A technical note on divergence of the Wald statistic

TL;DR

This paper investigates the divergence behavior of the Wald test statistic, linking it to eigenvalues of polynomial matrices and establishing the rate of divergence under certain conditions.

Contribution

It provides a mathematical connection between Wald statistic divergence and eigenvalues, and quantifies the divergence rate, enhancing understanding of Wald test limitations.

Findings

Divergence of Wald statistic linked to eigenvalues

Established divergence rate under specific conditions

Provides theoretical insights into Wald test behavior

Abstract

The Wald test statistic has been shown to diverge (Dufour et al, 2013, 2017) under some conditions. This note links the divergence to eigenvalues of a polynomial matrix and establishes the divergence rate.

Equations

$$H_0:g\left(\theta \right)=0,$$

$$\sqrt{T}{V}^{-\frac{1}{2}}\left({\hat{\theta }}_T-\bar{\theta }\right)$$

$$Z$$

$$\text{and }{\hat{V}}_T$$

$$W_T=T{g}^{\prime }({\hat{\theta }}_T){\left[\frac{\partial g}{\partial {\theta }^{\prime }}({\hat{\theta }}_T){\hat{V}}_T\frac{\partial g^{\prime }}{\partial \theta }({\hat{\theta }}_T)\right]}^{-1}g({\hat{\theta }}_T).$$

$$\left\{\begin{array}[]{ccc}\mathbf{xy}&=&0;\\ \mathbf{xw}&=&0;\\ \mathbf{yz}&=&0.\end{array}\right.$$

$$W=T\left(w^2+y^2\right)\frac{x^2+z^2}{w^2+x^2+y^2+z^2}.$$

$$z{\to }_{p}1;w{\to }_{p}1;T^{\frac{1}{2}}x{\to }_p{Z}_1;T^{\frac{1}{2}}y{\to }_p{Z}_2$$

$$W=\left(T+Z_2^2+o_p\left(1\right)\right)\frac{Z_1^2+T+o_p\left(1\right)}{2T+Z_1^2+Z_2^2+o_p\left(1\right)}.$$

$${W=T+O}_p\left(1\right).$$

$$g_l(\theta )=\sum _{\gamma =0}^{m_l}\sum _{j_1+...+j_p=\gamma }{c}_l(j_1,...,j_p,\bar{\theta })\prod _{k=1}^p{\left({\theta }_k-{\bar{\theta }}_k\right)}^{j_k}$$

$$c_l(0,...,0,\bar{\theta })=0$$

$$\left(\begin{array}[]{cc}y&0\\ y^{2}&0\end{array}\right)$$

$$g_l\left(\theta \right)={\bar{g}}_l\left(\theta -\bar{\theta }\right)+r_l\left(\theta -\bar{\theta }\right),$$

$$G\left(\theta \right)=\frac{\partial g}{\partial {\theta }^{\prime }}(\theta )$$

$$G_l\left(\theta \right)={\bar{G}}_l\left(\theta -\bar{\theta }\right)+R_l\left(\theta -\bar{\theta }\right),$$

$$G\left(\theta \right)=\bar{G}\left(\theta -\bar{\theta }\right)+R\left(\theta -\bar{\theta }\right).$$

$$\overline{SG}\left(x\right)=\left[\begin{array}[]{c}\left[\overline{SG}\left(x\right)\right]_{1}\\ \vdots\\ \left[\overline{SG}\left(x\right)\right]_{i}\\ \vdots\\ \left[\overline{SG}\left(x\right)\right]_{v}\end{array}\right],$$

$$G\left(\theta\right)=\left(\begin{array}[]{cccc}y&x&0&0\\ 1+\tilde{w}&0&x&0\\ 0&1+\tilde{z}&0&y\end{array}\right);$$

$$PG\left(\theta \right)$$

$$\overline{PG}\left(\theta \right)$$

$$T^{1/2}{\Delta }_{T}g({\hat{\theta }}_T)\underset{T\to \infty }{\stackrel{d}{\to }}\bar{g}\left(Y\right),$$

$${\Delta }_{T}G({\hat{\theta }}_T){\hat{V}}_{T}G({\hat{\theta }}_T{)}^{\prime }{\Delta }_T\underset{T\to \infty }{\stackrel{d}{\to }}\bar{G}\left(Y\right)V\bar{G}{\left(Y\right)}^{\prime },$$

$$W=T^{1/2}{\Delta }_{T}g({\hat{\theta }}_T{)}^{\prime }{\left[{\Delta }_{T}G({\hat{\theta }}_T){\hat{V}}_{T}G({\hat{\theta }}_T{)}^{\prime }{\Delta }_T\right]}^{-1}{T}^{1/2}{\Delta }_{T}g({\hat{\theta }}_T)$$

