Why Gaussian Dependence Has No Asymptotic Tail Dependence

Why Gaussian Tail Dependence Is a Limit Statement

A positive correlation can make large losses more likely to occur together. Yet for a non-degenerate bivariate normal pair, that association does not persist at the most extreme quantiles in the sense measured by asymptotic tail dependence. The distinction is important in risk analysis: a zero limiting coefficient does not say that joint exceedances are impossible at a particular threshold, or that the variables are independent.

This tutorial derives the result for both tails and explains its scope. The conclusion is aligned with the CAS Exam 9 (Risk Management for Actuaries) syllabus.

Definition: Let $X$ and $Y$ have continuous marginal distribution functions $F_X$ and $F_Y$. Their upper-tail dependence coefficient is $\lambda_U=\lim_{p\uparrow1}P(F_Y(Y)>p\mid F_X(X)>p)$, if the limit exists. Their lower-tail dependence coefficient is $\lambda_L=\lim_{p\downarrow0}P(F_Y(Y)\le p\mid F_X(X)\le p)$, if the limit exists. Each coefficient measures the limiting conditional probability of an extreme event in one variable, given a correspondingly extreme event in the other.

For a standard bivariate normal pair $(Z_1,Z_2)$, each marginal is standard normal. Write $\Phi$ for the standard normal distribution function and $\bar\Phi(z)=1-\Phi(z)$. Since the two marginals are identical and continuous, common upper quantiles correspond to events $Z_1>z$ and $Z_2>z$ as $z$ increases without bound. Common lower quantiles correspond to $Z_1<-z$ and $Z_2<-z$. Thus the coefficients can be written as $\lim_{z\to\infty}P(Z_2>z\mid Z_1>z)$ and $\lim_{z\to\infty}P(Z_2<-z\mid Z_1<-z)$, respectively.

The Gaussian Result and Its Derivation

Assume $(Z_1,Z_2)$ is jointly normal, both variables have mean zero and variance one, and their correlation is $\rho$, where $|\rho|<1$. This condition makes the covariance matrix positive definite, so the pair is non-degenerate. The key bound comes from adding the two variables: if both exceed $z$, their sum must exceed $2z$.

Theorem: For a non-degenerate bivariate normal pair with correlation $|\rho|<1$, $\lambda_U=0$ and $\lambda_L=0$. This does not imply independence: the pair can have nonzero correlation and positive probability of joint exceedance at every finite threshold.

Bound the joint upper tail.
If $Z_1>z$ and $Z_2>z$, then $Z_1+Z_2>2z$. The sum is normal with mean zero and variance $2(1+\rho)$. Therefore $P(Z_1>z,Z_2>z)\le\bar\Phi\!\left(z\sqrt{2/(1+\rho)}\right)$.

Divide by the marginal tail.
For $z>0$, $P(Z_1>z)=\bar\Phi(z)>0$. Hence $0\le P(Z_2>z\mid Z_1>z)\le \bar\Phi(az)/\bar\Phi(z)$, where $a=\sqrt{2/(1+\rho)}$.

Show that the bound vanishes.
Because $\rho<1$, $a>1$. The normal-tail asymptotic, $\bar\Phi(t)\sim\phi(t)/t$ as $t\to\infty$, gives $\bar\Phi(az)/\bar\Phi(z)\sim a^{-1}\exp[-(a^2-1)z^2/2]$, which tends to zero.

Apply the same argument to the lower tail.
The pair $(-Z_1,-Z_2)$ is also standard bivariate normal with correlation $\rho$. Its joint upper-tail coefficient is the original pair’s lower-tail coefficient, so that coefficient is zero as well.

For completeness, the normal-tail asymptotic used above follows from the standard Mills bounds $\frac{t}{t^2+1}\phi(t)<\bar\Phi(t)<\frac{\phi(t)}{t}$ for $t>0$: dividing each bound by $\phi(t)/t$ shows that the ratio tends to one. Also, $a^2-1=(1-\rho)/(1+\rho)>0$, so the exponential term in the ratio tends to zero. Since a conditional probability is nonnegative, the upper bound tending to zero proves that its limit is zero.

