10. Multivariate Time Series
本章主题：多变量时间序列。 §10.1 根、脉冲响应与协整：VAR(\(n\)) \((\mathbb I_m-\mathbf B(L))\mathbf y_t=\mathcal A\epsilon_t\)，合成冲击 \(\mathbf u_t=\mathcal A\epsilon_t\sim N(0,\Sigma)\)、\(\Sigma=\mathcal A\mathcal A'\)；堆叠为 VAR(1) \(\mathbf x_t=\mathcal B\mathbf x_{t-1}+\mathcal A\epsilon_t\)；特征多项式 \(p(\lambda)=\det(\lambda\mathbb I_{mn}-\mathcal B)\)，可对角化 \(\mathcal B=\mathbf V\mathbf D\mathbf V^{-1}\) 给出 Wold 分解；脉冲响应 \(\mathbf r_\mathbf a(k)=\mathcal B^k\mathbf a\)。§10.2 协整与误差修正：定义 10.3 秩 \(r\) 协整（各分量 \(I(1)\) 而某些线性组合 \(\beta'\mathbf y_t\) 平稳）；命题 10.1 协整矩阵 \(\beta=\)（\(\mathbf V^{-1}\) 前 \(r\) 行的转置）；脉冲响应拆为暂时部分 + 永久部分，\(\mathbf r_\mathbf a(\infty)=\nu_\star\mathbf a_\star\)；误差修正 \(\Delta\mathbf y_t=-\alpha\beta'\mathbf y_{t-1}+\mathbf u_t\)，\(\alpha=\nu(\mathbb I_r-\mathbf D_r)\)（命题 10.2 Engle–Granger 1987）。§10.3 \(\epsilon\) 冲击的脉冲响应识别：\(\epsilon_{t,j}\) 的脉冲响应是 \(\mathcal A\) 的第 \(j\) 列 \(\mathbf a_j\)；由 \(\Sigma=\mathcal A\mathcal A'\) 反解 \(\mathcal A\) 需 \(\frac{(m-1)m}2\) 个额外约束——Cholesky 分解（下三角，同期影响假设）、长期识别（Blanchard–Quah/Galí，暂时 vs 永久冲击）、符号约束。
Chapter theme: multivariate time series. §10.1 Roots, impulse responses, and cointegration: the VAR(\(n\)) \((\mathbb I_m-\mathbf B(L))\mathbf y_t=\mathcal A\epsilon_t\), with synthesized shocks \(\mathbf u_t=\mathcal A\epsilon_t\sim N(0,\Sigma)\), \(\Sigma=\mathcal A\mathcal A'\); stacking into a VAR(1) \(\mathbf x_t=\mathcal B\mathbf x_{t-1}+\mathcal A\epsilon_t\); characteristic polynomial \(p(\lambda)=\det(\lambda\mathbb I_{mn}-\mathcal B)\), with the diagonalization \(\mathcal B=\mathbf V\mathbf D\mathbf V^{-1}\) yielding the Wold decomposition; impulse response \(\mathbf r_\mathbf a(k)=\mathcal B^k\mathbf a\). §10.2 Cointegration and error correction: Definition 10.3 cointegration of rank \(r\) (components \(I(1)\) while some linear combinations \(\beta'\mathbf y_t\) are stationary); Proposition 10.1 the cointegrating matrix \(\beta=\) (transpose of the first \(r\) rows of \(\mathbf V^{-1}\)); impulse responses split into a transitory part + a permanent part, \(\mathbf r_\mathbf a(\infty)=\nu_\star\mathbf a_\star\); error correction \(\Delta\mathbf y_t=-\alpha\beta'\mathbf y_{t-1}+\mathbf u_t\), \(\alpha=\nu(\mathbb I_r-\mathbf D_r)\) (Proposition 10.2, Engle–Granger 1987). §10.3 Identifying impulse responses to \(\epsilon\) shocks: the impulse response to \(\epsilon_{t,j}\) is the \(j\)-th column \(\mathbf a_j\) of \(\mathcal A\); recovering \(\mathcal A\) from \(\Sigma=\mathcal A\mathcal A'\) needs \(\frac{(m-1)m}2\) extra restrictions — the Cholesky decomposition (lower triangular, contemporaneous-impact assumption), long-run identification (Blanchard–Quah/Galí, temporary vs permanent shocks), and sign restrictions.
