I replicated momentum-strategy returns for different markets and want to show that I have done it "right". I got research data for comparison and want to calculate the correlation between this data and mine. Which correlation coefficient is appropriate for stock returns? Pearson and Spearman both give similar results but I want to be able to explain why I have taken the specific one. So far I know that the interpretation for Pearson would be somewhat like : "The difference from the series' mean correlates". But is this what I would need in the case of stock returns?

Second question would be: Do I have to compare logarithmic returns since they are normally distributed or can I take the normal returns for measuring correlation?

  • $\begingroup$ "The difference from the series' mean correlates" can't be an interpretation of any correlation coefficient, except vacuously. Think that Pearson measures linearity of relationship and the others measure monotonicity. $\endgroup$ – Nick Cox Aug 22 at 16:18

Regardless of the field that you are working in, whenever the x and y variables show non-normal distribution or have ordinal scales; use cor(x, y, method = "spearman") or cor(x, y, method = "kendall").

Use pearson when your data is normally distributed. If the logarithmic transformation (or other transformation methods) results in unskewed and normally distributed variables, "pearson" is also a good choice. However, the interpretation of transformed variables is usually more difficult.

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    $\begingroup$ What's wrong with using the pearson correlation for just about everything? It makes sense whenever linear regression makes sense, which almost always makes sense as a first step in studying how variables depend on each other. And a correlation coefficient is never anything but such a first step in an analysis. $\endgroup$ – CloseToC Aug 22 at 16:04
  • $\begingroup$ I disagree. Suppose $x$ is uniformly distributed and $y = a + bx$ exactly for $b \ne 0$, so $y$ is uniformly distributed too. So Pearson correlation is invalid because neither variable is normally distributed? Not so, and similar comments for many other cases. This answer is the wrong way round. Pearson measures linearity of relationship and is useful whenever that's the reference. Sure, the usual inferential machinery, especially for producing P-values, requires circumspection if the distributions are some distance from bivariate normal, but we've had the bootstrap for 40 years now.... $\endgroup$ – Nick Cox Aug 22 at 16:12
  • $\begingroup$ (+1 to @CloseToC) $\endgroup$ – Nick Cox Aug 22 at 16:13

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