1
$\begingroup$

When building a model with many regressors, Is there any statistical test to examine unidirectional versus mutual influence of variables?

$\endgroup$
1
  • $\begingroup$ What about $t$-test and $F$-test to test that some coefficient(s) is/are equal to zero? Is there anything wrong with them? This is how we test for Granger causality, for example. $\endgroup$ Feb 24 '17 at 19:12
0
$\begingroup$

While your problem description is pretty brief, as far as I understand it you may be interested in Granger causality (Wikipedia). In a simple bivariate system $(x_t,y_t)$ you may ask whether $x_t$ Granger-causes $y_t$, denoted $x_t \xrightarrow{Granger} y_t$; or vice versa; or both. You may test that using $F$ test, for example. As an illustration, consider a VAR(1) model:

\begin{aligned} x_t &= \varphi_{11} x_{t-1} + \varphi_{12} y_{t-1} + \varepsilon_{1,t}, \\ y_t &= \varphi_{21} x_{t-1} + \varphi_{22} y_{t-1} + \varepsilon_{2,t}. \end{aligned}

You may test the following hypotheses:

  • $H_0\colon \ x_t \not\xrightarrow{Granger} y_t$ by testing $H_0\colon \ \varphi_{21}=0$.
  • $H_0\colon \ y_t \not\xrightarrow{Granger} x_t$ by testing $H_0\colon \ \varphi_{12}=0$.
  • $H_0\colon \ x_t \not\xrightarrow{Granger} y_t \ \text{and} \ y_t \not\xrightarrow{Granger} x_t$ by testing $H_0\colon \ \varphi_{21}=\varphi_{12}=0$.

This way you may conclude over whether the relation is unidirectional (and then which way) or bidirectional. This also works in a multivariate system.

$\endgroup$
1
  • $\begingroup$ Thanks much Richard for the explanation, I never used Granger test before. In my problem both x and y variables are caused by a third known variable (that we dont have data for that). The partial causal relation between x and y is unknown. I was testing the significancy of cross correlation of residuals of each fitted model with lag of other series to determine the causal relationship. It would be interesting to check the end result with Granger test if my approach is not fundamentally wrong :)) $\endgroup$ Feb 26 '17 at 2:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.