Archimedean Clayton copula entropy

Question

Question

I would like to derive the entropy of Archimedean parametric copulas (Clayton, Frank, or Gumbel), focusing here on the Clayton copula. Link to similar question on the t-copula.

The bivariate copula function, $C$, for the Clayton copula, with transformed marginals $u$ and $v$, and dependence parameter $\theta\in \mathbb{R}_{\geq 0}$, is

$$ C(u, v) = \bigg[ u^{-\theta} + v^{-\theta} -1 \bigg]^{-1/\theta} $$

Its copula density is the second mixed partial derivative of $C(u,v)$:

\begin{align} c\left(u,v ; \theta\right) = (1+\theta)(u \cdot v)^{-1-\theta}(u^{-\theta}+v^{-\theta}-1)^{-\frac{1}{\theta}-2} \end{align}

The differential entropy of a univariate variable's pdf, $f(x)$, is typically $$H(X)=-\int_{-\infty} ^{\infty} f(x) \ln f(x) dx$$

while any copula entropy can be estimated as

$$H(c(u,v))=-\int_{[0,1]^2} c(u,v) \ln c(u,v) \hspace{1mm} du \hspace{1mm} dv $$

How can we derive a closed-form analytical solution for the entropy of the Clayton copula density $c(u,v)$?

First attempt

\begin{align} H(c(u,v)) =& -\iint_{[0,1]^2} c\left(u, v ; \theta\right) \ln c\left(u, v ; \theta\right) \mathrm{d}u \, \mathrm{d}v \\ = & -\iint_{[0,1]^2} (1+\theta)\left(u v\right)^{-1-\theta}\left(u^{-\theta}+v^{-\theta}-1\right)^{-\frac{1}{\theta}-2}\\ \times & \ln\left[(1+\theta)\left(u v\right)^{-1-\theta}\left(u^{-\theta}+v^{-\theta}-1\right)^{-\frac{1}{\theta}-2}\right] \mathrm{d}u \, \mathrm{d}v \\ = & -(1+\theta) \iint_{[0,1]^2} \left(u v\right)^{-1-\theta}\left(u^{-\theta}+v^{-\theta}-1\right)^{-\frac{1}{\theta}-2}\\ \times & \left[\ln(1+\theta) - (1+\theta)\ln\left(u v\right) -\left(\frac{1}{\theta}+2\right)\ln\left(u^{-\theta}+v^{-\theta}-1\right)\right] \mathrm{d}u \, \mathrm{d}v \\ = & -(1+\theta) \iint_{[0,1]^2} \bigg[\ln(1+\theta) c - (1+\theta)\ln\left(u v\right) c -\left(\frac{1}{\theta}+2\right)\ln\left(u^{-\theta}+v^{-\theta}-1\right) c \bigg] \ \mathrm{d}u \, \mathrm{d}v \\ = & -(1+\theta) \ln(1+\theta) \iint_{[0,1]^2} c \ \mathrm{d}u \, \mathrm{d}v + (1+\theta)^2\iint_{[0,1]^2} c \ln\left(u v\right) \ \mathrm{d}u \, \mathrm{d}v \\ + & (1+\theta) \left(\frac{1}{\theta}+2\right) \iint_{[0,1]^2} c \ln\left(u^{-\theta}+v^{-\theta}-1\right) \ \mathrm{d}u \, \mathrm{d}v \\ \stackrel{\dagger}{=} & -(1+\theta) \ln(1+\theta) \int_0^1 c \bigg|_{v=0}^{v=1} \ \mathrm{d}u + (1+\theta)^2 \int_0^1 c \ln\left(u v\right) \bigg|_{v=0}^{v=1} \ \mathrm{d}u\\ + & (1+\theta) \left(\frac{1}{\theta}+2\right) \int_0^1 c \ln\left(u^{-\theta}+v^{-\theta}-1\right) \bigg|_{v=0}^{v=1} \ \mathrm{d}u \\ = & ? \end{align}

where $c = \left(u v\right)^{-1-\theta}\left(u^{-\theta}+v^{-\theta}-1\right)^{-\frac{1}{\theta}-2} $, and $\dagger$ is due to treating the double integrals as an iterated integral,

$$\iint_{[0,1]^2} c(u,v) \ \mathrm{d}u \, \mathrm{d}v = \int_0^1 c(u,v) \bigg|_{v=0}^{v=1} \mathrm{d}u $$

Why do I think an analytical solution of copula entropy can be found? Because there is one for the entropy of the Normal distribution's pdf, derived here.

Answer 1

The place for a comment was too short for the following, so this became an answer.

Well, removing a factor (namely the factor that makes live too short to compute...) simplifies things in a high measure, but we still have a mess... For the above i see only some first steps, but then things still get complicated. The $\theta$ is hard to type, there will be a $t$ instead. We have $$ \begin{aligned} H &=- \iint_{[0,1]^2} \frac {(1+t)(uv)^{-t}} {(u^{-t} + v^{-t} -1)^{2+1/t}}\;\ln \frac {(1+t)(uv)^{-t-1}} {(u^{-t} + v^{-t} -1)^{2+1/t}} \; \frac {du}u \; \frac {dv}v \\ &= \iint_{I^2} \frac {(1+t)UV} {(U + V -1)^{2+1/t}} \; \ln \frac {(1+t)(UV)^{-(1+t)/t}} {(U + V-1)^{2+1/t}} \; \frac 1{t^2} \; \frac {dU}U \; \frac {dV}V \\ &= \frac {1+t}{t^2} \iint_{I^2} \frac 1{(U + V -1)^{2+1/t}} \;\ln \frac {(1+t)(UV)^{-(1+t)/t}} {(U + V - 1)^{2+1/t}} \; dU\; dV\ . \end{aligned} $$ We substituted $U=u^{-t}$, $V=v^{-t}$, so $\frac {dU}U$ is $(-t)\frac {du}u$, and $\frac {dV}V$ is $(-t)\frac {dv}v$, so that we obtain a better looking expression.

The integral is now over $I^2$, where $I$ is $[1,\infty]$, because of the sign of $-t$ in $U=u^{-t}$.

Now under the logarithm we split four factors. And have to compute correspondingly four integrals.

• The integral in $\ln(t+1)$ is the simplest. It is in fact an integral in $W=(U+V)\ge 2$. We have working with $a=2+1/t$ $$\iint_{I^2} \frac 1{(U + V -1)^a} \; dU\; dV = \int_2^\infty\frac{W-2}{(W-1)^a}\; dW \\ =\frac 1{a^2-3a+2} =\frac 1{a-2}-\frac 1{a-1} = t-\frac t{1+t} \ . $$
• The integrals in $\ln U$ and $\ln V$ are equal, it is enough to compute only one of them. Again pass from $(U,V)$ to $(U,W)$, where $W=U+V$ formally. So we have to integrate something like $$ \iint_{\substack{1\le U<\infty\\2\le W<\infty\\1+U\le W}} \frac 1{(W -1)^a} \ln U \; dU\; dW $$ We can first integrate in $U$ from $1$ to $W-1$, but some $\log(W-2)$ will come into play and the work begins.
• The final integral can also be arranged in a better form, passing from $(U,V)$ to $(U,W)$ as above, but then we have to compute something like $$ \int_{2}^\infty \frac{W-2}{(W-1)^a}\ln(W-1)\;dW\ . $$

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Now try to apply integration by parts to get rid of the logarithmic term. For special values of $t$ (and $a$) this may be computed, but i will stop here.