In the first part of this section, we provide technical lemmas required to prove asymptotic properties of the estimator ${\stackrel{\u02c6}{\mathit{\theta}}}_{n}.$

Denote ${E}_{t}[\cdot ]$ as the expectation with respect to ${\mathit{Y}}_{t}=\{{\mathit{Y}}_{i,t},i\in {{L}}_{n}\},$ and $E[\cdot ]$ as the expectation of $\mathit{Y}=\{{\mathit{Y}}_{t},t=1,\dots ,T\}$. Let ${N}_{i,r}$ be the set of tiles in the neighborhood of tile $i$, with radius $r$. Specifically, for two locations $i$ and $j$, we say $j\in {N}_{i,r}$ if $\parallel i-j\parallel \le r.$ Thus, the neighborhood defined in Section 2 is of radius $1$, i.e. $\{j:j\sim i\}=\{j:j\in {N}_{i,1}\}$. Denote ${n}_{r}=\underset{i\in {{L}}_{n}}{max}|{N}_{i,r}|={r}^{2}+r+1$. Actually, for any tile $i$ that is not on the boundary of the image, $|{N}_{i,r}|={n}_{r}.$

In the remainder of this paper we use the following assumptions:

–

A.1: The parameter space $\mathbf{\Theta}$ is a compact subset of ${\mathbb{R}}^{p}$, and that ${\mathit{\theta}}_{0}$ is the unique maximiser of $\mathrm{\ell}(\mathit{\theta})=\underset{{n}_{{L}}\to \mathrm{\infty}}{lim}{\mathrm{\ell}}_{n}(\mathit{\theta}).$

–

A.2: The $({n}_{{C}}+1)\times {n}_{{L}}T$ matrix $(\mathrm{\nabla}{\mathit{v}}_{1,1},\mathrm{\nabla}{\mathit{v}}_{1,2},\dots ,\mathrm{\nabla}{\mathit{v}}_{1,T},\mathrm{\nabla}{\mathit{v}}_{2,1},\dots ,\mathrm{\nabla}{\mathit{v}}_{n,T})$ is full rank.

#### Lemma 1.

*Let ${Y}_{1},\dots ,{Y}_{n}$ be independent Poisson random variables with mean ${\lambda}_{1},\dots ,{\lambda}_{n}$ respectively, where $N$ is a finite positive integer. Then for any positive integer $h$*, $E\left[\underset{i=1,\dots ,n}{max}{Y}_{i}^{h}\right]\le {n}^{h}\underset{i=1,\dots ,n}{max}E\left[{Y}_{i}^{h}\right].$

#### Proof.

$\begin{array}{ll}E\left[{max}_{i=1,\dots ,n}{Y}_{i}^{h}\right]& \u2a7dE\left[{\left(\sum _{i=1}^{n}{Y}_{i}\right)}^{h}\right]\text{\hspace{0.17em}}\\ \u2a7d{n}^{h-1}E\left[\sum _{i=1}^{n}{Y}_{i}^{h}\right]\phantom{\rule{1em}{0ex}}\text{(convexity)}\\ \u2a7d{n}^{h}{max}_{i=1,\dots ,n}E\left[{Y}_{i}^{h}\right]\end{array}$

#### Lemma 2

*Denote ${\tilde{Y}}_{{N}_{i,r},t}=\underset{j\in {N}_{i,r},c\in {C}}{max}{Y}_{j,t}^{(c)}$, with corresponding observation ${\tilde{y}}_{{N}_{i,r},t}$ and conditional mean ${\tilde{\lambda}}_{{N}_{i,r},t}$, then*
$E\left[{\left({\tilde{Y}}_{{N}_{i,r},t}+1\right)}^{B}\right]\le {w}_{r,t}\sum _{k=0}^{{B}^{t}}{f}_{t}(k){e}^{k\tilde{\alpha}}{\left(1+{\tilde{y}}_{{N}_{i,r+t},0}\right)}^{Bk},\phantom{\rule{1em}{0ex}}t=1,2,\dots ,T$(8)

*where*
$\begin{array}{c}{f}_{t}(k)=\sum _{h=\u2308k/B\u2309}^{{B}^{t-1}}{e}^{\tilde{\alpha}h}g(k,Bh){f}_{t-1}(h),\phantom{\rule{1em}{0ex}}g(a,b)=\sum _{k=a}^{b}\left(\begin{array}{c}b\\ h\end{array}\right)\left\{\begin{array}{c}h\\ a\end{array}\right\},\\ {f}_{1}(k)=g(k,B)=\sum _{h=k}^{B}\left(\begin{array}{c}B\\ h\end{array}\right)\left\{\begin{array}{c}h\\ k\end{array}\right\},\phantom{\rule{1em}{0ex}}{w}_{r,t}=\prod _{k=0}^{t-1}{n}_{r+k}{2}^{{n}_{r+k}},\end{array}$

*the $\{\cdot \}$ denotes Stirling number of the second kind, $\tilde{\alpha}=\underset{c\in {C}}{max}{\alpha}^{(c)},B=\underset{c}{max}(\sum _{c\mathrm{\prime}\in {C}}{\beta}^{(c|c\mathrm{\prime})}){n}_{1}.$*