10.1 Roots, Impulse Responses, and Cointegration
10.1.1 VAR(\(n\))
考虑 \(\mathbf y_t\in\mathbb R^m\) 上的 VAR(\(n\))： $$\mathbf y_t=\sum_{j=1}^n\mathbf B_j\mathbf y_{t-j}+\mathcal A\epsilon_t,\quad \epsilon_t\sim N(0,\mathbb I_n),\quad t=1,\dots,T$$ $$\Rightarrow(\mathbb I_m-\mathbf B(L))\mathbf y_t=\mathcal A\epsilon_t$$ 其中 \(L^j\mathbf y_t=\mathbf y_{t-j}\)，\(\mathbf B(L)=\sum_{j=1}^n\mathbf B_j L^j\)。令合成冲击 $$\mathbf u_t=\mathcal A\epsilon_t,\quad \mathbf u_t\sim N(0,\Sigma),\quad \Sigma=\mathcal A\mathcal A'$$ \(\mathbf u_t\) 为一步预测误差。
注记 10.1 直观上：\(\epsilon_t\) 是一组假想的相互独立的冲击源，\(\mathbf u_t\) 是这些独立冲击经线性组合 \(\mathcal A\) 后产生的合成冲击。我们通常只能观测到合成冲击 \(\mathbf u_t\)，而非单个基本冲击 \(\epsilon_t\)——后者总是混合在一起出现。因此后面需要额外假设才能恢复每个 \(\epsilon_t\) 冲击的效应。
把 VAR(\(m\)) 堆叠成 VAR(1)。令 \(\mathbf x_t=(\mathbf y_t',\mathbf y_{t-1}',\dots,\mathbf y_{t-n+1}')'\)，则 $$\mathbf x_t=\mathcal B\mathbf x_{t-1}+\mathcal A\epsilon_t,\qquad \mathcal B=\begin{bmatrix}\mathbf B_1&\cdots&\mathbf B_{n-1}&\mathbf B_n\\\mathbb I_m&\cdots&\mathbf 0&\mathbf 0\\\vdots&\ddots&\vdots&\vdots\\\mathbf 0&\cdots&\mathbb I_m&\mathbf 0\end{bmatrix},\quad \mathcal A=\begin{bmatrix}\mathbf A\\\mathbf 0\\\vdots\\\mathbf 0\end{bmatrix}$$ 不失一般性，下文多数讨论基于 VAR(1)。
Consider a VAR(\(n\)) on \(\mathbf y_t\in\mathbb R^m\): $$\mathbf y_t=\sum_{j=1}^n\mathbf B_j\mathbf y_{t-j}+\mathcal A\epsilon_t,\quad \epsilon_t\sim N(0,\mathbb I_n),\quad t=1,\dots,T$$ $$\Rightarrow(\mathbb I_m-\mathbf B(L))\mathbf y_t=\mathcal A\epsilon_t$$ where \(L^j\mathbf y_t=\mathbf y_{t-j}\), \(\mathbf B(L)=\sum_{j=1}^n\mathbf B_j L^j\). Let the synthesized shock be $$\mathbf u_t=\mathcal A\epsilon_t,\quad \mathbf u_t\sim N(0,\Sigma),\quad \Sigma=\mathcal A\mathcal A'$$ with \(\mathbf u_t\) the one-step forecast error.
Remark 10.1 Intuitively: \(\epsilon_t\) is a vector of hypothetically independent sources of shocks, and \(\mathbf u_t\) is a vector of synthesized shocks produced by a linear combination \(\mathcal A\) of \(\epsilon_t\). We can usually observe only the synthesized shocks \(\mathbf u_t\), not the individual fundamental shocks \(\epsilon_t\) — those always come in combination. So later we need extra assumptions to recover the effect of each \(\epsilon_t\) shock.
Stack the VAR(\(m\)) into a VAR(1). With \(\mathbf x_t=(\mathbf y_t',\mathbf y_{t-1}',\dots,\mathbf y_{t-n+1}')'\), $$\mathbf x_t=\mathcal B\mathbf x_{t-1}+\mathcal A\epsilon_t,\qquad \mathcal B=\begin{bmatrix}\mathbf B_1&\cdots&\mathbf B_{n-1}&\mathbf B_n\\\mathbb I_m&\cdots&\mathbf 0&\mathbf 0\\\vdots&\ddots&\vdots&\vdots\\\mathbf 0&\cdots&\mathbb I_m&\mathbf 0\end{bmatrix},\quad \mathcal A=\begin{bmatrix}\mathbf A\\\mathbf 0\\\vdots\\\mathbf 0\end{bmatrix}$$ WLOG, most of what follows is based on the VAR(1).
10.1.2 Characteristic Polynomial and Roots
定义 10.1（VAR(\(n\)) 的特征多项式） 堆叠 VAR(\(n\)) 的特征多项式定义为 $$p(\lambda)=\det(\lambda\mathbb I_{mn}-\mathcal B)$$
若 \(\mathcal B\) 可对角化，则 \(\mathcal B=\mathbf V\mathbf D\mathbf V^{-1}\)，其中 \(\mathbf D=\operatorname{diag}(\lambda_1,\dots,\lambda_{nm})\) 是以 \(p(\lambda)\) 的根为对角元的对角矩阵。
例 10.1（VAR(\(n\)) 的 Wold 分解）。 对堆叠 \(\mathbf x_t=\mathcal B\mathbf x_{t-1}+\mathcal A\epsilon_t\)，若所有根 \(|\lambda_j|<1\) 且 \(\mathcal B\) 可对角化，则 $$\begin{aligned}\mathbf x_t&=\underbrace{\mathcal B^\infty\mathbf x_{-\infty}}_{\to0}+\sum_{j=0}^\infty\mathcal B^j\mathcal A\epsilon_{t-j}=\sum_{j=0}^\infty\mathcal B^j\mathcal A\epsilon_{t-j}\\&=\sum_{j=0}^\infty\mathbf V\mathbf D^j\mathbf V^{-1}\mathcal A\epsilon_{t-j}=\sum_{j=0}^\infty\mathbf V\operatorname{diag}(\lambda_1^j,\dots,\lambda_{nm}^j)\mathbf V^{-1}\mathcal A\epsilon_{t-j}\end{aligned}$$ 取最后一式的前 \(m\) 行即给出该 VAR(\(n\)) 的 Wold 分解。
Definition 10.1 (Characteristic polynomial of VAR(\(n\))) The characteristic polynomial of the stacked VAR(\(n\)) is $$p(\lambda)=\det(\lambda\mathbb I_{mn}-\mathcal B)$$
If \(\mathcal B\) is diagonalizable, then \(\mathcal B=\mathbf V\mathbf D\mathbf V^{-1}\), where \(\mathbf D=\operatorname{diag}(\lambda_1,\dots,\lambda_{nm})\) is the diagonal matrix carrying the roots of \(p(\lambda)\).
Example 10.1 (Wold decomposition of VAR(\(n\))). For the stacking \(\mathbf x_t=\mathcal B\mathbf x_{t-1}+\mathcal A\epsilon_t\), if all roots \(|\lambda_j|<1\) and \(\mathcal B\) is diagonalizable, then $$\begin{aligned}\mathbf x_t&=\underbrace{\mathcal B^\infty\mathbf x_{-\infty}}_{\to0}+\sum_{j=0}^\infty\mathcal B^j\mathcal A\epsilon_{t-j}=\sum_{j=0}^\infty\mathcal B^j\mathcal A\epsilon_{t-j}\\&=\sum_{j=0}^\infty\mathbf V\mathbf D^j\mathbf V^{-1}\mathcal A\epsilon_{t-j}=\sum_{j=0}^\infty\mathbf V\operatorname{diag}(\lambda_1^j,\dots,\lambda_{nm}^j)\mathbf V^{-1}\mathcal A\epsilon_{t-j}\end{aligned}$$ The first \(m\) rows of the last line give the Wold decomposition of this VAR(\(n\)).
