Solved – Help understanding copula version of Spearman rank correlation

copulacorrelationspearman-rho

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.

Best Answer

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 $C'(u,v) = \frac{\partial^2}{\partial u \partial v}C(u,v)$.

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

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.

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Solved – Estimation of critical values in Spearman rank correlation

There's no great mystery here.

The distribution of the Spearman correlation is discrete. (The set of ranks of $n$ values are discrete, so the Spearman correlation will necessarily be discrete.)
The relevant Wikipedia page lists several approximations for the distribution of the Spearman correlation coefficient; I think the t-approximation is the second listed. The resulting p-values are not exact in small samples (but are reasonably close considering it's a continuous approximation of a discrete distribution - see the plot).
Since the exact small sample distribution (under the null) for the Spearman correlation can be computed, exact small sample tables exist. The table you link to appears to be a table of critical correlations (yielding type I error no greater than the listed significance levels) derived from the exact distribution.

enter image description here

A continuity correction would probably help here.

Edit: actually, it looks like a suitable continuity correction helps quite a lot overall (however it's slightly less accurate for small portions of the range of $r_s$, so while most of the time it improves the approximation, sometimes it doesn't).

How to use the t-statistic that Wikipedia gives: you can get one- or two-tailed p-values from the cdf of the $t$ with $n-2$ d.f. just in the same way as you would for any t-statistic with given degrees of freedom.

You can get critical values from the inverse CDF of the $t$. As noted above, these will be approximate, but the approximation kicks in pretty quickly. You don't need very large $n$ before it's suitable for most purposes.

Solved – Relation between Spearman and Pearson correlation in Gaussian copula

A dependence measure is a parameter of a Gauss copula. Let denote by $\rho$ the Pearson correlation, $\rho_s$ the Spearman correlation, and $\tau_k$ the Kendell tau.

It's known $\rho \in [-1,1]$, and $$\rho_s = \frac{6}{\pi}\cdot arcsin(\frac{\rho}{2}) ,$$ $$\tau_k = \frac{2}{\pi}\cdot arcsin(\rho).$$

In the plot below one can see the dependencies between $\rho_s$, $\tau_k$ and $\rho$.

    rho <- seq(-1,1,0.01)     # Pearson correlation
    rho_s <- 6/pi*asin(rho/2) # Spearman correlation
    tau_k <- 2/pi*asin(rho)   # Kendell tau
    plot(rho, rho, type="n", xlab="rho", ylab="rho_s, tau_k")
    lines(rho, tau_k, col = "red")
    lines(rho, rho_s, col = "blue")
    legend("topleft", lty=c(1,1), col = c("red", "blue"), legend=c("tau_k", "rho_s"))

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Solved – Estimation of critical values in Spearman rank correlation

Solved – Relation between Spearman and Pearson correlation in Gaussian copula

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