Fréchet-Hoeffding Bounds for Copulas

Why Bound Dependence When It Is Unknown?

A pair of loss models specifies how likely each loss is to exceed a threshold, but does not by itself specify how often both losses exceed their thresholds together. That joint probability can materially affect aggregate loss, reinsurance attachment, and capital decisions. If the dependence structure is not known, the Fréchet–Hoeffding bounds give the sharpest possible limits using only the marginal probabilities.

A copula separates marginal distributions from dependence. For continuous losses $X$ and $Y$, define $U=F_X(X)$ and $V=F_Y(Y)$. Each variable is uniform on $[0,1]$, and their joint distribution is the copula $C$. The bounds below make no assumption of independence or of a particular parametric copula family.

Definition: A bivariate copula $C$ is a joint cumulative distribution function on $[0,1]^2$ whose two marginal distributions are uniform on $[0,1]$. In particular, $C(u,1)=u$, $C(1,v)=v$, and probabilities assigned to rectangles are nonnegative. The Fréchet–Hoeffding bounds are $$ W(u,v)=\max(u+v-1,0) \quad\text{and}\quad M(u,v)=\min(u,v). $$ For every bivariate copula and every $u,v\in[0,1]$, $W(u,v)\le C(u,v)\le M(u,v)$.

Deriving the Bounds

For a copula, $C(u,v)=P(U\le u,V\le v)$. Apply elementary probability bounds to the two events $A=\{U\le u\}$ and $B=\{V\le v\}$. Their probabilities are $u$ and $v$, because the margins are uniform.

Bound the intersection above.
The event $A\cap B$ is contained in each of $A$ and $B$. Thus $P(A\cap B)\le P(A)=u$, and $P(A\cap B)\le P(B)=v$. Together, these give $C(u,v)\le\min(u,v)$.

Bound the intersection below.
Since $P(A\cup B)\le1$ and $P(A\cup B)=u+v-P(A\cap B)$, it follows that $P(A\cap B)\ge u+v-1$. Probabilities are also nonnegative, so $C(u,v)\ge\max(u+v-1,0)$.

Conclude the sharp limits.
The inequalities hold for every pair of events with these marginal probabilities. The bivariate copulas $M$ and $W$ attain the respective limits, so neither bound can be improved without additional dependence information.

The attainability in Step 3 can be seen directly. If $Z$ is uniform on $[0,1]$, the pair $(U,V)=(Z,Z)$ has copula $M$: both variables are uniform and $P(U\le u,V\le v)=\min(u,v)$. The pair $(U,V)=(Z,1-Z)$ has copula $W$. For that pair, the events $Z\le u$ and $1-Z\le v$ overlap over an interval of length $\max(u+v-1,0)$. These constructions also show the extreme dependence represented by the bounds: perfect movement together for $M$, and perfect movement in opposite directions for $W$.

When Are the Bounds Copulas?

In two dimensions, both $W$ and $M$ are copulas. Their uniform margins follow from their boundary values: $W(u,1)=u$, $W(1,v)=v$, $M(u,1)=u$, and $M(1,v)=v$; both are zero when either argument is zero. Their rectangle probabilities are nonnegative, as is also guaranteed by the uniform-variable constructions above.

For $d$ uniform variables, the corresponding pointwise limits are $W_d(u_1,\ldots,u_d)=\max(\sum_{j=1}^d u_j-d+1,0)$ and $M_d(u_1,\ldots,u_d)=\min(u_1,\ldots,u_d)$. The upper limit $M_d$ is a copula in any dimension: setting all variables equal to one uniform $Z$ gives it. The lower limit $W_d$ is not, in general, a copula when $d\ge3$. A pointwise probability bound need not itself be a valid joint distribution.

For example, consider $W_3$ on the box $[0.5,0.9]^3$. At a corner with $k$ coordinates equal to $0.9$ and the rest equal to $0.5$, the sum of coordinates is $1.5+0.4k$, so $W_3$ is zero for $k=0,1$, $0.3$ for $k=2$, and $0.7$ for $k=3$. The probability a valid distribution assigns to this box would be its three-dimensional rectangle increment: $$ 0.7-3(0.3)+3(0)-0=-0.2. $$ A negative rectangle probability is impossible. Thus $W_3$ fails the required nonnegative-increment condition and is not a trivariate copula. The bivariate bounds remain valid for bivariate margins of a higher-dimensional model; the warning concerns treating the full lower limit as a valid joint model.

