[proofplan] The proof has two parts. First, we use only algebra of the backshift operator: because polynomials in $B$ commute, the factorization $\phi(z) = (1-z)\psi(z)$ gives $\phi(B)X_t = \psi(B)\Delta X_t$, so the differenced process satisfies the reduced autoregressive equation. Second, the root condition on $\psi$ implies that $1/\psi$ has a power-series expansion with absolutely summable coefficients, producing a causal linear-process solution driven by the white noise. The absolute summability of those coefficients gives weak stationarity, the finite convolution identity verifies the reduced equation in $L^2$, and coefficient comparison proves uniqueness in the explicitly stated class of causal absolutely summable linear-process solutions. [/proofplan]

[step:Use the polynomial factorization to remove the unit-root factor]For every real polynomial $q(z) = \sum_{k=0}^{m} q_k z^k$, define the backshift polynomial operator \begin{align*} q(B): \mathbb{R}^{\mathbb{Z}} &\to \mathbb{R}^{\mathbb{Z}} \\ (W_t)_{t\in\mathbb{Z}} &\mapsto \left(\sum_{k=0}^{m} q_k W_{t-k}\right)_{t\in\mathbb{Z}}. \end{align*} Write $\psi(z)=\sum_{k=0}^{p}\psi_k z^k$, where $p\in\mathbb{N}\cup\{0\}$ and $\psi_k\in\mathbb{R}$ for $0\leq k\leq p$. Since $B^iB^j = B^{i+j} = B^jB^i$ for all non-negative integers $i,j$, polynomial operators in $B$ commute. Therefore \begin{align*} \phi(B) = ((1-B)\psi(B)) = \psi(B)(1-B) = \psi(B)\Delta. \end{align*} Applying this identity to $(X_t)_{t \in \mathbb{Z}}$ gives, for every $t \in \mathbb{Z}$, \begin{align*} Z_t = \phi(B)X_t = \psi(B)\Delta X_t = \psi(B)Y_t. \end{align*} Thus the differenced process $(Y_t)_{t \in \mathbb{Z}}$ satisfies the reduced autoregressive equation \begin{align*} \psi(B)Y_t = Z_t. \end{align*}[/step]

[guided]We first isolate the purely algebraic point. If $q(z) = \sum_{k=0}^{m} q_k z^k$ is a polynomial, then $q(B)$ is the map \begin{align*} q(B): \mathbb{R}^{\mathbb{Z}} &\to \mathbb{R}^{\mathbb{Z}} \\ (W_t)_{t\in\mathbb{Z}} &\mapsto \left(\sum_{k=0}^{m} q_k W_{t-k}\right)_{t\in\mathbb{Z}}. \end{align*} Also write $\psi(z)=\sum_{k=0}^{p}\psi_k z^k$, with $p\in\mathbb{N}\cup\{0\}$ and $\psi_k\in\mathbb{R}$. The important fact is that the powers of $B$ commute: \begin{align*} B^iB^jW_t = W_{t-i-j} = B^jB^iW_t. \end{align*} Hence any two polynomials in $B$ commute. Now use the assumed factorization \begin{align*} \phi(z) = (1-z)\psi(z). \end{align*} Substituting $B$ for $z$ gives \begin{align*} \phi(B) = (1-B)\psi(B) = \psi(B)(1-B) = \psi(B)\Delta, \end{align*} where $\Delta = 1-B$. Since $Y_t = \Delta X_t$, the original equation becomes \begin{align*} Z_t = \phi(B)X_t = \psi(B)\Delta X_t = \psi(B)Y_t. \end{align*} So the unit-root factor $1-B$ has been absorbed by differencing, and the remaining dynamics are governed by $\psi(B)$.[/guided]