\frac{\bar\Phi(az)}{\bar\Phi(z)} \sim \frac{1}{a}\exp\!\left[-\frac{1-\rho}{2(1+\rho)}z^2\right] \longrightarrow 0 \qquad (|\rho|<1).

The strict condition matters. At $\rho=1$, the standardized variables are identical, so both coefficients equal one. That degenerate case is not covered by the proof: the sum-tail bound no longer has $a>1$. The theorem also does not classify other dependence models; its conclusion is specifically for a non-degenerate bivariate normal pair.

Worked Example: A Positive Finite-Threshold Co-Exceedance

Let $(Z_1,Z_2)$ be standard bivariate normal with $\rho=0.5$. At the finite threshold $1$, establish a positive lower bound for $P(Z_2>1\mid Z_1>1)$. Use the event $1<Z_1<1.1$ to obtain the bound.

Method. The conditional normal distribution is $Z_2\mid Z_1=x\sim N(\rho x,1-\rho^2)$. Here this is $N(0.5x,0.75)$, so $P(Z_2>1\mid Z_1=x)=\bar\Phi((1-0.5x)/\sqrt{0.75})$. For $x\ge1$, the argument of $\bar\Phi$ decreases as $x$ increases, so this conditional probability is at least its value at $x=1$.

At $x=1$, the standardized threshold is $(1-0.5)/\sqrt{0.75}=1/\sqrt{3}\approx0.577350$. Thus $P(Z_2>1\mid Z_1=1)=\bar\Phi(1/\sqrt{3})\approx0.281851$, which rounds to $0.2819$ to four decimals.

The probability of the conditioning interval is $P(1<Z_1<1.1)=\Phi(1.1)-\Phi(1)\approx0.86433394-0.84134475=0.02298919$. Consequently, the restricted event contributes approximately $0.02298919(0.281851)=0.0064795$ to $P(Z_1>1,Z_2>1)$. Dividing by $P(Z_1>1)=\bar\Phi(1)\approx0.15865525$, its contribution to the conditional probability is approximately $0.0064795/0.15865525=0.04084$.

To express a safely rounded lower bound, use the conservative values $P(1<Z_1<1.1)>0.02298$, $\bar\Phi(1/\sqrt{3})>0.2818$, and $\bar\Phi(1)<0.15866$. Their ratio gives $(0.02298)(0.2818)/0.15866\approx0.04081>0.0408$. Therefore $P(Z_2>1\mid Z_1>1)>0.0408$. This calculation verifies positive co-exceedance at this finite threshold; it does not contradict the zero limit as $z\to\infty$.

Check. The interval contribution to the joint probability is about $0.0064795$, and its conditional-probability contribution is about $0.04084$. The conservative lower bound $0.0408$ is rounded down, not asserted from multiplying rounded values as if they were exact.

Worked Example: The Exponential Rate of the Tail Bound

For a standard bivariate normal pair with $\rho=0.5$, find the exponential rate in the normal-sum upper bound and interpret it at $z=4$.

Method. In the bound, $a=\sqrt{2/(1+\rho)}=\sqrt{2/1.5}=\sqrt{4/3}\approx1.1547$. The ratio bounding the conditional tail is $\bar\Phi(az)/\bar\Phi(z)$. Its asymptotic form has prefactor $1/a=\sqrt{3}/2\approx0.8660$ and exponential factor $\exp[-(1-\rho)z^2/(2(1+\rho))]=\exp(-z^2/6)$.

At $z=4$, the exponent is $-16/6\approx-2.6667$, and the exponential factor is approximately $0.0695$. The asymptotic expression at this value is approximately $0.8660(0.0695)=0.0602$. This is an asymptotic approximation to the ratio bound, not an exact finite-threshold conditional probability.

As $z$ grows, the factor $\exp(-z^2/6)$ tends to zero, so the bound, and therefore the conditional joint-tail probability, tends to zero. The calculation illustrates the decay rate without treating an asymptotic approximation as an exact probability.

Worked Example: Standardizing Unequal Marginals

Suppose $X$ and $Y$ are jointly normal with means $80$ and $25$, standard deviations $12$ and $5$, and correlation $0.4$. Find their common $97.5$th-percentile thresholds and state the upper- and lower-tail dependence coefficients.

Step 1: Standardize. Define $Z_1=(X-80)/12$ and $Z_2=(Y-25)/5$. This pair is standard bivariate normal with correlation $0.4$, so $|\rho|<1$ and the theorem applies.