Using the Bounds for Loss Exceedances

For continuous loss variables, $P(X\le x,Y\le y)=C(F_X(x),F_Y(y))$. Write $u=F_X(x)$ and $v=F_Y(y)$. The joint probability of exceeding both thresholds is obtained by taking the complement of the union of the two non-exceedance events: $$ P(X>x,Y>y)=1-u-v+C(u,v). $$ Substituting the Fréchet–Hoeffding limits gives a useful form directly in terms of the marginal exceedance probabilities: $$ \max\bigl(P(X>x)+P(Y>y)-1,0\bigr) \le P(X>x,Y>y) \le \min\bigl(P(X>x),P(Y>y)\bigr). $$ The lower endpoint is often zero when the two marginal exceedance probabilities sum to less than one. These bounds require no assumption that the losses are independent. For discrete margins, the event-level probability inequalities still hold, although a copula representation of the joint distribution may not be unique.

Worked Example: Bounding Losses at Two Marginal Quantiles

Two hypothetical insurance portfolios have continuous losses $X\sim\operatorname{Uniform}(0,100{,}000)$ and $Y\sim\operatorname{Uniform}(0,200{,}000)$, with amounts in dollars. Dependence is unknown. Bound (a) the probability both losses are at most $80{,}000$ and $140{,}000$, respectively, and (b) the probability both exceed those thresholds.

Method. Evaluate the marginal distribution functions at the thresholds, apply the copula bounds to the joint non-exceedance event, then use the joint-survival formula for simultaneous exceedance.

For $X$, $u=F_X(80{,}000)=80{,}000/100{,}000=0.8$. For $Y$, $v=F_Y(140{,}000)=140{,}000/200{,}000=0.7$. Therefore, $$ \max(0.8+0.7-1,0)=0.5 \le P(X\le80{,}000,Y\le140{,}000) \le\min(0.8,0.7)=0.7. $$ The marginal exceedance probabilities are $1-u=0.2$ and $1-v=0.3$. Hence $$ \max(0.2+0.3-1,0)=0 \le P(X>80{,}000,Y>140{,}000) \le\min(0.2,0.3)=0.2. $$ Check: The joint-survival identity gives $1-0.8-0.7+C(0.8,0.7)$, which ranges from $1-1.5+0.5=0$ to $1-1.5+0.7=0.2$. The two answers are consistent. The common mistake is to use the joint-CDF lower bound $0.5$ as a lower bound on the joint exceedance probability; it applies to the opposite event.

Worked Example: Joint Exceedance With Exponential Losses

Two hypothetical lines have continuous exponential losses, measured in dollars: $X$ has mean $100{,}000$, so its rate is $1/100{,}000$ per dollar; $Y$ has mean $50{,}000$, so its rate is $1/50{,}000$ per dollar. With no dependence model specified, find bounds on the probability that both losses exceed $100{,}000$.

Method. Compute each exponential survival probability, then apply the intersection bounds for the two exceedance events.

For an exponential loss with mean $\theta$, $P(X>x)=e^{-x/\theta}$. Thus $$ p_X=P(X>100{,}000)=e^{-100{,}000/100{,}000}=e^{-1}=0.3679, $$ and $$ p_Y=P(Y>100{,}000)=e^{-100{,}000/50{,}000}=e^{-2}=0.1353. $$ The lower and upper limits for their intersection are $$ \max(p_X+p_Y-1,0)=\max(0.3679+0.1353-1,0)=0, $$ $$ \min(p_X,p_Y)=0.1353. $$ Therefore the simultaneous exceedance probability is between $0$ and $0.1353$, with probabilities rounded to four decimals. This is a wide range because the marginals alone do not identify dependence. Check: the upper limit cannot exceed the smaller marginal probability, $0.1353$; the lower limit is zero because the marginal probabilities sum to less than one. Independence would give $0.3679(0.1353)$, but that is only one possible value, not a bound that follows from the stated information.

