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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?

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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 $\vert\lambda_1\rvert<0$ and $\vert\lambda_2\rvert<0$. 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).

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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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