$${\bar{\Lambda }}_T=diag\left[{\bar{\lambda }}_{1T}\left({\hat{\theta }}_T\right),{\bar{\lambda }}_{2T}\left({\hat{\theta }}_T\right),...,{\bar{\lambda }}_{qT}\left({\hat{\theta }}_T\right)\right]$$

$$p_B\left(\lambda ;x,U\right)={\lambda }^q+{\Sigma }_{k=0}^q{a}_k\left(x,U\right){\lambda }^{q-k},$$

$$a_k\left(x,U\right)=D_{m_k\left(U\right)}\left(x,U\right)+{\tilde{R}}_k\left(x,U\right),$$

$$a_3\left(x,I\right)=w^2{x}^2{y}^2+2w{x}^2{y}^2+x^4{y}^2+x^2{y}^4+x^2{y}^2{z}^2+2x^2{y}^{2}z+2x^2{y}^2,$$

$$U=\left(\begin{array}[]{cccc}1&\sqrt{.98}&0&0\\ \sqrt{.98}&1&.1&.1\\ 0&.1&1&0\\ 0&.1&0&1\end{array}\right).$$

$$a_3\left(x,U\right)$$

$$P_k\left({\lambda }_1,...{\lambda }_q\right)={\Sigma }_{\left\{i_1,...,i_k\right\}\in I_q\left(k\right)}{\Pi }_{j=1}^k{\lambda }_{i_j}.$$

$$P_k\left[{\lambda }_1\left(x,U\right),...{\lambda }_q\left(x,U\right)\right]={\left(-1\right)}^k{a}_k\left(x,U\right)$$

Full text

1

The set-up and an example of divergence

Suppose that a $p\times 1$ parameter of interest $\bar{\theta}$ satisfies

[TABLE]

where $g\left(\theta\right)$ is a $q\times 1$ vector of differentiable functions; $g\left(\theta\right)=\left(g_{1}\left(\theta\right),...,g_{q}\left(\theta\right)\right)^{\prime};$ $q\leq p.$

Let $V$ be a symmetric positive definite matrix.

Assumption 1. In some open set $\Theta\subset R^{p}$ there is a random sequence $\hat{\theta}_{T}\in\Theta$ and $p\times p$ random matrix sequence, $\hat{V}_{T},$ such that as $T\rightarrow\infty$

[TABLE]

Define the usual Wald test statistic:

[TABLE]

For linear $g$ the statistic converges to a $\chi_{q}^{2}$ distribution, for other, e.g. polynomial restrictions the limit distribution may be not $\chi^{2}.$ The limit results for the statistic for testing general polynomial restrictions can be found in (Dufour et al., 2013, 2017); it is also established there that under some conditions the statistic may diverge when $q>1$. Below is an example of divergence.

Example 1. Restrictions for which the Wald statistic diverges.

Consider for $\theta=\left(\mathbf{x,y,z,w}\right)^{\prime}$ the set of restrictions, $H_{0}:$

[TABLE]

Then the Wald statistic for $\hat{\theta}_{T}=\left(x,y,z,w\right),$ assuming that the covariance matrix is identity $\hat{V}=V=I_{4},$ is

[TABLE]

Suppose that the true parameter value is $\bar{\theta}=(0,0,1,1);$ $H_{0}$ then holds. Suppose that the estimated parameter $\hat{\theta}_{T}=\left(x,y,z,w\right)$ as $T\rightarrow\infty$ is consistent and satisfies

[TABLE]

where $Z_{1},Z_{2}$ are independent standard normals. Then the Wald statistic can be expressed as

[TABLE]

This is

[TABLE]

As $T\rightarrow\infty$ the statistic diverges under $H_{0}.$

We shall assume that each $g_{l}\left(\theta\right)$ is a polynomial of order $m_{l}$ in the components of $\theta.$ Then for any $\bar{\theta}$ each polynomial component $g_{l}\left(\theta\right),$ can be written around $\bar{\theta}$ as

[TABLE]

with some coefficients $c(j_{1},...,j_{p},\bar{\theta}).$

If the value $\bar{\theta}$ satisfies the null hypothesis, then

[TABLE]

for each $l.$

A polynomial function is eiher identically zero or non-zero a.e. with respect to the Lebesgue measure. Consider a square matrix $G(y)$ of polynomials of variable $y\in\mathbb{R}^{p}.$ We say that the polynomial matrix $G(y)$ is non-singular if its determinant is a non-zero polynomial.

The rank of the $q\times p$ matrix $G(y)$ is the largest dimension of a square non-singular submatrix.

