We have $$y_t = a\mu_t + b\mu_{t-1} + c\mu_{t-2} + d\mu_{t-3}$$

and $\mu$ itself is an AR process, let's assume here an AR(2)

$$\mu_t = \phi_1 \mu_{t-1} + \phi_2 \mu_{t-2} + \epsilon_t$$

where $\epsilon_t$ is a white noise process with E($\epsilon_t$)=0 and $Var(\epsilon_t)=\sigma^2$.

As there will be covariance terms present, how can one solve (or compute numerically) the variance of $y_t$?

Note that this question is close to this one (which is still unanswered); the the goal here is however to find an approach that excludes any covariance terms between different lags of the $\mu$s, as the sum in the $y_t$ equations could get long.

My approach so far

  1. Rewrite the AR process in companion form: $$ \begin{bmatrix} \mu_{t} \\ \mu_{t-1} \end{bmatrix}= \begin{bmatrix} \phi_1 & \phi_2 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} \mu_{t-1} \\ \mu_{t-2} \end{bmatrix}+ \begin{bmatrix} \epsilon_{t} \\ 0 \end{bmatrix} $$

$$\pmb \mu_t = \pmb \Phi \pmb \mu_{t-1} + \pmb \epsilon_t$$

  1. Normally, I would now compute the variance of the $\mu$s by a solver for Lyapunov equations (in either R or Matlab). (Rewriting the AR as $\Sigma = \Phi \Sigma \Phi' + Q$, and then solving for $\Sigma$). Yet here, we have to deal with the covariances between the lags.

Ideally, I would now sub out the lags of the $\mu$s in the last equation, as in an AR(1). But can I do this? $$\pmb \mu_t = \sum_{i=0} \pmb \Phi^i \pmb \epsilon_{t-i}$$

  1. Once the $\mu$s are expressed in terms of $\epsilon$s only, one could calculate the variances of each $\mu$ term in the $y_t$ equation separately and ignore any covariances

Please note: The ultimate goal is to estimate $Var(y_t)$; parameters $a,b,c,d,\Phi$ and $\sigma$ are known.


2 Answers 2


Evidently $\operatorname{Var}(y_t)$ will involve covariances of the $\mu$ series at lags $0,$ $1,$ $2,$ and $3.$ A comment to the question you reference directs us to a thread that explains how to find those covariances. You should obtain the solution

$$\eqalign{ &\operatorname{Var}(\mu_t,\mu_t) &= \gamma_0 = \left(\frac{1-\phi_2}{1+\phi_2}\right)\frac{\sigma^2}{(1-\phi_2)^2-\phi_1^2} \\ &\operatorname{Var}(\mu_t,\mu_{t-1}) &=\gamma_1 = \frac{\phi_1}{1-\phi_2}\,\gamma_0 \\ &\operatorname{Var}(\mu_t,\mu_{t-2}) &=\gamma_2 = \phi_1 \gamma_1 + \phi_2 \gamma_0 \\ &\operatorname{Var}(\mu_t,\mu_{t-3}) &=\gamma_3 = \phi_1 \gamma_2 + \phi_2 \gamma_1. }$$

Because covariance is bilinear,

$$\eqalign{ \operatorname{Var}(y_t) &= \operatorname{Var}((a,b,c,d)(\mu_t,\mu_{t-1},\mu_{t-2},\mu_{t-3})^\prime) \\ &= \pmatrix{a&b&c&d}\Gamma \pmatrix{a\\b\\c\\d} }$$


$$\Gamma = (\gamma_{|i-j|}) = \pmatrix{\gamma_0 & \gamma_1 & \gamma_2 & \gamma_3 \\ \gamma_1 & \gamma_0 &\gamma_1 &\gamma_2 \\ \gamma_2 & \gamma_1 & \gamma_0 & \gamma_1 \\ \gamma_3 & \gamma_2 & \gamma_1 & \gamma_0}. $$


Writing the model using backshift operator polynomials, you have $$ y_t = \theta_3(B)\mu_t \tag{1} $$ and $$ \phi_2(B)\mu_t = \epsilon_t. \tag{2} $$ Applying $\phi_2(B)$ to both sides of (1) and using (2) yields $$ \phi_2(B)y_t = \theta_3(B)\phi_2(B)\mu_t = \theta_3(B)\epsilon_t $$ which shows that $y_t$ is ARMA(2,3) with autocovariance function that can be computed using standard methods, implemented in e.g. R-function ltsa:::tacvfARMA.

Note that if $a\neq 1$, you'll need to do some rescaling to make your model conform to the convention that the first coefficients of the the MA and AR-polynomial are 1.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.