Correlation Shifts Under Mixtures and Common Shocks

Why Correlation Can Shift When the Risk Environment Changes

A historical correlation is a summary of the observations that were pooled to calculate it. If those observations come from different regimes, the pooled value reflects both dependence within each regime and differences in the regimes’ average outcomes. Likewise, a shock that affects two risks at once can create covariance even when their other components are independent. These effects matter when interpreting a dependence estimate or applying it to a risk environment that may not resemble the historical mix.

This tutorial derives the moment calculations for mixtures and shared shocks. It concerns ordinary covariance and Pearson correlation, not tail-dependence measures. The methods are aligned with the CAS Exam 9 (Risk Management for Actuaries) syllabus.

Definition: Let \(R\) be a regime indicator taking values \(1,\ldots,k\), with \(P(R=r)=p_r\), where \(p_r\geq 0\) and \(\sum_{r=1}^k p_r=1\). A mixture distribution is the unconditional distribution of \(X,Y\) when their conditional distributions may differ by \(R\). Write \(\mu_{X,r}=E[X\mid R=r]\), \(\mu_{Y,r}=E[Y\mid R=r]\), \(\sigma^2_{X,r}=\operatorname{Var}(X\mid R=r)\), \(\sigma^2_{Y,r}=\operatorname{Var}(Y\mid R=r)\), and \(c_r=\operatorname{Cov}(X,Y\mid R=r)\). Assume finite second moments.

The Total Covariance and Total Variance Results

The unconditional covariance has two parts: the average conditional covariance and the covariance of the conditional means. The second part can be positive, negative, or zero. Therefore, even if every regime has positive within-regime covariance, the pooled covariance need not be positive.

\operatorname{Cov}(X,Y) = E[\operatorname{Cov}(X,Y\mid R)] + \operatorname{Cov}(E[X\mid R],E[Y\mid R]).

For a finite set of regimes, the same result and the corresponding variance formulas are:

\begin{aligned} \mu_X &= \sum_{r=1}^{k}p_r\mu_{X,r}, & \mu_Y &= \sum_{r=1}^{k}p_r\mu_{Y,r},\\ \operatorname{Cov}(X,Y) &= \sum_{r=1}^{k}p_rc_r +\sum_{r=1}^{k}p_r(\mu_{X,r}-\mu_X)(\mu_{Y,r}-\mu_Y),\\ \operatorname{Var}(X) &=\sum_{r=1}^{k}p_r\sigma^2_{X,r} +\sum_{r=1}^{k}p_r(\mu_{X,r}-\mu_X)^2,\\ \operatorname{Var}(Y) &=\sum_{r=1}^{k}p_r\sigma^2_{Y,r} +\sum_{r=1}^{k}p_r(\mu_{Y,r}-\mu_Y)^2. \end{aligned}

The pooled Pearson correlation is then calculated from the pooled moments:

\rho_{X,Y}= \frac{\operatorname{Cov}(X,Y)} {\sqrt{\operatorname{Var}(X)\operatorname{Var}(Y)}}.

This expression requires both unconditional variances to be positive. Notice that the conditional correlation \(\rho_r=c_r/(\sigma_{X,r}\sigma_{Y,r})\) does not, by itself, determine the pooled correlation. The conditional standard deviations, regime probabilities, and shifts in conditional means all matter.

Why the Decomposition Holds

The result follows by separating each observation into its deviation from its regime mean and the regime mean’s deviation from the overall mean. This also explains why regime shifts contribute to pooled dependence.

Split each centered variable.
Write \(X-\mu_X=(X-E[X\mid R])+(E[X\mid R]-\mu_X)\), and use the analogous split for \(Y-\mu_Y\).

Multiply and take expectations.
The product contains four terms. The product of the two within-regime deviations has expectation \(E[\operatorname{Cov}(X,Y\mid R)]\); the product of the two regime-mean deviations has expectation \(\operatorname{Cov}(E[X\mid R],E[Y\mid R])\).

