Rotated Archimedean Copulas for Opposite-Tail Association

Why Rotate an Archimedean Copula?

Some risks can move in opposite directions in the extremes: for example, one loss may be unusually large when another is unusually small. A standard Archimedean copula can represent dependence in the lower-lower or upper-upper corners, but its dependence may not match this opposite-tail pattern. Rotating the copula changes which corners carry its dependence while leaving the uniform marginal distributions unchanged.

The rotation is applied to the copula of the probability ranks, not directly to the dollar losses. If $X$ and $Y$ have continuous marginal distributions $F_X$ and $F_Y$, their probability integral transforms $U=F_X(X)$ and $V=F_Y(Y)$ are uniform on $[0,1]$. A large value of $U$ means a large loss relative to the distribution of $X$; a small value of $V$ means a small loss relative to the distribution of $Y$.

Definition: Given a bivariate copula $C$, define its 90-degree and 270-degree rotations, using the convention in this tutorial, by

C^{90}(u,v)=v-C(1-u,v), \qquad C^{270}(u,v)=u-C(u,1-v).

The 90-degree rotation corresponds to replacing the first uniform variable $U_0$ by $1-U_0$; the 270-degree rotation corresponds to replacing the second uniform variable $V_0$ by $1-V_0$. Thus both constructions are copulas, with uniform margins.

A rotation does not mean simply evaluating the base copula at a complemented argument. The subtraction is essential: it changes a joint upper-tail event into the required joint distribution function while preserving the margins. The stated convention also matters because some sources use different orientation labels; the formulas, rather than the label alone, determine the event.

Connecting Rotations to Opposite-Tail Events

For a pair $(U,V)$ with copula $C^{90}$, the probability that the first rank is above $u$ while the second is at most $v$ is

P(U>u,V\le v)=v-C^{90}(u,v)=C(1-u,v).

The first equality follows by subtracting $P(U\le u,V\le v)$ from $P(V\le v)=v$. For the 270-degree rotation, the corresponding event with a small first rank and large second rank is

P(U\le u,V>v)=u-C^{270}(u,v)=C(u,1-v).

These identities are useful because they reduce an opposite-tail calculation under a rotated copula to a base-copula calculation in the corner to which the base copula assigns association. For example, lower-lower association in a base copula becomes upper-left association under a 90-degree rotation and lower-right association under a 270-degree rotation. The exact strength and corner behavior depend on the selected base copula and its parameter.

For an Archimedean copula with a decreasing generator $\phi$, the base copula in two dimensions is

C(u,v)=\phi^{-1}\bigl(\phi(u)+\phi(v)\bigr).

The generator and parameter must be in a range that makes this a valid copula. Rotations can be applied to any valid bivariate copula, including these Archimedean examples. The formulas below state the specific base models and parameter restrictions being used.

Worked Example: A 90-Degree Rotated Clayton Opposite-Tail Event

Let $C$ be the Clayton copula with parameter $\theta=2$, $C(u,v)=(u^{-2}+v^{-2}-1)^{-1/2}$. Under its 90-degree rotation, calculate $P(U>0.9,V\le0.1)$ and compare it with independence.

Method. Use the 90-degree opposite-tail identity, which gives $P(U>u,V\le v)=C(1-u,v)$. The event therefore maps to base-copula arguments $1-0.9=0.1$ and $0.1$.

Calculation. The two reciprocal-square terms are $0.1^{-2}=100$ and $0.1^{-2}=100$. Thus

P(U>0.9,V\le0.1) =C(0.1,0.1) =(100+100-1)^{-1/2} =\frac{1}{\sqrt{199}} \approx 0.0709.

Under independence, the event probability is $P(U>0.9)P(V\le0.1)=0.1(0.1)=0.0100$. The rotated Clayton model assigns a probability about seven times as large. The result is consistent with opposite-tail association: a very high first rank and a very low second rank occur together more often than under independence.

Check and common mistake. The base-copula arguments must be $(1-0.9,0.1)=(0.1,0.1)$, not $(0.9,0.1)$. Also, for $\theta=2$, the denominator inside the square root is $199$, not a value based on $\theta=1$. The probability is $0.0709$, not $0.0526$; the latter does not follow from the stated parameter and event.

Constructing and Using the Rotated Copula

The rotated formulas provide more than a shortcut for one corner probability: they define the full joint distribution of the ranks. For a 90-degree rotation, one can calculate $C^{90}(u,v)$ from $v-C(1-u,v)$; for a 270-degree rotation, use $u-C(u,1-v)$. Keeping the subtraction and its order intact is important for probabilities inside rectangles and for checking whether a proposed model has the intended orientation.