#### Proof

$\begin{array}{ll}{\lambda}_{i,t}^{(c)}& =exp\left[{\alpha}^{(c)}+\sum _{c\mathrm{\prime}\in {C}}{\beta}^{(c|c\mathrm{\prime})}\sum _{j\in {N}_{i,1}}log({y}_{j,t-1}^{(c\mathrm{\prime})}+1)\right]\u2a7d{e}^{\tilde{\alpha}}{\left({\tilde{y}}_{{N}_{i,1},t-1}+1\right)}^{B}.\end{array}$(9)

Similarly, for any $c\in {C}$, we have ${\lambda}_{{N}_{i,r},t}^{(c)}\le {e}^{\tilde{\alpha}}{\left({\tilde{y}}_{{N}_{i,r+1},t-1}+1\right)}^{B},$ since $\{j\mathrm{\prime}\in {N}_{j,1};j\in {N}_{i,r},i\in {{L}}_{n},r>0\}=\{j\in {N}_{i,r+1};i\in {{L}}_{n},r>0\}$.

Next, we proceed by induction.

For $T=1$, by the conditional independence assumption and Lemma 1, we have $\begin{array}{rl}& {E}_{T-1}\left[{E}_{T}\left({\left({\tilde{Y}}_{{N}_{i,r},T}+1\right)}^{B}|{\mathit{Y}}_{T-1}\right)\right]={E}_{T-1}\left[\sum _{h=0}^{B}\left(\genfrac{}{}{0ex}{}{B}{h}\right){E}_{T}\left(\underset{j\in {N}_{i,r},c\in {C}}{max}{{Y}_{i,T}^{(c)}}^{h}|{\mathit{Y}}_{T-1}\right)\right]\\ & <{n}_{r}{2}^{{n}_{r}}{E}_{T-1}[\sum _{k=0}^{B}\sum _{h=k}^{B}\left(\genfrac{}{}{0ex}{}{B}{h}\right)\left\{\genfrac{}{}{0pt}{}{h}{k}\right\}{\tilde{\lambda}}_{{N}_{i,r},T}^{k}]\le {w}_{r,1}\sum _{k=0}^{B}{f}_{1}(k){e}^{k\tilde{\alpha}}{E}_{T-1}\left[{\left(1+{\tilde{Y}}_{{N}_{i,r+1},T-1}\right)}^{Bk}\right].\end{array}$

Since $T-1=0$ and ${Y}_{t}$ has constant entries at time point $0$, ${E}_{T-1}\left[{\left(1+{\tilde{Y}}_{{N}_{i,r+1},T-1}\right)}^{Bk}\right]={\left(1+{\tilde{y}}_{{N}_{i,r+1},0}\right)}^{Bk}.$

Suppose eq. (8) is true for $T=t$, then for $T=t+1$, we have $\begin{array}{rl}& {E}_{T-t-1}{E}_{T-t}{E}_{T-t+1}\dots {E}_{T}\left[{\left({\tilde{Y}}_{{N}_{i,r},T}+1\right)}^{B}|{\mathbf{Y}}_{T-1},\dots ,{\mathbf{Y}}_{T-t-1}\right]\\ & \phantom{\rule{2em}{0ex}}\u2a7d{E}_{T-t-1}\left\{{w}_{r,t}\sum _{k=0}^{{B}^{t}}{f}_{t}(k){e}^{k\tilde{\alpha}}{E}_{T-t}\left[{\left(1+{\tilde{Y}}_{{N}_{i,r+t},T-t}\right)}^{Bk}|{\mathbf{Y}}_{T-t-1}\right]\right\}\\ & \phantom{\rule{2em}{0ex}}={w}_{r,t}\sum _{k=0}^{{B}^{t}}{f}_{t}(k){e}^{k\tilde{\alpha}}\left\{\sum _{k\mathrm{\prime}=0}^{Bk}\left(\begin{array}{c}Bk\\ k\mathrm{\prime}\end{array}\right){E}_{T-t-1}\left[{E}_{T-t}\left({\tilde{Y}}_{{N}_{i,r+t},T-t}^{k\mathrm{\prime}}|{\mathbf{Y}}_{T-t-1}\right)\right]\right\}\\ & \phantom{\rule{2em}{0ex}}\u2a7d{w}_{r,t}\sum _{k=0}^{{B}^{t}}{f}_{t}(k){e}^{k\tilde{\alpha}}\left\{\sum _{k\mathrm{\prime}=0}^{Bk}\left(\begin{array}{c}Bk\\ k\mathrm{\prime}\end{array}\right){E}_{T-t-1}\left[{n}_{r+t}{2}^{{n}_{r+t}}{max}_{j\in {N}_{i,r+t},c\in {C}}{E}_{T-t}\left(Y{{j,T-t}_{(c)}^{}}^{k\mathrm{\prime}}|{\mathbf{Y}}_{T-t-1}\right)\right]\right\}\\ & \phantom{\rule{2em}{0ex}}={w}_{r,t+1}\sum _{k=0}^{{B}^{t}}{f}_{t}(k){e}^{k\tilde{\alpha}}\left[\sum _{k\mathrm{\prime}\mathrm{\prime}=0}^{Bk}\sum _{k\mathrm{\prime}=k\mathrm{\prime}\mathrm{\prime}}^{Bk}\left(\begin{array}{c}Bk\\ k\mathrm{\prime}\end{array}\right)\left\{\begin{array}{c}k\mathrm{\prime}\\ k\mathrm{\prime}\mathrm{\prime}\end{array}\right\}{E}_{T-t-1}\left({\tilde{\lambda}}_{{N}_{i,r+t},T-t}^{k\mathrm{\prime}\mathrm{\prime}}\right)\right]\text{\hspace{0.17em}}\\ & \phantom{\rule{2em}{0ex}}\u2a7d{w}_{r,t+1}\sum _{k\mathrm{\prime}\mathrm{\prime}=0}^{{B}^{t+1}}\sum _{k=\u2308k\mathrm{\prime}\mathrm{\prime}/B\u2309}^{{B}^{t}}{f}_{t}(k){e}^{k\tilde{\alpha}}g(k\mathrm{\prime}\mathrm{\prime},Bk){e}^{k\mathrm{\prime}\mathrm{\prime}\tilde{\alpha}}{E}_{T-t-1}\left[{\left(1+{\tilde{Y}}_{{N}_{i,r+t+1},T-t-1}\right)}^{Bk\mathrm{\prime}\mathrm{\prime}}\right]\\ & \phantom{\rule{2em}{0ex}}={w}_{r,t+1}\sum _{k\mathrm{\prime}\mathrm{\prime}=0}^{{B}^{t+1}}{f}_{t+1}(k\mathrm{\prime}\mathrm{\prime}){e}^{k\mathrm{\prime}\mathrm{\prime}\stackrel{\u02c9}{\alpha}}{\left(1+{\tilde{y}}_{{N}_{i,r+t+1},0}\right)}^{Bk\mathrm{\prime}\mathrm{\prime}}\end{array}$

