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Linear Processes Basics Composition of 2 line processes Autoregressive processes of $AR(1)$ An alternative perspective inversion of operators <- Go Back

401-4623-00: Time Series Analysis

Section 5

Linear Processes

Swiss Federal Institute of Technology Zurich

Eidgenössische Technische Hochschule Zürich

Last Edit Date: 01/09/2024

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Linear Processes Basics

A time series $\{X_t\}_t$ is said to be a linear process if it admits: $X_t = \sum_{j = -\infty}^{\infty} \psi_j Z_{t-j}, t\in \mathbb{Z}$ with $Z_t \sim WN(0, \sigma^2)$ and $\{\psi_j\}_j$ is a (determinstic) and an absolutely convergence sequence:

$$\sum_{j = -\infty}^{\infty} |\psi_j| < \infty.$$

Alternative notaion: Let $\psi(Z) = \sum_{j = -\infty}^{\infty} \psi_j Z^j$ for $Z$ in the convergence domain of $\sum_{j = -\infty}^{\infty} \psi_j Z^j$.

Recall that $B$, the backward shift operator is defined as

$$ \begin{align} BX_t &= X_{t-1}\\ B^jX_t &= X_{t-j} \end{align} $$

For some stationary time series $\{Y_t\}_t$,

$$ \begin{align} \psi (B) Y_t &= \left( \sum_{j=-infty}^{\infty} \psi_j B^j \right) Y_t \\ &= \sum_{j=-\infty}^{\infty} \psi_j B^j Y_t \\ &= \sum_{j=-\infty}^{\infty} \psi_j Y_{t-j} \end{align} $$

This means that the linear process can be rewritten as

$$X_t = \psi (B) Z_t.$$

Remark: The condition $\sum_{j=-\infty}^{\infty} |\psi_j| < \infty$ ensures that $X_t$ is finite with probability 1.

Recall that $X_t = \sum_{j = -\infty}^{\infty} \psi_j Z_{t-j}$. Let $Y_t = \sum_{j=-\infty}^{\infty} |\psi_j| |Z_{t-j}| \ge 0$, we have

$$ \begin{align} \mathbb{E}[Y_t] &= \mathbb{E}\left[ \sum_{j=-\infty}^{\infty} |\psi_j| |Z_{t-j}| \right]\\ &= \sum_{j=-\infty}^{\infty} |\psi_j| \mathbb{E}\left[ |Z_{t-j}| \right] \end{align} $$

by Fubini's theorem (Fubini's theorem states that a double integral can be computed by iterated integrals if the function is Lebesgue integrable on a rectangle)

$$ \begin{align} \mathbb{E}[|Z_{t-j}|] &\le \sqrt{\mathbb{E}[Z_{t-j}^2]}\\ &= \sqrt{\mathrm{Var}(Z_{t-j})}\\ &= \sqrt{\sigma^2} = \sigma < \infty \end{align} $$

so

$$\sum_{j=-\infty}^{\infty} |\psi_j| \mathbb{E}[|Z_{t-j}|] \le \sigma \sum_{j=-\infty}^{\infty} |\psi_j| < \infty.$$

Hence, $\mathbb{E}[Y_t]$ is finite.

This implies that $\mathbb{P}(Y_t < \infty) = 1$

$$ \begin{align} \mathbb{P}(Y_t = \infty) &= \mathbb{P}\left( \lim_{M\rightarrow \infty} \bigcap_{m=1}^M \left\{ Y_t > m \right\} \right) \\ &= \mathbb{P}\left( \lim_{M\rightarrow \infty} \left\{ Y_t > M \right\} \right)\\ &= \lim_{M\rightarrow \infty} \mathbb{P} \left( \{Y_t > M\} \right)~~~~~\text{(Monotone convergence theorem)}\\ &\le \lim_{M\rightarrow \infty} \frac{\mathbb{E}[Y_t]}{M} = 0~~~~~\text{(Markov inequality)} \end{align} $$

Hence, $\sum_{j=-\infty}^\infty \psi_j Z_{t-j}$ is absolutely convergent, meaning that $X_t$ is well-defined with probability 1.