10.1.3 Impulse Responses
定义 10.2（脉冲响应） 设 \(\mathbf a\) 为某个 \(m\) 维向量。\(\mathbf y_k\) 对冲击向量 \(\mathbf a\) 的脉冲响应 \(\mathbf r_\mathbf a(k)\) 是给定 \(\mathbf u_0=\mathbf a\) 时的预测修正： $$\mathbf r_\mathbf a(k)=E[\mathbf y_k\mid \mathbf u_0=\mathbf a,\ \mathbf y_t:t<0]-E[\mathbf y_k\mid \mathbf y_t:t<0]$$ 对 \(k\ge0\)。
Definition 10.2 (Impulse response) Let \(\mathbf a\) be an \(m\)-dimensional vector. The impulse response \(\mathbf r_\mathbf a(k)\) of \(\mathbf y_k\) to the shock vector \(\mathbf a\) is the forecast revision given \(\mathbf u_0=\mathbf a\): $$\mathbf r_\mathbf a(k)=E[\mathbf y_k\mid \mathbf u_0=\mathbf a,\ \mathbf y_t:t<0]-E[\mathbf y_k\mid \mathbf y_t:t<0]$$ for \(k\ge0\).
10.1.4 Computing Impulse Responses by Recursive Substitution
与单变量类似，可用递归代入计算 \(k\) 步脉冲响应。对 VAR(1) \(\mathbf y_t=\mathbf B\mathbf y_{t-1}+\mathbf u_t\)，在 0 期施加冲击 \(\mathbf a\)、其余期冲击为零：
| \(k\) | \(\mathbf r_\mathbf a(k)\) | \(\mathbf r_\mathbf a(k-1)\) |
|---|---|---|
| 0 | \(\mathbf a\) | \(\mathbf 0\) |
| 1 | \(\mathbf B\mathbf a\) | \(\mathbf a\) |
| 2 | \(\mathbf B^2\mathbf a\) | \(\mathbf B\mathbf a\) |
| 3 | \(\mathbf B^3\mathbf a\) | \(\mathbf B^2\mathbf a\) |
一般地，对 VAR(1)， $$\mathbf r_\mathbf a(k)=\mathbf B^k\mathbf a$$
数值例 1。 设 \(\mathbf B=\begin{bmatrix}0.5&0\\0&1\end{bmatrix}\)、\(\mathbf a=\begin{bmatrix}1\\1\end{bmatrix}\)，则 $$\mathbf r_\mathbf a(k)=\mathbf B^k\mathbf a=\begin{bmatrix}0.5^k&0\\0&1\end{bmatrix}\begin{bmatrix}1\\1\end{bmatrix}=\begin{bmatrix}0.5^k\\1\end{bmatrix}$$ 画出脉冲响应函数：第二分量 \(y(2)\) 对应单位根，冲击永久持续（水平线停在 1）；第一分量 \(y(1)\) 模长小于 1，冲击衰减至零。
数值例 2。 设 \(\mathbf B=\begin{bmatrix}0.68&-0.24\\-0.24&0.82\end{bmatrix}\)、\(\mathbf a=\begin{bmatrix}1\\1\end{bmatrix}\)，对角化得 $$\mathbf r_\mathbf a(k)=\mathbf B^k\mathbf a=\begin{bmatrix}0.8&-0.6\\0.6&0.8\end{bmatrix}\begin{bmatrix}0.5^k&0\\0&1\end{bmatrix}\begin{bmatrix}0.8&0.6\\-0.6&0.8\end{bmatrix}\begin{bmatrix}1\\1\end{bmatrix}\xrightarrow{k\to\infty}\begin{bmatrix}-0.12\\0.16\end{bmatrix}$$ 即便存在一个单位根，系统也可能收敛到一个非零稳态（取决于对应的特征向量方向）：脉冲响应中可与单位根特征向量正交的成分会衰减掉，只留下沿单位根方向的永久成分。
As in the univariate case, the \(k\)-step impulse response can be computed by recursive substitution. For a VAR(1) \(\mathbf y_t=\mathbf B\mathbf y_{t-1}+\mathbf u_t\), apply a shock \(\mathbf a\) at date 0 and zero shocks otherwise:
| \(k\) | \(\mathbf r_\mathbf a(k)\) | \(\mathbf r_\mathbf a(k-1)\) |
|---|---|---|
| 0 | \(\mathbf a\) | \(\mathbf 0\) |
| 1 | \(\mathbf B\mathbf a\) | \(\mathbf a\) |
| 2 | \(\mathbf B^2\mathbf a\) | \(\mathbf B\mathbf a\) |
| 3 | \(\mathbf B^3\mathbf a\) | \(\mathbf B^2\mathbf a\) |
Generally, for a VAR(1), $$\mathbf r_\mathbf a(k)=\mathbf B^k\mathbf a$$
Numerical Example 1. Let \(\mathbf B=\begin{bmatrix}0.5&0\\0&1\end{bmatrix}\), \(\mathbf a=\begin{bmatrix}1\\1\end{bmatrix}\); then $$\mathbf r_\mathbf a(k)=\mathbf B^k\mathbf a=\begin{bmatrix}0.5^k&0\\0&1\end{bmatrix}\begin{bmatrix}1\\1\end{bmatrix}=\begin{bmatrix}0.5^k\\1\end{bmatrix}$$ Plotting the impulse-response functions: the second component \(y(2)\) corresponds to the unit root and the shock persists indefinitely (a flat line at 1); the first component \(y(1)\) has modulus below 1 and the shock dies out to zero.