Unlike matrices of constants for polynomial matrices the rows may be linearly independent vectors of polynomial functions, while the matrix may have defficient rank. For example, in the matrix

[TABLE]

the two rows are given by independent vectors of polynomials, but the rank of this matrix of polynomials is one.

Assumption 2. The $q\times 1$ function $g\left(\theta\right)$ is a vector of polynomial functions; the matrix of polynomials $G\left(\theta\right)=\frac{\partial g}{\partial\theta^{\prime}}\left(\theta\right)$ is of rank $q.$

This does not exclude the possibility of reduced rank at some particular point or on a low dimensional space.

Under the stated assumptions for $g\left(\bar{\theta}\right)=0$ the standard asymptotic $\chi_{q}^{2}$ distribution holds for $W_{T}$ as long as $G\left(\bar{\theta}\right)=\frac{\partial g}{\partial\theta^{\prime}}(\bar{\theta})$ is a (numerical) matrix of rank $q.$

Each restriction $g_{l}\left(\theta\right)$ can be represented as a sum

[TABLE]

where $\bar{g}_{l}\left(.\right)$ denotes the lowest degree non-zero homogeneous polynomial and has degree $\bar{\gamma}_{l}+1,$ the degrees of all non-zero monomials in $r_{l}\left(.\right)$ are >$$\bar{\gamma}_{l}+1. We ascribe the degree of homogeneity $\infty$ to a function that is identically zero.

Correspondingly to $\left(\ref{decomp g}\right)$, in the matrix

[TABLE]

for each row write

[TABLE]

the degree of any non-zero homogenious polynomial in the row vector, $\bar{G}_{l}\left(\theta-\bar{\theta}\right),$ is $\bar{\gamma}_{l}$; any non-zero monomial in $R_{l}\left(\theta-\bar{\theta}\right)$ has degree higher than $\gamma_{l}.$ Then collecting the lowest degree homogeneous polynomials in each row we have

[TABLE]

2 **The property of full rank reached at lowest degrees (FRALD)

and FRALD-T**

**Definition (FRALD). **If the matrix $\bar{G}\left(\theta-\bar{\theta}\right)$ of lowest degree polynomials for $g\left(\theta\right)$ is of full rank $q$ we say that the Full Rank at Lower Degrees (FRALD) property is satisfied for $g\left(.\right)$ and $\bar{\theta}.$

Examples in Dufour et al (2017) illustrate the possibilities that the FRALD property may hold at some points $\bar{\theta},$ but not others, and that even if FRALD property does not hold for $g\left(.\right)$ at $\bar{\theta},$ it may hold for $Sg\left(.\right),$ where $S$ is a non-degenerate numerical matrix.

Recall that the distribution of the Wald statistic is invariant with respect to non-degenerate linear transformation of the restrictions.

**Definition (FRALD-T). **There exists some numerical non-degenerate matrix $S$ such that FRALD holds for $Sg\left(\theta\right)$ at $\bar{\theta}.$

If FRALD-T holds for $g,$ then for some $S$ FRALD holds for $Sg,$ meaning that $\overline{SG}\left(.\right)$ is a full rank matrix of polynomials.

It is shown in Dufour et al (2017) that for polynomial $g\left(x\right)$ with a full rank matrix $G\left(x\right)$ there always exists a non-degenerate numerical matrix $S,$ such that $SG\left(x\right)$ has the property that $\overline{SG}\left(x\right)$ has all the rows represented by linearly independent vectors of polynomials (each row contains non-zero homogeneous polynomials); these rows could be stacked by a permutation in an ”eschelon form”, with the degrees of the non-zero homogeneous polynomials in non-decreasing order.

The eschelon form is given by

[TABLE]

where $\left[\overline{SG}\left(x\right)\right]_{i}$ has dimension $n_{i}\times q,$ all non-zero polynomials in $\left[\overline{SG}\left(x\right)\right]_{i}$ have degree $\bar{s}_{i}$ for $i=1,\ldots,\,\nu$, with $0\leq\bar{s}_{1}<\cdots<\bar{s}_{i}<\cdots<\bar{s}_{\nu},$ and all the rows of $\overline{SG}\left(x\right)$ are linearly independent functions. Once any $S$ that provides such a structure is found, the rank of $\overline{SG}\left(\theta\right)$ is either $q,$ and FRALD-T holds, or is less than $q,$ in which case this property is violated. An algorithm to find $S$ is provided in Dufour et al (2017).

Example 2 (Example 1 continued). FRALD-T does not hold.