[step:Invert the stable autoregressive polynomial by an absolutely summable power series]Because every zero of $\psi$ lies outside the closed unit disk, there exists a number $r > 1$ such that $\psi(z) \neq 0$ whenever $|z| < r$. Therefore the reciprocal map \begin{align*} R_\psi: \{z \in \mathbb{C}: |z|<r\} &\to \mathbb{C} \\ z &\mapsto \frac{1}{\psi(z)} \end{align*} is holomorphic and has a Taylor expansion \begin{align*} \frac{1}{\psi(z)} = \sum_{j=0}^{\infty} \pi_j z^j \end{align*} converging absolutely for every $|z| < r$. Choose $\rho$ with $1 < \rho < r$. By [Cauchy's coefficient estimate](/page/Cauchy%20Coefficient%20Estimate) applied on the circle $|z|=\rho$, there is a finite constant \begin{align*} M_\rho := \sup_{|z|=\rho} \left|\frac{1}{\psi(z)}\right| < \infty \end{align*} such that \begin{align*} |\pi_j| \leq M_\rho \rho^{-j} \end{align*} for every $j \geq 0$. Hence \begin{align*} \sum_{j=0}^{\infty} |\pi_j| \leq M_\rho \sum_{j=0}^{\infty} \rho^{-j} = M_\rho \frac{\rho}{\rho-1} < \infty. \end{align*}[/step]

[guided]The root condition is exactly what makes the reduced autoregressive equation stable. Since all zeros of $\psi$ lie outside the closed unit disk, the polynomial $\psi$ has no zero on some larger open disk around the origin. More precisely, there exists $r > 1$ such that \begin{align*} \psi(z) \neq 0 \end{align*} whenever $|z| < r$. Therefore the reciprocal map \begin{align*} R_\psi: \{z \in \mathbb{C}: |z|<r\} &\to \mathbb{C} \\ z &\mapsto \frac{1}{\psi(z)} \end{align*} is holomorphic. Its Taylor series at $0$ has the form \begin{align*} \frac{1}{\psi(z)} = \sum_{j=0}^{\infty} \pi_j z^j, \end{align*} and this series converges throughout the disk of radius $r$. We need more than convergence at each point: we need absolute summability of the coefficient sequence $(\pi_j)_{j=0}^{\infty}$. Choose a radius $\rho$ satisfying $1 < \rho < r$. Since the circle $\{z \in \mathbb{C}: |z|=\rho\}$ is compact and $1/\psi$ is continuous there, the constant \begin{align*} M_\rho := \sup_{|z|=\rho} \left|\frac{1}{\psi(z)}\right| \end{align*} is finite. [Cauchy's coefficient estimate](/page/Cauchy%20Coefficient%20Estimate) gives \begin{align*} |\pi_j| \leq M_\rho \rho^{-j} \end{align*} for every $j \geq 0$. Since $\rho > 1$, the geometric series is summable, and therefore \begin{align*} \sum_{j=0}^{\infty} |\pi_j| \leq M_\rho \sum_{j=0}^{\infty} \rho^{-j} = M_\rho \frac{\rho}{\rho-1} < \infty. \end{align*} This absolute summability is the analytic form of the autoregressive stationarity condition.[/guided]

