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If I were to use Spearman's rho as a measure of correlation, must the data I'm working with be normally distributed/homoscedastic?

Also, if I am plotting the data against time, must the data be time-independent?

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Spearman's rho is nonparametric. There are no distributional assumptions. Interpreting the Pearson correlation as strength of linearity and tests for statistical significance rely on bivariate normality. Homoskedasticity never enters in for any correlation. Remember it only has to do with the bivariate pair (X,Y) that you are calculating correlations for. Regression doesn't enter in unless in the case of Pearson you want the slope of the regression of X on Y. When you say you are plotting with time is time a variable included in the correlation? If you are just looking at pairs that are from a time-dependent sequence, there is no need to make any assumptions about time. If you are correlating X with time T then Spearman's rho is used to see if there is a relationship between X and T and then clearly you don't want to assume T is independent of X.

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    $\begingroup$ (+1) Homoskedasticity is an assumption in the sense that $X_1,\ldots,X_n$ are assumed to be i.i.d. (and thus have the same variance). It's possible that this is what the OP meant. $\endgroup$
    – MånsT
    Jul 26, 2012 at 17:03
  • $\begingroup$ @MansT That would have a bearing on the Pearson correlation since the Xs and Ys are both assumed to have a constant variance and the formula is Cov(X,Y)/[√Var(X) √Var(Y)]. While IID is assumed in the nonparametric framework too you do not see constant variance having a big impact on the Spearman correlation. But yes the assumption is that the pairs (X$_i$, Y$_i$) are independent and each component has constant variance. $\endgroup$ Jul 26, 2012 at 18:10

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