[proofplan] The proof is the standard likelihood expansion argument. Consistency places the maximiser in an interior neighbourhood of the true parameter, so the score equation holds at $\hat{\vartheta}_n$. Taylor expansion of the score around $\vartheta_0$ expresses $\sqrt n(\hat{\vartheta}_n-\vartheta_0)$ as the inverse observed information times the normalised score. The ARMA hypotheses enter through the stated standard moment and regularity conditions, which supply consistency, score asymptotic normality, and observed-information convergence. These assumed score and Hessian limits then give the claimed normal limit. [/proofplan]

[step:Use consistency to place the estimator in the regular neighbourhood] Let $U \subset \Theta$ be the neighbourhood of $\vartheta_0$ from the regularity hypotheses. The ARMA causality, invertibility, identifiability, moment, and no-common-zero assumptions are used here through those likelihood regularity hypotheses: they give consistency, twice continuous differentiability of the log-likelihood on $U$ with probability tending to one, validity of the first-order likelihood equation at any interior maximum, the score central limit theorem, and local uniform observed-information convergence in shrinking neighbourhoods of $\vartheta_0$. Thus the proof below is the abstract likelihood argument after the ARMA-specific regularity facts have been assumed. Since $\vartheta_0$ is an interior point of $\Theta$, choose $r>0$ such that \begin{align*} B(\vartheta_0,r) := \{\vartheta \in \mathbb{R}^{p+q} : |\vartheta-\vartheta_0| < r\} \subset U \subset \Theta, \end{align*} where $|\cdot|$ denotes the Euclidean norm on $\mathbb{R}^{p+q}$. Because $\hat{\vartheta}_n \xrightarrow{\mathbb{P}} \vartheta_0$, we have \begin{align*} \mathbb{P}\bigl(\hat{\vartheta}_n \in B(\vartheta_0,r)\bigr) \to 1. \end{align*} Let $E_n$ denote the event that $\hat{\vartheta}_n \in B(\vartheta_0,r)$ and that $\ell_n$ is twice continuously differentiable on $U$. By consistency and the differentiability regularity hypothesis, $\mathbb{P}(E_n) \to 1$. On $E_n$, $\hat{\vartheta}_n$ is an interior maximiser of $\ell_n$ on the differentiability neighbourhood $U$. Therefore the first-order likelihood equation holds: \begin{align*} \nabla \ell_n(\hat{\vartheta}_n)=0. \end{align*} [/step]

[step:Expand the score around the true parameter using the integral Taylor formula]Define the score map \begin{align*} S_n: U &\to \mathbb{R}^{p+q},\\ \vartheta &\mapsto \nabla \ell_n(\vartheta), \end{align*} and define the observed Hessian matrix map as the Jacobian of the score, \begin{align*} H_n: U &\to \mathbb{R}^{(p+q)\times(p+q)},\\ \vartheta &\mapsto JS_{n,\vartheta}. \end{align*} Equivalently, the $(i,j)$ entry of $H_n(\vartheta)$ is $\partial_{\vartheta_j}S_{n,i}(\vartheta)$ for $1 \le i,j \le p+q$. Let $\mathcal{L}^1$ denote one-dimensional Lebesgue measure on $[0,1]$. On $E_n$, the line segment from $\vartheta_0$ to $\hat{\vartheta}_n$ is contained in $B(\vartheta_0,r) \subset U$, and $S_n$ is continuously differentiable on that segment. Define the pathwise score curve \begin{align*} g_n: [0,1] &\to \mathbb{R}^{p+q},\\ t &\mapsto S_n\bigl(\vartheta_0+t(\hat{\vartheta}_n-\vartheta_0)\bigr). \end{align*} By the [Fundamental Theorem of Calculus](/theorems/632) applied componentwise to $g_n$, whose components are continuously differentiable on $[0,1]$, we obtain \begin{align*} S_n(\hat{\vartheta}_n)-S_n(\vartheta_0) = \int_0^1 H_n\bigl(\vartheta_0+t(\hat{\vartheta}_n-\vartheta_0)\bigr)(\hat{\vartheta}_n-\vartheta_0)\,d\mathcal{L}^1(t). \end{align*} Define the averaged observed Hessian matrix \begin{align*} \bar{H}_n: E_n &\to \mathbb{R}^{(p+q)\times(p+q)},\\ \omega &\mapsto \int_0^1 H_n\bigl(\vartheta_0+t(\hat{\vartheta}_n(\omega)-\vartheta_0)\bigr)\,d\mathcal{L}^1(t). \end{align*} Since $S_n(\hat{\vartheta}_n)=0$ on $E_n$, we obtain \begin{align*} 0 = S_n(\vartheta_0)+\bar{H}_n(\hat{\vartheta}_n-\vartheta_0). \end{align*} Multiplying by $n^{-1/2}$ and rearranging gives \begin{align*} -\frac{1}{n}\bar{H}_n\sqrt n(\hat{\vartheta}_n-\vartheta_0) = \frac{1}{\sqrt n}S_n(\vartheta_0). \end{align*}[/step]

