqrt{(nWz)^\prime (nWz)}$ is not necessarily equal to $1$. To standardize, we must first subtract the mean of $w = nWz$,

$$\bar{w} = \frac{1}{n}\mathbf{1}^\prime n W z = \mathbf{1}^\prime W z,$$

from each component of $n W z$. Let $V$ be the matrix for this transformation,

$$V:z \to Vz = nWz - \mathbf{1}\bar w = \left(n\mathbb{I}_n - \mathbf{1}\mathbf{1}^\prime\right) W z.$$

(The vector $\mathbf{1}$ is the column vector $(1,1,\ldots,1)^\prime$, so the matrix $\mathbf{1}\mathbf{1}^\prime$ is the $n\times n$ matrix all of whose entries are ones.)

By standardizing $Wz$ in this manner--subtracting $\bar w$ and dividing by the length of the resulting centered vector--we get an honest correlation coefficient whose size cannot exceed $1$ (by virtue of the Cauchy-Schwarz Inequality):

$$1 \ge |\rho(z, Wz)| = \frac{|z^\prime Vz|}{|Vz|}.$$

This can be related to $I$ because $\mathbf{1}^\prime z = 0$, whence $z^\prime V z = z^\prime W z$. Clearing the denominator in the preceding inequality gives

$$|I| = |z^\prime (nW z)| = n|z^\prime V z| \le |V z|.$$

Equality will hold if and only if $Vz \ne 0$ and $Vz$ is parallel to $z$: that is, when $z$ is an eigenvector of $V$ with nonzero eigenvalue. When such an eigenvector exists (and it almost always will), this shows that

The size of $I$ is bounded by the largest eigenvalue (in absolute value) of $V = (n\mathbb{I} - \mathbf{1}\mathbf{1}^\prime)W$ and it can attain this bound.

Whenever we can find a basis in which $V$ is diagonal the largest absolute entry of $V$ is its largest eigenvalue.

In the question the unit-sum-normalized version of the weights matrix is

$$W = \pmatrix{1&0&0&0 \\ 0&0&0&0 \\ 0&0&0&0 \\ 0&0&0&0}$$

whence

$$V = \pmatrix{ 3&0&0&0 \\ -1&0&0&0 \\ -1&0&0&0 \\ -1&0&0&0 }.$$

It is simple to compute that the largest value attained by $|Vz|$ among all centered unit vectors $z$ (that is, $\mathbf{1}^\prime z = 0$ and $|z|=1$) is $3$, uniquely attained at $z=\pm(3,-1,-1,-1)/\sqrt{12}$. Consequently, all we can hope in general for this particular weights matrix $W$ is that $|I| \le 3$.


Resolution of the Original Question

Ordinarily, $W$ has nonnegative entries that originally are row standardized to sum to unity. This automatically makes the sum of each row in the unit-sum-normalized version of $W$ equal to $1/n$. Consequently $n W \mathbf{1} = \mathbf{1}$, making $\mathbf{1}$ an eigenvector of $n W$ with eigenvalue $1$.

It is also well-known that in this case if $W$ is symmetric, all other eigenvectors $z$ will be orthogonal to $\mathbf{1}$--that is, $z$ will be zero-centered--and will have eigenvalue $\lambda$ between $0$ and $1$. Considering any such unit $z$,

$$z^\prime V z = z^\prime\left(n\mathbb{I}_n - \mathbf{1}\mathbf{1}^\prime\right)W z = z^\prime (nW z) = z^\prime (\lambda z) = \lambda z^\prime z = \lambda \le 1.$$

In this ordinary application of Moran's I we may conclude that $\max{|I|} = 1$, as expected. But when $W$ is not symmetric and row-standardized--or cannot possibly be row-standardized because one or more rows are entirely zero--then the maximum has to be computed using the general result attained previously.

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That is true because you assume no variation(weight) on$X_i,i=2,\ldots,4$. In your case you have, \begin{align} I &=\frac{N}{1} \frac{(X_1-\bar{X})(X_1-\bar{X})}{\sum_i(X_i-\bar{X})^2}\\ &=\frac{(X_1-\bar{X})(X_1-\bar{X})}{\hat{\sigma}_X}\\ &=\frac{(X_1-\bar{X})^2}{\hat{\sigma}_X} \end{align}

In this case $\mu_x=0$ and $\sigma_x=3$ then, $I=\frac{9}{3}=3$. As @Andy W mentioned (see) , Moran's $|I|$ can be greater that one in some cases.

However, I am not sure your configuration about weight matrix $W$ is true because you assumed zero weights on some variables (that means you can remove them from analysis and just do the analysis on the rest).

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