#### Lemma 3.

*Given Assumption A.1, for any finite constant $a,b\ge 0$ and $\theta \in \mathrm{\Theta},$
$E({{\lambda}_{i,t}^{(c)}}^{a}{{S}_{i,t-1}^{(c\mathrm{\prime})}}^{b})<\mathrm{\infty},\phantom{\rule{1em}{0ex}}\mathrm{\forall}c,c\mathrm{\prime}\in {C},i\in {{L}}_{n},t=1,\dots ,T.$*

#### Proof.

By the definition of ${f}_{t}(k)$ given in Lemma 2, we know that ${f}_{t}(k)$ is bounded for all bounded $t$ under assumption A.1. Thus, Lemma 2 implies $\begin{array}{rl}E({{\lambda}_{i,t}^{(c)}}^{a}{{S}_{i,t-1}^{(c\mathrm{\prime})}}^{b})& =E\left[{\left(\sum _{j\in {N}_{i,1}}log(1+{Y}_{j,t-1}^{(c\mathrm{\prime})})\right)}^{b}{{\lambda}_{i,t}^{(c)}}^{a}\right]\\ & \le E\left[(1+{\tilde{Y}}_{{N}_{i,1},t-1}{)}^{bB}{{\lambda}_{i,t}^{(c)}}^{a}\right]\le E\left[{e}^{a\stackrel{\u02c9}{\alpha}}(1+{\tilde{Y}}_{{N}_{i,1},t-1}{)}^{(a+b)B}\right]\\ & \le {e}^{a\tilde{\alpha}}{w}_{1,t}\sum _{k=0}^{{B}^{t}}{f}_{t}(k){e}^{k\tilde{\alpha}}{\left(1+{\tilde{y}}_{{N}_{i,1+t},0}\right)}^{Bk}<\mathrm{\infty}.\end{array}$

For simplicity, define the distance between tile $i$ and $j$ as $d(i,j)=r$ if $r-1<\parallel i-j\parallel \le r.$

#### Lemma 4.

*For any* $i\in {{L}}_{n},{t}_{1}=1,\dots ,T,$$\text{Cov}({Y}_{i,{t}_{1}},{Y}_{j,{t}_{2}})=0,\phantom{\rule{1em}{0ex}}\text{for\hspace{0.17em}}\mathrm{\forall}j\in {{L}}_{n},{t}_{2}=1,\dots ,T,\text{if\hspace{0.17em}}d(i,j)>{t}_{1}+{t}_{2}.$

*and*
$|(j,{t}_{2}):\text{Cov}({Y}_{j,{t}_{2}},{Y}_{i,{t}_{1}})\ne 0;j\in {{L}}_{n},{t}_{2}=1,\dots ,T,|\le T(8{T}^{2}+4T+1)$

#### Proof.

Let ${N}_{i,t}^{\ast}=\{j:\text{Cov}({Y}_{j,0},{Y}_{i,t})\ne 0;j\in {L}\}$ be the collection of counts in tiles at time $0$ that are correlated with the count in tile $i$ at time $t$ (${Y}_{i,t}$). Due to the neighborhood structure in the autoregressive term described in Section 2, one can easily tell that ${N}_{i,t}^{\ast}$ is a neighbourhood around tile $i$, with the radius equal to $t$.

