# Solved – On the Autocorrelation Matrix of an ARMA(2,2) to derive the Yule Walker Equations

For an AR(2) I can get the Yule-Walker equations: \$\$begin{cases} rho_1=alpha_1+alpha_2rho_1 \ rho_2=alpha_1rho_1+alpha_2 \ rho_k=alpha_1rho_{k-1}+alpha_2rho_{k-2} end{cases}\$\$ starting from the Autocorrelation Matrix.
But how can I manage the MA part to get the complete autocorrelation matrix in the case of an ARMA(2,2)?

Contents

Let's define a general ARMA model of orders \$(p,q)\$ as follows:

\$\$ psi_t equiv sum_{i=0}^p alpha_i, y_{t-i} = sum_{i=0}^q theta_i, epsilon_{t-i} ,, mbox{ with } epsilon_t sim NID,(0, sigma^2_epsilon) ,. \$\$

where \$alpha_0\$ and \$theta_0\$ are normalised to \$1\$.

You can check that multiplying \$psi_t\$ by \$psi_{t-tau}\$ and taking expectations in both sides of the equation yields:

begin{equation} sum_{i=0}^p sum_{j=0}^p alpha_i alpha_j gamma_{tau+j-i} = sigma^2_epsilon sum_{j=0}^{q-tau} theta_j theta_{j+tau} ,, end{equation}

where \$gamma_i\$ is the autocovariance of order \$i\$.

The mapping between the autocovariances and the parameters in an ARMA model is not as rewarding as in an AR model. The above equation does not return a system of equations that can be easily solved to obtain an estimate of the parameters by the method of moments. The Yule-Walker equations are instead easy to solve and return an estimate of the AR coefficients.

Although it is not straightforward, the method of moments can still be applied for an ARMA model by means of a two-steps procedure: the first step uses the Yule-Walker equations and the second step is based on the equation given above. If your question goes in this direction I could give you further details about it.

Edit

The following is an extract from pp. 545-546 in D.S.G. Pollock (1999) A handbook of time series analysis, signal processing and dynamics, Academic Press (changed notation \$theta\$ is \$mu\$ in the original source):

1)

begin{eqnarray} begin{array}{lcl} E(psi_tpsi_{t-tau}) &=& Eleft{ left( sum_i theta_i epsilon_{t-i} right) left( sum_j theta_j epsilon_{t-tau-j} right) right} \ &=& sum_i sum_j theta_i theta_j E(epsilon_{t-i} epsilon_{t-tau-j}) \ &=& sigma^2_epsilon sum_j theta_j theta_{j+tau} ,. end{array} end{eqnarray}

2)

begin{eqnarray} begin{array}{lcl} E(psi_tpsi_{t-tau}) &=& Eleft{ left( sum_i alpha_i y_{t-i} right) left( sum_j alpha_j y_{t-tau-j} right) right} \ &=& sum_i sum_j alpha_i alpha_j E(y_{t-i} y_{t-tau-j}) \ &=& sum_i sum_j alpha_i alpha_j gamma_{tau+j-i} ,. end{array} end{eqnarray}

Putting (1) and (2) together:

\$\$ sum_isum_j alpha_ialpha_jgamma_{tau+j-i} = sigma^2_epsilon sum_j theta_j theta_{j+tau} ,. \$\$

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