[step:Construct the causal stationary solution of the reduced equation]Let $(\Omega,\mathcal{F},\mathbb{P})$ denote the probability space on which the white noise process $(Z_t)_{t \in \mathbb{Z}}$ is defined, and let $L^2(\Omega,\mathcal{F},\mathbb{P})$ denote the Hilbert space of square-integrable real-valued random variables modulo equality $\mathbb{P}$-a.s., with norm $\|U\|_{L^2}:=(\mathbb{E}[|U|^2])^{1/2}$. Write $\sigma^2:=\operatorname{Var}(Z_0)$; by the zero-mean white-noise hypothesis, $\mathbb{E}[Z_t]=0$ and $\operatorname{Cov}(Z_s,Z_u)=0$ for $s\neq u$. In this proof, a causal absolutely summable linear process means a process $(W_t)_{t\in\mathbb{Z}}$ for which there exists a sequence $(c_j)_{j=0}^{\infty}\in\ell^1$ such that \begin{align*} W_t=\sum_{j=0}^{\infty}c_jZ_{t-j} \end{align*} in $L^2(\Omega,\mathcal{F},\mathbb{P})$ for every $t\in\mathbb{Z}$. Define the process $(\widetilde{Y}_t)_{t \in \mathbb{Z}}$ by \begin{align*} \widetilde{Y}_t := \sum_{j=0}^{\infty} \pi_j Z_{t-j}. \end{align*} Since $(\pi_j)_{j=0}^{\infty} \in \ell^1 \subset \ell^2$ and $\operatorname{Var}(Z_t)=\sigma^2<\infty$, the partial sums form a Cauchy sequence in $L^2(\Omega,\mathcal{F},\mathbb{P})$, because for $N>M$, \begin{align*} \mathbb{E}\left[\left|\sum_{j=M+1}^{N}\pi_j Z_{t-j}\right|^2\right] = \sigma^2 \sum_{j=M+1}^{N}|\pi_j|^2 \leq \sigma^2 \left(\sum_{j=M+1}^{N}|\pi_j|\right)^2. \end{align*} Thus the series converges in $L^2(\Omega,\mathcal{F},\mathbb{P})$ for each $t \in \mathbb{Z}$. The process is causal because $\widetilde{Y}_t$ is a function of $Z_t,Z_{t-1},Z_{t-2},\dots$ only. The expectation functional is continuous on $L^2(\Omega,\mathcal{F},\mathbb{P})$: by the [Cauchy-Schwarz inequality](/page/Cauchy-Schwarz%20Inequality), \begin{align*} |\mathbb{E}[U]| \leq \mathbb{E}[|U|] \leq \|U\|_{L^2} \end{align*} for every $U\in L^2(\Omega,\mathcal{F},\mathbb{P})$. Therefore the mean may be computed by passing from finite partial sums to the $L^2$ limit: \begin{align*} \mathbb{E}[\widetilde{Y}_t] = \sum_{j=0}^{\infty} \pi_j \mathbb{E}[Z_{t-j}] = 0. \end{align*} For $h \in \mathbb{Z}$, using the white-noise covariance relation \begin{align*} \operatorname{Cov}(Z_s,Z_u) = \begin{cases} \sigma^2, & s=u,\\ 0, & s \neq u, \end{cases} \end{align*} we obtain the covariance by first computing finite partial sums and then passing to the $L^2$ limit. This passage is valid because if $U_N\to U$ and $V_N\to V$ in $L^2(\Omega,\mathcal{F},\mathbb{P})$, then the [Cauchy-Schwarz inequality](/page/Cauchy-Schwarz%20Inequality) gives \begin{align*} |\operatorname{Cov}(U_N,V_N)-\operatorname{Cov}(U,V)| \leq \|U_N-U\|_{L^2}\|V_N\|_{L^2}+\|U\|_{L^2}\|V_N-V\|_{L^2}, \end{align*} after expanding around $U$ and $V$. Hence \begin{align*} \operatorname{Cov}(\widetilde{Y}_{t+h},\widetilde{Y}_t) &= \operatorname{Cov}\left(\sum_{i=0}^{\infty}\pi_i Z_{t+h-i}, \sum_{j=0}^{\infty}\pi_j Z_{t-j}\right) \\ &= \sigma^2 \sum_{\substack{i,j \geq 0 \\ t+h-i=t-j}} \pi_i\pi_j \\ &= \sigma^2 \sum_{\substack{i,j \geq 0 \\ j=i-h}} \pi_i\pi_j. \end{align*} This depends only on $h$, not on $t$. Hence $(\widetilde{Y}_t)_{t \in \mathbb{Z}}$ is weakly stationary.[/step]

