What is $\rho$ in case of $Cov(X,Y) = \rho \cdot SD(Y) \cdot SD(X)$

Question

$\rho$ has been used as a symbol for rank correlation especially in the sense of Spearman $\rho$. What is the interpretation in the case indicated in the the question, where the covariance equals $\rho$ multiplied by the standard deviation $X$ multiplied by standard deviation $Y$. Suppose we are working on two random variables. The variables are independent as well as interdependent. For example, r is a composite of population correlation (in psychometric sense) and sampling error.

Answer 1

If the random variables $X$ and $Y$ are independent as you claim they are, then their covariance $\text{cov}(X,Y)$ equals $0$ and therefore so does the Pearson correlation coefficient $\displaystyle \rho = \frac{\text{cov}(X,Y)}{\sigma_X\sigma_Y}$ equal $0$.

However, given independent samples $\{(X_i,Y_i) \colon 1 \leq i \leq n\}$ of independent random variables $X$ and $Y$, the sample covariance and the sample Pearson correlation coefficient are not necessarily identically $0$, though it is every statistician's fondest hope that both these statistics will be small in magnitude, especially when $n$ is large.

Answer 2

It is (by definition) the Pearson correlation coefficient. Whether the two variables are independent or "interactive" has no bearing on this definition.

Answer 3

It seems that there is a little confusion in the use of terms. Here, $\rho$ is a parametric estimate i.e. $E(r)$. Mathai and Rathie (1977) have given the formula of $\rho$ in terms of $E(r)$ i.e. "linear correlation coefficient" and in any case it does not connote the sample correlation coefficient i.e. $r$. Moreover, we should interpret statistical independence between two variables as if there is no common moderator for the two random variables. And, generally, there is a real relationship (it may be small) between two independent explanatory variables that can be measured by covariance.