Due to the condition that ${Y}_{t}$ has constant entries at time $0$, we have $\text{Cov}({Y}_{i,{t}_{1}},{Y}_{j,{t}_{2}})=0$ if ${N}_{i,{t}_{1}}^{\ast}\cap {N}_{j,{t}_{2}}^{\ast}=\mathrm{\varnothing},$ which is true when $d(i,j)>{t}_{1}+{t}_{2}.$

For any $(i,{t}_{1})\in {D}_{n}$, $\{(j,{t}_{2}):{N}_{i,{t}_{1}}^{\ast}\cap {N}_{j,{t}_{2}}^{\ast}\ne \mathrm{\varnothing}\}$ is a neighborhood around tile $i$, with a radius ${t}_{1}+{t}_{2}$.

Since ${n}_{r}=2{r}^{2}+2r+1,$ we have $|(j,{t}_{2}):{N}_{i,{t}_{1}}^{\ast}\cap {N}_{j,{t}_{2}}^{\ast}\ne \mathrm{\varnothing}|\le T|j:{N}_{i,T}^{\ast}\cap {N}_{j,T}^{\ast}\ne \mathrm{\varnothing}|=T{N}_{2T}=T(8{T}^{2}+4T+1).$

In the second part of this section, we study the asymptotic properties of the estimator ${\stackrel{\u02c6}{\theta}}_{n}$.

#### Proposition 1

*(Existence and uniqueness) If assumption A.3 holds, then there exist unique maximizer of ${\mathrm{\ell}}_{n}(\mathit{\theta})$, denoted by ${\stackrel{\u02c6}{\mathit{\theta}}}_{n}$*.

#### Proof.

First, since $\mathbf{\Theta}$ is compact and ${\mathrm{\ell}}_{n}(\mathit{\theta})$ is continuous, at least one maximiser of ${\mathrm{\ell}}_{n}(\mathit{\theta})$ exist. Next, we wish to prove that the maximiser is unique. The $p\times p$ Hessian matrix of $-{\mathrm{\ell}}_{n}(\mathit{\theta})$ can be written as a block matrix ${\mathit{H}}_{n}(\mathit{\theta})=-{\mathrm{\nabla}}^{2}{\mathrm{\ell}}_{n}(\mathit{\theta})=\left(\begin{array}{cccc}{\mathit{H}}_{n}^{(1)}(\mathit{\theta})& \mathbf{0}& \cdots & \mathbf{0}\\ \mathbf{0}& {\mathit{H}}_{n}^{(2)}(\mathit{\theta})& \cdots & \mathbf{0}\\ \vdots & \vdots & \ddots & \vdots \\ \mathbf{0}& \mathbf{0}& \cdots & {\mathit{H}}_{n}^{({n}_{{C}})}(\mathit{\theta})\end{array}\right),$

where ${\mathit{H}}_{n}^{(c)}(\mathit{\theta})=\sum _{i\in {{L}}_{n}}\sum _{t=1}^{T}exp\left[{\mathit{v}}_{i,t}^{(c)}(\mathit{\theta})\right]\left[\mathrm{\nabla}{\mathit{v}}_{i,t}\right]{\left[\mathrm{\nabla}{\mathit{v}}_{i,t}\right]}^{\mathrm{\top}}$ is a $({n}_{{C}}+1)\times ({n}_{{C}}+1)$ matrix. Matrix $\left[\mathrm{\nabla}{\mathit{v}}_{i,t}\right]{\left[\mathrm{\nabla}{\mathit{v}}_{i,t}\right]}^{\mathrm{\top}}$ is positive semidefinite with rank 1. By Assumption A.2, $\sum _{i\in {{L}}_{n}}\sum _{t=1}^{T}\left[\mathrm{\nabla}{\mathit{v}}_{i,t}\right]{\left[\mathrm{\nabla}{\mathit{v}}_{i,t}\right]}^{\mathrm{\top}}$ is full rank, which means ${\mathit{H}}_{n}^{(c)}(\mathit{\theta})$ is positive definite for all $c\in {C}$ and $\mathit{\theta}\in \mathbf{\Theta}$, since $exp\left[{\mathit{v}}_{i,t}^{(c)}(\mathit{\theta})\right]>0.$ This shows that $-{\mathrm{\ell}}_{n}(\mathit{\theta})$ is strictly convex, which implies ${\stackrel{\u02c6}{\mathit{\theta}}}_{n}$ is unique.

#### Proposition 2

*[Consistency] If the regularity assumption A.1 holds, then ${\stackrel{\u02c6}{\mathit{\theta}}}_{n}\stackrel{p}{\to}{\mathit{\theta}}_{0}$ with probability tending 1, as ${n}_{{L}}\to \mathrm{\infty}$*.

#### Proof.

We proceed by verifying the conditions of Theorem 2 in [31]. First we show that the score functions are ${{L}}_{p}$-Uniform Integrable for $p<3$, i.e. $\underset{n\to \mathrm{\infty}}{lim}\underset{\begin{array}{c}i\in {{L}}_{n}\\ t=1,\dots ,T\end{array}}{sup}\underset{\mathit{\theta}\in \mathbf{\Theta}}{sup}E[{\mathit{u}}_{i,t}^{p}(\mathit{\theta})I({\mathit{u}}_{i,t}(\mathit{\theta})>k)]\to \mathbf{0},\phantom{\rule{1em}{0ex}}\text{as\hspace{0.17em}}k\to \mathrm{\infty}.$(10)