Moving Average Processes

If $\psi_j = 0$, $\forall j < 0$, then $X_t = \sum_{j=0}^{\infty} \psi_j Z_{t-j}$. In this case, $\{X_t\}_t$ is called a Moving Average Process:

$$\{X_t\}_t \sim MA(\infty).$$

Let $\{X_t\}_t$ be stationary with $\mathbb{E}[Y_t] = 0$ and ACVF to be $\gamma_y$. Also, let $\{\psi_j\}_j$ be a sequence such that

$$\sum_{j=-\infty}^{\infty} |\psi_j| < \infty.$$

Then, $X_t = \sum_{j=-\infty}^{\infty} \psi_j Y_{t-j}$ is stationary with $\mu_x(t) = \mathbb{E}[X_t] = 0$ and $\gamma_x(h) = \sum_{j=-\infty}^{\infty} \sum_{k=-\infty}^{\infty} \psi_j \psi_k \gamma_y(h + k -j)$.

In particular, if $\{Y_t\}_t \sim WN(0, \sigma^2)$, then $\gamma_x(h) = \sigma^2 \sum_{j=-\infty}^{\infty} \psi_j \psi_{j+h}$.

Proof: Consider $W_t = \sum_{j = -\infty} |\psi_j| |Y_{t-j}|$ and

$$ \begin{align} \mathbb{E}[W_t] &= \sum_{j=-\infty}^{\infty} |\psi_j| \mathbb{E}[|Y_t-j|] \\ &\le \sum_{j=-\infty}^{\infty} |\psi_j| \sqrt{\gamma_y(0)}~~~~~\text{(Jensen's inequality)}\\ &= \sqrt{\gamma_y(0)} \sum_{j=-\infty}^{\infty} |\psi_j| < \infty \end{align} $$

This implies that $W_t$ is finite with probability 1, meaning that $X_t$ is finite with probability 1.

We can also conclude that $\mathbb{E}[|X_t|] \le \infty$ (as $\le \mathbb{E}[W_t]$). Hence, $\mathbb{E}[X_t]$ is finite. Then

$$ \begin{align} \mathbb{E}[X_t] &= \sum_{j=-\infty}^{\infty} \psi_j \underbrace{\mathbb{E}[Y_{t-j}]}_{0}~~~~~\text{(Fubini's theorem)}\\ &= 0 \end{align} $$

Fix $h \in \mathbb{Z}$

$$ \begin{align} |X_t X_{t+h}| &= \left| \left( \sum_{j=-\infty}^{\infty} \psi_j Y_{t-j} \right) \left( \sum_{j=-\infty}^{\infty} \psi_j Y_{t+h-j} \right)\right| \\ &= \left| \sum_{j,k=-\infty}^{\infty} \psi_j \psi_k Y_{t-j} Y_{t+h-k} \right| \\ &\le \sum_{j,k=-\infty}^{\infty} \left| \psi_j \right| \left| \psi_k \right| \left| Y_{t-j} \right| \left| Y_{t+h-k} \right| \end{align} $$

Now we have

$$ \begin{align} \mathbb{E}\left[ \sum_{j,k=-\infty}^{\infty} |\psi_j| |\psi_k| |Y_{t-j}| |Y_{t+h-k}| \right] &= \sum_{j,k=-\infty}^{\infty} |\psi_j| |\psi_k| \mathbb{E}[|Y_{t-j}| |Y_{t+h-k}|]\\ &\le \sum_{j,k=-\infty}^{\infty} |\psi_j| |\psi_k| \sqrt{\gamma_y(0)} \sqrt{\gamma_y(0)}\\ &= \gamma_y(0) \sum_{j,k=-\infty}^{\infty} |\psi_j| |\psi_k|\\ &= \gamma_y(0) \left( \sum_{j=-\infty}^{\infty} |\psi_j| \right)^2 < \infty \end{align} $$

This implies that $\mathbb{E}[|X_t X_{t+h}|] \le \infty$, meaning that $\mathbb{E}[X_t X_{t+h}] < \infty$.

$$ \begin{align} \gamma_x(t, t+h) &= \mathrm{Cov}(X_t, X_{t+h})\\ &= \mathbb{E}[X_t X_{t+h}] - 0 \cdot 0\\ &= \mathbb{E}[X_t X_{t+h}] \end{align} $$

