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Factor in the information from the question "Does causation imply correlation?", "Mathematical Definition of Causality", and "How to formally tell is one time series affects another".

Let $y = f(x)$, where $f$ is continuous and differentiable. Does this guarantee that $x$ will Granger-cause $y$?

Given that $f$ can be horribly nonlinear, it seems that there are a limited number of tests and methods for conditioning the data in such a way that the granger causality test can be applied effectively. Even something like $y=e^x + \epsilon$, where $\epsilon$ is white noise, can cause problems with estimating the order of integration (thus testing for cointegration, and then Granger-causality according to the Toda and Yamamoto method as described in Dave Giles' blog post).

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  • $\begingroup$ Could you clarify the sense in which a function $f$ could be considered as related to Granger causality, which concerns predictability in a time series? $\endgroup$
    – whuber
    Sep 16 '20 at 16:52
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    $\begingroup$ See also my comment under the answer. $\endgroup$ Sep 16 '20 at 17:08
  • $\begingroup$ @whuber - Take the equation for force, mass, and acceleration F = ma, where mass is constant. I'm under the impression that if F = ma, then F should granger-cause a (because it actually causes a). Is that misguided? $\endgroup$ Sep 16 '20 at 17:16
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    $\begingroup$ @whuber - Thinking about this harder, I will post a separate question. $\endgroup$ Sep 16 '20 at 17:26
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No.

Consider $x(t) \sim \operatorname{Uniform}(-1,1)$. For any function $f$, including $f(x)=x$ and $f(x)=0$ it is obvious to see that past values of $x$ don't help in predicting $y$.

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    $\begingroup$ Granger causality requires past values of $x$ to help in predicting $y$ once the predictive information contained in past values of $y$ has been accounted for. Forgetting the latter clause seems to be a common mistake. $\endgroup$ Sep 16 '20 at 17:07
  • $\begingroup$ @RichardHardy - I'll replace my previous question with this one: doesn't that imply that for test for Granger-Causality, y must be a function of at least 2 variables with (each with their own auto-regressions)? $\endgroup$ Sep 16 '20 at 17:23

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