The general form of each entry of ${\mathit{u}}_{i,t}(\mathit{\theta})$ is $({\lambda}_{i,t}^{(c)}-{y}_{i,t}^{(c)}){S}_{i,t-1}^{c\mathrm{\prime}}$, take $p=3$, we have $\begin{array}{c}E[(({\lambda}_{i,t}^{(c)}-{y}_{i,t}^{(c)}){S}_{i,t-1}^{c\mathrm{\prime}}{)}^{3}]\\ ={E}_{1}\dots {E}_{t-2}{E}_{t-1}\left[{E}_{t}\left[{(({\lambda}_{i,t}^{(c)}-{Y}_{i,t}^{(c)}){S}_{i,t-1}^{c\mathrm{\prime}})}^{3}|{\mathbf{Y}}_{t-1}\right]|{\mathbf{Y}}_{t-2}\right]\dots \\ ={E}_{1}\dots {E}_{t-2}{E}_{t-1}\left[S{{i,t-1}_{c\mathrm{\prime}}^{}}^{3}[\lambda {{i,t}_{(c)}^{}}^{3}-3\lambda {{i,t}_{(c)}^{}}^{2}{E}_{t}\left[{Y}_{i,t}^{(c)}|{\mathbf{Y}}_{t-1}\right]+3{\lambda}_{i,t}^{(c)}{E}_{t}\left[Y{{i,t}_{(c)}^{}}^{2}|{\mathbf{Y}}_{t-1}\right]+{E}_{t}\left[Y{{i,t}_{(c)}^{}}^{3}|{\mathbf{Y}}_{t-1}\right]]|{\mathbf{Y}}_{t-2}\right]\dots \\ ={E}_{1}\dots {E}_{t-2}{E}_{t-1}\left[S{{i,t-1}_{c\mathrm{\prime}}^{}}^{3}(2\lambda {{i,t}_{(c)}^{}}^{3}+6\lambda {{i,t}_{(c)}^{}}^{2}+{\lambda}_{i,t}^{(c)})|{\mathbf{Y}}_{t-2}\right]\dots ,\end{array}$(a)

which is finite by lemma 3. This gives us the ${{L}}_{3}-$boundedness of ${\mathit{u}}_{i,t}(\mathit{\theta})$, i.e. $\underset{n\to \mathrm{\infty}}{lim}\underset{\begin{array}{c}i\in {{L}}_{n}\\ t=1,\dots ,T\end{array}}{sup}\underset{\mathit{\theta}\in \mathbf{\Theta}}{sup}E[{{\mathit{u}}_{i,t}^{(c)}(\mathit{\theta})}^{3}]<\mathrm{\infty},$

which implies ${{L}}_{p}$-Uniform Integrability, for $p<3$.

Second, we show the stochastic equicontinuity of ${\mathit{u}}_{i,t}(y;\mathit{\theta})$, i.e. $\underset{n\to \mathrm{\infty}}{lim}\underset{\begin{array}{c}i\in {{L}}_{n}\\ t=1,\dots ,T\end{array}}{sup}P\left(\underset{\begin{array}{c}\mathit{\theta},\mathit{\theta}\mathrm{\prime}\in \mathbf{\Theta}\\ \parallel \mathit{\theta}-\mathit{\theta}\mathrm{\prime}\parallel <\delta \end{array}}{sup}|{\mathit{u}}_{i,t}(\mathit{\theta})-{\mathit{u}}_{i,t}(\mathit{\theta}\mathrm{\prime})|>\u03f5\right)=\mathbf{0}.$

The $\mathrm{\nabla}{\mathit{u}}_{i,t}(\mathit{\theta})$ is a $p\times p$ matrix, with each column being either $\frac{\mathrm{\partial}{\mathit{\gamma}}_{i,t}(\mathit{\theta})}{\mathrm{\partial}{\beta}^{(c|c\mathrm{\prime})}}\otimes \mathrm{\nabla}{\mathit{v}}_{i,t}$ or $\frac{\mathrm{\partial}{\mathit{\gamma}}_{i,t}(\mathit{\theta})}{\mathrm{\partial}{\alpha}^{(c)}}\otimes \mathrm{\nabla}{\mathit{v}}_{i,t}$, and $\frac{\mathrm{\partial}{\mathit{\gamma}}_{i,t}(\mathit{\theta})}{\mathrm{\partial}{\beta}^{(c|c\mathrm{\prime})}}=(0,\dots ,0,{\lambda}_{i,t}^{(c)}{S}_{i,t}^{(c)},0,\dots ),\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\frac{\mathrm{\partial}{\mathit{\gamma}}_{i,t}(\mathit{\theta})}{\mathrm{\partial}{\alpha}^{(c)}}=(0,\dots ,0,{\lambda}_{i,t}^{(c)},0,\dots ).$

Thus, the non-zero entries of $E\underset{\mathit{\theta}\in \mathbf{\Theta}}{sup}\left[\mathrm{\nabla}{\mathit{u}}_{i,t}(\mathit{\theta})\right]$ have the general form: $E\underset{\mathit{\theta}\in \mathbf{\Theta}}{sup}\left[{\lambda}_{i,t}^{(c)}{S}_{i,t}^{(c)}{S}_{i,t}^{(c\mathrm{\prime})}\right]$, which are bounded by an equivalent analogous to Lemma 3.