Numerical Example 2. Let \(\mathbf B=\begin{bmatrix}0.68&-0.24\\-0.24&0.82\end{bmatrix}\), \(\mathbf a=\begin{bmatrix}1\\1\end{bmatrix}\); diagonalizing, $$\mathbf r_\mathbf a(k)=\mathbf B^k\mathbf a=\begin{bmatrix}0.8&-0.6\\0.6&0.8\end{bmatrix}\begin{bmatrix}0.5^k&0\\0&1\end{bmatrix}\begin{bmatrix}0.8&0.6\\-0.6&0.8\end{bmatrix}\begin{bmatrix}1\\1\end{bmatrix}\xrightarrow{k\to\infty}\begin{bmatrix}-0.12\\0.16\end{bmatrix}$$ Even with a unit root, the system can converge to a nonzero steady state (depending on the corresponding eigenvector direction): the part of the impulse response orthogonal to the unit-root eigenvector dies out, leaving only the permanent component along the unit-root direction.
10.2 Cointegration and Error Correction
10.2.1 Cointegration
定义 10.3（协整） 向量时间序列 \(\mathbf y_t\) 称为秩 \(r\) 协整，若 \(\mathbf y_t\) 的每个分量单独看都非平稳（如 \(I(1)\)），而其分量的某些线性组合 \(\beta'\mathbf y_t\) 平稳，其中 \(\beta\) 是秩 \(r\ge1\) 的 \(m\times r\) 协整矩阵。
直观例子。 实际 GDP 与实际消费各自都是 \(I(1)\)（随机趋势上行），但二者之差（对数产出减对数消费）围绕一个稳定水平波动——这条差序列是平稳的，体现两者共享同一长期随机趋势。
Definition 10.3 (Cointegration) A vector time series \(\mathbf y_t\) is cointegrated of rank \(r\) if each component of \(\mathbf y_t\) taken individually is non-stationary (e.g. \(I(1)\)), while some linear combinations of its components, \(\beta'\mathbf y_t\), are stationary, where \(\beta\) is an \(m\times r\) cointegrating matrix of rank \(r\ge1\).
Intuitive example. Real GDP and real consumption are each \(I(1)\) (drifting along a stochastic trend), but their difference (log output minus log consumption) fluctuates around a stable level — this difference series is stationary, reflecting that the two share the same long-run stochastic trend.
10.2.2 The Cointegrating Matrix
命题 10.1（协整矩阵） 对 VAR(1) \(\mathbf y_t=\mathbf B\mathbf y_{t-1}+\mathbf u_t\)，设平稳根 \(|z_1|<1,\dots,|z_r|<1\)、单位根 \(z_{r+1}=\dots=z_m=1\)。若 \(\mathbf B\) 可对角化 \(\mathbf B=\mathbf V\mathbf D\mathbf V^{-1}\)，则一个协整矩阵 \(\beta\) 由 \(\mathbf V^{-1}\) 前 \(r\) 行的转置给出。
证明 用分块矩阵。记 \(\mathbf V=\begin{bmatrix}\nu&\nu_\star\end{bmatrix}\)、\(\mathbf V^{-1}=\begin{bmatrix}\beta'\\(\beta_\star)'\end{bmatrix}\)，其中 \(\nu\) 为 \(m\times r\)、\(\nu_\star\) 为 \(m\times(m-r)\)，\(\beta'\) 为 \(r\times m\)、\((\beta_\star)'\) 为 \((m-r)\times m\)。动态写为 $$\mathbf y_t=\begin{bmatrix}\nu&\nu_\star\end{bmatrix}\begin{bmatrix}\mathbf D_r&\mathbf 0\\\mathbf 0&\mathbb I_{m-r}\end{bmatrix}\begin{bmatrix}\beta'\\(\beta_\star)'\end{bmatrix}\mathbf y_{t-1}+\mathbf u_t$$ 其中 \(\mathbf D_r=\operatorname{diag}(z_1,\dots,z_r)\)。检验 \(\beta'\mathbf y_t\) 的协方差平稳性： $$\begin{aligned}\beta'\mathbf y_t&=\begin{bmatrix}\beta'\nu&\beta'\nu_\star\end{bmatrix}\begin{bmatrix}\mathbf D_r&\mathbf 0\\\mathbf 0&\mathbb I_{m-r}\end{bmatrix}\begin{bmatrix}\beta'\\(\beta_\star)'\end{bmatrix}\mathbf y_{t-1}+\beta'\mathbf u_t\\&=(\beta'\nu\mathbf D_r\beta'+\beta'\nu_\star\beta_\star')\mathbf y_{t-1}+\beta'\mathbf u_t\end{aligned}$$ 由 \(\mathbf V^{-1}\mathbf V=\mathbb I_m\) 得 \(\begin{bmatrix}\beta'\\(\beta_\star)'\end{bmatrix}\begin{bmatrix}\nu&\nu_\star\end{bmatrix}=\begin{bmatrix}\mathbb I_r&\mathbf 0\\\mathbf 0&\mathbb I_{m-r}\end{bmatrix}\)，即 \(\beta'\nu=\mathbb I_r\)、\(\beta'\nu_\star=\mathbf 0\)。代入得 $$\beta'\mathbf y_t=(\mathbb I_r\mathbf D_r\beta'+\mathbf 0\cdot\beta_\star')\mathbf y_{t-1}+\beta'\mathbf u_t=\mathbf D_r\beta'\mathbf y_{t-1}+\beta'\mathbf u_t$$ 记 \(\mathbf x_t=\beta'\mathbf y_t\)、\(\mathbf e_t=\beta'\mathbf u_t\)，则 \(\mathbf x_t=\mathbf D_r\mathbf x_t+\mathbf e_t\)（应作 \(\mathbf x_t=\mathbf D_r\mathbf x_{t-1}+\mathbf e_t\)）。由于 \(\mathbf D_r\) 的根 \(|z_1|,\dots,|z_r|<1\)，\(\mathbf x_t\) 协方差平稳。\(\blacksquare\)
Proposition 10.1 (Cointegrating matrix) For a VAR(1) \(\mathbf y_t=\mathbf B\mathbf y_{t-1}+\mathbf u_t\), let the stationary roots be \(|z_1|<1,\dots,|z_r|<1\) and the unit roots \(z_{r+1}=\dots=z_m=1\). If \(\mathbf B\) is diagonalizable \(\mathbf B=\mathbf V\mathbf D\mathbf V^{-1}\), then a cointegrating matrix \(\beta\) is given by the transpose of the first \(r\) rows of \(\mathbf V^{-1}\).
