For a bivariate process $(\textbf{X},\textbf{Y})=( (X_t, Y_t)^\top, t\in\mathbb{Z})$, we say that the process $\textbf{X}$ Sims-causes the process $\textbf{Y}$ (notation $\textbf{X}\overset{Sims}{\to}\textbf{Y}$), if $$ (Y_{t+s}, s\geq 1)\not\perp X_t \mid X_{past(t-1)}, Y_{past(t)}, Z_{past(t)}, \text{ for all }t\in\mathbb{Z}, $$ where $past(t) = (t, t-1, \dots)$, $\perp$ means independence and $Z$ are all revelant (causallty sufficient) variables.

Consider my new definition: we say that the process $\textbf{X}$ N-causes the process $\textbf{Y}$ (notation $\textbf{X}\overset{N}{\to}\textbf{Y}$), if $$ (Y_{t+s}, s\geq 1)\not\perp X_t \mid Y_{past(t)}, Z_{past(t)}, \text{ for all }t\in\mathbb{Z}. $$

The difference is that we omit $\textbf{X}$ in the conditioning set.

Can you give me an example when $\textbf{X}\overset{N}{\to}\textbf{Y}$ but $\textbf{X}\overset{Sims}{\not\to}\textbf{Y}$? Or vice-versa? I feel that in the vast majority of random processes, these definitions should be equivalent. Can you find a condition under which the definitions are equivalent?

  • $\begingroup$ Where does the process $\mathbf{Z}$ come in? $\endgroup$ Sep 12, 2023 at 7:23
  • $\begingroup$ $Z$ are possible confounders. In the most general form, $Z$ should contain all variables in the universe. But for the sake of argument, say that there is at most one common confounder. For example, if $(\textbf{X},\textbf{Y},\textbf{Z})$ follow some statistical model such as VAR model, we assume that neither component of $Z$ is hidden. That is, $Y_{t+1}$ is a function of $(X_{past(t)}, Y_{past(t)}, Z_{past(t)})$. How does $Z$ come in in the definition? We condition on $Z_{past(t)}$, because if we wont, we can not distinguish between dependence due to $X\to Y$ or $Y\leftarrow Z\to X$. $\endgroup$ Sep 12, 2023 at 11:17
  • $\begingroup$ In that case, the definitions should be applied to $(\textbf{X},\textbf{Y},\textbf{Z})$ instead. $\endgroup$ Sep 12, 2023 at 11:44
  • $\begingroup$ Whatt do you mean by 'applied'? The definiton relies on $Z$, but it is not specified explicitely $\endgroup$ Sep 14, 2023 at 10:55
  • $\begingroup$ The definition as stated, is about trivariate processes, not bivariate processes. $\endgroup$ Sep 14, 2023 at 11:21

1 Answer 1


Let $\mathbf{X}$ be conditionally iid on $\{0,1\}$ with distribution $(P,1-P)$, with $P$ being a nontrivial random variable. Let $Y_t=\limsup_{T\to\infty} T^{-1}\sum_{i=1}^TX_{t-i}$. Then $\textbf{X}\overset{N}{\to}\textbf{Y}$ but not $\textbf{X}\overset{\text{Sims}}{\to}\textbf{Y}$.

Here is an example that shows that the other direction need not hold either:

Let $\mathbf{X}$ be iid uniform on $\{0,1\}$ and let $Y_{t+1}=X_t+X_{t-1}\mod 2$. Then $\textbf{X}\overset{\text{Sims}}{\to}\textbf{Y}$ but not $\textbf{X}\overset{N}{\to}\textbf{Y}$.

  • $\begingroup$ The $\mathbf{X}$ variable is normally known as Bernoulli. $\endgroup$ Sep 14, 2023 at 5:05
  • $\begingroup$ Yeah, this counterexample seems valid, thank you for it. But it only holds because $Y_{t}$ is independent with $X_s$ for any $s,t$ which seems as an extremely untraditional situation that cant hold for example in continuous processes. However, I mainly asked this question because of the other implication, which is more interesting (that I feel it should hold). Can $\textbf{X}\overset{N}{\to} \textbf{Y}$ but $\textbf{X}\overset{Sims}{\not\to} \textbf{Y}$? $\endgroup$ Sep 14, 2023 at 10:49
  • $\begingroup$ That direction need not hold either; see my updated answer. $\endgroup$ Sep 14, 2023 at 11:44
  • $\begingroup$ Thank you for the update. However, I am not sure if I understand. Won't it be just $Y_t = constant$ due to the law of large numbers? $\endgroup$ Sep 14, 2023 at 15:24
  • $\begingroup$ @AlbertParadek It will be constant conditional on the realization of $P$. But since $P$ is random, the value of this constant is not known. $X_t$ is slightly informative, and the complete history is perfectly informative. $\endgroup$ Sep 14, 2023 at 15:59

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