By Fubini's theorem, we comput

$$ \begin{align} \mathbb{E}[X_t X_{t+h}] &= \sum_{j,k=-\infty}^{\infty} \psi_j \psi_k \mathbb{E}[Y_{t-j}Y_{t+h-k}]\\ \gamma_y(t, t+h) &= \sum_{j,k=-\infty}^{\infty} \psi_j \psi_k \gamma_y(t+h-k-(t-j))\\ &= \sum_{j,k=-\infty}^{\infty} \psi_j \psi_k \gamma_y(h-k+j) \end{align} $$

Suppose $\{Y_t\} \sim WN(0, \sigma^2)$. In this case, we have that $\gamma_y(h-k+j) = \sigma^2 \mathbb{1}\{h-k+j=0\} = \sigma^2 \mathbb{1}\{k=h+j\}$.

This means that

$$\gamma_x(h) = \sigma^2 \sum_{j=-\infty}^{\infty} \psi_j \psi_{h+j}.$$

Example

Consider $X_t = Z_t + \theta Z_{t-1}$ with $\theta \in \mathbb{R}$ and $\{Z_t\} \sim WN(0, \sigma^2)$. Note that $\{X_t\} \sim MA(1)$, a special case of linear processes.

Given $X_t = \psi(B) Z_t$ with $\psi(B) = \sum_{j=-\infty}^{\infty} \psi_j B^j$, where

$$ \psi_j = \begin{cases} 1 & j = 0\\ \theta & j = 1\\ 0 & \text{otherwise} \end{cases}. $$

Since $\gamma_x$ is even, we conclude that

$$ \gamma_x(h) = \begin{cases} \sigma^2 (1 + \theta^2) & h = 0\\ \sigma^2 \theta & |h| = 1\\ 0 & \text{otherwise} \end{cases} $$

Composition of 2 line processes

Let $\{\alpha_j\}_j$ and $\{\beta_j\}_j$ be 2 real sequences such that $\sum_{j=-\infty}^{\infty} |\alpha_j| < \infty$ and $\sum_{j=-\infty}^{\infty} |\beta_j| < \infty$. Also, let $\{Y_t\}_t$ be some stationary time series.

Consider $X_t = \alpha(B) Y_t = \sum_{j=-\infty}^{\infty} \alpha_j Y_{t-j}$ and $W_t = \beta(B) X_t = \sum_{j=-\infty}^{\infty} \beta_j X_{t - j}$.

How can we write $W_t$ as a function of $Y_t$?

$$\underbrace{Y_t \stackrel{\alpha(B)}{\rightarrow} X_t \stackrel{\beta(B)}{\rightarrow} W_t}_{?}$$

We can consider write like this

$$W_t = \left( \underbrace{\beta(B) \cdot \alpha(B)}_{\psi{B}} \right) Y_t$$

where $\psi_j = \sum_{k=-\infty}^{\infty} \beta_k \alpha_{j-k}$.

Indeed, $W_t = \sum_{j=-\infty}^{\infty} \beta_j \alpha_{t - j} = \sum_{j=-\infty}^{\infty}\beta_j \sum_{k=-\infty}^{\infty} \alpha_k Y_{t-j-k}$.

Now, note that $\sum_{j,k=-\infty}^{\infty} \beta_j \alpha_k Y_{t-j-k}$ is finite with probability 1.

In fact

$$ \begin{align} \mathbb{E}\left[ \sum_{j,k=-\infty}^{\infty} |\beta_j| |\alpha_k| |Y_{t-j-k}| \right] &= \sum_{j,k=-\infty}^{\infty} |\beta_j| |\alpha_j| \mathbb{E} [|Y_{t-j-k}|]~~~~~\text{(Fubini's theorem)}\\ &\le \sqrt{\gamma_y(0)} \sum_{j,k=-\infty} |\beta_j| |\alpha_k|\\ &= \sqrt{\gamma_y(0)} \left( \sum_{j=-\infty}^{\infty} |\beta_j| \right) \left( \sum_{k=-\infty}^{\infty} |\alpha_k| \right) < \infty \end{align} $$

