I was preparing for CFA and encountered this question, which is quite puzzling.

To use autoregressive model, it has to be covariance stationary (same mean, covariance). If a model's residual is not auto-correlated, then the model is well-specified(covariance stationary). However, random walk model's error term is uncorrelated, but it is NOT covariance stationary. This seems quite contradictory to me, and the textbook does not explain it clearly.

Anyone has any ideas how this thing works?

  • $\begingroup$ Could you elaborate on what you mean by "random walk model's error term"? $\endgroup$ – whuber Mar 6 '18 at 0:19

There is a little bit of confusion here.

The AR(p) is weakly stationary by definition:

$$x_t=c+\phi_1 x_{t-1}+\varepsilon_t$$ $$\varepsilon_t\sim\mathcal{N}(0,\sigma)$$

with $|\phi_1| \le 1$.

Under these hypothesis you can prove that both mean, variance and autocovariance do not depend on time, hence it is a weakly stationary process.

So that was the theoretical part. Now there is the model estimation. If you fit an AR(p) to your data and that is the true data generating process, you should find (statistical evidence) that residuals are not autocorrelated (as well as rejecting hypothesis of unit root). If that's the case then you can use that model for prediction.

In my opinion the first sentence of your question is referred to this aspect.

Stationarity DOES NOT imply absence of autocorrelation

In fact AR(1) is a stationary but autocorrelated process. In the theoretical model the errors are $i.i.d.$

The random walk process is not stationary despite having $i.i.d$ errors.

see Autocorrelation vs Non-stationary

  • $\begingroup$ The variance $\sigma^{2}_{x_{t}}$ manifestly depends on time in the random walk of your first equation. The variance of the pertubation $\varepsilon_{t}$ does not vary by time. You might want to edit "mean, variance and autocovariance do not depend on time" to address this. $\endgroup$ – Alexis Mar 6 '18 at 2:59
  • $\begingroup$ I am sorry @Alexis but my first equation is that of an AR(1), with $ |\phi_1| \leq 1$, hence I think what I wrote is correct. Am I missing something? $\endgroup$ – gioxc88 Mar 6 '18 at 7:41
  • $\begingroup$ If $\phi=1$ and $c=0$, then $\sigma^{2}_{x_{t}}=t$, which is a function of $t$. $\endgroup$ – Alexis Mar 6 '18 at 22:53
  • $\begingroup$ Yes of course I know, but I specified it is AR(1) and AR(1) cannot have $\phi_1 = 1$ by definition otherwise it would be a random walk. $\endgroup$ – gioxc88 Mar 7 '18 at 14:06
  • 2
    $\begingroup$ It is this fact: that $\sigma^{2}_{x_{t}}$ is a function of t while $\sigma^{2}_{\varepsilon}$ is not that makes differencing so useful in permitting us to make inference about change in x. Of course, there's more nuance to it, since $|\phi|<1$ means that variance eventually stabilizes for some value of $t$... but, per my original comment, this is precisely what I was asking for: that you make your assertions more explicit. $\endgroup$ – Alexis Mar 7 '18 at 16:57

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