24.6 Summary of Copulas

Table 24.1: Summary of Copulas

Copula	Shape	Dependency	\(\tau\)
Frank’s	Symmetric	Light tails	Complicated has an integral
Gumbel	Asymmetric	More weight in the right. Higher tail than Frank	\(1 - \dfrac{1}{a}\)
HRT	Asymmetric	Less tail on the left but high on the right	\(\dfrac{1}{2a + 1}\)
Normal	Symmetric	Higher tail than Frank	\(\dfrac{2\mathrm{arcsin}(a)}{\pi}\)

Key things to know:

• Density of the copula is the 2^nd derivative w.r.t. both \(u\) and \(v\)

• Conditional probability of \(y\) is the derivative of the copula w.r.t. \(u\) (Or \(x\) and \(v\))

• How to simulate a joint distribution

Not important to memorize formulas

24.6.1 Frank’s Copula

Properties:

• Small tail dependencies compared to Gumbel and HRT

• \(\tau\) for Frank is not closed form

Frank Copula

\[\begin{equation} C(u,v) = - \dfrac{1}{a} \ln \left(\ 1 + \dfrac{g_u g_v}{g_1} \right) \tag{24.6} \end{equation}\] \[\begin{equation} g_z = e^{-az} -1 \tag{24.7} \end{equation}\]

Conditional distribution

\[\begin{equation} \Pr(Y \leq y \mid X = x) = C_1(u,v) = \dfrac{g_u g_v + g_v}{g_u g_v + g_1} \tag{24.8} \end{equation}\]

Density

\[\begin{equation} c(u,v) = - \dfrac{a g_1(1 + g_{u+v})}{(g_u g_v + g_1)^2} \tag{24.9} \end{equation}\]

Kendall’s \(\tau\)

\[\begin{equation} \tau(a) = 1 - \dfrac{4}{a} + \dfrac{4}{a^2} \int \limits_0^a \dfrac{t}{e^t - 1} dt \tag{24.10} \end{equation}\]

\(\tau\) is negative when \(a\) is negative

24.6.1.1 Simulation

Need the conditional distribution \(P(Y \leq y \mid X = x)\)

Given \(\tau\) we can solve for \(a\) which is used in the conditional formula

1. For a given \(\tau\), we solve for \(a\) using (24.10)

2. Simulate 2 r.v.

\[u,p \sim U[0,1]\]

3. Use \(u\) to determine \(x\) and use \(p\) to find \(y \mid x\)

\[x = F^{-1}(u)\]

4. Use the conditional distribution to find \(y\) by setting \(p = P(Y \leq y \mid X = x) = C_1(u,v)\) and solve for \(v\)

5. We know the formula for \(p\) and can do some algebra to get \(v\) after some algebra

\[\begin{equation} v = - \dfrac{1}{a} \ln \left( 1 + \dfrac{p g_1}{1 + g_u(1-p)} \right) \tag{24.11} \end{equation}\]

6. Plug in \(g_1\) and \(g_u\) (based of formula (24.7)) in to equation (24.11) to get \(v\)

7. Use the \((u,v)\) we get from above and then find the values in the 2 distⁿ associated with the percentiles

24.6.2 Gumbel Copula

Properties:

• Asymmetric

• more probability in the tails than Frank’s

Gumbel Copula

\[\begin{equation} C(u,v) = \exp\left\{ -\left[(-\ln(u))^a + (- \ln (v))^a \right]^{1/a} \right\} \tag{24.12} \end{equation}\]

Conditional distribution

\[\begin{equation} C_1(u,v) = C(u,v) \left[(-\ln(u))^a + (- \ln (v))^a \right]^{(-1 + 1/a)} \cdot (-\ln(u))^{a-1} \cdot \dfrac{1}{u} \tag{24.13} \end{equation}\]

Density

• Super complicated…

Kendall’s \(\tau\)

\[\begin{equation} \tau = 1 - \frac{1}{a} \tag{24.14} \end{equation}\]

\(v\) can’t be solve from \(p=C_1(u,v)\) \(\Rightarrow\) Need to simulate to some other ways

24.6.2.1 Simulation

We can’t solve \(v\) like how we did with Frank

1. For a given \(\tau\), we solve for \(a\) using (24.14)

2. Simulate 2 r.v.:

\[p,r \sim U[0,1]\]

3. Numerically solve for \(s\) that satisfies

\[\begin{equation} s \ln(s) = a(s-p) \:\: : \:\: 0<s<1 \tag{24.15} \end{equation}\]

4. Use \(s\) we can get:

\[\begin{equation} (u,v) = \left( \exp\left[ \ln(s)r^{\frac{1}{a}} \right], \exp\left[ \ln(s)(1-r)^{\frac{1}{a}} \right] \right) \tag{24.16} \end{equation}\]

24.6.3 Heavy Right Tail Copula

Properties

• Less correlation in the left tail but high in the right tail

HRT Copula

\[\begin{equation} C(u,v) = [u + v - 1] + \left[ (1-u)^{-1/a} + (1 - v)^{-1/a} -1 \right]^{-a} \:\: : \:\: a > 0 \tag{24.17} \end{equation}\]

Conditional distribution

\[\begin{equation} C_1(u,v) = 1 - \left[ (1-u)^{-1/a} + (1 - v)^{-1/a} -1 \right]^{-a-1} \times (1-u)^{-1-1/a} \tag{24.18} \end{equation}\]

Density

\[\begin{equation} c(u,v) = (1 + \dfrac{1}{a}) - \left[ (1-u)^{-1/a} + (1 - v)^{-1/a} -1 \right]^{-a-2} \times [(1-u)(1-v)]^{-1-1/a} \tag{24.19} \end{equation}\]

