How to write an AR(2) stationary process in the Wold representation 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?
 A: 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 $\theta_1 = \theta_2 = 0$.
A: 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).