Worked Example: Bounding the Probability of Any Trigger

A hypothetical insurer monitors two continuous annual losses. The first is exponential with mean $100{,}000$ dollars and triggers above $100{,}000$ dollars. The second is exponential with mean $50{,}000$ dollars and triggers above $100{,}000$ dollars. Find bounds on the probability that at least one trigger occurs and on the probability neither occurs.

Method. Bound the intersection of the two trigger events first. Use inclusion–exclusion to bound their union; take its complement for the probability of no triggers.

The trigger probabilities, using the given means as the exponential parameters, are $$ p_1=e^{-100{,}000/100{,}000}=e^{-1}=0.3679, \qquad p_2=e^{-100{,}000/50{,}000}=e^{-2}=0.1353. $$ Let $I$ be the probability both triggers occur. The bounds are $$ \max(p_1+p_2-1,0)=0 \le I\le\min(p_1,p_2)=0.1353. $$ The probability of at least one trigger is $p_1+p_2-I$. It is largest when $I$ is smallest and smallest when $I$ is largest: $$ 0.3679+0.1353-0.1353=0.3679 \le P(\text{at least one trigger}) \le0.3679+0.1353-0=0.5032. $$ Taking complements gives $$ 0.4968\le P(\text{neither trigger})\le0.6321. $$ All probabilities are rounded to four decimals. Check: the probability of at least one trigger is at least the larger marginal trigger probability, $0.3679$, and cannot exceed the sum, $0.5032$. The bounds satisfy both checks.

Exam Traps and Interpretation

The bounds are sharp but can be broad. They describe what is possible given only the marginals, not what is most likely. A chosen dependence model can give a more precise estimate, but that estimate depends on the model and its assumptions. The bounds are useful as a dependence-agnostic benchmark and as a check on calculations made under a particular copula.

Using the CDF bound for a survival event. $W(u,v)$ and $M(u,v)$ bound $P(X\le x,Y\le y)$, not $P(X>x,Y>y)$. For simultaneous exceedances, use $1-u-v+C(u,v)$, or apply the bounds directly to the two exceedance probabilities.
Assuming the upper limit is a product. The product of marginal probabilities is the intersection probability only under independence. The general upper bound is the smaller marginal probability.
Calling the multivariate lower limit a copula. In dimension two, $W$ is a copula. Its direct higher-dimensional analogue is not generally a valid copula, as the negative rectangle increment demonstrates.
Confusing a bound with a model. The limits constrain possible probabilities; they do not specify a complete dependence structure between losses.

Key takeaway: For any two events with probabilities $p$ and $q$, their intersection lies between $\max(p+q-1,0)$ and $\min(p,q)$. Applying this fact to marginal loss events yields sharp bivariate joint-probability bounds without assuming a dependence model. Both bivariate Fréchet–Hoeffding bounds are copulas, but the lower limit does not generally extend to a copula in three or more dimensions.

Check Your Understanding

For each problem, find the requested bounds without assuming independence.

A bivariate copula is evaluated at $u=0.25$ and $v=0.60$. Find the lower and upper bounds for $P(U\le u,V\le v)$. (Answer: $0$ to $0.25$.)
Two losses have marginal exceedance probabilities $0.30$ and $0.25$ at specified thresholds. Bound the probability that both exceed their thresholds. (Answer: $0$ to $0.25$.)
At two thresholds, continuous losses have $F_X(x)=0.90$ and $F_Y(y)=0.85$. Find bounds for both losses being at most their thresholds, then for both exceeding. (Answer: joint non-exceedance $0.75$ to $0.85$; joint exceedance $0$ to $0.10$.)
Two trigger events have probabilities $0.20$ and $0.40$. Bound the probability at least one occurs. (Answer: $0.40$ to $0.60$.)
For $W_3(u_1,u_2,u_3)=\max(u_1+u_2+u_3-2,0)$, calculate the rectangle increment on $[0.5,0.9]^3$ and state what it implies. (Answer: $-0.20$; $W_3$ is not a valid trivariate copula.)

What You'll Learn