Take for $\theta^{\prime}=\left(x;y;w;z\right)$ the function g\left(\theta\right)=\left(\begin{array}[]{c}xy\\ xw\\ yz\end{array}\right). With $\bar{\theta}^{\prime}=\left(0,0,1,1\right)$ denote $\theta^{\prime}=\left(x;y;1+\tilde{w};1+\tilde{z}\right)$ with $\sqrt{T}\left(x;y;\tilde{w};\tilde{z}\right)^{\prime}$ converging to $N\left(0,I_{4}\right).$ Then

[TABLE]

by applying a transformation (here permutation), $P,$ to the rows of this matrix we get

[TABLE]

The matrix $\overline{PG}\left(\theta\right)$ has independent polynomial row vectors, and the rows are stacked so that the degrees of ”leading” polynomials do not decline from row to row (eschelon form). The rank of the matrix $\overline{PG}\left(\theta\right)$ is not full in an eschelon form, no linear transformation applied to $G$ can remedy this rank defficiency. So FRALD-T does not hold for this example.

In Dufour et al (2014, 2017) the limit distribution for the Wald statistic was established for $\bar{\theta}$ when the FRALD-T property holds.

The example 1 here of the case where the statistic was shown to diverge does not satisfy FRALD-T. The next section demonstrates the mechanism whereby the violation of the FRALD-T property leads to divergence of the statistic.

3 Divergence of the Wald statistic where FRALD-T does not hold

Assume that for $\bar{\theta}$ for which the null is satisfied, $g\left(\bar{\theta}\right)=0,$ the FRALD-T property does not hold. Without loss of generality we may assume that $G\left(x\right)=\frac{\partial g}{\partial x^{\prime}}$ is such that the eschelon form $\left(\ref{eschelon}\right)$ applies to $\bar{G}\left(x\right)$ (so that $S$ in FRALD-T and in $\left(\ref{eschelon}\right)$ is identity).

Denote by $Y$ the Gaussian limit $\sqrt{T}(\hat{\theta}-\bar{\theta})\rightarrow_{d}Y\sim N\left(0,V\right).$ Define $\Delta_{T}:=\mathrm{diag}[T^{s_{1}/2}I_{n_{1}},\ldots,\,T^{s_{q}/2}I_{n_{v}}].$ With the scaling $\Delta_{T}$ we get

[TABLE]

[TABLE]

where $\bar{G}\left(Y\right)V\bar{G}\left(Y\right)^{\prime}$ has rank $r<q$ when FRALD-T does not hold. Consequently, inverting the consistent estimator $\left[\Delta_{T}\,G(\hat{\theta}_{T})\hat{V}_{T}G(\hat{\theta}_{T})^{\prime}\Delta_{T}\right]$ for

[TABLE]

will lead to an explosion as $T\rightarrow\infty.$

We next examine the matrix $\bar{\Sigma}_{T}\left(\hat{\theta}_{T},V_{T}\right)=\left[\Delta_{T}\,G(\hat{\theta}_{T})\hat{V}_{T}G(\hat{\theta}_{T})^{\prime}\Delta_{T}\right]$ and its limit eigenvalues which provide the key ingredient to prove the divergence of the Wald statistic.

Denote by $\bar{\lambda}_{1T}\left(\hat{\theta}_{T}\right),\bar{\lambda}_{2T}\left(\hat{\theta}_{T}\right),...,\bar{\lambda}_{qT}\left(\hat{\theta}_{T}\right)$ the eigenvalues of the matrix $\bar{\Sigma}_{T}\left(\hat{\theta}_{T},V_{T}\right),$ arranged in decreasing order: $\bar{\lambda}_{1T}\left(\hat{\theta}_{T}\right)\geq\bar{\lambda}_{2T}\left(\hat{\theta}_{T}\right)\geq...\geq\bar{\lambda}_{qT}\left(\hat{\theta}_{T}\right)$ and denote by

[TABLE]

the $q\times q$ diagonal matrix of these eigenvalues.

We prove several auxilliary results about eigenvalues of non-random polynomial matrices (proofs are in the next section).

Start with $B\left(x,U\right)=G(x)UG(x)^{\prime},$ where $G\left(x\right)$ is a non-zero matrix of polynomial functions and define the characteristic polynomial, $p\left(\lambda;B\left(x,U\right)\right)=\det\left[\lambda I_{p}-B\left(x,U\right)\right]$.

The next proposition describes a polynomial representation for the coefficients of $p\left(\lambda;B\left(x,U\right)\right)$ as a polynomial in $\lambda$. Denote by $F_{p}$ the set of all real symmetric positive-definite matrices.