Step 2: Find the upper thresholds. The standard normal $97.5$th percentile is $z_{0.975}\approx1.959964$. Thus $x_{0.975}=80+12(1.959964)=103.519568$, or $\$103.52$, and $y_{0.975}=25+5(1.959964)=34.799820$, or $\$34.80$. The two exceedance events are exactly $Z_1>z_{0.975}$ and $Z_2>z_{0.975}$.

Step 3: State the limiting coefficients. The standardization preserves the dependence structure, and the resulting Gaussian pair has $|\rho|<1$. Therefore $\lambda_U=0$ and $\lambda_L=0$. Unequal means and scales change the dollar thresholds, not the conclusion about asymptotic tail dependence.

Check. Both thresholds use the same marginal percentile, not the same dollar amount. The conclusion follows from the correlation and joint normality assumptions, not from comparing the numerical sizes of the thresholds.

What Zero Tail Dependence Does—and Does Not—Mean

A zero coefficient says that the conditional probability of a matching extreme event tends to zero as the marginal quantile approaches its endpoint. It does not say that the joint probability is zero at every finite threshold. For a non-degenerate bivariate normal distribution, the joint density is positive throughout the plane, so any finite rectangle such as $(z,\infty)\times(z,\infty)$ has positive probability.

Nor does the result say the variables are independent. Independence is the special case $\rho=0$; when $\rho\ne0$, the pair is dependent, even though both asymptotic coefficients are zero. The distinction matters when interpreting a Gaussian dependence assumption for risk aggregation: an asymptotic coefficient is a limiting property, not a complete summary of co-exceedance probabilities at operational stress thresholds.

Common Exam Trap

Wrong calculation: “The pair has correlation $0.5$, so its upper-tail dependence coefficient is $0.5$.” Correlation is not the definition of a tail dependence coefficient; it summarizes linear association across the distribution.

Correction: For a non-degenerate bivariate normal pair, apply the sum-tail bound. When $|\rho|<1$, the resulting conditional-probability bound tends to zero, so $\lambda_U=\lambda_L=0$, including when $\rho=0.5$. At finite thresholds, however, the co-exceedance probability can still be positive, as the worked calculation demonstrates.

Key takeaway: For a non-degenerate bivariate normal pair, $|\rho|<1$ implies zero upper- and lower-tail dependence. The proof uses the thinner normal tail of the sum to bound joint exceedance. Do not confuse this limiting result with independence or with zero co-exceedance at a finite threshold.

Check Your Understanding

Work through these problems before checking each final answer.

A standard bivariate normal pair has correlation $0.25$. What are its upper- and lower-tail dependence coefficients, and what is the exponential coefficient multiplying $z^2$ in the normal-sum tail ratio? (Answer: $\lambda_U=\lambda_L=0$; the ratio has exponential factor $\exp(-0.3z^2)$.)
A standard bivariate normal pair has correlation $-0.4$. Calculate $a=\sqrt{2/(1+\rho)}$ in the upper-tail bound and state the upper-tail dependence coefficient. (Answer: $a=\sqrt{10/3}\approx1.8257$, and $\lambda_U=0$.)
Let $X,Y$ be jointly normal with means $60,30$, standard deviations $8,4$, and correlation $0.3$. Find the common 97.5th-percentile thresholds, using $z_{0.975}=1.959964$, and state $\lambda_U$. (Answer: $75.68$ and $37.84$, respectively; $\lambda_U=0$.)
For a standard bivariate normal pair with $\rho=0.5$, calculate $P(Z_2>1\mid Z_1=1)$ to four decimals. Does this value equal $P(Z_2>1\mid Z_1>1)$? (Answer: $\bar\Phi(1/\sqrt{3})\approx0.2819$; no, conditioning on $Z_1=1$ is not the same as conditioning on $Z_1>1$.)
If $\rho=1$ for standard normal marginals, identify the dependence between the variables and their upper-tail coefficient. Why is this case not covered by the theorem for $|\rho|<1$? (Answer: $Z_2=Z_1$ almost surely and $\lambda_U=1$; the pair is degenerate and does not satisfy $|\rho|<1$.)

What You'll Learn