Thirdly, we check $\alpha -$mixing conditions. Let $U$ and $V$ be two subsets of ${D}_{n}$, and let $\sigma (U)=\sigma \{{Y}_{i,t};(i,t)\in U\}$ be the $\sigma -$algebra generated by random variables ${Y}_{i,t},(i,t)\in U$.

Define $\alpha (U,V)=sup\{|P(A\cap B)-P(A)P(B)|;A\in \sigma (U),B\in \sigma (V)\}.$

Then the $\alpha -$mixing coefficient for the random field $\{{Y}_{i,t},i\in {{L}}_{n},t=1,\dots ,T\}$ is defined as $\alpha (k,l,m)=sup\{\alpha (U,V),|U|\le k,|V|\le l,d(U,V)\ge m\}.$

Following Bai et al. [40], in an $a-$dimensional space, we need (a) $\mathrm{\exists}\delta >0s.t.\sum _{m=1}^{\mathrm{\infty}}{m}^{a-1}\alpha (1,1,m{)}^{\delta /(2+\delta )}<\mathrm{\infty},$ (b) For $k+l\le 4,\sum _{m=1}^{\mathrm{\infty}}{m}^{a-1}\alpha (k,l,m)<\mathrm{\infty},$ (c) $\mathrm{\exists}\u03f5>0\phantom{\rule{1em}{0ex}}s.t.\phantom{\rule{thickmathspace}{0ex}}\alpha (1,\mathrm{\infty},m)={O}({m}^{-a-\u03f5}),$ where $k,l,m\in \mathbb{N}$ and $d(U,V)=min\{\parallel i-j\parallel :i\in U,j\in V\}$ is the distance between sets $U$ and $V$.

For any fixed ${i}_{1},\dots ,{i}_{k}\in {{L}}_{n},k<\mathrm{\infty}$ and ${t}_{1}=0,\dots ,T$,

consider $U=\{{Y}_{i,{t}_{1}}={y}_{i,{t}_{1}},\dots ,{Y}_{{i}_{k},{t}_{1}}={y}_{{i}_{k},{t}_{1}}\}$ and $V=\{{Y}_{j,{t}_{2}}={y}_{j,{t}_{2}};j\in {{L}}_{n},{t}_{2}=0,\dots ,T\}$, then $|U|=k$ and $|V|\to \mathrm{\infty}$ as $n\to \mathrm{\infty}.$ By Lemma 4, we have $P({Y}_{i,{t}_{1}}={y}_{i,{t}_{1}},{Y}_{j,{t}_{2}}={y}_{i,{t}_{1}})-P({Y}_{i,{t}_{1}}={y}_{i,{t}_{1}})P({Y}_{j,{t}_{2}}={y}_{j,{t}_{2}})=0,$ if $d(i,j)>{t}_{1}+{t}_{2}$. Thus, $\alpha (U,V)=0$ for any $|U|=k$, provided that $d(U,V)>2T$, that is, $\alpha (k,\mathrm{\infty},m)=0$ if $m>2T$.

This implies all three mixing conditions.

Finally, by Theorem 3 in Jenish and Prucha [31], Uniform Integrability in eq. (10) and mixing condition (a) ensure that the score functions ${\mathit{u}}_{i,t}(\mathit{y};\mathit{\theta})$ satisfy a point wise law of large numbers in the sense that $\frac{1}{{n}_{{L}}}\sum _{i\in {{L}}_{n}}\sum _{t=1}^{T}\underset{\mathit{\theta}\in \mathbf{\Theta}}{sup}({\mathit{u}}_{i,t}(\mathit{y},\mathit{\theta})-E{\mathit{u}}_{i,t}(\mathit{y};\mathit{\theta}))\stackrel{p}{\to}\mathbf{0},\text{as\hspace{0.17em}}{n}_{{L}}\to \mathrm{\infty},$

for all $\mathit{\theta}\in \mathbf{\Theta}.$

#### Proposition 3.

*If the regularity assumptions A.1 and A.2 hold, we have*
$\sqrt{{n}_{{L}}}{\mathit{V}}_{n}(\mathit{\theta}{)}^{-1/2}({\stackrel{\u02c6}{\mathit{\theta}}}_{n}-{\mathit{\theta}}_{0})$
*converges in distribution to a*
$p-$*variate Normal with zero mean vector and identity variance, as*
${n}_{{L}}\to \mathrm{\infty}$.

#### Proof.

First, we show the uniform law of large numbers for $\mathrm{\nabla}{\mathit{u}}_{n}(\mathit{\theta})$: $\underset{\mathit{\theta}}{sup}\u2225\mathrm{\nabla}{\mathit{u}}_{n}(\mathit{\theta})-E\left[\mathrm{\nabla}{\mathit{u}}_{n}(\mathit{\theta})\right]\u2225\stackrel{p}{\to}\mathbf{0},\phantom{\rule{1em}{0ex}}\text{as\hspace{0.17em}}{n}_{{L}}\to \mathrm{\infty},$(11)

