# Why is Gaussian Copula's Tail Dependence Zero?

I know that the Gaussian copula has a zero tail dependence (tail independence) due to the exponential behaviour at the tail. I am just wondering if there is a rigorous proof for this?

For simplicity sake, a bivariate Gaussian copula $C(X,Y)$ would do. In particular, there are no perfect correlations between $X$ and $Y$.

• What definition of tail dependence do you have? – Glen_b Nov 13 '16 at 5:37
• @Glen_b Limit as q->1 P[F(X) > q | F(Y) > q], where F is the standard normal marginal CDF. – NicTamHK Nov 13 '16 at 5:40
• In what sense do you assume that the faster that exponential decaying normal distributions' tails are "exponential"? Where do you get that from? – Carl Nov 13 '16 at 5:50
• @Carl I'm just assuming that might be the reason why the tail dependence is zero, since the normal PDF/CDF involves the exponential factor. – NicTamHK Nov 13 '16 at 5:52
• @NicTamHK Tail heaviness is usually adjudicated by survival function ratios. See, stats.stackexchange.com/questions/86429/… as an example. With respect to Cupolas, tail heaviness may be irrelevant. Why do you think it relevant? – Carl Nov 13 '16 at 6:03

Consider a bivariate Gaussian copula $C(\cdot)$.

Because of the radial symmetry of a Gaussian copula we can consider just the lower tail dependence. We know that the lower tail dependence for this copula is:

\begin{align} \lambda&=\lim_{\,\,q\to 0^{+}} \frac{\partial C(q,q)}{\partial q}\\ &=\lim_{\,\,q\to 0^{+}} \text{Pr}(U_{2}\leq q\,|\,U_{1}=q)+ \lim_{\,\,q\to 0^{+}} \text{Pr}(U_{1}\leq q\,|\,U_{2}=q) \end{align}

Since a Gaussian copula is exchangeable, it follows that:

$$\lambda=2\lim_{\,\,q\to 0^{+}}\text{Pr}(U_{2}\leq q\,|\,U_{1}=q)$$

Now, let:

$$(X_{1},X_{2}):=\Big(\Phi^{-1}(U_{1}),\,\Phi^{-1}(U_{2})\Big)$$

This means that $(X_{1},X_{2})$ has a bivariate normal distribution with standard marginals and correlation $\rho$. Now:

\begin{align} \lambda&=2\lim_{\,\,q\to 0^{+}}\text{Pr}(\Phi^{-1}(U_{2})\leq \Phi^{-1}(q)\,|\,\Phi^{-1}(U_{1})=\Phi^{-1}(q))\\ &=2\lim_{x\to -\infty}\text{Pr}(X_{2}\leq x\,|\, X_{1}=x) \end{align}

Finally, we know that $X_{2}\,|\,X_{1}\sim N(\rho x,1-\rho^{2})$, so:

$$\lambda=2\lim_{x\to -\infty}\Phi\Bigg(x\sqrt{\frac{(1-\rho)}{(1 +\rho)}}\Bigg)=0$$

• I was wondering why in the partial derivative of the Gaussian copula there are two terms in the sum: $\lambda=\lim_{\,\,q\to 0^{+}} \text{Pr}(U_{2}\leq q\,|\,U_{1}=q)+ \lim_{\,\,q\to 0^{+}} \text{Pr}(U_{1}\leq q\,|\,U_{2}=q)$ – AlexMe Mar 5 '19 at 10:22
• @AlexMe because both random variables have lower tails – epp Mar 5 '19 at 13:21
• Thanks @StatsPlease, I'm just confused with how can one obtain the above mentioned sum as the partial derivative of the double integral (which is the form of the Gaussian copula). – AlexMe Mar 5 '19 at 13:55
• @AlexMe Might be worth raising another question related to this. – epp Mar 5 '19 at 22:05

For a non-technical, intuitive view of what the tail index is telling you, we can look at simulation and compute sample estimates of the quantity $P[F(Y) > q | F(X) > q]$ as $q$ increases.

Here the original correlation is $0.96$, but as we get further into the upper tail of $X$ the correlation decreases, and the proportion where $F(Y) > q$ given $F(X) > q$ decreases.

• I think a formal proof of this observed convergence is given in my answer. – epp Jan 24 '17 at 6:56
• Indeed, but often people looking at this also seek intuition about what's going on. I attempted to supply some. – Glen_b Jan 24 '17 at 7:37