Determining the shape of the relationship between $y$ and $x$ after passing the Granger causality test - Cross Validated most recent 30 from stats.stackexchange.com 2019-07-17T17:03:25Z https://stats.stackexchange.com/feeds/question/245783 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://stats.stackexchange.com/q/245783 2 Determining the shape of the relationship between $y$ and $x$ after passing the Granger causality test Nirmal https://stats.stackexchange.com/users/138526 2016-11-14T05:58:30Z 2016-11-14T18:25:50Z <p>Let's say you have two time series, and you have already established a Granger-causal relationship between them. E.g. when testing if $X$ Granger-causes $Y$ (with a lag of 1), we calculated a $p$-value of 0.01. The results of our test mean that the $\beta X_{t-1}$ term in the equation </p> <p>$$Y_t = \alpha Y_{t-1} + \beta X_{t-1} + \varepsilon$$ </p> <p>provides statistically significant information about $Y_t$, compared to the original equation</p> <p>$$Y_t = \alpha Y_{t-1} + \varepsilon.$$</p> <p>Here's my question. The equation $Y_t = \alpha Y_{t-1} + \beta X_{t-1} + \varepsilon$ has an $R^2$ value of 0.2. This means that the data does not fit well to this regression line. Is there a better way to determine a function to calculate $Y_t$?</p> https://stats.stackexchange.com/questions/245783/-/245842#245842 0 Answer by Richard Hardy for Determining the shape of the relationship between $y$ and $x$ after passing the Granger causality test Richard Hardy https://stats.stackexchange.com/users/53690 2016-11-14T14:05:25Z 2016-11-14T18:25:50Z <p>By testing for Granger causality you have established that there is a statistically significant <em>linear</em> relationship. You cannot tell whether there are other forms of relationships just from this information.</p> <p>Also, an $R^2=0.20$ may be considered either low or high depending on the context. For example, if the noise is much stronger than the signal in the true data generating process (as if often the cases with financial returns, for example), you would not expect a high $R^2$ from a good model. (Actually, a high $R^2$ would then be a sign of overfitting.)</p> <p>Given this, the remaining question is how to determine a good model for $y_t$ based on $x_t$. This is a very general question, probably too general to have a concise yet useful answer.</p>