[proofplan] We centre the process and use the autoregressive equation to exhibit an explicit linear predictor of $Y_t$ using only the first $p$ lags. For any larger lag order $k>p$, the corresponding residual is exactly the innovation $Z_t$, which is orthogonal to every variable in the larger lag space by causality. The nonsingularity of the lag covariance matrix forces the projection coefficients onto the $k$ lag variables to be unique, so the unique coefficient of the extra lag $Y_{t-k}$ must be zero. [/proofplan]

[step:Introduce the finite lag space and the candidate predictor using only the first $p$ lags] Fix $k \in \mathbb{N}$ with $k > p$ and fix $t \in \mathbb{Z}$. Define the finite-dimensional subspace $V_k \subset L^2(\Omega,\mathcal{F},\mathbb{P})$ by \begin{align*} V_k := \operatorname{span}\{Y_{t-1},Y_{t-2},\dots,Y_{t-k}\}. \end{align*} Define the candidate predictor $P_k \in V_k$ by \begin{align*} P_k := \sum_{i=1}^{p}\phi_i Y_{t-i}. \end{align*} This belongs to $V_k$ because $k>p$, so each lag $Y_{t-i}$ with $1 \leq i \leq p$ is among the generators of $V_k$. By the centred autoregressive equation, \begin{align*} Y_t - P_k = Z_t. \end{align*} [/step]

[step:Show that the innovation residual is orthogonal to the entire lag space]For every $j \in \{1,\dots,k\}$, the time index $t-j$ satisfies $t-j<t$. The causal innovation orthogonality hypothesis therefore gives \begin{align*} \mathbb{E}[Z_tY_{t-j}] = 0. \end{align*} Since $Y_t-P_k=Z_t$, it follows that \begin{align*} \mathbb{E}[(Y_t-P_k)Y_{t-j}] = 0 \end{align*} for every $j \in \{1,\dots,k\}$. Thus $Y_t-P_k$ is orthogonal in $L^2(\Omega,\mathcal{F},\mathbb{P})$ to each generator of $V_k$, and hence to every element of $V_k$ by linearity of expectation.[/step]

[guided]We need to verify that $P_k$ is not merely a plausible autoregressive predictor, but actually satisfies the defining orthogonality condition for the centred $L^2$ projection onto the whole lag space $V_k$. The residual after subtracting $P_k$ is \begin{align*} Y_t - P_k = Z_t. \end{align*} Now take an arbitrary generator $Y_{t-j}$ of $V_k$, where $j \in \{1,\dots,k\}$. Since $j \geq 1$, the variable $Y_{t-j}$ is a past centred value relative to time $t$. The causal AR hypothesis says precisely that the innovation $Z_t$ is uncorrelated with every centred past variable. Therefore \begin{align*} \mathbb{E}[Z_tY_{t-j}] = 0. \end{align*} Substituting $Z_t=Y_t-P_k$ gives \begin{align*} \mathbb{E}[(Y_t-P_k)Y_{t-j}] = 0. \end{align*} This holds for every generator $Y_{t-1},\dots,Y_{t-k}$. If $W \in V_k$, then there are scalars $c_1,\dots,c_k \in \mathbb{R}$ such that \begin{align*} W = \sum_{j=1}^{k} c_jY_{t-j}. \end{align*} Using linearity of expectation and the orthogonality just proved, \begin{align*} \mathbb{E}[(Y_t-P_k)W] = \sum_{j=1}^{k} c_j\mathbb{E}[(Y_t-P_k)Y_{t-j}] = 0. \end{align*} Hence the residual $Y_t-P_k$ is orthogonal to all of $V_k$. This is exactly the normal equation condition characterising the centred $L^2$ projection onto $V_k$.[/guided]

[step:Use nonsingularity to identify the unique projection coefficients] The space $V_k$ is finite-dimensional, hence closed in the [Hilbert space](/page/Hilbert%20Space) $L^2(\Omega,\mathcal{F},\mathbb{P})$. By the Hilbert-space [projection theorem](/theorems/1985), the orthogonality condition from the preceding step characterises the centred $L^2$ projection onto $V_k$, so \begin{align*} \operatorname{Proj}_{V_k}Y_t = P_k. \end{align*} Writing $P_k$ in the $k$-lag generators gives the coefficient vector $b_k=(b_{k,1},\dots,b_{k,k}) \in \mathbb{R}^k$ defined by $b_{k,j}:=\phi_j$ for $1 \leq j \leq p$ and $b_{k,j}:=0$ for $p<j\leq k$. If another coefficient vector $c=(c_1,\dots,c_k)\in\mathbb{R}^k$ represents the same projection, then the difference $d=c-b_k \in \mathbb{R}^k$ satisfies \begin{align*} \sum_{j=1}^{k} d_jY_{t-j}=0 \end{align*} in $L^2(\Omega,\mathcal{F},\mathbb{P})$. Taking the square in $L^2$ gives \begin{align*} 0 = \mathbb{E}\left[\left(\sum_{j=1}^{k} d_jY_{t-j}\right)^2\right] = d^\top \Gamma_k d. \end{align*} Because $\Gamma_k$ is nonsingular and is a covariance matrix, it is positive definite under the stated nonsingularity hypothesis. Hence $d=0$, so the coefficient vector is unique and equals $b_k$. [/step]

[step:Read off the coefficient of the $k$th lag] Since $k>p$, the definition of $b_k$ gives $b_{k,k}=0$. The uniqueness of projection coefficients gives $a_{k,k}=b_{k,k}$. Therefore \begin{align*} a_{k,k}=0. \end{align*} Because $k>p$ was arbitrary, the partial autocorrelation is zero at every lag $k>p$. [/step]