# Does $y=f(x)$ implies $X$ Granger-causes $Y$?

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).

• 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?
– whuber
Sep 16 '20 at 16:52
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 $$y$$.
• 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. Sep 16 '20 at 17:07