Proposition 1. Let $x\in R^{p}$ and $G\left(x\right)$ be a $q\times p$ non-zero matrix of polynomial functions in $x$ such that rank$\left[G\left(x\right)\right]=q$ a.e., $U\in F_{p}$, $B\left(x,U\right)=G(x)UG(x)^{\prime}$ and $p_{B}\left(\lambda;x,U\right)=\det\left[\lambda I_{p}-B\left(x,U\right)\right]$ is the characteristic polynomial. Then $p_{B}\left(\lambda;x,U\right)$ can be written as

[TABLE]

where the coefficients $a_{k}\left(x,U\right)$ have the following polynomial expansions

[TABLE]

where $D_{m_{k}\left(U\right)}\left(x,U\right)$ is a homogeneous in $x$ polynomial of degree $m_{k}\left(U\right),$ and $\tilde{R}_{k}\left(x,U\right)$ is a sum of polynomials with any non-zero mononomials of degree strictly greater than $m_{k}\left(U\right).$ Further, if $m_{k}=\underset{Q\in F_{p}}{\min}m_{k}\left(Q\right),$ then $m_{k}\left(U\right)=m_{k}$ for almost all $U\in F_{p}.$

Example 3 Restrictions of Example 1 but with a covariance matrix $U$ for which $m_{k}\left(U\right)>m_{k}.$

In the example 1 we had divergence at the rate $T$ when the matrix $U$ was identity. The characteristic polynomial for the same restrictions, $\mathit{g}\left(.\right)$ of example 1 with a covarince matrix $U,$ possibly different from $I,$ has as the $q-th$ (here for $q=3)\,$ coefficient the determinant of $G(x)UG(x)^{\prime}.$ With $U=I$ the determinant of $G(x)IG(x)^{\prime}$ provides

[TABLE]

and we note that the lowest degree monomial is $x^{2}y^{2}.$ It can be verified that for these restrictions and $\bar{\theta}$ we get $m_{3}=4$ (so that there can be no $U$ for which the degree could be smaller) and by Proposition 1 $m_{3}\left(U\right)=m_{3}=4$  for almost every $U\in F_{q}.$ However, below we provide $U$ for which $a_{3}\left(x,U\right)$ is such that $m_{3}\left(U\right)>4.$* Consider*

[TABLE]

For this $U$ we get

[TABLE]

with the lowest degree of monomial $m_{3}\left(U\right)=6>m_{3}=4.$

Since the coefficients of a polynomial represent symmetric polynomials in the roots (e.g., Horn and Johnson, 1985, Section 1.2), elementary symmetric polynomials in the eigenvalues can be expressed as polynomial functions in $x.$ Denote by $I_{q}\left(k\right)$ the set of all combinations of $k$ integers out of $\left\{1,...,q\right\}.$ Denote by $P_{k}\left(\lambda_{1},...\lambda_{q}\right)$ the $k-th$ elementary symmetric polynomial in $\lambda_{1},...\lambda_{q}:$

[TABLE]

**Corollary to Proposition 1. **For the eigenvalues $\lambda_{i}\left(x,U\right),$ $i=1,...,q,$  that are the solutions of the characteristic polynomial, we have that

[TABLE]

*and thus the representation $\left(\ref{Dm(k)}\right)$ applies. *

In the next proposition we apply scaling to the argument $x$ by considering $x=T^{-1/2}y$ and exploit the polynomial terms from $\left(\ref{Dm(k)}\right)$ with lowest degree of homogeneity in $\left(\ref{sym polyn}\right)$ to establish the rates for the eigenvalues of a scaled polynomial matrix. Recall that from the convergence result $\left(\ref{Delta2}\right)$ the matrix scaling $\Delta_{T}$ is associated with $G\left(.\right).$ We show that when rank of $\bar{G}\left(.\right)$ is less than $q$ (in violation of the FRALD-T condition) some eigenvalues will be converging to zero and additional scaling can be applied to have the eigenvalues converge to continuous limit functions. This additional scaling will provide the divergence rate.

Proposition 2. Under the conditions of Proposition 1 consider the scaled matrix for $y\in R^{p}:$

[TABLE]

and its eigenvalues $\bar{\lambda}_{l}^{\left(T\right)}\left(y,U\right)=\bar{\lambda}_{l}^{\left(T\right)}\left(T^{-1/2}y,U\right),$ $l=1,...,q$ in descending order. Then for some non-negative integers $\beta_{l}=\beta_{l}\left(U\right)$ that satisfy

[TABLE]

we have that

[TABLE]

where $\lambda_{l}\left(y,U\right)$ are continuous a.e. non-zero functions.

Thus we see that for eigenvalues beyond $r$ additional non-trivial scaling provides convergence to a continuous a.e. non-zero function.