[guided]We expand the score, not the likelihood itself, because the estimator is characterised by the first-order condition. Define the score map \begin{align*} S_n: U &\to \mathbb{R}^{p+q},\\ \vartheta &\mapsto \nabla \ell_n(\vartheta), \end{align*} and define the observed Hessian matrix map as the Jacobian of the score, \begin{align*} H_n: U &\to \mathbb{R}^{(p+q)\times(p+q)},\\ \vartheta &\mapsto JS_{n,\vartheta}. \end{align*} Thus the $(i,j)$ entry of $H_n(\vartheta)$ is $\partial_{\vartheta_j}S_{n,i}(\vartheta)$ for $1 \le i,j \le p+q$. Let $\mathcal{L}^1$ denote one-dimensional Lebesgue measure on $[0,1]$. This avoids using a scalar one-dimensional mean value theorem in a vector setting. In several parameters, a componentwise Taylor theorem need not produce one common intermediate point for every component of the score, so we use the integral Taylor formula instead. On $E_n$, the estimator belongs to $B(\vartheta_0,r)$, and this ball is convex as a Euclidean ball in $\mathbb{R}^{p+q}$. Therefore for every $t \in [0,1]$, the parameter \begin{align*} \vartheta_{n,t}:=\vartheta_0+t(\hat{\vartheta}_n-\vartheta_0) \end{align*} belongs to $B(\vartheta_0,r) \subset U$. The same event $E_n$ also gives twice continuous differentiability of $\ell_n$ on $U$, hence continuous differentiability of $S_n$ on the whole segment $\{\vartheta_{n,t}:0\le t\le 1\}$. These hypotheses allow us to reduce the vector-valued expansion to one-dimensional calculus along the line segment. Define the pathwise score curve \begin{align*} g_n: [0,1] &\to \mathbb{R}^{p+q},\\ t &\mapsto S_n(\vartheta_{n,t}). \end{align*} Because $S_n$ is continuously differentiable on the segment, every component of $g_n$ is continuously differentiable on $[0,1]$. By the [Fundamental Theorem of Calculus](/theorems/632) applied componentwise to $g_n$, \begin{align*} S_n(\hat{\vartheta}_n)-S_n(\vartheta_0) = g_n(1)-g_n(0) = \int_0^1 H_n(\vartheta_{n,t})(\hat{\vartheta}_n-\vartheta_0)\,d\mathcal{L}^1(t). \end{align*} Since the vector $\hat{\vartheta}_n-\vartheta_0$ is independent of the integration variable $t$, we may factor it to the right of the matrix integral. Define the averaged observed Hessian matrix \begin{align*} \bar{H}_n: E_n &\to \mathbb{R}^{(p+q)\times(p+q)},\\ \omega &\mapsto \int_0^1 H_n\bigl(\vartheta_0+t(\hat{\vartheta}_n(\omega)-\vartheta_0)\bigr)\,d\mathcal{L}^1(t). \end{align*} Then the Taylor identity becomes \begin{align*} S_n(\hat{\vartheta}_n) = S_n(\vartheta_0)+\bar{H}_n(\hat{\vartheta}_n-\vartheta_0). \end{align*} Because $\hat{\vartheta}_n$ is an interior maximiser on $E_n$, the first-order likelihood equation gives $S_n(\hat{\vartheta}_n)=0$. Hence \begin{align*} 0 = S_n(\vartheta_0)+\bar{H}_n(\hat{\vartheta}_n-\vartheta_0). \end{align*} Multiplying by $n^{-1/2}$ and moving the Hessian term to the left gives \begin{align*} -\frac{1}{n}\bar{H}_n\sqrt n(\hat{\vartheta}_n-\vartheta_0) = \frac{1}{\sqrt n}S_n(\vartheta_0). \end{align*} This identity is the mechanism of the proof: the left side contains the averaged observed information, while the right side is the normalised score.[/guided]