where ${\mathit{u}}_{n}(\mathit{\theta})=\mathrm{\nabla}{\mathrm{\ell}}_{n}(\mathit{\theta})/{n}_{{L}}$ as defined in Section 2. Note that $\begin{array}{rl}\text{Var}\left(\mathrm{\nabla}{\mathit{u}}_{n}(\mathit{\theta})\right)& =\frac{1}{{n}_{{L}}^{2}}\text{Var}\left(\sum _{i=1}^{n}\sum _{t=1}^{T}\mathrm{\nabla}{\mathit{u}}_{i,t}(\mathit{\theta})\right)\\ & =\frac{1}{{n}_{{L}}^{2}}\sum _{i\in {{L}}_{n}}\sum _{t=1}^{T}\text{Var}\left(\mathrm{\nabla}{\mathit{u}}_{i,t}(\mathit{\theta})\right)\\ & +\frac{1}{{n}_{{L}}^{2}}\sum _{i\in {{L}}_{n}}\sum _{{t}_{1}=1}^{T}\sum _{\begin{array}{c}j\in {{L}}_{n}j\ne i\end{array}}\sum _{\begin{array}{c}{t}_{2}=1{t}_{2}\ne {t}_{1}\end{array}}^{T}\text{Cov}\left(\mathrm{\nabla}{\mathit{u}}_{i,{t}_{1}}(\mathit{\theta}),\mathrm{\nabla}{\mathit{u}}_{j,{t}_{2}}(\mathit{\theta})\right)\end{array}$(12)

The first term in eq. (12) is ${O}({n}_{{L}}^{-1})$, since $\text{Var}\left(\mathrm{\nabla}{\mathit{u}}_{i,t}(\mathit{\theta})\right)\le {\left[E\left(\mathrm{\nabla}{\mathit{u}}_{i,t}(\mathit{\theta})\right)\right]}^{2}$, which is shown to be finite in the proof of Proposition 2.

For the second term in eq. (12), by Lemma 2 we have $\begin{array}{rl}& \frac{1}{{n}_{{L}}^{2}}\sum _{i\in {{L}}_{n}}\sum _{{t}_{1}=1}^{T}\sum _{\begin{array}{c}j\in {{L}}_{n}j\ne i\end{array}}\sum _{\begin{array}{c}{t}_{2}=1{t}_{2}\ne {t}_{1}\end{array}}^{T}\text{Cov}\left(\mathrm{\nabla}{\mathit{u}}_{i,{t}_{1}}(\mathit{\theta}),\mathrm{\nabla}{\mathit{u}}_{j,{t}_{2}}(\mathit{\theta})\right)\\ & \le \frac{1}{{n}_{{L}}^{2}}\sum _{i\in {{L}}_{n}}\sum _{{t}_{1}=1}^{T}T(8{T}^{2}+4T+1)\underset{\begin{array}{c}j:d(i,j)\le 2T{t}_{2}\ne {t}_{1}\end{array}}{max}\text{Cov}\left(\mathrm{\nabla}{\mathit{u}}_{i,{t}_{1}}(\mathit{\theta}),\mathrm{\nabla}{\mathit{u}}_{j,{t}_{2}}(\mathit{\theta})\right),\end{array}$

where $\text{Cov}\left(\mathrm{\nabla}{\mathit{u}}_{i,{t}_{1}}(\mathit{\theta}),\mathrm{\nabla}{\mathit{u}}_{j,{t}_{2}}(\mathit{\theta})\right)\le E\left(\mathrm{\nabla}{\mathit{u}}_{i,{t}_{1}}(\mathit{\theta}),\mathrm{\nabla}{\mathit{u}}_{j,{t}_{2}}(\mathit{\theta})\right)\le E{\left(\mathrm{\nabla}{\mathit{u}}_{i,{t}_{1}}(\mathit{\theta})\right)}^{2}+E{\left(\mathrm{\nabla}{\mathit{u}}_{i,{t}_{2}}(\mathit{\theta})\right)}^{2}$ is finite by Lemma 2. Thus, the second term in eq. (12) is also of order ${O}({n}_{{L}}^{-1})$ element wise, which means $\text{Var}\left(\mathrm{\nabla}{\mathit{u}}_{n}(\mathit{\theta})\right)\to \mathbf{\text{0}}$ as $n\to \mathrm{\infty}.$ Therefore, eq. (11) follows by Chebyshev’s inequality.

Second, ${\mathit{V}}_{n}(\mathit{\theta})=1/{n}_{{L}}\text{Var}\left(\sum _{i\in {{L}}_{n}}\sum _{t=1}^{T}{\mathit{u}}_{it}(\mathit{\theta})\right)=-1/{n}_{{L}}E\left(\sum _{i\in {{L}}_{n}}\sum _{t=1}^{T}{\mathit{u}}_{it}(\mathit{\theta})\right)=1/{n}_{{L}}{\mathit{H}}_{n}(\mathit{\theta})$, which is shown to be positive definite under Assumption A.2 in Proposition 1. Thus, together with uniform Integrability in eq. (10) and the mixing conditions, by Theorem 1 in [31], we have $\sqrt{{n}_{{L}}}{\mathit{V}}_{n}(\mathit{\theta}{)}^{-1/2}{\mathit{u}}_{n}(\mathit{\theta})\to N(\mathbf{0},{\mathit{I}}_{p})$(13)

