I managed to write an AR(1) process in the Wold representation with help from the geometric series.

I am having trouble with a stationary AR(2). How could I do?


2 Answers 2


Let $X_t$ be a zero-mean covariance-stationary time series such that $$X_t = \varphi_1 X_{t-1} + \varphi_2 X_{t-2} + \varepsilon_t$$ where $\varepsilon_t$ is white noise.

Using $L$ to mean the lag (backshift) operator, the above can be expressed as $$(1-\varphi_1L - \varphi_2L^2)X_t=\varepsilon_t . \tag{1}$$

Since $X_t$ is a covariance-stationary AR(2) process, the roots of its characteristic polynomial $(1-\varphi_1 z - \varphi_2 z^2) = 0$ must lie outside the unit circle. Thus, Equation (1) can be written as $$(1-\lambda_1 L)(1-\lambda_2 L)X_t=\varepsilon_t $$ where $\vert \lambda_1 \rvert<1$ and $\vert \lambda_2 \rvert<1$. The last two inequalities are true of a covariance-stationary AR(2) process since then the roots of the characteristic polynomial, $z^*_1=1/ \lambda_1$ and $z^*_2=1/ \lambda_2$, will lie outside the unit circle. Therefore, $$X_t= \frac{1}{(1-\lambda_1 L)} \frac{1}{(1- \lambda_2 L)} \varepsilon_t .$$

Expand the two fractions on the right-hand side in the equation above using the geometric series. and you'll have the Wold decomposition of an AR(2).


The Wold representation is an infinite weighted sum of the current and past innovations $\epsilon_t$:

$$ X_t = \psi_0 \epsilon_t + \psi_1 \epsilon_{t-1} + \psi_1 \epsilon_{t-2} + ... = \sum_{i=0}^\infty \psi_i \epsilon_{t-i} $$

If you were interested in obtaining the values of the first weights, $\psi_i$, for a given ARMA process, you can proceed as follows (more details are given in Brockwell and Davis (1991) Time Series: Theory and Methods §3.3):

By definition, we have:

$$ X_t = \psi(L) \epsilon_t \quad \hbox{ and } \quad \phi(L) X_t = \theta(L) \epsilon_t \rightarrow X_t = \frac{\theta(L)}{\phi(L)} \epsilon_t $$

where $\psi(L)$ is the infinite polynomial, $\phi(L)$ is the autoregressive polynomial and $\theta(L)$ is the moving average polynomial. Your question is about an AR process but for generality I would consider an ARMA process, which may also be of interest for this question.

Thus, we can write:

$$ \psi(L) \epsilon_t = X_t = \frac{\theta(L)}{\phi(L)} \epsilon_t \rightarrow \psi(L) \phi(L) \epsilon_t = \theta(L) \epsilon_t $$

The values of $\psi_i$ can be obtained equating the coefficients related to the same lags, $L^i$, from both sides of the last equation, $\psi(L) \phi(L) = \theta(L)$.

Example: Let's take the following ARMA(2,2) process:

$$ X_t = 0.4 X_{t-1} + 0.2 X_{t-1} + \epsilon_t + 0.3 \epsilon_{t-1} - 0.4 \epsilon_{t-2} $$

You can check that the values $\psi_i$ can be obtained recursively (normalizing $\psi_0=1$, $\phi_0=0$ and $\theta_0=0$):

\begin{eqnarray} \begin{array}{l} \psi_1 = \theta_1 + \phi_1 = 0.3 + 0.4 = 0.7 \\ \psi_2 = \theta_2 + \phi_2 + \phi_1 \psi_1 = -0.4 + 0.2 + 0.7\times0.4 = 0.08 \\ \psi_3 = \phi_1 \psi_2 + \phi_2 \psi_1 = 0.4 \times 0.08 + 0.2 \times 0.7 = 0.172 \\ \psi_4 = \phi_1 \psi_3 + \phi_2 \psi_2 = 0.4 \times 0.172 + 0.2 \times 0.08 = 0.0848 \\ \psi_5 = \phi_1 \psi4 + \phi_2 \psi_3 = 0.4 \times 0.0848 + 0.2 \times 0.172 = 0.0683 \\ \psi_6 = ... = 0.0443 \\ ... \end{array} \end{eqnarray}

For an AR(2) process you can simply set $\theta_1 = \theta_2 = 0$.


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