The next proposition shows that convergence with these rates to a continuous (but now in some exceptional cases possibly zero) function is preserved when $U$ is replaced with a sequence $U_{T},$ of matrices from $F_{p}$ such that $U_{T}\rightarrow U.$

Proposition 3. Under the conditions of Proposition 2 consider a sequence $U_{T}\in F_{p},$ such that $U_{T}\rightarrow U$. Then

(a) if $m_{l}\left(U\right)=m_{l}$ for $l=1.,,,q$ we have that

[TABLE]

(b) if for some $k\in\left\{1,...,q\right\}$ it holds that $m_{l}\left(U\right)=m_{l}$ for $l<k$ and $m_{k}<m_{k}\left(U\right),$ then

[TABLE]

Recall that case (a) will hold for almost all $U$ by Proposition 1. Example 3 illustrates part (b) of Proposition 2: there $m_{3}=4,$ so $\beta_{3}=2,$ but $m_{3}\left(U\right)=6>4$ and thus $\lambda_{3}^{\left(T\right)}\left(T^{-1/2}y,U_{T}\right)$ scaled up by $T^{2}$ goes to zero. It is possible that for  $U_{T}$ itself $m_{3}\left(U_{T}\right)=6;$ in that case $T^{3}\lambda_{3}^{\left(T\right)}\left(T^{-1/2}y,U_{T}\right)$ converges to a non-zero limit. Alternatively, if $U_{T}$ has $m_{3}\left(U_{T}\right)=m_{3}$ for almost all $U_{T}$ but $U_{T}$ converges to $U$ sufficiently fast the rate could still be as high as $T^{3}.$ However, in case (b) to get a precise rate we also need to consider the convergence rate for $U_{T}.$

The next proposition applies the deterministic properties to provide limits for eigenvalues of the random matrix $\bar{\Sigma}_{T}\left(\hat{\theta}_{T},\hat{V}_{T}\right),$ in the diagonal eigenvalue matrix $\bar{\Lambda}_{T}\left(\hat{\theta}_{T}\right)$ of $\left(\ref{diag}\right)$. Without loss of generality consider $\bar{\theta}=0.$

Proposition 4. Suppose that assumptions 1,2 hold at $\bar{\theta}=0.$ Then there is a sequence of integers $\beta_{l},l=1,...,q,$ which depends on $G$ and $V,$ such that

[TABLE]

with all $\lambda_{l}\left(y\right)$ continuous non-negative functions a.e..

We see that if FRALD-T were not violated, no additional scaling would be required, but once it is violated the extra scaling is captured by $\bar{\beta}_{l}\geq 1$ for $r<l\leq q$ that determines the rate of explosion of the Wald statistic. The Theorem below shows this.

Theorem. Under the conditions of Proposition 4 if FRALD-T property does not hold, i.e. $r<q,$ then we have for $\bar{\beta}\geq 1$ that

[TABLE]

where $\mu_{T}\left(\hat{\theta}_{T},V_{T}\right)\underset{d}{\rightarrow}\mu\left(Y\right),$ a continuous positive a.e. function.

We thus see that if FRALD-T is violated the rate of the exlosion is at least $T$ (as in example 1 here), but could be stronger even with the same restrictions (as could be in example 3).

4 Proofs

Proof of Proposition 1.

First, consider the polynomial expansion for $p_{B}\left(\lambda;x,U\right)$, given e.g. in Harville (2008, Corollary 13,7,4). Denote by $I_{q}\left(k\right)$ the set of all combinations of $k$ integers out of $\left\{1,...,q\right\};$ denote for $\left\{i_{1},...,i_{r}\right\}\in I_{q}\left(r\right)$ by $B^{\left\{i_{1},...,i_{r}\right\}}$ a minor of the matrix $B$ obtained by striking out all the rows and columns numbered $i_{1},...,i_{r}.$ Then

[TABLE]

Since all the components of the $B\left(x,U\right)$ matrix are polynomials in $x$ it follows that the determinants of the minors are also polynomials in $x.$ Then, given $U,$ denote by $D_{m_{k}\left(U\right)}\left(x,U\right)$ the homogeneous polynomial in $a_{k}\left(x,U\right)$ of the lowest degree, denoted $m_{k}\left(U\right),$ to obtain the polynomial expansion of the Proposition.

Next, note that $a_{k}\left(x,U\right)$ is also a polynomial function in the components of the matrix $U.$ By varying $U$ over $F_{q}$ we can find the minimum possible $m_{k}\left(U\right),$ denoted $m_{k}.$ Thus there is some matrix, $Q\in F_{q}$ such that for $a_{k}\left(x,Q\right)$ we have that $m_{k}\left(Q\right)=m_{k}.$ This implies that in $a_{k}\left(x,Q\right)$ there is a homogeneous polynomial of degree $m_{k}$ in $x$ that is non-zero, thus has at least one non-zero coefficient on a monomial term of degree $m_{k}.$ Since this coefficient is a polynomial function of the components of $Q,$ considering this polynomial over the corresponding components of all $U\in F_{q}$ we note that it is non-zero a.e.. This implies that $m_{k}\left(U\right)=m_{k}$ for almost all $U\in F_{q}.$

$\blacksquare$

Proof of Proposition 2.