[guided]We now build the stationary causal solution explicitly. Let $(\Omega,\mathcal{F},\mathbb{P})$ be the probability space on which the white noise process $(Z_t)_{t \in \mathbb{Z}}$ is defined. The notation $L^2(\Omega,\mathcal{F},\mathbb{P})$ means the Hilbert space of square-integrable real-valued random variables modulo equality $\mathbb{P}$-a.s., equipped with the norm $\|U\|_{L^2}:=(\mathbb{E}[|U|^2])^{1/2}$. Write $\sigma^2:=\operatorname{Var}(Z_0)$. The zero-mean white-noise hypothesis says that $\mathbb{E}[Z_t]=0$ for every $t\in\mathbb{Z}$ and that $\operatorname{Cov}(Z_s,Z_u)=0$ whenever $s\neq u$. Here causal has a precise meaning: a process $(W_t)_{t\in\mathbb{Z}}$ is a causal absolutely summable linear process if there is a sequence $(c_j)_{j=0}^{\infty}\in\ell^1$ such that \begin{align*} W_t=\sum_{j=0}^{\infty}c_jZ_{t-j} \end{align*} in $L^2(\Omega,\mathcal{F},\mathbb{P})$ for every $t\in\mathbb{Z}$. Define \begin{align*} \widetilde{Y}_t := \sum_{j=0}^{\infty} \pi_j Z_{t-j}. \end{align*} This is causal in the stated sense because the coefficient sequence $(\pi_j)_{j=0}^{\infty}$ is in $\ell^1$ and only present and past shocks appear: the terms are $Z_t,Z_{t-1},Z_{t-2},\dots$. The coefficient sequence is absolutely summable, so in particular it is square summable. To verify convergence, consider the difference of two partial sums with $N>M$. The white-noise covariance relation gives \begin{align*} \mathbb{E}\left[\left|\sum_{j=M+1}^{N}\pi_j Z_{t-j}\right|^2\right] = \sigma^2 \sum_{j=M+1}^{N}|\pi_j|^2 \leq \sigma^2 \left(\sum_{j=M+1}^{N}|\pi_j|\right)^2. \end{align*} The right-hand side tends to $0$ as $M,N \to \infty$ because $(\pi_j)_{j=0}^{\infty}\in\ell^1$. Hence the series converges in $L^2(\Omega,\mathcal{F},\mathbb{P})$ for each fixed $t$. Its mean is computed by first using finite partial sums and then passing to the $L^2$ limit. This passage is valid because expectation is continuous on $L^2(\Omega,\mathcal{F},\mathbb{P})$: by the [Cauchy-Schwarz inequality](/page/Cauchy-Schwarz%20Inequality), \begin{align*} |\mathbb{E}[U]| \leq \mathbb{E}[|U|] \leq \|U\|_{L^2} \end{align*} for every $U\in L^2(\Omega,\mathcal{F},\mathbb{P})$. Therefore linearity of expectation for the finite sums and $\mathbb{E}[Z_t]=0$ give \begin{align*} \mathbb{E}[\widetilde{Y}_t] = \sum_{j=0}^{\infty} \pi_j \mathbb{E}[Z_{t-j}] = 0. \end{align*} To verify weak stationarity, we compute the covariance at lag $h \in \mathbb{Z}$. The white-noise covariance rule is \begin{align*} \operatorname{Cov}(Z_s,Z_u) = \begin{cases} \sigma^2, & s=u,\\ 0, & s \neq u. \end{cases} \end{align*} We first compute the covariance for finite partial sums and then let the truncation level tend to infinity. This is legitimate because covariance is continuous with respect to $L^2$ convergence: if $U_N\to U$ and $V_N\to V$ in $L^2(\Omega,\mathcal{F},\mathbb{P})$, then