2
$\begingroup$

In my professor's notes, it is written that if the variable $y$ is explained with an ARMA($p$,$q$) model, then $y_t$ (i.e. $y$ at time t) depends on the most recent $p$ lags of its own value and the most recent $q$ lags of a white noise process.

I understand the fact that $y_t$ depends on the past $p$ values of $y$, but I don't quite understand the $q$ lags of a white noise process. My understanding of a white noise process is that it is a sequence of random variables over time, where each random variable in the sequence is i.i.d. and has a mean of zero.

Can someone explain what it means to depend on $q$ lags of a white noise process? I think it means that at the last $q$ points in time, the value taken by the particular white noise process explains the value of $y$, but I also don't understand the difference between depending on a white noise process and simply having an error term. Is it that the last $q$ values of the white noise process explain the size of the error term for $y_t$?

$\endgroup$
2
$\begingroup$

An AR(1) model looks like this: $y_{t} = \rho_{1} y_{t-1} + \epsilon_{t}$

Where $\epsilon_{t}$ is your innovation/error term/white noise process. Note that $y_{t}$ depends on $\epsilon_{t}$ but not $\epsilon_{t-1}$, $\epsilon_{t-2}$, etc. The error term is not correlated across time.

An ARMA(1,1) model looks like this: $y_{t} = \rho_{1} y_{t-1} + \epsilon_{t} + \theta_{1} \epsilon_{t-1}$

An ARMA(1,2) model looks like this: $y_{t} = \rho_{1} y_{t-1} + \epsilon_{t} + \theta_{1} \epsilon_{t-1} + \theta_{2} \epsilon_{t-2}$

For the ARMA process, $y_{t}$ and $y_{t-1}$ will be correlated not just because of the $\rho_{1} y_{t-1}$ term but because of the $\theta_{1} \epsilon_{t-1}$ term. The errors do not disappear immediately, they die out over time. $y_{t}$ depends directly on not just one realization of the white noise process, but on the $q$ past realizations.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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