For every $T$ consider $T^{-1/2}y$ in place of $x$ and $M_{T}\left(y,U\right)$ in place of $B\left(x,U\right)$ in Proposition 1. Then for the corresponding characteristic polynomial, $p_{M_{T}}\left(\lambda;y,U\right),$ the expansion similar to $\left(\ref{charact polyn}\right)$ will provide coeffficients

[TABLE]

Note that we have that $\det\left[M_{T}\left(y,U\right)^{\left\{i_{1},...,i_{k}\right\}}\right]\rightarrow\det\left[\bar{G}\left(y\right)V\bar{G}\left(y\right)^{\prime\left\{i_{1},...,i_{k}\right\}}\right],$ that is a non-zero constant for $k\leq r$ and zero for $k>r.$ Therefore the coefficients can be represented as

[TABLE]

where $\tilde{m}_{k}\left(U\right)$ is zero for $k=1,...,r.$ But for $k=r+1,...,q$ there is some $\gamma_{k}\geq 1$ such that $D_{\tilde{m}_{k}\left(U\right)}\left(y,U;T\right)=T^{-\gamma_{k}}\bar{R}_{k}\left(y\right),$ where $\gamma_{k}=\frac{1}{2}\tilde{m}_{k}\left(U\right)$ and $\bar{R}_{k}\left(y\right)$ is a (positive a.e.) homogeneous polynomial in $y$ of degree $\gamma_{k}.$ So altogether we can write

[TABLE]

where $\tilde{R}_{k}\left(.\right)$ is a polynomial that can contain non-zero monomials only of degree strictly higher than $\gamma_{k}.$ Then apply the representation $\left(\ref{sym polyn}\right)$ to the corresponding coefficients to write for every $k=1,...,q$

[TABLE]

The proof is by induction on $k.$

For $k=1$ consider the largest eigenvalue $\lambda_{1}^{\left(T\right)}(T^{-1/2}y).$ Note that $r\geq 1,$ so that $\gamma_{1}$ is always zero. Since $P_{1}\left[.\right]$ is the sum of all eigenvalues we have by replacing all the $q$ eigenvalues by the largest, $\lambda_{1}^{\left(T\right)}(T^{-1/2}y),$ that

[TABLE]

then since $\bar{R}_{1}\left(y\right)>0$ a.e. the limit of $T^{\beta_{1}}\lambda_{1}^{\left(T\right)}(T^{-1/2}y)$ (with $\beta_{1}=0)$ is positive a.e..

Suppose that for $k^{\prime}\geq 1$ all $T^{\beta_{l}}\lambda_{l}^{\left(T\right)}(T^{-1/2}y)$ for $l\leq k^{\prime}$ converge to continuous positive a.e. functions.

Then by replacing in the symmetric polynomial $P_{k^{\prime}+1}\left[\lambda^{\left(T\right)}(T^{-1/2}y)\right]$ all the terms by the largest, $\lambda_{k^{\prime}+1}^{\left(T\right)}(T^{-1/2}y),$ and multipying by the rate, $T^{\gamma_{k^{\prime}+1}},$ we can write that

[TABLE]

Since the expression in the last line has a limit that is non-zero a.e., so does the expression in the second line; by the induction hypothesis $\left[\Pi_{l\leq k^{\prime}}T^{\beta_{l}}\lambda_{l}^{\left(T\right)}(T^{-1/2}y)\right]$ converges to a continuous positive a.e. function. Thus for $\beta_{k^{\prime}+1}=\left(\gamma_{k^{\prime}+1}-\Sigma_{l\leq k^{\prime}}\beta_{l}\right)$ the function $T^{\beta_{k^{\prime}+1}}\lambda_{k^{\prime}+1}^{\left(T\right)}(T^{-1/2}y)$ converges to a continuous positive a.e. function.

From the derivation it follows that $\beta_{l}=\frac{1}{2}\tilde{m}_{l}\left(U\right)=0$ for $l=1,...,r;$ $\beta_{r+1}=\gamma_{r+1}=\frac{1}{2}\tilde{m}_{r+1}\left(U\right)\geq 1;$ and generally $\beta_{k}=\gamma_{k}-\gamma_{k-1}=\frac{1}{2}\left(\tilde{m}_{k}\left(U\right)-\tilde{m}_{k-1}\left(U\right)\right)\geq 1$ for $k>r.\blacksquare$

Proof of Proposition 3.