Using Fubini's theorem, we can write

$$ \begin{align} W_t &= \sum_{j,k=-\infty}^{\infty} \beta_j \alpha_k Y_{t-j-k} \\ &= \sum_{j=-\infty}^{\infty} \sum_{k=-\infty}^{\infty} \beta_j \alpha_k Y_{t-j-k}\\ &= \sum_{j=-\infty}^{\infty} \sum_{k=-\infty}^{\infty} \beta_j \alpha_{k'-j} Y_{t-k'}~~~~~\text{(replacing } k = k' - j \text{)}\\ &= \sum_{k'=-\infty}^{\infty} \sum_{j=-\infty}^{\infty} \beta_j \alpha_{k'-j} Y_{t-k'}\\ &= \sum_{k=-\infty}^{\infty} \left( \sum_{j=-\infty}^{\infty} \beta_j \alpha_{k-j} \right) Y_{t-k}\\ &= \sum_{k=-\infty}^{\infty} \psi_k Y_{t-k} \end{align} $$

As a result we can have

$$W_t = \psi(B)Y_t$$

where $\psi(B) = \sum_{j=-\infty}\psi_j B^j$ and $\psi_j = \sum_{k=-\infty}^{\infty}\beta_k \alpha_{j-k}$.

Note that $\psi_j = \sum_{k=-\infty}^{\infty} \alpha_j \beta_{k-j}$. This implies that $\alpha(B) \cdots \beta(B) = \beta(B) \cdot \alpha(B)$.

Autoregressive processes of $AR(1)$

Let $\phi \in \mathbb{R} \{-1, 1\}$. An autoregressive process of order (1) ($AR(1)$) is a stationary solution to the equations:

$$X_t = \phi X_{t-1} + Z_t$$

where $\{Z_t\}_t \sim WN(0, \sigma^2)$.

We can rewrite it as

$$ \begin{align} &\Leftrightarrow X_t - \phi X_{t-1} = Z_t\\ &\Leftrightarrow X_t - \phi B X_t = Z_t\\ &\Leftrightarrow \phi(B)X_t = Z_t \end{align} $$

where $\phi(B) = 1 - \phi B$.

1. First case - $|\phi| < 1$

   We show that the $MA(\infty)$ process given by $\sum_{j=0}^{\infty} \phi^j Z_{t-j}$ is the unique stationary time series satisfying $X_t = \phi X_{t-1} + Z_t$ almost surely.

   Consider $X_t = \sum_{j=0}^{\infty} \phi^j Z_{t-j} = \psi(B) Z_t$, where $\psi(B) = \sum_{j=-\infty}^{\infty} \psi_j B^j$ and $\psi_j = \begin{cases} \phi^j & j \ge 0\\ 0 & j < 0 \end{cases}$.

   $$\sum_{j = -\infty}^{\infty} |\psi_j| = \sum_{j=0}^{\infty} |\phi^j| = \frac{1}{1 - |\phi|} < \infty$$

   Using the previous proposition, we see that $\{X_t\}_t$ is stationary.

   $$ \begin{align} X_t - \phi X_{t-1} &= \sum_{j=0}^{\infty} \phi^j Z_{t-j} - \phi \sum_{j=0}^{\infty} \phi^j Z_{t-1-j}\\ &= \sum_{j=0}^{\infty} \phi^j Z_{t-j} - \phi \sum_{j=1}^{\infty} \phi^{j-1} Z_{t-j}\\ &= \sum_{j=0}^{\infty} \phi^j Z_{t-j} - \sum_{j=1}^{\infty} \phi^{j} Z_{t-j}\\ &= \phi^0 Z_{t-0} = Z_t \end{align} $$

   Hence, $X_t = \sum_{j=0}^{\infty} \phi^j Z_{t-j}$ is a stationary solution of $X_t = \phi X_{t-1} + Z_t$.

