# Spearman $\rho$ as a function of Pearson $r$

It is common to talk about the linear correlation, Pearson's $$r$$, between two random variables $$\{(x_1,y_1),(x_2,y_2),\ldots,(x_n,y_n)\}$$ as having two components: a) the copula and b) the marginal distributions of $$x$$ and $$y$$. In contrast, the rank correlation, Spearman's $$\rho$$, depends only on the copula.

As a reminder, the calculation of $$r$$ and $$\rho$$ are identical except that in the case of $$\rho$$, the variables are first transformed to ordered ranks $$R(.) = 1,2,\ldots, n$$.

$$r = \frac{\operatorname{Cov}(x_i,y_i)}{\sqrt{\operatorname{Var}(x_i)}\sqrt{\operatorname{Var}(y_i)}} \quad, \quad \rho = \frac{\operatorname{Cov}(R[x_i],R[y_i])}{\sqrt{\operatorname{Var}(R[x_i])}\sqrt{\operatorname{Var}(R[y_i])}} .$$

Ranks follow a uniform distribution by definition. Let's further assume a normal bivariate distribution for nontransformed $$x$$ and $$y$$.

Now my question: I want to express $$\rho$$ as a transformation of $$r$$. It seems that this should be a fairly simple function that somehow involves a mapping between the normal to a uniform.

My intuition of $$\rho$$ is that it is a weighted function of $$r$$ with large weights placed on small distances in the middle of the distribution; again, this would seem proportional to the normal distribution itself.

Is anyone able to work out the specific expression?

• Unless you can quantify what you mean by "approximately" normal, there is no possible correct answer. The reason is that in many common, meaningful definition of "approximate," a mixture of a bivariate normal and any other distribution (with tiny weight) is "close" to bivariate normal, but by placing the center of the contaminating distribution far from the center of the normal distribution you can make $r$ attain any value in $(-1,1).$ This operation scarcely changes the distribution of Spearman's $\rho,$ proving there's no relationship between $r$ and $\rho$ whatsoever. – whuber Aug 25 '19 at 13:01
• I appreciate this and am fine with an expression that holds in the normal case. (Hence my last sentence: Assume normality if that helps.) Thankful for any help. – pengzell Aug 25 '19 at 13:04
• Do you just want to compare expectations or do you want to derive the joint distribution of $(r,\rho)$ for every $n$? – whuber Aug 25 '19 at 13:12
• I would be happy with expectations! I see that my original question was perhaps unnecessarily wordy... – pengzell Aug 25 '19 at 13:16
• To aid with intuition somewhat, it might be worth noting the close connection between (suitably standardized) ranks and a marginal transformation to standard uniform (in particular, with increasing sample size). – Glen_b Aug 25 '19 at 23:18

I think I found the answer. In Pearson's "On further methods of determining correlation" (1907) he derives the expression: $$r=2 \sin \Big(\frac{\pi}{6}\rho\Big),$$ which implies, $$\rho= \frac{6}{\pi} \sin^{-1} \frac{r}{2}.$$ This made me realize that for the bivariate normal case, $$r$$ and $$\rho$$ are, as a rule of thumb, near identical. I would still be curious if anyone has an answer for more exotic distributions but that is a separate question.