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I have a question about the properties of white noise in a time series context. Specifically, I want to know:

If we assume that the error term $u_t$ in a time series model is white noise, does this guarantee that the lagged variable $X_{t-1}$ is uncorrelated with the current error term $u_t$?

In other words, does the white noise assumption for the error term imply that $\text{Cov}(X_{t-1}, u_t) = 0$?

Richard Hardy
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Newbie
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2 Answers2

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Generally, no. The concept of white noise refers only to a single process such as $\{u_t\}$ and its internal structure (how the different elements of $\{u_t\}$ relate to each other). It does not refer to any other variables or processes. Knowing that $\{u_t\}$ is white noise tells us nothing about the relation between $\{u_t\}$ and some other process $\{X_t\}$.

However, if you are interested in autoregressive processes and such (as you probably are, though you do not quite spell that out in the original version of the question), see Jarle Tufto's answer for a special case where indeed $\text{Cov}(X_{t-1}, u_t) = 0$.

Richard Hardy
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  • I'm having a problem since I saw the example of an AR(1) process in the Wikipedia article on autoregressive models. When deriving the variance, it seems to consider X_(t-1) and u_t as uncorrelated. Am I missing something? Autoregressive model example – Newbie Mar 15 '24 at 08:27
  • But if this other process is defined in terms of $u_{t-k}$ with $k \geq 1$, for example $$X_{t-1} = \sum_{k=1}^\infty a^{k-1} u_{t-k} $$ , doesn't this make it possible to say something about the correlation between $u_t$ and that other process. – Sextus Empiricus Mar 15 '24 at 08:40
  • @SextusEmpiricus, your comment and Jarle Tufto's answer consider some special cases where it is entirely possible that $\text{Cov}(X_{t-1}, u_t) = 0$. My answer is about the general case. – Richard Hardy Mar 15 '24 at 08:58
  • As always with downvotes, I would appreciate a constructive comment to follow along. If you see a mistake in my answer, it would be helpful if you pointed it out. – Richard Hardy Mar 15 '24 at 09:00
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    Indeed, it is based on filling in what the OP means with $X_t$. The question is not well written and misses to explain what $X_t$ means. We could for instance have the situation where $X_t$ is white noise and we define $u_{t} := X_{t-1}$ (such that $u_t$ is also white noise). White noise is not neccesarily independent from everything in the past. – Sextus Empiricus Mar 15 '24 at 09:21
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Yes, if $\{X_t\}$ is a causal stationary solution $X_t=\sum_{j=0}^\infty\psi_i u_t$ of the model equation $\phi(B)X_t = \theta(B)u_t$ , then $X_t$ is clearly independent of $u_{t+1}$ since $\{u_t\}$ is white noise. If $\{X_t\}$ is a non-causal solution, the answer is no. This post clearly explains the difference between the model equation and its causal and non-causal solutions.

Richard Hardy
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Jarle Tufto
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  • I'm having a problem since I saw the example of an AR(1) process in the Wikipedia article on autoregressive models. When deriving the variance, it seems to consider X_(t-1) and u_t as uncorrelated when it only assumes that u_t is a white noise. Am I missing something? Autoregressive model example – Newbie Mar 15 '24 at 08:35
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    @Newbie Yes, usually only causal solutions are of interest and then $X_{t-1}$ and $u_t$ are indeed uncorrelated. – Jarle Tufto Mar 15 '24 at 08:43
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    The formula describing the process might have a non-causal solution, but it doesn't mean that this solution is actually what happens. If we define $X_{t} := \varphi X_{t-1} + u_t$ with $\varphi \geq 1$ and initialize $X_{0} = 0$ then you get a random walk, and no AR process, and you do not get this non-causal solution. – Sextus Empiricus Mar 15 '24 at 08:44