   Suppose there exists another stationary solution, say $\{Y_t\}_t$. Then

   $$ \begin{align} X_t - Y_t &= (\phi X_{t-1} + Z_t) - (\phi Y_{t-1} + Z_t)\\ &= \phi (X_{t-1} - Y_{t-1})\\ &= \phi^{k} (X_{t-k} - Y_{t-k})\\ \Rightarrow |X_t - Y_t| &= |\phi|^k |X_{t-k} - Y_{t-k}| \end{align} $$

   As a result, the expectation can be computed as

   $$ \begin{align} \mathbb{E}[|X_t - Y_t|] &= |\phi|^k \mathbb{E}[|X_{t-k} - Y_{t-k}|]\\ &\le |\phi|^k (\mathbb{E}[|X_{t-k}|] + \mathbb{E}[|Y_{t-k}|])\\ &\le |\phi|^k (\sqrt{\gamma_x(0)} + \gamma_y(0))\\ &= 0~~~~~\text{(as } k \text{ goes to infinity, it approaches 0)} \end{align} $$

   This is because $\mathbb{E}[X_t] = \mathbb{E}[Y_t] = 0$, $\mathbb{E}[X_t^2] = \gamma_x(0)$, and $\mathbb{E}[Y_t^2] = \gamma_y(0)$.

   As a result, we have $X_t = Y_t$ almost surely.

   Note that in this case, the $AR$ model is automatically causal, meaning that $\mathrm{Cov}(X_s, Z_t) = 0, \forall s < t$. Indeed, $X_t$ depends only on the present and past terms of $\{Z_t\}_t$.

2. Second case - $|\phi| > 1$

   Note that $\sum_{j=0}^{\infty} |\phi|^j = \infty$. However we can write

   $$X_{t+1} = \phi X_t + Z_{t+1} \Leftrightarrow X_t = \frac{1}{\phi} X_{t+1} + \frac{1}{\phi}Z_{t+1}$$

   where $\phi \neq 0$ always.

   We can now show that $-\sum_{j=1}^{\infty} \frac{1}{\phi^j} Z_{t+j}$ is the unique stationary solution of $X_t = \phi X_{t-1} + Z_t$ almost surely.

   Consider $X_t = -\sum_{j=1}^{\infty} \frac{1}{\phi^j} Z_{t+j}$. Note that $X_t = \psi(B) Z_t$ where $\psi_j = \begin{cases} \phi^j & j \le -1\\ 0 & j > -1 \end{cases}$.

   $$ \begin{align} &\sum_{j=-\infty} |\psi_j| = \sum_{j=-\infty}^{-1} |\phi|^j = \sum_{j=1}^{\infty} |\phi|^{0j}\\ = & -1 + \sum_{j=0}^{\infty} |\phi|^{-1} = -1 + \frac{1}{1 - \frac{1}{|\phi|}} = \frac{1}{|\phi| - 1} < \infty \end{align} $$

   Using the first proposition, we see that $\{X_t\}_t$ is stationary.

   $$ \begin{align} X_t - \phi X_{t-1} &= -\sum_{j=1}^{\infty} \phi^{-j} Z_{t+j} - \phi \sum_{j=1}^{\infty} \phi^{-j} Z_{t-1+j}\\ &= -\sum_{j=1}^{\infty} \phi^{-j} Z_{t+j} - \phi \sum_{j=0}^{\infty} \phi^{-j-1} Z_{t+j}\\ &= -\sum_{j=1}^{\infty} \phi^{-j} Z_{t+j} - \sum_{j=0}^{\infty} \phi^{-j} Z_{t+j}\\ &= \phi^0 Z_{t+0} = Z_t \end{align} $$

   Suppose there exists a stationary solution $\{Y_t\}_t$. Then

   $$ \begin{align} X_t - Y_t &= \left(\frac{1}{\phi} X_{t+1} + Z_{t+1}\right) - \left(\frac{1}{\phi} Y_{t+1} + Z_{t+1}\right) \\ &= \frac{1}{\phi} (X_{t+1} - Y_{t+1})\\ &= \frac{1}{\phi^k} (X_{t+k} - Y_{t+k}) \end{align} $$

   As for the expectation, we have

   $$\mathbb{E}[|X_t - Y_t|] \le |\phi|^{-k} \left( \sqrt{\gamma_x(0)} + \sqrt{\gamma_y(0)} \right)$$

   approaches to 0 as $k$ goes to infinity.

   As a result $X_t = Y_t$ almost surely.

   Note that the $AR$ model is non-causal in this case since $X_t$ depends on future terms of $\{Z_t\}_t$.