Show the cross terms vanish.
For example, conditional on \(R\), \(E[X-E[X\mid R]\mid R]=0\), while \(E[Y\mid R]-\mu_Y\) is fixed. Thus \(E[(X-E[X\mid R])(E[Y\mid R]-\mu_Y)]=0\). The other cross term vanishes in the same way.

Apply the same argument to a variance.
Set \(Y=X\) in the covariance identity. This gives \(\operatorname{Var}(X)=E[\operatorname{Var}(X\mid R)]+\operatorname{Var}(E[X\mid R])\), which yields the finite-regime variance formula.

Worked Example: A Mixture Reverses the Within-Regime Relationship

A hypothetical portfolio analyst models annual changes in two risk measures, \(X\) and \(Y\), in thousands of dollars. Regime A and regime B are equally likely. Conditional on regime A, \((X,Y)\) is bivariate normal with means \((10,20)\), variances \(4\) and \(9\), and covariance \(1\). Conditional on regime B, it is bivariate normal with means \((20,10)\), the same variances, and covariance \(5\). Each covariance matrix is valid: its determinant is positive (\(4\cdot9-1^2=35\) in A and \(4\cdot9-5^2=11\) in B). Find the conditional and pooled correlations.

Method. Calculate conditional correlations from the conditional covariance matrices, then use the total covariance and variance formulas for the pooled correlation.

In regime A, \(\rho_A=1/\sqrt{4\cdot9}=1/6\), or approximately \(0.1667\). In regime B, \(\rho_B=5/\sqrt{4\cdot9}=5/6\), or approximately \(0.8333\). The regime means are \(\mu_X=(10+20)/2=15\) and \(\mu_Y=(20+10)/2=15\).

The average within-regime covariance is \((1+5)/2=3\). The between-regime covariance is \[ \frac12(10-15)(20-15)+\frac12(20-15)(10-15) =\frac12(-25)+\frac12(-25)=-25. \] Thus the pooled covariance is \(3-25=-22\).

For \(X\), the average conditional variance is \(4\), and the between-regime variance of the means is \(\frac12(10-15)^2+\frac12(20-15)^2=25\). Hence \(\operatorname{Var}(X)=29\). For \(Y\), the corresponding terms are \(9\) and \(25\), so \(\operatorname{Var}(Y)=34\). Therefore, \[ \rho_{X,Y}=\frac{-22}{\sqrt{29\cdot34}} =\frac{-22}{\sqrt{986}} \approx -0.7006. \] The calculation is checked by noting that the pooled covariance is negative and the denominator is positive, so the pooled correlation must be negative. The method avoids the common mistake of averaging the conditional correlations: their simple average is \(0.5000\), not the pooled correlation.

Shared Shocks and Correlation

A common shock is a random component that enters two variables. The variables need not respond to it equally: the coefficients \(a\) and \(b\) represent their respective loadings. Suppose \(X=X_0+aZ\) and \(Y=Y_0+bZ\), where \(Z\) has finite variance and is independent of the pair \((X_0,Y_0)\). No independence between \(X_0\) and \(Y_0\) is required. Expanding variances and covariance gives:

\begin{aligned} \operatorname{Cov}(X,Y) &=\operatorname{Cov}(X_0,Y_0)+ab\operatorname{Var}(Z),\\ \operatorname{Var}(X) &=\operatorname{Var}(X_0)+a^2\operatorname{Var}(Z),\\ \operatorname{Var}(Y) &=\operatorname{Var}(Y_0)+b^2\operatorname{Var}(Z). \end{aligned}

For instance, expanding the covariance gives \(\operatorname{Cov}(X_0,Y_0)+b\operatorname{Cov}(X_0,Z)+a\operatorname{Cov}(Z,Y_0)+ab\operatorname{Var}(Z)\). Independence makes the two cross-covariances zero, leaving the stated result. When \(a\) and \(b\) have the same sign, the shock adds positive covariance; when their signs differ, it adds negative covariance. The correlation still depends on the changes to both variances, so it does not generally change by the same amount as the covariance. If \(Z\) is not independent of the baseline pair, the omitted cross-covariances must also be included.