expanding the covariance difference and applying the [Cauchy-Schwarz inequality](/page/Cauchy-Schwarz%20Inequality) gives \begin{align*} |\operatorname{Cov}(U_N,V_N)-\operatorname{Cov}(U,V)| \leq \|U_N-U\|_{L^2}\|V_N\|_{L^2}+\|U\|_{L^2}\|V_N-V\|_{L^2}, \end{align*} which tends to $0$ when the approximating sequences are $L^2$ convergent and hence bounded in $L^2$. Therefore \begin{align*} \operatorname{Cov}(\widetilde{Y}_{t+h},\widetilde{Y}_t) &= \operatorname{Cov}\left(\sum_{i=0}^{\infty}\pi_i Z_{t+h-i}, \sum_{j=0}^{\infty}\pi_j Z_{t-j}\right) \\ &= \sum_{i=0}^{\infty}\sum_{j=0}^{\infty}\pi_i\pi_j \operatorname{Cov}(Z_{t+h-i},Z_{t-j}) \\ &= \sigma^2 \sum_{\substack{i,j \geq 0 \\ t+h-i=t-j}} \pi_i\pi_j \\ &= \sigma^2 \sum_{\substack{i,j \geq 0 \\ j=i-h}} \pi_i\pi_j. \end{align*} The final expression depends on the lag $h$ but not on the time index $t$. Thus the mean is constant and the autocovariance depends only on the lag, which is precisely weak stationarity.[/guided]

[step:Verify that the causal solution satisfies the reduced autoregressive equation]Using the notation $\psi(z)=\sum_{k=0}^{p}\psi_k z^k$ fixed above, the identity \begin{align*} \psi(z)\sum_{j=0}^{\infty}\pi_j z^j = 1 \end{align*} holds for $|z|<r$. Since the series is absolutely convergent on $|z|\leq 1$, multiplication by the finite polynomial $\psi$ is justified term-by-term, and comparison of coefficients gives \begin{align*} \sum_{k=0}^{\min\{p,m\}}\psi_k\pi_{m-k} = \begin{cases} 1, & m=0,\\ 0, & m\geq 1, \end{cases} \end{align*} where $\pi_j:=0$ for $j<0$. For $N\in\mathbb{N}$, define the truncated process $\widetilde{Y}_{t,N}:=\sum_{j=0}^{N}\pi_j Z_{t-j}$. Because $\widetilde{Y}_{t,N}\to\widetilde{Y}_t$ in $L^2(\Omega,\mathcal{F},\mathbb{P})$ for every $t$ and $\psi(B)$ is the finite linear combination $\sum_{k=0}^{p}\psi_kB^k$, we have \begin{align*} \psi(B)\widetilde{Y}_{t,N}\to \psi(B)\widetilde{Y}_t \end{align*} in $L^2(\Omega,\mathcal{F},\mathbb{P})$. For each finite $N$, rearranging the finite sums gives \begin{align*} \psi(B)\widetilde{Y}_{t,N} &= \sum_{m=0}^{N+p}\left(\sum_{k=0}^{\min\{p,m\}}\psi_k\pi_{m-k}\right)Z_{t-m}, \end{align*} where $\pi_j:=0$ for $j<0$ and for $j>N$ in this finite truncation. The coefficient identity gives the main term $Z_t$ and leaves only the boundary tail \begin{align*} R_{t,N}:=\sum_{m=N+1}^{N+p}\left(\sum_{k=m-N}^{p}\psi_k\pi_{m-k}\right)Z_{t-m}. \end{align*} Thus \begin{align*} \psi(B)\widetilde{Y}_{t,N}=Z_t+R_{t,N}. \end{align*} Using orthogonality of the white-noise variables and then the triangle inequality in $\ell^1$, we have \begin{align*} \mathbb{E}[|R_{t,N}|^2] &=\sigma^2\sum_{m=N+1}^{N+p}\left|\sum_{k=m-N}^{p}\psi_k\pi_{m-k}\right|^2 \\ &\leq \sigma^2\left(\sum_{m=N+1}^{N+p}\sum_{k=m-N}^{p}|\psi_k|\,|\pi_{m-k}|\right)^2 \\ &\leq \sigma^2\left(\sum_{k=0}^{p}|\psi_k|\right)^2\left(\sum_{j=N+1-p}^{N}|\pi_j|\right)^2. \end{align*} The last expression tends to $0$ as $N\to\infty$ because $(\pi_j)_{j=0}^{\infty}\in\ell^1$. Since also $\psi(B)\widetilde{Y}_{t,N}\to\psi(B)\widetilde{Y}_t$ in $L^2(\Omega,\mathcal{F},\mathbb{P})$, it follows that \begin{align*} \psi(B)\widetilde{Y}_t=Z_t \end{align*} in $L^2(\Omega,\mathcal{F},\mathbb{P})$ for every $t\in\mathbb{Z}$. Now let $(V_t)_{t\in\mathbb{Z}}$ be any causal absolutely summable linear-process solution of the reduced equation in the sense defined above. Thus there is a coefficient sequence $(a_j)_{j=0}^{\infty}\in\ell^1$ such that \begin{align*} V_t=\sum_{j=0}^{\infty}a_j Z_{t-j} \end{align*} in $L^2(\Omega,\mathcal{F},\mathbb{P})$ and $\psi(B)V_t=Z_t$. If $\sigma^2=0$, then $Z_t=0$ in $L^2(\Omega,\mathcal{F},\mathbb{P})$ for every $t$, so every causal linear process driven by $(Z_t)$ is the zero process in $L^2$; hence $V_t=\widetilde{Y}_t=0$ for every $t$. Assume now that $\sigma^2>0$. Define the absolutely summable sequence $(c_m)_{m=0}^{\infty}$ by \begin{align*} c_m:=\sum_{k=0}^{\min\{p,m\}}\psi_k a_{m-k}, \end{align*} where $a_j:=0$ for $j<0$. Since $\psi$ is a finite polynomial and $(a_j)_{j=0}^{\infty}\in\ell^1$, applying $\psi(B)$ to $V$ gives \begin{align*} \psi(B)V_t=\sum_{m=0}^{\infty}c_m Z_{t-m} \end{align*} in $L^2(\Omega,\mathcal{F},\mathbb{P})$. Because $\psi(B)V_t=Z_t$, the difference sequence $(d_m)_{m=0}^{\infty}$ defined by $d_0:=c_0-1$ and $d_m:=c_m$ for $m\geq1$ satisfies \begin{align*} \sum_{m=0}^{\infty}d_m Z_{t-m}=0 \end{align*} in $L^2(\Omega,\mathcal{F},\mathbb{P})$. For a fixed $\ell\geq0$, take covariance with $Z_{t-\ell}$. The covariance is continuous under the $L^2$ limit and the white-noise covariance relation gives \begin{align*} 0=\operatorname{Cov}\left(\sum_{m=0}^{\infty}d_m Z_{t-m},Z_{t-\ell}\right)=\sigma^2 d_\ell. \end{align*} Since $\sigma^2>0$, $d_\ell=0$ for every $\ell\geq0$. Therefore \begin{align*} \sum_{m=0}^{\infty}\left(\sum_{k=0}^{\min\{p,m\}}\psi_k a_{m-k}\right)z^m=1 \end{align*} for $|z|<1$. Since $\psi(z)\neq 0$ for $|z|<1$, this implies \begin{align*} \sum_{m=0}^{\infty}a_m z^m=\frac{1}{\psi(z)}=\sum_{m=0}^{\infty}\pi_m z^m \end{align*} for $|z|<1$. Uniqueness of Taylor coefficients gives $a_m=\pi_m$ for every $m\geq0$, so $V_t=\widetilde{Y}_t$ in $L^2(\Omega,\mathcal{F},\mathbb{P})$ for every $t\in\mathbb{Z}$. Thus the reduced equation has the unique causal absolutely summable linear-process solution \begin{align*} \widetilde{Y}_t = \sum_{j=0}^{\infty} \pi_j Z_{t-j}. \end{align*} The preceding covariance computation proves that this solution is weakly stationary. The first step proves separately that the differenced process $Y_t=\Delta X_t$ satisfies $\psi(B)Y_t=Z_t$; when $Y$ is taken in the stated causal absolutely summable linear-process class, it is therefore this unique stationary reduced ARMA process.[/step]

