Clayton Copula Solution: Statistical Analysis and Likelihood

This document offers a comprehensive solution to a Clayton Copula assignment, addressing two key questions. The first question involves deriving the density function of the Clayton Copula, presenting detailed steps and calculations. It starts with the copula function and proceeds to calculate the first and second derivatives with respect to u and v. The solution then proceeds to derive the likelihood function, providing the product of the copula density functions and the log-likelihood function. The second question focuses on determining the left tail dependence coefficient, using the copula function to calculate the limit as u approaches 0. The solution provides the formula and the step-by-step calculation to arrive at the final answer for the tail dependence coefficient. References to relevant academic papers are included to support the analysis.

Running head: CLAYTON COPULA SOLUTION 1
Clayton Copula Solution
Name
Institution

CLAYTON COPULA SOLUTION 2
Clayton Copula Solution
Question 1
The Clayton Copula
C ( u , v ) = ( u−θ + v−θ−1 )
−1
θ for θ>0
1) The density function is derived as (Oakes, 2005);
c ( u , v ) = ∂2 C ( u , v )
∂ u ∂ v
But, ∂2 C ( u , v )
∂u ∂ v = ∂
∂ v ( ∂ C ( u , v )
∂u )
Now, ∂C ( u , v )
∂ u = ∂
∂u {( u−θ+ v−θ −1 )
−1
θ }
Let y = ( u−θ+v−θ −1 )
−1
θ = t
−1
θ ………………………………. (i)
Also, let t = u−θ + v−θ−1 ……………………………………. (ii)
Using Product rule
∂ y
∂u = ∂ y
∂t . ∂ t
∂u
From equation (ii), ∂ t
∂u =−θu−θ−1
And from (i), ∂ y
∂ t =−1
θ t
−1
θ −1
=−1
θ ( u−θ+v−θ −1 )
−1
θ −1
Then,
∂ y
∂u =−1
θ ( u−θ + v−θ−1 )
−1
θ −1
(−θu−θ−1 )
∂ y
∂u =u−θ−1 ( u−θ +v−θ−1 )
−1
θ −1
= ∂C ( u , v )
∂ u
The second derivative with respect to v
∂
∂ v (u−θ−1 ( u−θ +v−θ−1 )
−1
θ −1
)

CLAYTON COPULA SOLUTION 3
Similarly,
let y = u−θ −1 ( u−θ+ v−θ−1 )
−1
θ −1 = u−θ −1 t
−1
θ −1……………………… (iii)
Also, let t = u−θ + v−θ−1 ……………………………………. (iv)
From equation (iv), ∂ t
∂ v =−θv−θ−1
And from (i),
∂ y
∂ t =( −1
θ −1)u
−θ−1
t
−1
θ −2
=( −1
θ −1)u
−θ −1
( u−θ + v−θ−1 )
−1
θ −2
Then,
∂ y
∂ v =( −1
θ −1)u
−θ−1
( u−θ + v−θ−1 )
−1
θ −2
(−θv−θ−1 )
c ( u , v ) = ∂2 C ( u , v )
∂ u ∂ v = ∂ y
∂u =(θ+1)(uv )−(θ+1 ) ( u−θ +v−θ −1 )
−2 θ+1
θ
2) Likelihood, L(θ) = k∏
i=1
n
c ( u , v ) where k is any positive constant chosen
conveniently.
∏
i=1
n
c ( u , v )=∏
i=1
n
{(θ+1)(uv )−(θ +1) ( u−θ +v−θ−1 )
−2 θ +1
θ }
∏
i=1
n
c ( u , v )= ( θ+1 )n
(∏
i=1
n
uv )−(θ+1 )
[∏
i=1
n
( u−θ+ v−θ−1 ) ]−2 θ+1
θ
Choose k =∏
i=1
n
uv
L ( θ )= ( θ+ 1 )n
(∏
i=1
n
uv )−θ
[∏
i=1
n
( u−θ + v−θ−1 ) ]−2θ +1
θ
Then, loglikelihood l ( θ ) =log L(θ)
l ( θ )=log {( θ+1 )n
(∏i=1
n
uv )−θ
[∏i=1
n
( u−θ+ v−θ −1 ) ]−2 θ +1
θ
}

CLAYTON COPULA SOLUTION 4
l ( θ )=nlog ( θ+1 )−θ ∑
i=1
n
log (uv ) −2 θ+1
θ ∑
i =1
n
log ( u−θ + v−θ−1 )
Question 2
The left tail dependence coefficient is defined as:
λL ( X , Y )= lim
u → 0+¿ {P (Y ≤ FY
−1 (u ) , X ≤ F X
−1 ( u ) )}¿
¿
λL ( X , Y )= lim
u → 0+¿ C (u ,u )
u ¿
¿
From question (1);
C ( u , v ) = ( u−θ + v−θ−1 )
−1
θ
Implying (Dakovic & Czado, 2011);
C ( u ,u )
u = ( 2u−θ−1 )
−1
θ
u = ( 2−uθ )
−1
θ ( u−θ )
−1
θ
u = ( 2−uθ )
−1
θ u
u
C ( u ,u )
u = ( 2−uθ )
−1
θ u
u = 1
( 2−uθ )
1
θ
Then,
λL ( X , Y )= lim
u → 0+¿ C (u ,u )
u = lim
u→0 +¿ 1
(2−uθ )
1
θ
=2
−1
θ ¿
¿¿
¿
λL ( X , Y )=2
−1
θ

CLAYTON COPULA SOLUTION 5
References
Dakovic, R., & Czado, C. (2011). Comparing point and interval estimates in the bivariate t-
copula model with application to financial data. Statistical Papers, 52(3), 709-731.
Oakes, D. (2005). On the preservation of copula structure under truncation. The Canadian
Journal of Statistics/La revue canadienne de statistique, 33(3), 465-468.