Population Spearman's correlation coefficient For Pearson correlation coefficient one has a "population" formula in terms of the expectations /  variances / covariances, and a "sample" estimate formula based on actual observations. However, for Spearman's correlation, I have only seen it given in terms of the "sample" formula, never the population one. 
It seems that in order to have a population formula for Spearman's correlation, one has to at least fix the sample size. Assuming absolutely continuous densities $f_X$, $f_Y$, and $f_{XY}$ (both joint and marginal) for random variables $X$ and $Y$, and denoting by $\mathbf X_n = \{X_1,X_2,...X_n\}$ a sample of size $n$, the probability of any two $X$ in the sample being the same is 0 and hence the ranks are always given by the set of integers from $1$ to $n$, which has the same "variance" as the uniform discreet distribution given by 
$$ \mathrm {Var}(\mathrm{rank}(X)) = \frac{n^2-1}{12}$$
and similarly for $Y$. So, this is independent of the marginal distributions. We now need to find the expression for the covariance of ranked samples. If we denote by $\mathbf X_{(n)} = \{X_{(1)},X_{(2)},...X_{(n)}\}$ the corresponding order statistics in the increasing order (and similarly for $Y$), then the density function for the order statistics are given by 
$$f_{\mathbf X_{(n)}} = n!f_X(x_1)f_X(x_2) \cdots f_X(x_n) \mathbb I_{x_1 \le x_2} \mathbb I_{x_2 \le x_3} \cdots \mathbb I_{x_{n-1} \le x_n}$$
and
$$f_{\mathbf Y_{(n)}} = n!f_Y(y_1)f_Y(y_2) \cdots f_Y(y_n) \mathbb I_{y_1 \le y_2} \mathbb I_{y_2 \le y_3} \cdots \mathbb I_{y_{n-1} \le y_n}$$
However, it is not clear to me how to express the joint distribution of ranked variables in order to calculate their covariance.
 A: It depends on the ranking criteria of $(X_1,Y_1), (X_2, Y_2), \ldots, (X_n, Y_n)$.
Cases:


*

*$(X_i, Y_i) \leq (X_j, Y_j) \iff X_i \leq X_j$ then,


$$f_{(X,Y)_{(n)}} = n! f_{X,Y}(x_1,y_1)f_{X,Y}(x_2,y_2) \ldots f_{X,Y}(x_n, y_n)\mathbb{I}_{x_1 \leq x_2 \leq x_3 \ldots \leq x_n}$$


*$(X_i, Y_i) \leq (X_j, Y_j) \iff Y_i \leq Y_j$ then,


$$f_{(X,Y)_{(n)}} = n! f_{X,Y}(x_1,y_1)f_{X,Y}(x_2,y_2) \ldots f_{X,Y}(x_n, y_n)\mathbb{I}_{y_1 \leq y_2 \leq y_3 \ldots \leq y_n}$$


*$(X_i, Y_i) \equiv (\text{ith smallest }_{j=1}^{n} \ X_j, \ \text{ith smallest }_{j=1}^{n} \ Y_j)$ then,


$$f_{(X,Y)_{(n)}} = n!n! f_{X,Y}(x_1,y_1)f_{X,Y}(x_2,y_2) \ldots f_{X,Y}(x_n, y_n){I}_{x_1 \leq x_2 \leq x_3 \ldots \leq x_n}\mathbb{I}_{y_1 \leq y_2 \leq y_3 \ldots \leq y_n}$$
In each case, you multiply $f_{X,Y}(x_1,y_1)f_{X,Y}(x_2,y_2) \ldots f_{X,Y}(x_n, y_n)$ with number of possible ways of assigning a realization of the sample to the random variables in the sample such that the ranking criteria is satisfied i.e. if the realization of the sample is $x_1, x_2, \ldots, x_n, y_1, y_2, \ldots y_n$, then multiply with the number of ways you can assign these $x_i$'s to $X_j$'s and $y_k$'s to $Y_l$'s such that the ranking criteria is satisfied. You also multiple with a proper indicator function which is one if the given realization of the sample satisfies the ranking criteria otherwise zero.
