1

If $X$ causes $Y$,$X \rightarrow Y$, as defined in Granger causality, is it possible that $Y \rightarrow X $? How can we prove if it is not possible? That is the relationship is one-directional.

rando
  • 236
  • 1
    To be more precise with the notation, Granger causality is a relationship between $X_{t-1}$ and $Y_t$. Are you asking if it's also possible that $Y_{t-1}$ and $X_t$ be related at the same time (it is, of course)? Or are you suggesting a different notion of causality in which it is possible for a variable to "cause" another backwards in time? – Chris Haug Nov 25 '20 at 20:12
  • @ChrisHaug, the notation you are suggesting seems unorthodox. $X\xrightarrow{\text{Granger}}Y$ means the history of $X$ is helpful in predicting $Y$ beyond the history of $Y$. In my understanding, you could say the notation $X\xrightarrow{\text{Granger}}Y$ could be equivalently written as $X_t\xrightarrow{\text{Granger}}Y_t$ but not $X_{t-1}\xrightarrow{\text{Granger}}Y_t$. The latter would be equivalent to $\text{lag}(X,1)\xrightarrow{\text{Granger}}Y$. In short, $X_{t-1}\xrightarrow{\text{Granger}}Y_t$ might be a confusing notation and not equivalent to $X\xrightarrow{\text{Granger}}Y$. – Richard Hardy Nov 25 '20 at 20:49
  • @RichardHardy Right, but I didn't write any of those arrows, so I don't think what I wrote was confusing. I just wanted to highlight the time indices because it makes it immediately clear that it can go both ways. I feel that the $X\xrightarrow{\text{Granger}}Y$ notation obscures this. – Chris Haug Nov 25 '20 at 22:03
  • @ChrisHaug, maybe it is only me that got confused. Instead of saying Granger causality is* a relationship between $X_{t−1}$ and $Y_t$* I would try to phrase it otherwise. For me it is a relationship between $X_t$ and $Y_t$ characterized by a statistical relationship between $X_{t−p}$ for $p>0$ and $Y_t$ conditional on $Y_{t-q}$ for $q>0$. Though not elegant, I hope it is understandable. – Richard Hardy Nov 26 '20 at 06:28

1 Answers1

2

Bidirectional Granger causality is definitely possible. Take a simple bivariate VAR(1) model for variables $X$ and $Y$ with all coefficients being nonzero: \begin{aligned} X_t&=\varphi_{10}+\varphi_{11}X_{t-1}+\varphi_{12}Y_{t-1}+\varepsilon_{1,t},\\ Y_t&=\varphi_{20}+\varphi_{21}X_{t-1}+\varphi_{22}Y_{t-1}+\varepsilon_{2,t}. \end{aligned} In it, $X\xrightarrow{\text{Granger}}Y$ as well as $Y\xrightarrow{\text{Granger}}X$. The former holds because $Y_t$ is a linear function of $X_{t-1}$ even once $Y_{t-1}$ and all older lags of $Y_t$ are accounted for. The latter holds by symmetry.

You can make $X\xrightarrow{\text{Granger}}Y$ disappear by setting $\varphi_{21}=0$.
You can make $Y\xrightarrow{\text{Granger}}X$ disappear by setting $\varphi_{12}=0$.

Richard Hardy
  • 67,272