[guided]The power series was chosen so that it is the reciprocal of $\psi$. Using the notation $\psi(z)=\sum_{k=0}^{p}\psi_k z^k$ fixed above, the reciprocal identity is \begin{align*} \psi(z)\sum_{j=0}^{\infty}\pi_j z^j = 1 \end{align*} for $|z|<r$. Because $\psi$ is a finite polynomial and $(\pi_j)_{j=0}^{\infty}\in\ell^1$, multiplying the finite polynomial into the absolutely convergent series is legitimate. Comparing Taylor coefficients gives \begin{align*} \sum_{k=0}^{\min\{p,m\}}\psi_k\pi_{m-k} = \begin{cases} 1, & m=0,\\ 0, & m\geq 1, \end{cases} \end{align*} where $\pi_j:=0$ for $j<0$. We now translate this coefficient identity back into the stochastic equation using finite truncations. For $N\in\mathbb{N}$, set \begin{align*} \widetilde{Y}_{t,N}:=\sum_{j=0}^{N}\pi_jZ_{t-j}. \end{align*} The construction above gives $\widetilde{Y}_{t,N}\to\widetilde{Y}_t$ in $L^2(\Omega,\mathcal{F},\mathbb{P})$ for every $t$. Since $\psi(B)=\sum_{k=0}^{p}\psi_kB^k$ is a finite sum of shifts, applying $\psi(B)$ preserves $L^2$ convergence, so \begin{align*} \psi(B)\widetilde{Y}_{t,N}\to\psi(B)\widetilde{Y}_t \end{align*} in $L^2(\Omega,\mathcal{F},\mathbb{P})$. For each fixed $N$, all sums are finite, and therefore \begin{align*} \psi(B)\widetilde{Y}_{t,N} &= \sum_{m=0}^{N+p}\left(\sum_{k=0}^{\min\{p,m\}}\psi_k\pi_{m-k}\right)Z_{t-m}, \end{align*} where $\pi_j:=0$ for $j<0$ and for $j>N$ in this finite truncation. The coefficient identity kills all coefficients except the coefficient of $Z_t$, but the finite truncation creates boundary terms near the cutoff. More precisely, \begin{align*} \psi(B)\widetilde{Y}_{t,N}=Z_t+R_{t,N}, \end{align*} where \begin{align*} R_{t,N}:=\sum_{m=N+1}^{N+p}\left(\sum_{k=m-N}^{p}\psi_k\pi_{m-k}\right)Z_{t-m}. \end{align*} The tail $R_{t,N}$ vanishes in $L^2$. Indeed, orthogonality of the white-noise variables and the triangle inequality give \begin{align*} \mathbb{E}[|R_{t,N}|^2] &=\sigma^2\sum_{m=N+1}^{N+p}\left|\sum_{k=m-N}^{p}\psi_k\pi_{m-k}\right|^2 \\ &\leq \sigma^2\left(\sum_{m=N+1}^{N+p}\sum_{k=m-N}^{p}|\psi_k|\,|\pi_{m-k}|\right)^2 \\ &\leq \sigma^2\left(\sum_{k=0}^{p}|\psi_k|\right)^2\left(\sum_{j=N+1-p}^{N}|\pi_j|\right)^2. \end{align*} Because $(\pi_j)_{j=0}^{\infty}\in\ell^1$, the final tail tends to $0$ as $N\to\infty$. Since $\psi(B)\widetilde{Y}_{t,N}\to\psi(B)\widetilde{Y}_t$ in $L^2(\Omega,\mathcal{F},\mathbb{P})$, we conclude \begin{align*} \psi(B)\widetilde{Y}_t=Z_t. \end{align*} This is the precise reason the infinite filter may be convolved with the finite autoregressive polynomial: the convolution is first performed at the finite level, the boundary tail is estimated, and only then the $L^2$ limit is taken. It remains to explain uniqueness in the stated causal absolutely summable linear-process class without appealing to