Let say we have random variables $X$ and $Y$, and we take their rank transforms to be $g(X)$ and $g(Y)$. The Spearman rank correlation coeffiicient can be considered to be $$R[g(X),g(Y)] = \frac{\mathbb{E}\left[ (g(X) - \mathbb{E}[g(X)])(g(Y) - \mathbb{E}[g(Y)]) \right]}{\sqrt{\mathbb{E}[(g(X) - \mathbb{E}[g(X)])]}\sqrt{\mathbb{E}[(g(Y) - \mathbb{E}[g(Y)])]}}$$

where $R$ is the Pearson product-moment correlation coefficient (PPMCC). It is also known from the algebra of random variables that $X \perp \!\!\! \perp Y \implies R[X,Y] = 0$. In other words, if two random variables are independent, then their PPMCC will be zero.

From the above, we can understand that a non-zero Spearman rank correlation coefficient implies that the rank-transformed variables $g(X)$ and $g(Y)$ are mutually dependent.


Can we further deduce that $X$ and $Y$ are mutually dependent if their Spearman rank correlation is non-zero? I.e. that $R[g(X),g(Y)] \neq 0 \implies \lnot (X \perp \!\!\! \perp Y)$?


Perhaps the rank transform being a monotonic transform could be useful if combined with the notion copulas. Specifically, the copula that maps the marginal cumulative distribution functions to the joint cumulative distribution function is invariant to monotonic transformations of the random variables. Thus ranking should not change the copula. If the copula was not the independence copula, then it is a copula that describes some sort of dependence. I see a potential argument here, but I have not closed the gap.


1 Answer 1


Yes, it does.

Here is Proposition 7.1.3(i) from Athreya and Lahiri 2006.

Let $(\Omega, \mathcal{F}, P)$ be a probability space and let $\{ X_j \}_{j=1}^n$, $2 \leq n < \infty$, be a collection of random variables on $(\Omega, \mathcal{F}, P)$. Then $\{ X_j \}_{j=1}^n$ are independent if-and-only-if $$\mathbb{E} \left[ \prod_{j=1}^n f_j(X_j) \right] = \prod_{j=1}^n \mathbb{E} \left[ f_j(X_j) \right]$$ for all bounded Borel-measurable functions $f_j: \mathbb{R} \mapsto \mathbb{R}, \forall j \in \{ 1, \cdots, n \}$.

The Spearman rank correlation can be written in the form of this result. By a contra-positive argument with this result, a non-zero Spearman correlation implies statistical dependence.


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