Kendall’s \(\tau\)

\[\begin{equation} \tau(a) = \dfrac{1}{2a + 1} \tag{24.20} \end{equation}\]

24.6.3.1 Simulation

We can solve \(v\) from \(p=C_1(u,v)\) so we can simulate it same way as Frank’s

24.6.3.2 Joint Burr Distribution

Joint distⁿ where marginal distⁿ are Burr and the conditionals are also Burr

• Join the 2 marginal Burr using the HRT copula

• The 2 marginal distⁿ needs to have the same \(a\) parameter as the HRT copula

Start with 2 Burr distribution:

\[F(x) = 1 - \left[ 1 + \left( \dfrac{x}{b} \right)^p \right]^{-a}\]

\[G(y) = 1 - \left[ 1 + \left( \dfrac{y}{d} \right)^q \right]^{-a}\]

Join with HRT with the same parameters \(a\) by plugging \(F(x)\) and \(G(y)\) into \(C(F(x), G(y))\)

\[H(x,y) = 1 - \left[ 1 + \left( \dfrac{x}{b} \right)^p \right]^{-a} - \left[ 1 + \left( \dfrac{y}{d} \right)^q \right]^{-a} + \left[ 1 + \left( \dfrac{x}{b} \right)^p + \left( \dfrac{y}{d} \right)^q \right]^{-a}\]

With condition distribution

\[G_{(Y \mid X)}(y \mid x) = 1 - \left[ 1 + \left( \dfrac{y}{d_x} \right)^q \right]^{-(a+1)}\]

Where

\[d_x = d \left[ 1 + \left( \dfrac{x}{b} \right)^{\frac{p}{q}} \right]\]

Analogous to the joint normal, in that the marginal and conditional distⁿ are the same

• \(a\) is to control correlation

• \(p\) & \(q\) is used to fit the tail

• \(b\) & \(d\) are used to set the scales of the distⁿ

24.6.4 Normal Copula

Joins 2 distⁿ using correlations from the bi-variate normal

Properties

• More dependencies in the tail then Frank

• Symmetrical

The copula take \(u\) and \(v\) from any distⁿ and converts them to standard normal variables and calculates the probability under the joint normal distⁿ with parameter \(a\)

Definitions

\(N(x; \: m,v) =\) Normal distribution with mean \(m\) and variance \(v\)

Therefore

\[P(X \leq x) = N(x; \: m,v) \:\:\:\:\: \mathbb{R}^2 \rightarrow [0,1]\]

Use the following shorthand for standard normal

\[N(x) = N(x; \: 0,1)\]

\(B(x,y; \: a)\) Bivariate standard normal distribution with Pearson correlation \(a\)

\[P(X \leq x, Y \leq y) = B(x, y; \:a) \:\:\:\:\: \mathbb{R}^2 \rightarrow [0,1]\]

\(p(u)\) is the inverse of the standard normal \(\Phi^{-1}(u)\)

\[p(u) \:\:\:\:\: [0,1] \rightarrow \mathbb{R}\]

\[N(p(u)) = u\]

e.g. \(p(0,95) = 1.645\)

Normal Copula

\[\begin{equation} C(u,v) = B[p(u), p(v); \: a] \tag{24.21} \end{equation}\]

Conditional distribution

\[\begin{equation} C_1(u,v) = N(p(v); a p(u), 1-a^2) \tag{24.22} \end{equation}\]

Density

\[\begin{equation} c(u,v) = \dfrac{\exp\left( \dfrac{[a^2 p(u)^2 - 2a p(u) p(v) + a^2 p(v)^2]}{2(1-a^2)} \right)}{\sqrt{(1-a^2)}} \tag{24.23} \end{equation}\]

Kendall’s \(\tau\)

\[\begin{equation} \tau(a) = \dfrac{2 \arcsin(a)}{\pi} \tag{24.24} \end{equation}\] \[\begin{equation} a = \sin \left( \tau \times \dfrac{\pi}{2} \right) \tag{24.25} \end{equation}\]

24.6.4.1 Simulation

Normal copula is simple to simulate from and also easy to generalizes to multiple dimensions

1. Solve for \(a\) with the \(\tau\) using (24.25)

2. Simulate 2 r.v.

\[u,r \sim U[0,1]\]

3. Get the standard normal value of the 2 percentiles

\[x = p(u)\]

\[y = p(r)\]

4. Join the 2 normal value with

\[z = ax + y \sqrt{1 - a^2}\]

5. Convert \(z\) to a percentile with the normal function to get \(v\)

\[ v = N(z)\]

24.6.5 Partial Perfect Correlation Copula

Krep’s Partial Perfect Correlation Copula

Sort of artificial, more useful for simulating evens then for description of actual outcomes

Generates dependencies by sometimes simulating \(u\) & \(v\) where they are independent and sometimes 100% dependent by setting \(v=u\)

• Level of dependency is controlled by

\[q(u,p): [0,1]^2 \rightarrow [0,1]\]

• \(q(u,p)\) needs to be symmetric

Definition

Simulate PPC by drawing \(u\), \(p\) & \(w\) from \(U[0,1]\)

\(v = \begin{cases} p &:q(u,p) < w & Independent \\ u &:q(u,p) \geq w & Dependent \\ \end{cases}\)

Partial Perfect Power

\(q(u,p) = (up)^a\)

More concentrate at the top right, since if either \(u\) or \(v\) is large then the other will likely be at the same percentile \(\Rightarrow\) perfectly correlated