I am struggling to get to the general formula of the general linear process autocovariance.

If $Y_t = \mu + \sum_{k=0}^\infty \omega_k e_{t-k}$ where $e \sim WN(0,\sigma_e^2)$ (a.k.a. the general linear process), then what is $\gamma_k$?

I know we're working towards $\gamma_k = (\omega_0\omega_k + \omega_1 \omega_{k+1} + \omega_2\omega_{k+2} + \dots)\sigma_e^2$ but I can't get past the first step.

e.g. $$ \gamma_k = \mathrm{Cov}(\mu + \omega_0 e_t + \omega_1 e_{t-1} + \dots, \mu + \omega_0 e_{t-k} + \omega_1 e_{t-k-1} + \dots) $$

But since $e$ is uncorrelated with any previous white noise terms, doesn't all the terms fall away except if $k=0$?


1 Answer 1


Without caring about too much rigor, a heuristic calculation can go as follows (suppose $k \geq 0$): \begin{align*} & \gamma_k = \operatorname{Cov}(Y_t, Y_{t - k}) = E((Y_t - \mu)(Y_{t - k} - \mu)) \\ =& E\left[\sum_{m = 0}^\infty \omega_me_{t - m} \times \sum_{n = 0}^\infty \omega_ne_{t - k - n}\right] \\ =& \sum_{m = 0}^\infty\sum_{n = 0}^\infty \omega_m\omega_nE[e_{t - m}e_{t - k - n}] \\ =& \sigma_e^2\sum_{m = k}^\infty \omega_m\omega_{m - k} \tag{1}\label{1} \\ =& \sigma_e^2\sum_{j = 0}^\infty \omega_j\omega_{j + k}. \tag{2}\label{2} \end{align*}

In $\eqref{1}$, we used the condition $\{e_t\} \sim WN(0, \sigma_e^2)$ so that $E[e_ie_j] = \delta_{ij}\sigma_e^2$, that is, $E[e_{t - m}e_{t - k - n}]$ is non-zero if and only if $t - m = t - k - n$, hence the double-sum can be reduced to a single-sum.

For a more formal proof that deals with the interchanging of infinite sum and the expectation operator (note that we need to impose conditions on coefficients $\{\omega_k\}$ in order $\eqref{2}$ holds, typically this condition is called absolutely summable, i.e., $\sum_{k = 0}^\infty |\omega_k| < \infty$), refer to Proposition 3.1.2 and Theorem 3.2.1 in Time Series: Theory and Methods (2nd edition) by P. J. Brockwell and R. A. Davis.


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