Proof Use block matrices. Write \(\mathbf V=\begin{bmatrix}\nu&\nu_\star\end{bmatrix}\), \(\mathbf V^{-1}=\begin{bmatrix}\beta'\\(\beta_\star)'\end{bmatrix}\), where \(\nu\) is \(m\times r\), \(\nu_\star\) is \(m\times(m-r)\), \(\beta'\) is \(r\times m\), \((\beta_\star)'\) is \((m-r)\times m\). The dynamics read $$\mathbf y_t=\begin{bmatrix}\nu&\nu_\star\end{bmatrix}\begin{bmatrix}\mathbf D_r&\mathbf 0\\\mathbf 0&\mathbb I_{m-r}\end{bmatrix}\begin{bmatrix}\beta'\\(\beta_\star)'\end{bmatrix}\mathbf y_{t-1}+\mathbf u_t$$ with \(\mathbf D_r=\operatorname{diag}(z_1,\dots,z_r)\). Check covariance stationarity of \(\beta'\mathbf y_t\): $$\begin{aligned}\beta'\mathbf y_t&=\begin{bmatrix}\beta'\nu&\beta'\nu_\star\end{bmatrix}\begin{bmatrix}\mathbf D_r&\mathbf 0\\\mathbf 0&\mathbb I_{m-r}\end{bmatrix}\begin{bmatrix}\beta'\\(\beta_\star)'\end{bmatrix}\mathbf y_{t-1}+\beta'\mathbf u_t\\&=(\beta'\nu\mathbf D_r\beta'+\beta'\nu_\star\beta_\star')\mathbf y_{t-1}+\beta'\mathbf u_t\end{aligned}$$ From \(\mathbf V^{-1}\mathbf V=\mathbb I_m\), \(\begin{bmatrix}\beta'\\(\beta_\star)'\end{bmatrix}\begin{bmatrix}\nu&\nu_\star\end{bmatrix}=\begin{bmatrix}\mathbb I_r&\mathbf 0\\\mathbf 0&\mathbb I_{m-r}\end{bmatrix}\), so \(\beta'\nu=\mathbb I_r\), \(\beta'\nu_\star=\mathbf 0\). Substituting, $$\beta'\mathbf y_t=(\mathbb I_r\mathbf D_r\beta'+\mathbf 0\cdot\beta_\star')\mathbf y_{t-1}+\beta'\mathbf u_t=\mathbf D_r\beta'\mathbf y_{t-1}+\beta'\mathbf u_t$$ Denote \(\mathbf x_t=\beta'\mathbf y_t\), \(\mathbf e_t=\beta'\mathbf u_t\), so \(\mathbf x_t=\mathbf D_r\mathbf x_{t-1}+\mathbf e_t\). Since the roots of \(\mathbf D_r\) satisfy \(|z_1|,\dots,|z_r|<1\), \(\mathbf x_t\) is covariance stationary. \(\blacksquare\)
10.2.3 Splitting the Impulse Responses
回忆脉冲响应 \(\mathbf r_\mathbf a(k)=\mathbf B^k\mathbf a\)。用 \(\mathbf B=\mathbf V\mathbf D\mathbf V^{-1}\) 的分块形式： $$\mathbf r_\mathbf a(k)=\begin{bmatrix}\nu&\nu_\star\end{bmatrix}\begin{bmatrix}\mathbf D_r^k&\mathbf 0\\\mathbf 0&\mathbb I_{m-r}\end{bmatrix}\begin{bmatrix}\beta'\\(\beta_\star)'\end{bmatrix}\mathbf a=\big(\nu\mathbf D_r^k\beta'+\nu_\star(\beta_\star)'\big)\mathbf a$$ 定义 \(\mathbf a_r=\beta'\mathbf a\in\mathbb R^r\)、\(\mathbf a_\star=(\beta_\star)'\mathbf a\in\mathbb R^{m-r}\)，则 $$\mathbf r_\mathbf a(k)=\nu\mathbf D_r^k\mathbf a_r+\nu_\star\mathbf a_\star$$ 第一项是暂时部分（随 \(k\) 衰减），第二项是永久部分（不随 \(k\) 变）。由于 \(\lim_{k\to\infty}\mathbf D_r^k=\mathbf 0\)， $$\mathbf r_\mathbf a(\infty)=\lim_{k\to\infty}\mathbf r_\mathbf a(k)=\nu_\star\mathbf a_\star$$