[step:Show the averaged observed information converges to Fisher information] For $t \in [0,1]$, define the random parameter \begin{align*} \vartheta_{n,t}:=\vartheta_0+t(\hat{\vartheta}_n-\vartheta_0). \end{align*} For a matrix $A \in \mathbb{R}^{(p+q)\times(p+q)}$, let $\|A\|_{\mathrm{op}}$ denote the operator norm induced by the Euclidean norm on $\mathbb{R}^{p+q}$: \begin{align*} \|A\|_{\mathrm{op}}:=\sup\{|Av|:v\in\mathbb{R}^{p+q},\ |v|=1\}. \end{align*} Since \begin{align*} |\vartheta_{n,t}-\vartheta_0| = t|\hat{\vartheta}_n-\vartheta_0| \le |\hat{\vartheta}_n-\vartheta_0|, \end{align*} the whole segment $\{\vartheta_{n,t}:0\le t\le 1\}$ lies in the random ball $B(\vartheta_0,|\hat{\vartheta}_n-\vartheta_0|)$. Because $\hat{\vartheta}_n \xrightarrow{\mathbb{P}} \vartheta_0$, for every fixed $\delta>0$, \begin{align*} \mathbb{P}\bigl(|\hat{\vartheta}_n-\vartheta_0|>\delta\bigr) \to 0. \end{align*} The assumed local uniform observed-Hessian convergence says that \begin{align*} \sup_{\vartheta \in B(\vartheta_0,\delta)}\left\| -\frac{1}{n}H_n(\vartheta)-I(\vartheta_0)\right\|_{\mathrm{op}} \xrightarrow{\mathbb{P}} 0 \end{align*} as first $n\to\infty$ and then $\delta\downarrow 0$. Hence, for every $\varepsilon>0$ and $\eta>0$, choose $\delta>0$ small enough for the local uniform convergence bound and then use consistency to obtain \begin{align*} \mathbb{P}\left( \sup_{0\le t\le 1}\left\| -\frac{1}{n}H_n(\vartheta_{n,t})-I(\vartheta_0)\right\|_{\mathrm{op}}>\varepsilon \right) &\le \mathbb{P}\bigl(|\hat{\vartheta}_n-\vartheta_0|>\delta\bigr)\\ &\quad+ \mathbb{P}\left( \sup_{\vartheta \in B(\vartheta_0,\delta)}\left\| -\frac{1}{n}H_n(\vartheta)-I(\vartheta_0)\right\|_{\mathrm{op}}>\varepsilon \right)\\ &<\eta \end{align*} for all sufficiently large $n$. Therefore \begin{align*} \sup_{0\le t\le 1}\left\| -\frac{1}{n}H_n(\vartheta_{n,t})-I(\vartheta_0)\right\|_{\mathrm{op}} \xrightarrow{\mathbb{P}} 0. \end{align*} Using the triangle inequality for the operator norm and the definition of $\bar{H}_n$, we obtain \begin{align*} \left\|-\frac{1}{n}\bar{H}_n-I(\vartheta_0)\right\|_{\mathrm{op}} &= \left\|\int_0^1\left(-\frac{1}{n}H_n(\vartheta_{n,t})-I(\vartheta_0)\right)\,d\mathcal{L}^1(t)\right\|_{\mathrm{op}}\\ &\le \int_0^1\left\|-\frac{1}{n}H_n(\vartheta_{n,t})-I(\vartheta_0)\right\|_{\mathrm{op}}\,d\mathcal{L}^1(t)\\ &\le \sup_{0\le t\le 1}\left\| -\frac{1}{n}H_n(\vartheta_{n,t})-I(\vartheta_0)\right\|_{\mathrm{op}}, \end{align*} so \begin{align*} -\frac{1}{n}\bar{H}_n \xrightarrow{\mathbb{P}} I(\vartheta_0). \end{align*} Because $I(\vartheta_0)$ is nonsingular, choose $\rho>0$ such that every matrix $A \in \mathbb{R}^{(p+q)\times(p+q)}$ with $\|A-I(\vartheta_0)\|_{\mathrm{op}}<\rho$ is nonsingular. The convergence \begin{align*} -\frac{1}{n}\bar{H}_n \xrightarrow{\mathbb{P}} I(\vartheta_0) \end{align*} therefore implies that $-n^{-1}\bar{H}_n$ is nonsingular with probability tending to one. On this event, the adjugate formula for matrix inversion and continuity of determinant and cofactors give continuity of the inversion map at $I(\vartheta_0)$, hence \begin{align*} \left(-\frac{1}{n}\bar{H}_n\right)^{-1} \xrightarrow{\mathbb{P}} I(\vartheta_0)^{-1}. \end{align*} [/step]

[step:Combine the score limit with the information limit] From the Taylor identity, \begin{align*} \sqrt n(\hat{\vartheta}_n-\vartheta_0) = \left(-\frac{1}{n}\bar{H}_n\right)^{-1} \frac{1}{\sqrt n}S_n(\vartheta_0) \end{align*} with probability tending to one. The score regularity assumption gives \begin{align*} \frac{1}{\sqrt n}S_n(\vartheta_0) \xrightarrow{d} \mathcal{N}(0,I(\vartheta_0)). \end{align*} Together with \begin{align*} \left(-\frac{1}{n}\bar{H}_n\right)^{-1} \xrightarrow{\mathbb{P}} I(\vartheta_0)^{-1}, \end{align*} the [Slutsky Theorem](/page/Slutsky%20Theorem) yields \begin{align*} \sqrt n(\hat{\vartheta}_n-\vartheta_0) \xrightarrow{d} I(\vartheta_0)^{-1}Z, \end{align*} where $Z$ is a random vector with distribution $\mathcal{N}(0,I(\vartheta_0))$. Therefore \begin{align*} I(\vartheta_0)^{-1}Z \sim \mathcal{N}\left(0,I(\vartheta_0)^{-1}I(\vartheta_0)I(\vartheta_0)^{-1}\right) = \mathcal{N}(0,I(\vartheta_0)^{-1}). \end{align*} This proves \begin{align*} \sqrt n(\hat{\vartheta}_n-\vartheta_0) \xrightarrow{d} \mathcal{N}(0,I(\vartheta_0)^{-1}). \end{align*} [/step]