An alternative perspective inversion of operators

Consider an $AR(1)$ process,

$$X_t - \phi X_{t-1} = Z_t.$$

Let $|\phi| < 1$, we can write the equation as

$$\Phi(B)X_t = Z_t$$

where $\Phi(z) = 1 - \phi_z$ for $z \in \mathbb{C}$. We have shown that $X_t - \phi X_{t-1} = Z_t$ is uniquely solved by

$$X_t = \sum_{j=0}^{\infty} \phi^j Z_{t-j} = \chi(B)Z_t$$

for $\chi(z) = \sum_{j=0}^{\infty} \phi^j z^j$ for $|\phi_z| < 1$ as if the $AR(1)$ equations in $X_t - \phi X_{t-1} = Z_t$ are solved by

• $X_t = (1 - \phi B)^{-1} Z_t$

• $(1 - \phi B)^{-1} = \frac{1}{1 - \phi B} = \sum_{j=0} \phi^j B^j$

• $X_t = \chi(B) Z_t = \sum_{j=0}^{\infty} \phi^j Z_{t-j}$

which is the only solution as seen previously.

Question: How can we justifiy these manipulations?

Answer: Let $\{\alpha_t\}_t$ and $\{\beta_t\}_t$ be such that $\sum |\alpha_t| < \infty$ and $\sum |\beta_t| < \infty$. For any stationary time series $\{Y_t\}_t$, $\{\psi(B) Y_t\}_t$ is also stationary for $\psi(B) = \alpha(B) \circ \beta(B)$, where $\alpha (B) = \sum_{j=0}^{\infty} \phi^j B^j$ and $\beta(B) = \sum_{j=-\infty}^{\infty} \beta_j B^j$.

Now let $\alpha(B) = \chi(B) = \sum_{j=0}^{\infty} \phi^j B^j$ and $\beta(B) = \Phi(B) = 1 - \phi B$.

It is clear that the requirements satisfied

• $\sum_{j=-\infty}^{\infty} |\alpha_j| = \sum_{j=0}^{\infty} |\phi|^j < \infty$ and $\sum_{j=-\infty}^\infty |\beta_j| = 1 + |\phi| < \infty$

• $\underbrace{\beta(B)}_{\Phi(B)} X_t = Z_t \Rightarrow (\underbrace{\alpha(B) \circ \beta(B)}_{\psi(B)}) X_t = \alpha(B) Z_t = \chi(B) Z_t = \sum_{j=0}^{\infty}\phi^j Z_{t-j}$.

Recall that $\psi_j = \sum_{k=-\infty} \beta_k \alpha_{j-k}$ where $\alpha_j = \begin{cases} \phi^j & j \ge 0 \\ 0 & \text{otherwise} \end{cases}$ and $\beta_j = \begin{cases} 1 & j=0 \\ -\phi & j=1 \\ 0 & \text{otherwise} \end{cases}$, we have

$$ \psi_j = \alpha_j - \phi \alpha_{j-1} = \begin{cases} 0 & j = 0\\ 1 & j = 1\\ \phi^j - \phi \phi^{j-1} = 0 & \text{otherwise} \end{cases} $$

As a result

$$\psi(B) = \sum_{j=-\infty}^{\infty} \psi_j B^j = B^0 = I.$$

Moreover we have

$$IX_t = \sum_{j=0}^{\infty} \phi^j Z_{t-1} \Leftrightarrow X_t = \sum_{j=0}^{\infty} \phi^j Z_{t-j}.$$

It is also possible to directly show that $\psi(B) = I$. Generally, we can use these manipulations as such:

Let $\{Y_t\}_t$ be a stationary time series and $\{\psi_j\}_j$ a real sequence with $\sum |\phi_j| < \infty$.

Put $\psi(z) = \sum_{j=-\infty}^{\infty} \psi_j z^j$, $\forall z \in$ convergence fomain. If $\frac{1}{\psi(z)} = \sum_{j=-\infty}^{\infty} \chi_j z^j$ for $\{X_j\}_j$ such that a real sequence $\sum_{j=-\infty}^{\infty} |X_j| < \infty$ and $z \in$ suitable convergence domain.

Then the stationary solution, $\{X_t\}_t$ of $\psi(B)X_t = Y_t$ is given by

$$X_t = \chi(B) Y_t.$$

In fact, we can show that $\chi(B) \circ \psi(B) = I$.

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