Recall the impulse response \(\mathbf r_\mathbf a(k)=\mathbf B^k\mathbf a\). Using the block form of \(\mathbf B=\mathbf V\mathbf D\mathbf V^{-1}\): $$\mathbf r_\mathbf a(k)=\begin{bmatrix}\nu&\nu_\star\end{bmatrix}\begin{bmatrix}\mathbf D_r^k&\mathbf 0\\\mathbf 0&\mathbb I_{m-r}\end{bmatrix}\begin{bmatrix}\beta'\\(\beta_\star)'\end{bmatrix}\mathbf a=\big(\nu\mathbf D_r^k\beta'+\nu_\star(\beta_\star)'\big)\mathbf a$$ Defining \(\mathbf a_r=\beta'\mathbf a\in\mathbb R^r\), \(\mathbf a_\star=(\beta_\star)'\mathbf a\in\mathbb R^{m-r}\), $$\mathbf r_\mathbf a(k)=\nu\mathbf D_r^k\mathbf a_r+\nu_\star\mathbf a_\star$$ The first term is the transitory part (decays with \(k\)), the second is the permanent part (constant in \(k\)). Since \(\lim_{k\to\infty}\mathbf D_r^k=\mathbf 0\), $$\mathbf r_\mathbf a(\infty)=\lim_{k\to\infty}\mathbf r_\mathbf a(k)=\nu_\star\mathbf a_\star$$
10.2.4 Error Correction
考虑一阶差分。由 \(\mathbf B=\nu\mathbf D_r\beta'+\nu_\star(\beta_\star)'\)（带单位根块）展开： $$\begin{aligned}\Delta\mathbf y_t&=(\mathbf B-\mathbb I_m)\mathbf y_{t-1}+\mathbf u_t\\&=\Big(\begin{bmatrix}\nu&\nu_\star\end{bmatrix}\begin{bmatrix}\mathbf D_r-\mathbb I_r&\mathbf 0\\\mathbf 0&\mathbf 0\end{bmatrix}\begin{bmatrix}\beta'\\(\beta_\star)'\end{bmatrix}\Big)\mathbf y_{t-1}+\mathbf u_t\\&=\nu(\mathbf D_r-\mathbb I_r)\beta'\mathbf y_{t-1}+\mathbf u_t\end{aligned}$$ 定义 \(\alpha=\nu(\mathbb I_r-\mathbf D_r)\)（即 \(\alpha\beta'=\mathbb I_m-\mathbf B\)），则 VAR(1) 可写成误差修正表示： $$\Delta\mathbf y_t=-\alpha\beta'\mathbf y_{t-1}+\mathbf u_t$$
直观。 \(\mathbf y_t\) 一期到下期的变化由两部分驱动：(1) 新冲击 \(\mathbf u_t\)；(2) 协整的平稳成分 \(\beta'\mathbf y_{t-1}\)——当系统偏离长期均衡（\(\beta'\mathbf y_{t-1}\ne0\)）时，\(-\alpha\beta'\mathbf y_{t-1}\) 把它拉回，\(\alpha\) 是调整速度。
命题 10.2（Engle–Granger 1987） 一个秩 \(r\) 协整的 VAR(\(k\)) 可写成误差修正表示 $$A^\star(L)\Delta\mathbf y_t=-\alpha\beta'\mathbf y_{t-1}+\mathbf u_t$$ 其中 \(A^\star(L)\) 只有稳定根。
Consider first differences. Expanding \(\mathbf B=\nu\mathbf D_r\beta'+\nu_\star(\beta_\star)'\) (with the unit-root block): $$\begin{aligned}\Delta\mathbf y_t&=(\mathbf B-\mathbb I_m)\mathbf y_{t-1}+\mathbf u_t\\&=\Big(\begin{bmatrix}\nu&\nu_\star\end{bmatrix}\begin{bmatrix}\mathbf D_r-\mathbb I_r&\mathbf 0\\\mathbf 0&\mathbf 0\end{bmatrix}\begin{bmatrix}\beta'\\(\beta_\star)'\end{bmatrix}\Big)\mathbf y_{t-1}+\mathbf u_t\\&=\nu(\mathbf D_r-\mathbb I_r)\beta'\mathbf y_{t-1}+\mathbf u_t\end{aligned}$$ Defining \(\alpha=\nu(\mathbb I_r-\mathbf D_r)\) (so \(\alpha\beta'=\mathbb I_m-\mathbf B\)), the VAR(1) can be written as the error-correction representation: $$\Delta\mathbf y_t=-\alpha\beta'\mathbf y_{t-1}+\mathbf u_t$$
Intuition. The change in \(\mathbf y_t\) from one period to the next is driven by two parts: (1) the new shock \(\mathbf u_t\); (2) the cointegrated stationary component \(\beta'\mathbf y_{t-1}\) — when the system departs from long-run equilibrium (\(\beta'\mathbf y_{t-1}\ne0\)), \(-\alpha\beta'\mathbf y_{t-1}\) pulls it back, with \(\alpha\) the speed of adjustment.
Proposition 10.2 (Engle–Granger 1987) A VAR(\(k\)) that is cointegrated of rank \(r\) can be given an error-correction representation $$A^\star(L)\Delta\mathbf y_t=-\alpha\beta'\mathbf y_{t-1}+\mathbf u_t$$ where \(A^\star(L)\) has only stable roots.