Choose the corner and orientation.
State which risk is expected to be high and which low. Select 90 degrees for the event $U$ high, $V$ low under the convention here; select 270 degrees for $U$ low, $V$ high.

Specify the base copula.
Write its formula and parameter, including the conditions under which the formula defines a valid copula.

Transform the arguments.
Use $1-u$ in the first argument for a 90-degree rotation, or $1-v$ in the second argument for a 270-degree rotation.

Calculate and interpret.
Evaluate the rotated CDF or the appropriate opposite-tail identity, then compare with independence if that helps assess the effect.

Worked Example: A 270-Degree Rotated Gumbel Event

Let $C$ be a Gumbel copula with parameter $\theta=2$, $C(u,v)=\exp\{- [(-\ln u)^2+(-\ln v)^2]^{1/2}\}$, for $0<u,v\le1$. Under the 270-degree rotation, calculate $P(U\le0.2,V>0.8)$, then compare it with independence.

Method. For the 270-degree rotation, $P(U\le u,V>v)=C(u,1-v)$. Here $1-v=1-0.8=0.2$, so both base-copula arguments are $0.2$.

Calculation. Since $-\ln(0.2)=\ln 5$, the sum of the squared logarithms is $2(\ln 5)^2$, and its square root is $\sqrt{2}\ln 5$. Therefore

P(U\le0.2,V>0.8) =C(0.2,0.2) =\exp(-\sqrt{2}\ln 5) =5^{-\sqrt{2}} \approx 0.1027.

Under independence, the probability is $0.2(1-0.8)=0.0400$. The rotated model assigns more probability to this lower-right corner. The calculation uses the base copula at $(0.2,0.2)$, not at $(0.2,0.8)$.

Check and common mistake. The event specifies $V>0.8$, so the 270-degree identity complements the second threshold: $1-0.8=0.2$. Using $C(0.2,0.8)$ would correspond to the wrong corner and would not calculate the requested event.

Rectangle Probabilities and Loss Marginals

Many questions specify an interval of ranks rather than a single tail event. For any copula $C_R$, the probability of a rectangle $(a,b]\times(c,d]$ is obtained by inclusion-exclusion:

P(a<U\le b,\ c<V\le d) =C_R(b,d)-C_R(a,d)-C_R(b,c)+C_R(a,c).

This follows by subtracting the two lower-left rectangles that extend beyond the desired interval and adding back their overlap. For an event that includes an entire lower tail, such as $a<U\le b$ and $V\le d$, the probability simplifies to $C_R(b,d)-C_R(a,d)$.

Worked Example: A Rank Interval Under a Rotated Clayton Copula

Let $C$ be a Clayton copula with $\theta=1$, so $C(s,t)=(s^{-1}+t^{-1}-1)^{-1}$. Under its 90-degree rotation, calculate $P(0.4<U\le0.8,V\le0.3)$.

Method. The event contains a lower tail in $V$, so subtract two rotated-CDF values: $C^{90}(0.8,0.3)-C^{90}(0.4,0.3)$. The rotated formula is $C^{90}(u,v)=v-C(1-u,v)$.

At $(u,v)=(0.8,0.3)$, the base arguments are $(0.2,0.3)$. Thus $C(0.2,0.3)=(5+10/3-1)^{-1}=(22/3)^{-1}=3/22$, and $C^{90}(0.8,0.3)=3/10-3/22=9/55$.

At $(u,v)=(0.4,0.3)$, the base arguments are $(0.6,0.3)$. Thus $C(0.6,0.3)=(5/3+10/3-1)^{-1}=4^{-1}=1/4$, and $C^{90}(0.4,0.3)=3/10-1/4=1/20$.

Subtracting gives

P(0.4<U\le0.8,V\le0.3) =\frac{9}{55}-\frac{1}{20} =\frac{25}{220} =\frac{5}{44} \approx 0.1136.

Check and common mistake. The rotated CDF increases from $1/20=0.0500$ to $9/55\approx0.1636$, so their difference is positive and lies below the full $V\le0.3$ probability. Subtracting base-copula values in the opposite order would give a negative number, signaling an error.

To translate rank events into losses, use each loss’s own marginal distribution. Under continuous marginals, $U=F_X(X)$ and $V=F_Y(Y)$, so a rank threshold $U>u$ is equivalent to $X>F_X^{-1}(u)$, and $V\le v$ is equivalent to $Y\le F_Y^{-1}(v)$. The copula describes the dependence between the ranks; the marginals determine the dollar thresholds.

