I've come across this question in my class which I'm struggling with.
If we have the copula $C(u_1,u_2)=\phi^{-1}(\phi(u_1)+\phi(u_2))$ where $\phi(u) = (-\ln (u))^{\gamma}$ for ${\gamma} \ge 1$, express $C(u_1,u_2)$ as explicitly as possible.
$\begingroup$
$\endgroup$
3
-
$\begingroup$ Note that to get (e.g.) $\phi$ you just need a prior backslash e.g. \phi. $\endgroup$– Nick CoxCommented Mar 3, 2015 at 13:46
-
$\begingroup$ Self-study questions come with an obligation to show precisely what you have tried so far. $\endgroup$– Nick CoxCommented Mar 3, 2015 at 13:47
-
$\begingroup$ Try looking up the 'Gumbel-Hougaard' copula. $\endgroup$– Jordan CollinsCommented Mar 3, 2015 at 14:32
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
3
The copula that you are looking for is the Gumbel-Hougaard copula.
First find $\phi^{-1}(u) = \exp(-u^{1/γ})$.
Then use this to express $C(u_1, u_2)$ explicitly:
\begin{align} C(u_1, u_2) &= \phi^{-1}(\phi(u_1) + \phi(u_2)) \\ &= \exp(-[\phi(u_1) + \phi(u_2)]^{1/γ}) \\ &= \exp(-[(-\ln(u_1))^γ + (-\ln(u_2))^γ]^{1/γ}) \end{align}
This is the Gumbel-Hougaard copula which appeared in:
- Gumbel, E.J. (1960): "Bivariate exponential distributions." Journal of the American Statistical Association 55, 698–707.
A derivation of it can be found in:
- Hougaard, P.(1986): "A class of multivariate failure time distributions." Biometrika 73, 671–678.
-
$\begingroup$ Thank you for updating this. I took the liberty of formatting your answer using the $\LaTeX$ markdown the site supports. Please ensure it still says what you want it to. $\endgroup$ Commented Mar 3, 2015 at 20:35
-
$\begingroup$ Hi Jordan, welcome to CrossValidated. This is quite a good answer (so +1 for your effort), but I think for future questions like this it would be useful to take a read through the guidelines in the
self-studytag wiki, in particular the section 'Answering self-study questions'. $\endgroup$– Glen_bCommented Mar 4, 2015 at 6:29 -
$\begingroup$ Thanks - my original post (a hint to a self-studier) was relegated to a comment so I was under the impression that a full answer was more appropriate. Tricky business. $\endgroup$ Commented Mar 14, 2016 at 14:12