10.3 Identifying the Impulse Responses to \(\epsilon\) Shocks
在真实经济中，\(\epsilon\) 冲击可代表货币政策变化、财政政策变化、战争爆发等。我们关心的是某个 \(\epsilon_{t,j}\) 冲击（且仅它）发生后对 \(\mathbf y_t\) 的影响。可惜 \(\epsilon_{t,1},\dots,\epsilon_{t,m}\) 在每期都混合在一起出现（即 \(\mathbf u_t=\mathcal A\epsilon_t\)），无法在真实经济中单独实验。
利用我们能观测的合成冲击 \(\mathbf u_t\) 加上对 \(\mathbf A\) 的合理假设，可反推出 \(\mathbf A\)。回忆 $$\mathbf u_t=\mathbf A\epsilon_t,\quad \Sigma=\mathbf A\mathbf A'=\begin{bmatrix}\mid&\mid&&\mid\\\mathbf a_1&\mathbf a_2&\cdots&\mathbf a_m\\\mid&\mid&&\mid\end{bmatrix}\begin{bmatrix}\epsilon_{t,1}\\\vdots\\\epsilon_{t,m}\end{bmatrix}=\mathbf a_1\epsilon_{t,1}+\dots+\mathbf a_m\epsilon_{t,m}$$ 注意 \(\epsilon_{t,j}\) 的脉冲是 \(\mathbf u_t=\mathbf a_j\) 的脉冲，其中 \(\mathbf a_j\) 是 \(\mathbf A\) 的第 \(j\) 列。故只要有 \(\mathbf A\) 的良好候选，就能计算单独 \(\epsilon_{t,j}\) 的脉冲响应。
约束计数。 设 \(\mathbf A\) 为 \(k\times m\)（下设 \(k\ge m\)，这里假设 \(\mathbf A\) 为方阵），共 \(m^2\) 个元素。\(\Sigma=\mathbf A\mathbf A'\)（对称）给出 \(\frac{m(m+1)}2\) 个方程，故还需至少 $$m^2-\frac{m(m+1)}2=\frac{(m-1)m}2$$ 个额外约束才能由合理假设定出 \(\mathbf A\)。
In the real economy, \(\epsilon\) shocks may represent monetary policy changes, fiscal policy changes, the outbreak of war, and so on. We are interested in the effect on \(\mathbf y_t\) of one \(\epsilon_{t,j}\) shock (and only it) happening. Unfortunately \(\epsilon_{t,1},\dots,\epsilon_{t,m}\) always happen together in any period \(t\) (i.e. \(\mathbf u_t=\mathcal A\epsilon_t\)), so we cannot experiment around in the real economy.
Using the synthesized shock \(\mathbf u_t\) we can observe, plus reasonable assumptions on \(\mathbf A\), we can back out \(\mathbf A\). Recall $$\mathbf u_t=\mathbf A\epsilon_t,\quad \Sigma=\mathbf A\mathbf A'=\begin{bmatrix}\mid&\mid&&\mid\\\mathbf a_1&\mathbf a_2&\cdots&\mathbf a_m\\\mid&\mid&&\mid\end{bmatrix}\begin{bmatrix}\epsilon_{t,1}\\\vdots\\\epsilon_{t,m}\end{bmatrix}=\mathbf a_1\epsilon_{t,1}+\dots+\mathbf a_m\epsilon_{t,m}$$ Notice the impulse of \(\epsilon_{t,j}\) is the impulse response to \(\mathbf u_t=\mathbf a_j\) where \(\mathbf a_j\) is the \(j\)-th column of \(\mathbf A\). So with a good candidate for \(\mathbf A\) we can compute the impulse response to \(\epsilon_{t,j}\) alone.
Counting restrictions. Let \(\mathbf A\) be \(k\times m\) (below assume \(k\ge m\); here we take \(\mathbf A\) square), with \(m^2\) elements. The symmetric \(\Sigma=\mathbf A\mathbf A'\) gives \(\frac{m(m+1)}2\) equations, so we still need at least $$m^2-\frac{m(m+1)}2=\frac{(m-1)m}2$$ extra restrictions to pin down \(\mathbf A\) from reasonable assumptions.
10.3.1 Identification via the Cholesky Decomposition
一个合理假设是 \(\mathbf A\) 为下三角（Cholesky 分解）。上三角的 \(\frac{(m-1)m}2\) 个元素为零，恰好给出所需数目的额外约束： $$\mathbf A=\begin{bmatrix}\star&0&0&0&0\\\star&\star&0&0&0\\\star&\star&\star&0&0\\\vdots&\vdots&\vdots&\ddots&\vdots\\\star&\star&\star&\cdots&\star\end{bmatrix}$$ 其中可约定对角元非负（必要时翻转 \(\epsilon_{t,j}\) 的经济定义符号即可）。
- 该识别假设冲击 \(\epsilon_{t,j}\) 对变量 \(y_{t,i}\)（\(i
- 在此假设下，\(\mathbf A\) 可由 \(\Sigma=\mathbf A\mathbf A'\) 的估计直接解出。
- 但下三角假设很强，除非有很好的经济故事支持 \(\mathbf y_t\) 各分量与 \(\epsilon_{t,j}\) 之间的这种关系，否则可能受到批评。
One reasonable assumption is that \(\mathbf A\) is lower triangular (Cholesky decomposition). The \(\frac{(m-1)m}2\) upper-triangular elements being zero gives exactly the required number of extra restrictions: $$\mathbf A=\begin{bmatrix}\star&0&0&0&0\\\star&\star&0&0&0\\\star&\star&\star&0&0\\\vdots&\vdots&\vdots&\ddots&\vdots\\\star&\star&\star&\cdots&\star\end{bmatrix}$$ where we adopt a sign convention that diagonal elements are non-negative (achieved by flipping the economic definition of \(\epsilon_{t,j}\) if necessary).
- This identification assumes shock \(\epsilon_{t,j}\) has no contemporaneous impact on variables \(y_{t,i}\) for \(i
- Under this assumption, \(\mathbf A\) can be easily solved from our estimate of \(\Sigma=\mathbf A\mathbf A'\).
- But the lower-triangular assumption is strong and may be criticized unless we have a very good economic story for such relationships among the elements of \(\mathbf y_t\) and the shocks \(\epsilon_{t,j}\).
