0
$\begingroup$

Consider I have an AR(p)-process $ (X_{t})_{t\in Z } $, for example the following AR(1)-process:

$$X_{t}=\alpha_{0} + \alpha_{1}X_{t-1}+\epsilon_{t}$$ where $ \epsilon_{t}\sim D(0,\sigma^2) $ is an uncorrelated, zero-mean, finite variance process (White Noise) with distribution function $D$ (e.g. normal distribution).

Given the information set $\Phi_{t-1}$, the conditional distribution of $X_t$ is $$X_t|\Phi_{t-1}\sim D(\alpha_{0} + \alpha_{1}X_{t-1},\sigma^2).$$

  1. How do we know that $\epsilon_{t}$ is independent of all past values {$X_{t-1},X_{t-2}...$}? The assumption that $\epsilon_{t}$ is White Noise does not imply its independence of past $X_t$.

So I guess "$\epsilon_{t}$ is independent of {$X_{t-1},X_{t-2}...$}" is just another assumption for AR-processes?

  1. Saying that $\epsilon_t$ is independent of the infinite set {$X_{t-1},X_{t-2}...$} means that $\epsilon_t$ is independent of every subset of {$X_{t-1},X_{t-2}...$}?
$\endgroup$
  • $\begingroup$ Why would the white noise be dependent on the process? $\endgroup$ – Quality Jun 15 '18 at 1:05
  • $\begingroup$ @Quality If the $\epsilon_t$ is dependent on previous $\epsilon$'s one would expect it to be dependent on previous values of the process, no? $\endgroup$ – spaceisdarkgreen Jun 15 '18 at 1:07
  • $\begingroup$ The definition of white noise is that they ARE independent $\endgroup$ – Quality Jun 15 '18 at 1:25
  • $\begingroup$ @Quality They explicitly say "uncorrelated" and not independent and this is also the stipulation of many definitions of white noise I've seen. (It's also common to see it require only pairwise independence between the $\epsilon$'s... I confess I've never been sure why all the fuss. I guess it's easier to statistically check the weaker assumptions?) $\endgroup$ – spaceisdarkgreen Jun 15 '18 at 1:27
  • $\begingroup$ In Hamilton: Time Series Analysis, he also refers to the uncorrelated White Noise in the definition of the AR(1)-process $\endgroup$ – Abc123 Jun 15 '18 at 1:40
1
$\begingroup$
  1. How do we know that $\epsilon_{t}$ is independent of all past values {$X_{t-1},X_{t-2}...$}? The assumption that $\epsilon_{t}$ is White Noise does not imply its independence of past $X_t$.

"White noise" is sometimes (not always) taken to imply that terms are independent, not just uncorrelated. If $\epsilon_t$ is independent of all previous $\epsilon_{t-n}$ then it must also be independent of any function of those variables.

But $X_{t-n}$ can be expressed as an infinite sum involving only $\alpha_{0}$. $\alpha_{1}$, and $\epsilon_{t-n}, \epsilon_{t-n-1}, \epsilon_{t-n-2}, ...$ and hence $\epsilon_t$ must be independent of $X_{t-n}$ for positive $n$.

If we apply only the "uncorrelated" interpretation, then it's definitely not true that $\epsilon_t$ is independent of past X-values.

(Trivial example: choose any distribution for $\epsilon$ that is uncorrelated but not independent, and set $\alpha_0 = \alpha_1 = 0$.)

$\endgroup$
  • $\begingroup$ @spaceisdarkgreen Thanks, I missed that little detail in the OP. Better now? $\endgroup$ – Geoffrey Brent Jun 15 '18 at 2:17
  • $\begingroup$ The white noise is here gaussian. For a jointly gaussian process, uncorrelatedness implies independence. Hence, the white noise should be assumed to be independent. (Strictly speaking, it would be possible for the process to be marginally gaussian, but not jointly gaussian - hence uncorrelated would not necessarily mean independent - but I'd bet the author meant jointly gaussian). $\endgroup$ – leonbloy Jun 15 '18 at 3:24
  • $\begingroup$ I didn‘t want to restrict the process to be Gaussian, the normal distribution was just an example. But you‘re right, if it‘s jointly Gaussian, then uncorrelatedness would imply independence $\endgroup$ – Abc123 Jun 15 '18 at 8:55
  • $\begingroup$ But I have one more question: Is it equivalent to say $\epsilon_t$ are independent of each other <=> $\epsilon_t$ is independent of $\Phi_{t-1}$? $\endgroup$ – Abc123 Jun 15 '18 at 12:00
  • $\begingroup$ @Abc123 What exactly is $\Phi_{t-1}$? I don't think you actually defined it. $\endgroup$ – Geoffrey Brent Jun 15 '18 at 13:09

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.