From what I'm reading it seems like a nonstationary ARIMA model can have a nonzero constant term. I'm not understanding how this can happen. Suppose we have an AR(1) model where $\phi_1=1$. If p is the number of autoregressive terms, the constant, C would be defined as:

$$ C=\mu \left(1-\sum_i^p\phi_i\right) $$

Then wouldn't the constant equal zero since $$ C=\mu(1-\phi_1) $$ then becomes

$$ C=\mu(1-1) $$ which equals $$ C=0 $$

  • $\begingroup$ Hi: $\phi_{i}$ in an AR(whatever) model can never be one because you then have a unit root and need to difference the series before you can continue. This is because ARIMA models need to be stationary before they can be estimated, simulated etc. If you don't difference, the process explodes in the sense that it becomes a random walk. $\endgroup$ – mlofton Jan 6 '20 at 2:28
  • $\begingroup$ Please have a quick look at my edits. In particular Latex avoids the need to insert images. $\endgroup$ – Christoph Hanck Jan 6 '20 at 13:13
  • $\begingroup$ @ChristophHanck Thank you. This is my first time using Latex. $\endgroup$ – Michael Howell Jan 7 '20 at 2:21

You can define the following process: $$ Y_t=c+Y_{t-1}+u_t $$ Assuming an initial condition $Y_0=0$ for simplicity, recursive substitution yields $$ Y_t=c\cdot t+\sum_{s=1}^tu_s, $$ a so-called random walk with drift.

  • $\begingroup$ Thank you that's a great way to think of it. But how do you calculate the constant, C? It seems like in a random walk the process breaks down. $\endgroup$ – Michael Howell Jan 7 '20 at 2:19
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    $\begingroup$ I am not sure I understand what you mean by "calculate". Estimate from data? In that case you could for example (which of course presumes that you are sure about the unit coefficient on the lag of $Y_t$) take first differences to get $\Delta Y_t=c+u_t$ and estimate $c$ by simply the sample average of the $Y_t$. $\endgroup$ – Christoph Hanck Jan 7 '20 at 5:36
  • $\begingroup$ Thank you that does help. But still, wouldn't that be inconsistent with the above definition of the constant? $\endgroup$ – Michael Howell Jan 8 '20 at 1:41
  • $\begingroup$ Ah OK. What you define as $C$ is related to the expectation of a stationary AR process, which does not depend on $t$. And indeed, what I define as $c$ is not, because the expected value of that process is not $C$, but $ct$. $\endgroup$ – Christoph Hanck Jan 8 '20 at 5:19
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    $\begingroup$ Yes, I think that is a suitable way to see it. $\endgroup$ – Christoph Hanck Jan 9 '20 at 4:52

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