10.3.2 Long-Run Identification (Blanchard–Quah, Galí, etc.)
有时我们有很好的经济故事，能断定某些 \(\epsilon_{t,j}\) 冲击是暂时的，而另一些 \(\epsilon_{t,j'}\) 冲击是永久的。如前所述，对协整 VAR(1)， $$\mathbf y_t=\begin{bmatrix}\nu&\nu_\star\end{bmatrix}\begin{bmatrix}\mathbf D_r&\mathbf 0\\\mathbf 0&\mathbb I_{m-r}\end{bmatrix}\begin{bmatrix}\beta'\\(\beta_\star)'\end{bmatrix}\mathbf y_{t-1}+\mathbf A\epsilon_t$$ \(\epsilon_{t,j}\) 单独的 \(k\) 步脉冲响应是 \(\mathbf u_t=\mathbf a_j\) 的脉冲响应，\(\mathbf a_j\) 为 \(\mathbf A\) 第 \(j\) 列。由 §10.2.3， $$\mathbf r_{\mathbf a_j}(k)=\nu\mathbf D_r^k\beta'\mathbf a_j+\nu_\star(\beta_\star)'\mathbf a_j$$ 其长期（永久）部分为 \(\mathbf r_{\mathbf a_j}(\infty)=\nu_\star(\beta_\star)'\mathbf a_j\)。
- 对任意暂时冲击 \(\epsilon_{t,j}\)，可假设 \(\nu_\star(\beta_\star)'\mathbf a_j=\mathbf 0\)（无长期影响）——这成为对 \(\mathbf A\) 的额外约束。
- 对任意永久冲击 \(\epsilon_{t,j'}\)，可假设 \(\nu_\star(\beta_\star)'\mathbf a_{j'}\ne\mathbf 0\)（有长期影响）——也成为额外约束。
- 这些假设单独不足以完全定出 \(\mathbf A\)，但至少缩小候选范围，与其他约束结合可把 \(\mathbf A\) 限制到较小范围。
Sometimes we have a good economic story to say some \(\epsilon_{t,j}\) shocks are temporary while other \(\epsilon_{t,j'}\) shocks are permanent. As discussed, for a cointegrated VAR(1), $$\mathbf y_t=\begin{bmatrix}\nu&\nu_\star\end{bmatrix}\begin{bmatrix}\mathbf D_r&\mathbf 0\\\mathbf 0&\mathbb I_{m-r}\end{bmatrix}\begin{bmatrix}\beta'\\(\beta_\star)'\end{bmatrix}\mathbf y_{t-1}+\mathbf A\epsilon_t$$ The \(k\)-lag impulse response to \(\epsilon_{t,j}\) alone is the impulse response to \(\mathbf u_t=\mathbf a_j\), where \(\mathbf a_j\) is the \(j\)-th column of \(\mathbf A\). By §10.2.3, $$\mathbf r_{\mathbf a_j}(k)=\nu\mathbf D_r^k\beta'\mathbf a_j+\nu_\star(\beta_\star)'\mathbf a_j$$ whose long-run (permanent) part is \(\mathbf r_{\mathbf a_j}(\infty)=\nu_\star(\beta_\star)'\mathbf a_j\).
- For any temporary shock \(\epsilon_{t,j}\), we can assume \(\nu_\star(\beta_\star)'\mathbf a_j=\mathbf 0\) (no long-run impact) — an additional restriction on \(\mathbf A\).
- For any permanent shock \(\epsilon_{t,j'}\), we can assume \(\nu_\star(\beta_\star)'\mathbf a_{j'}\ne\mathbf 0\) (a long-run impact) — also an additional restriction.
- Such assumptions alone are not sufficient to pin down \(\mathbf A\), but they at least shrink the scope, and combined with other restrictions can restrict our candidates for \(\mathbf A\) to a smaller scope.
10.3.3 Identification with Sign Restrictions
考虑同期冲击 $$\mathbf u_t=\mathbf a_1\epsilon_{t,1}+\dots+\mathbf a_m\epsilon_{t,m}$$ 若对 \(\epsilon_{t,j}\) 施加正冲击（\(\epsilon_{t,j}=1\)，其余 \(\epsilon_{t,i}=0\ \forall i\ne j\)），应导致 \(\mathbf y_t\) 的同期变化为 \(\mathbf a_j\)。若有很好的经济故事或直觉判断 \(\mathbf a_j\) 各元素的符号，就更有把握。
例。 设 \(\epsilon_{t,j}\) 是货币政策变化、正的 \(\epsilon_{t,j}\) 表示货币宽松；又设 \(\mathbf y_t=(y_{t,1},y_{t,2},\dots)'\)，\(y_{t,1}\) 为实际 GDP 增长、\(y_{t,2}\) 为利率。经济上我们相信货币宽松应提高 GDP 增长、降低利率，故对 \(\mathbf a_j=(a_1,a_2,\dots)'\) 可合理地令 \(a_1>0\)、\(a_2<0\)。
这类符号约束又给 \(\mathbf A\) 增加限制。它们单独仍不足以定出 \(\mathbf A\)，但能缩小候选范围。本节所有技巧可组合使用，配合其他合理假设把候选范围缩到较小，再选一个作为先验、用贝叶斯更新去逼近真实的 \(\mathbf A\)。
Consider the contemporaneous shock $$\mathbf u_t=\mathbf a_1\epsilon_{t,1}+\dots+\mathbf a_m\epsilon_{t,m}$$ A positive shock to \(\epsilon_{t,j}\) (\(\epsilon_{t,j}=1\) with all \(\epsilon_{t,i}=0\ \forall i\ne j\)) should lead to a contemporaneous change in \(\mathbf y_t\) equal to \(\mathbf a_j\). With a good economic story or intuition about the signs of the elements of \(\mathbf a_j\), we gain more confidence.
Example. Suppose \(\epsilon_{t,j}\) is a monetary policy change and positive \(\epsilon_{t,j}\) means monetary stimulus; also suppose \(\mathbf y_t=(y_{t,1},y_{t,2},\dots)'\) where \(y_{t,1}\) is real GDP growth and \(y_{t,2}\) is the interest rate. Economically we believe monetary stimulus should increase GDP growth and decrease the interest rate, so for \(\mathbf a_j=(a_1,a_2,\dots)'\) it is fair to say \(a_1>0\) and \(a_2<0\).
Such sign restrictions add more restrictions on \(\mathbf A\). They are again not sufficient to pin down \(\mathbf A\), but they shrink the scope of candidates. All the tricks in this section can be used in combination, along with other reasonable assumptions, to reduce the candidates for \(\mathbf A\) to a fairly small scope; then pick one as our prior to do Bayesian updating to approximate the real \(\mathbf A\).