Help understanding copula version of Spearman rank correlation

I'm reading through this article (http://www.sciencedirect.com/science/article/pii/S0047259X06000662) where they have a population version of the Spearman rank correlation. I'm having a little bit of trouble understanding the author's derivation, which I reproduce as follows:

"For a two-dimensional random vector $X$ $=(X_{1},X_{2})'$ with distribution function $F$, univariate marginal distributions $F_{X_{1}}$, $F_{X_{2}}$ and a copula $C$ is defined by

$\rho_{S}=\frac{cov(F_{X_{1}}(X_{1}), F_{X_{2}}(X_{2}))}{\sqrt{var(F_{X_{1}}(X_{1}))}\sqrt{var(F_{X_{2}}(X_{2}))}}=$

$= \frac{cov(U_{1},U_{2})}{\sqrt{var(U_{1})}\sqrt{var(U_{2})}}=$

$=\frac{\int_{0}^{1}\int_{0}^{1}uv dC(u,v)-(\frac{1}{2})^{2}}{\sqrt{\frac{1}{12}}\sqrt{\frac{1}{12}}}=$

$=12\int_{0}^{1}\int_{0}^{1}C(u,v)dudv-3$

I understand every step with the exception of the last one. Where did the 3 come from? And why were they able to change the integrand from $uvdC(u,v)$ to $C(u,v)dudv$?

I'm trying to teach myself some basic copula theory so there are a few holes in my knowledge here and there.

• The second but last line is incorrect, the denominator should be $\sqrt{\frac{1}{12}}\sqrt{\frac{1}{12}}$ instead of $\sqrt{\frac{1}{2}}\sqrt{\frac{1}{2}}$. Aug 24, 2015 at 19:12
• Oh! You're totally right. Sorry about that, my mistake. Do you have any idea about my question regarding the last step? I can see that this is a very common way to define the Spearman rank correlation but nobody really explains where the 3 and the change of integrand came from.
Aug 24, 2015 at 19:22
• Maybe integrating $\int\int uv \text{d}C(u,v)$ by parts (two times, in both integrating variables) can show that it is equal to $\int\int C(u,v)\text{d}u\text{v}$ Aug 24, 2015 at 19:31
• But the "3" is obvious: $\frac{(\frac12)^2}{\sqrt{\frac{1}{12}}\sqrt{\frac{1}{12}}}=\frac{12}{4}=3$. Aug 24, 2015 at 22:06
• D'oh! It has been staring at me all the time. You're right! Thanks! Any ideas about the change of integrand? Do you think it's like StijnDeVuyst said?
Aug 25, 2015 at 0:00

I came to your question after looking for the same proof and not finding it anywhere. Following StijnDeVuyst suggestion, I managed to prove it using integration by parts. I will omit the integration limits, just keep in mind they exist.

$$\int \int uv \ dC(u, v) = \int\int uv C'(u,v) \ dudv,$$ where $$C'(u,v) = \frac{\partial^2}{\partial u \partial v}C(u,v)$$.

$$$$\tag{1} \label{main} \int\int uv C'(u,v) \ dudv = \int v \left(\int u C'(u,v) du \right) dv.$$$$

Solving the inner integral \begin{align*} \int_0^1 u C'(u,v) du &= u\frac{\partial}{\partial v}C(u,v) \big\rvert_{u=0}^1 - \int_0^1 \frac{\partial}{\partial v} C(u,v) \ du\\ &= 1-\int_0^1 \frac{\partial}{\partial v} C(u,v) \ du\\ \end{align*}

where the first equality is integration by parts on $$u$$ and the last equality comes from the fact that $$C(1, v) = v$$.

Substituting in \ref{main} we have \begin{align*} \int \int uvC'(u,v) \ dudv &= \int v \left(1-\int \frac{\partial}{\partial v} C(u,v) \ du \right)dv\\ &= \int_0^1 v \ dv - \int\int v \frac{\partial}{\partial v}C(u,v) \ dudv\\ &= \frac{1}{2}-\int\left(\int v \frac{\partial}{\partial v} C(u,v) \ dv \right) du. \tag{2}\label{segunda} \end{align*}

Solving the inner integral with integration by parts on $$v$$ we get $$\int_0^1 v \frac{\partial}{\partial v} C(u,v) \ dv = u - \int_0^1C(u,v) \ dv$$

and substituting in \ref{segunda} we have (now including the integration limits) \begin{align*} \int_0^1\int_0^1 uv \ dC(u,v) &= \int_0^1 \int_0^1 uvC'(u,v) \ dudv\\ &= \frac{1}{2} - \int_0^1 u \ du + \int_0^1 \int_0^1 C(u,v) \ dv du\\ &= \int_0^1 \int_0^1 C(u,v) \ du dv \end{align*}

which proves the identity \begin{align*} \rho_S &= 12\int_0^1 \int_0^1 uv \ dC(u,v) - 3\\ &= 12\int_0^1\int_0^1 C(u,v) \ dudv - 3. \end{align*}

This proof assumes that the copula $$C$$ has density given by $$\frac{\partial^2}{\partial u\partial v}C(u,v)$$. When this is not the case, check Theorem 5.1.1 and Corollary 5.1.2 from An Introduction to Copulas by Roger B. Nelsen.