an external criterion. Let $(V_t)_{t\in\mathbb{Z}}$ be another such solution. Then, by definition of this solution class, there is a sequence $(a_j)_{j=0}^{\infty}\in\ell^1$ such that \begin{align*} V_t=\sum_{j=0}^{\infty}a_j Z_{t-j} \end{align*} in $L^2(\Omega,\mathcal{F},\mathbb{P})$ and $\psi(B)V_t=Z_t$. There is one degenerate case to separate. If $\sigma^2=0$, then $Z_t=0$ in $L^2(\Omega,\mathcal{F},\mathbb{P})$ for every $t$, so every causal linear process driven by $(Z_t)$ is the zero process in $L^2$. In that case $V_t=\widetilde{Y}_t=0$ for every $t$. Assume now that $\sigma^2>0$. Define \begin{align*} c_m:=\sum_{k=0}^{\min\{p,m\}}\psi_k a_{m-k}, \end{align*} with $a_j:=0$ for $j<0$. The sequence $(c_m)_{m=0}^{\infty}$ is absolutely summable because it is the convolution of the finite sequence $(\psi_0,\dots,\psi_p)$ with the $\ell^1$ sequence $(a_j)_{j=0}^{\infty}$. Hence applying $\psi(B)$ to $V$ gives \begin{align*} \psi(B)V_t=\sum_{m=0}^{\infty}c_mZ_{t-m} \end{align*} in $L^2(\Omega,\mathcal{F},\mathbb{P})$. Since $\psi(B)V_t=Z_t$, define $d_0:=c_0-1$ and $d_m:=c_m$ for $m\geq1$; then \begin{align*} \sum_{m=0}^{\infty}d_mZ_{t-m}=0 \end{align*} in $L^2(\Omega,\mathcal{F},\mathbb{P})$. To identify the coefficients, fix $\ell\geq0$ and take covariance with $Z_{t-\ell}$. Continuity of covariance under $L^2$ convergence justifies passing through the infinite sum, and the white-noise covariance relation yields \begin{align*} 0=\operatorname{Cov}\left(\sum_{m=0}^{\infty}d_m Z_{t-m},Z_{t-\ell}\right)=\sigma^2 d_\ell. \end{align*} Because $\sigma^2>0$, this gives $d_\ell=0$. Since $\ell$ was arbitrary, $c_0=1$ and $c_m=0$ for every $m\geq1$. Therefore \begin{align*} \sum_{m=0}^{\infty}\left(\sum_{k=0}^{\min\{p,m\}}\psi_k a_{m-k}\right)z^m=1 \end{align*} for $|z|<1$. Since the root condition gives $\psi(z)\neq0$ on $|z|<1$, division by $\psi(z)$ is valid there, and hence \begin{align*} \sum_{m=0}^{\infty}a_m z^m=\frac{1}{\psi(z)}=\sum_{m=0}^{\infty}\pi_m z^m \end{align*} for $|z|<1$. Uniqueness of Taylor coefficients now forces $a_m=\pi_m$ for every $m\geq0$. Thus any causal absolutely summable linear-process solution agrees with $\widetilde{Y}$ in $L^2(\Omega,\mathcal{F},\mathbb{P})$ at every time. We have therefore verified the reduced ARMA description: the equation $\psi(B)Y_t=Z_t$ has the unique causal absolutely summable linear-process solution \begin{align*} \widetilde{Y}_t = \sum_{j=0}^{\infty} \pi_j Z_{t-j}. \end{align*} The covariance computation above shows that this solution is weakly stationary. The first step proved separately that the algebraic differencing operation sends $\phi(B)X_t=Z_t$ to $\psi(B)\Delta X_t=Z_t$. Thus differencing cancels the simple unit-root factor $1-B$, and, in the stated causal linear-process class, the stationary reduced dynamics are governed by $\psi(B)$.[/guided]