Worked Example: Translating Opposite Ranks Into Hypothetical Loss Thresholds

Consider two hypothetical annual losses. Let $X$ be exponential with mean $\$100{,}000$, so $F_X(x)=1-\exp(-x/100{,}000)$ for $x\ge0$. Let $Y$ be exponential with mean $\$50{,}000$, so $F_Y(y)=1-\exp(-y/50{,}000)$ for $y\ge0$. Suppose their ranks follow a 90-degree rotated Clayton copula with $\theta=1$. Calculate $P(X>F_X^{-1}(0.8),Y\le F_Y^{-1}(0.2))$, and compare it with independence.

Step 1: Find the loss thresholds. For an exponential distribution with mean $m$, solving $F(x)=q$ gives $x=-m\ln(1-q)$. Therefore

F_X^{-1}(0.8)=-100{,}000\ln(0.2) =100{,}000\ln 5 \approx \$160{,}943.79,

and

F_Y^{-1}(0.2)=-50{,}000\ln(0.8) \approx \$11{,}157.18.

The loss event is therefore a large $X$ loss, above $\$160{,}943.79$, together with a small $Y$ loss, at most $\$11{,}157.18$.

Step 2: Calculate the dependence probability. The 90-degree identity gives $P(U>0.8,V\le0.2)=C(0.2,0.2)$. With $\theta=1$,

C(0.2,0.2) =(5+5-1)^{-1} =\frac{1}{9} \approx 0.1111.

Under independence the probability is $(1-0.8)(0.2)=0.0400$. The rotated model therefore assigns greater probability to the specified opposite-tail loss scenario.

Check and common mistake. The exponential quantile formula uses $1-q$ inside the logarithm. The threshold for the 80th percentile of $X$ is not $-100{,}000\ln(0.8)$; that expression is the 20th percentile. The event probability uses the rank thresholds $0.8$ and $0.2$, not the dollar amounts themselves.

Interpretation, Assumptions, and Exam Traps

The useful feature of a rotated copula is its ability to place dependence in an opposite-tail corner. It does not imply that every large loss in one risk is paired with a small loss in the other, nor does a single corner probability describe all dependence in the model. The copula is a full joint distribution; event probabilities must be calculated at the thresholds relevant to the decision.

These rank-to-loss interpretations assume continuous marginals so the probability integral transforms are uniform without adjustment. With discrete or mixed marginals, the ordinary transform need not be uniform, and the copula representation may not be unique; additional methods are needed. Also, selecting a rotated family and parameter is a modeling choice, not a conclusion established by the rotation formula.

Using the wrong orientation. Under the convention here, 90 degrees targets $U$ high and $V$ low; 270 degrees targets $U$ low and $V$ high.
Dropping the subtraction. $C^{90}(u,v)$ is $v-C(1-u,v)$, not just $C(1-u,v)$. The latter is the probability $P(U>u,V\le v)$ under the rotated model.
Mixing ranks and losses. A rank such as $0.9$ is a percentile, not a loss of 90 dollars or 90 percent of a policy limit. Apply the marginal quantile function to obtain the loss threshold.
Assuming all opposite-tail events are identical. Which risk is large and which is small matters. Check the event wording before choosing a rotation and transforming an argument.

Key takeaway: Define the rotation by its formula, map the event to the correct base-copula corner, and use the marginals only after the rank probability is clear. This approach to opposite-tail association is aligned with the CAS Exam 9 (Risk Management for Actuaries) syllabus.

Check Your Understanding

For each problem, identify the rotation and transform the event before evaluating the base copula.

A 90-degree rotated Clayton copula has $\theta=1.5$. Calculate $P(U>0.8,V\le0.2)$, using $C(u,v)=(u^{-\theta}+v^{-\theta}-1)^{-1/\theta}$. (Answer: $C(0.2,0.2)=(2(0.2^{-1.5})-1)^{-2/3}\approx0.1299$.)
A 270-degree rotated Gumbel copula has $\theta=2$. Calculate $P(U\le0.25,V>0.75)$, using $C(u,v)=\exp\{- [(-\ln u)^2+(-\ln v)^2]^{1/2}\}$. (Answer: $C(0.25,0.25)=4^{-\sqrt{2}}\approx0.1408$.)
A 90-degree rotated Clayton copula has $\theta=1$. Calculate $P(U\le0.6,V\le0.3)$. (Answer: $0.3-C(0.4,0.3)=0.3-6/29=27/290\approx0.0931$.)
A 90-degree rotated Clayton copula has $\theta=1$. Calculate $P(0.5<U\le0.9,V\le0.2)$. (Answer: $C(0.5,0.2)-C(0.1,0.2)=1/6-1/14=2/21\approx0.0952$.)
A 90-degree rotated Clayton copula has $\theta=2$. Compare $P(U>0.95,V\le0.05)$ with independence. (Answer: $C(0.05,0.05)=1/\sqrt{799}\approx0.0354$; independence gives $0.05(0.05)=0.0025$.)

What You'll Learn