Finally, by Taylor’s expansion, $\begin{array}{c}{\mathbf{u}}_{n}({\stackrel{\u02c6}{\theta}}_{n})=\mathbf{0}={\mathbf{u}}_{n}({\theta}_{0})+\mathrm{\nabla}{\mathbf{u}}_{n}({\theta}_{0})({\stackrel{\u02c6}{\theta}}_{n}-{\theta}_{0})+\frac{1}{2}{\mathrm{\nabla}}^{2}{\mathbf{u}}_{n}({\theta}_{0}){({\tilde{\theta}}_{n}-{\theta}_{0})}^{2}\\ \Rightarrow \mathbf{0}=\sqrt{{n}_{{L}}}{\mathbf{V}}_{n}{({\theta}_{0})}^{-1/2}{\mathbf{u}}_{n}({\theta}_{0})+\sqrt{{n}_{{L}}}{\mathbf{V}}_{n}{(\theta )}^{-1/2}\mathrm{\nabla}{\mathbf{u}}_{n}({\theta}_{0})({\stackrel{\u02c6}{\theta}}_{n}-{\theta}_{0})+\\ \frac{1}{2}\sqrt{{n}_{{L}}}{\mathbf{V}}_{n}{({\theta}_{0})}^{-1/2}{\mathrm{\nabla}}^{2}{\mathbf{u}}_{n}({\tilde{\theta}}_{n}){({\stackrel{\u02c6}{\theta}}_{n}-{\theta}_{0})}^{2},\end{array}$(14)

where ${\tilde{\mathit{\theta}}}_{n}$ is a vector with elements between ${\stackrel{\u02c6}{\mathit{\theta}}}_{n}$ and ${\mathit{\theta}}_{0}.$ Since ${\stackrel{\u02c6}{\mathit{\theta}}}_{n}={\mathit{\theta}}_{0}+{o}_{p}(\mathbf{1})$ by Proposition 2, we have $({\tilde{\mathit{\theta}}}_{n}-{\mathit{\theta}}_{0}{)}^{2}=({\stackrel{\u02c6}{\mathit{\theta}}}_{n}-{\mathit{\theta}}_{0}){o}_{p}(\mathbf{1}).$ The second derivative ${\mathrm{\nabla}}^{2}{\mathit{u}}_{n}(\mathit{\theta})$ is a $p\times p\times p$ matrix, with entries being either $0$ or ${\lambda}_{it}^{(c)}{S}_{i,t-1}^{({c}_{1})}{S}_{i,t-1}^{({c}_{2})}{S}_{i,t-1}^{({c}_{3})}$, where $i=1,\dots ,n,$ and $t=1,\dots ,T,$ and $c,{c}_{1},{c}_{2},{c}_{3}\in {C}.$ Due to the structure of ${\lambda}_{it}^{(c)}$ and ${S}_{i,t-1}^{(c)}$ in Section 2, all non-zero elements in ${\mathrm{\nabla}}^{2}{\mathit{u}}_{n}(\mathit{\theta})$ are monotone with respect to $\mathit{\theta}.$ Thus, there exists ${\mathit{\theta}}_{s}\in \mathbf{\Theta}$ such that ${\mathrm{\nabla}}^{2}{\mathit{u}}_{n}({\mathit{\theta}}_{s})\ge {\mathrm{\nabla}}^{2}{\mathit{u}}_{n}(\mathit{\theta})$ for all $\mathit{\theta}\in \mathbf{\Theta}.$ Therefore, we have $E\underset{\mathit{\theta}\in \mathbf{\Theta}}{sup}{\mathrm{\nabla}}^{2}{\mathit{u}}_{n}(\mathit{\theta})=\underset{\mathit{\theta}\in \mathbf{\Theta}}{sup}E{\mathrm{\nabla}}^{2}{\mathit{u}}_{n}(\mathit{\theta})$,

which can be shown to be finite by an equivalent analogous to Lemma 3.

Thus, eq. (13) can be written as $\begin{array}{rl}\mathbf{0}=& \sqrt{{n}_{{L}}}{\mathit{V}}_{n}({\mathit{\theta}}_{0}{)}^{-1/2}{\mathit{u}}_{n}({\mathit{\theta}}_{0})+\sqrt{{n}_{{L}}}{\mathit{V}}_{n}(\mathit{\theta}{)}^{-1/2}\left(\mathrm{\nabla}{\mathit{u}}_{n}({\mathit{\theta}}_{0})+{o}_{p}(\mathbf{1})\right)({\stackrel{\u02c6}{\mathit{\theta}}}_{n}-{\mathit{\theta}}_{0}),\end{array}$

By eq. (11), $\mathrm{\nabla}{\mathit{u}}_{n}({\mathit{\theta}}_{0})\stackrel{p}{\to}E\left[\mathrm{\nabla}{\mathit{u}}_{n}({\mathit{\theta}}_{0})\right]=-{\mathit{V}}_{n}({\mathit{\theta}}_{0})$, since ${\mathrm{\ell}}_{n}(\mathit{\theta})$ is the full likelihood. Therefore, by eqs. (12) and (13), we have $\sqrt{{n}_{{L}}}{\mathit{V}}_{n}({\mathit{\theta}}_{0}{)}^{1/2}({\stackrel{\u02c6}{\mathit{\theta}}}_{n}-{\mathit{\theta}}_{0})\stackrel{d}{\to}N(\mathbf{0},{\mathit{I}}_{p}).$

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