Under the condition in (a) the degrees of homogeneity $m_{k}\left(U\right)$ for the coefficients of the characteristic polynomial of $M_{T}\left(y,G,U\right)$ and $m_{k}\left(U_{T}\right)$ for the corresponding coefficient in $M_{T}\left(y,G,U_{T}\right)$ has to be the same for large enough $T$ and thus by the proof of Proposition 2 we conclude that $T^{\beta_{l}}\bar{\lambda}_{l}^{\left(T\right)}\left(y,U_{T}\right)$ have the same positive a.e. limit as $T^{\beta_{l}}\bar{\lambda}_{l}^{\left(T\right)}\left(y,U\right).$

Under the condition in (b) we can write

[TABLE]

where $\left[\Pi_{i\leq k}T^{\beta_{l}}\lambda_{l}^{\left(T\right)}\left(T^{-1/2}y,U_{T}\right)\right]$ converges to a function that is positve a.e., but by the condition for $k$ the left-hand side converges to zero. Thus $\left[T^{(m_{k+1}-m_{k})}\lambda_{k+1}^{\left(T\right)}\left(T^{-1/2}y,U_{T}\right)\right]$ converges to zero and so does $\left[T^{(m_{k+1}-m_{k})}\lambda_{l}^{\left(T\right)}\left(T^{-1/2}y,U_{T}\right)\right]$ for any $l>k.\blacksquare$

Proof of Proposition 4.

Consider the scaling matrix $\tilde{\Delta}$ and the scaled matrix $\hat{\Sigma}_{T}\left(\hat{\theta}_{T},\hat{V}_{T}\right)=\tilde{\Delta}\Delta_{T}G\left(\hat{\theta}_{T}\right)\hat{V}_{T}G\left(\hat{\theta}_{T}\right)^{\prime}\Delta_{T}\tilde{\Delta}.$ By Assumption 1 since $T^{\frac{1}{2}}\left(\hat{\theta}_{T}\right)\underset{d}{\rightarrow}Y$ that is absolutely continuous by Proposition 2 the eigenvalues of $\tilde{\Delta}\Delta_{T}G\left(\hat{\theta}_{T}\right)VG\left(\hat{\theta}_{T}\right)^{\prime}\Delta_{T}\tilde{\Delta}$ converge in distribution to continuous functions in $Y,$ some of which are non-zero a.e.. Additionally, for any sequence of $\hat{V}_{T}$ we can select a subsequence $\hat{V}_{T^{\prime}}$ that converges a.s. to $V$ and by Proposition 3 we have the same result for the limits of eigenvalues of $\tilde{\Delta}\Delta_{T}G\left(\hat{\theta}_{T}\right)\hat{V}_{T}G\left(\hat{\theta}_{T}\right)^{\prime}\Delta_{T}\tilde{\Delta}.\blacksquare$

Proof of the Theorem.

Consider now the matrix $\hat{\Sigma}_{T}\left(\hat{\theta}_{T},\hat{V}_{T}\right)=\tilde{\Delta}\Delta_{T}G\left(\hat{\theta}_{T}\right)\hat{V}_{T}G\left(\hat{\theta}_{T}\right)^{\prime}\Delta_{T}\tilde{\Delta}$ as defined in Proposition 4. The eigenvalues of the scaled matrix, $\hat{\Sigma}_{T}\left(\hat{\theta}_{T},\hat{V}_{T}\right),$ denoted $\tilde{\lambda}_{iT},$ $i=1,...,q$ by Proposition 4 converge in distribution

[TABLE]

Then rewrite the Wald statistic as

[TABLE]

Since $\tilde{\lambda}_{1T}^{-1},...,\tilde{\lambda}_{qT}^{-1}$ are the eigenvalues of the non-negative definite matrix $\left[\hat{\Sigma}_{T}\left(\hat{\theta}_{T},\hat{V}_{T}\right)\right]^{-1},$ for any vector $\xi$ we have

[TABLE]

thus

[TABLE]

We have that

[TABLE]

with all components of the vector function $\bar{g}\left(Y\right)$ non-zero a.e. for absolutely continuous $Y.$ Then

[TABLE]

Define

[TABLE]

then

[TABLE]

By Proposition 4, continuity of the eigenvalue function, and of the maximum of continuous functions

[TABLE]

with the limit functions non-zero a.e.. Also,

[TABLE]

which is a piece-wise polynomial continuous function. The ratio

[TABLE]

exists and is non-zero a.e. and

[TABLE]

When the FRALD-T condition is violated $\bar{\beta}\geq 1$ and the Wald statistic diverges to $+\infty.$

$\blacksquare$