Referring to the answer here https://stats.stackexchange.com/a/66617

It is written that $\rho_s(X_1,X_2) = \rho(F_1(X_1),F_2(X_2))$

My Questions are :-

  • Is that forumla correct? Because I am not able to find it in other resources.
  • If it is correct, then how does it relate with the formula given in the wikipedia, $\rho_s(X_1,X_2) = \rho(rank(X_1),rank(X_2))$.
  • In general how to find the spearman's correlation coefficient "theoretically", i.e without any observed sample. To be specific let : $ \begin{pmatrix}X_{1}\\ X_{2}\\ \end{pmatrix} \sim \mathcal{N} \begin{bmatrix} \begin{pmatrix} 0\\ 0\\ \end{pmatrix}\!\!,& \begin{pmatrix} 1 & \rho\\ \rho & 1 \end{pmatrix} \end{bmatrix} $

How to compute the Spearman's Correlation Coefficient between $X_1$ and $X_2$?

  • 1
    $\begingroup$ Not sure about Spearman's correlation, but $\rho (F_1(X_1),F_2(X_2))$ is sometimes called the grade correlation between $X_1$ and $X_2$, which is derived for bivariate normal $(X_1,X_2)$ here and here. $\endgroup$ Feb 28, 2019 at 11:13
  • $\begingroup$ @StubbornAtom Got your point. But now the question is, whether grade correlation and spearman's correlation exactly the same thing? if so we have to prove that $\rho(F_1(X_1),F_2(X_2))=\rho(rank(X_1),rank(X_2))$, which by my gut feeling is most probably not true. If they're not the same thing my original questions still remains unanswered, how to compute spearman's correlation coefficient for bivariate normal distribution? $\endgroup$ Feb 28, 2019 at 11:35
  • $\begingroup$ @StubbornAtom Hey! I just now realized that $\rho(rank(X_1),rank(X_2))$ doesn't make any sense unless we are talking about an observed sample. So can we safely conclude that the concept of Spearman's Correlation Coefficient only exists for observed samples. And a natural extension of this concept for the Theoretical computation is the Grade's Correlation coefficient. Why I say it could be a natural extension is because both the formulas, i.e $\rho(rank(X_1),rank(X_2))$ and $\rho(F_1(X_1),F_2(X_2))$ seem to give very close results for large samples. $\endgroup$ Feb 28, 2019 at 12:18


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