Worked Example: A Shared Shock Creates Positive Correlation

Two hypothetical annual risk changes, in thousands of dollars, are \(X=X_0+Z\) and \(Y=Y_0+2Z\). The components \(X_0,Y_0,Z\) are mutually independent, normally distributed, and have mean zero with variances \(4,9,4\), respectively. Find the correlation before and after the shared shock is included.

Method. Use independence to calculate the baseline moments and the common-shock contributions. Before adding \(Z\), \(X_0\) and \(Y_0\) are independent, so their covariance and correlation are both zero.

After adding the shock, \(\operatorname{Cov}(X,Y)=0+(1)(2)(4)=8\). The variances are \(\operatorname{Var}(X)=4+1^2(4)=8\) and \(\operatorname{Var}(Y)=9+2^2(4)=25\). Thus, \[ \rho_{X,Y}=\frac{8}{\sqrt{8\cdot25}} =\frac{8}{\sqrt{200}} \approx 0.5657. \] As a check, the result is positive because both variables load positively on the same shock; its magnitude is less than one because each variable also contains its own independent variation. This method avoids counting the shared component in the covariance while forgetting to include it in each variance.

Worked Example: A Rare Common Event Creates Pooled Covariance

In a hypothetical model, \(H\) indicates a severe weather event in a year, with \(P(H=1)=0.10\) and \(P(H=0)=0.90\). Let \(H\) be independent of \(\epsilon_X,\epsilon_Y\), which are mutually independent normal variables with mean zero and variances \(1\) and \(4\). Define the annual changes \(X=\epsilon_X+2H\) and \(Y=\epsilon_Y+3H\), in thousands of dollars. Find the conditional and pooled correlations.

Method. First condition on the event indicator, then calculate the pooled covariance and variances. The event probability gives \(\operatorname{Var}(H)=0.10(0.90)=0.09\).

Given either value of \(H\), the event-related terms are constants. The remaining errors are independent, so the conditional covariance is zero and the conditional correlation is zero in both states. Unconditionally, \(E[X]=2(0.10)=0.2\) and \(E[Y]=3(0.10)=0.3\). The only covariance contribution is from the shared event: \[ \operatorname{Cov}(X,Y)=2\cdot3\cdot0.09=0.54. \] The variances are \(\operatorname{Var}(X)=1+2^2(0.09)=1.36\) and \(\operatorname{Var}(Y)=4+3^2(0.09)=4.81\). Therefore, \[ \rho_{X,Y}=\frac{0.54}{\sqrt{1.36\cdot4.81}} =\frac{0.54}{\sqrt{6.5416}} \approx 0.2111. \] The result is checked by the positive common-event contribution and positive denominator. This example shows that zero conditional correlation does not imply zero pooled correlation: years with the event have higher conditional means for both variables.

Interpreting a Historical Correlation

A historical correlation generally summarizes the sample’s particular mixture of regimes and shared influences. A change in regime frequency can therefore change the pooled correlation, even if the conditional distributions within each regime stay fixed. A different shock frequency or different shock loadings can have the same effect. Correlation calculated in a pooled sample should not automatically be treated as a stable property of the underlying risks.

A useful interpretation separates three questions:

Within-regime dependence: What are the conditional covariances or correlations in each relevant regime?
Regime-mean movement: Do the conditional means tend to rise together or move in opposite directions?
Common components: Is there a shock that enters both risks, and what are the signs and sizes of the loadings?

These questions help explain how a pooled estimate arose. They do not imply that correlation alone captures the full joint loss behavior. Nor do they guarantee that historical regimes or shock frequencies will persist. In applications, the model assumptions and the period represented by the data should be made explicit.

A Common Exam Trap

Common mistake: Calculate the pooled correlation as \(\sum_r p_r\rho_r\). This is generally wrong because correlations are standardized by regime-specific standard deviations, and it omits the covariance and variance contributions from shifts in regime means. In the first example, this shortcut gives \(\frac12(0.1667+0.8333)=0.5000\), whereas the correct pooled correlation is \(-0.7006\). The correction is to pool covariance and variances using the total-moment formulas first, then standardize once using the pooled variances.

Another trap is to say that a common shock increases correlation whenever it increases covariance. The correlation’s denominator also changes. For opposite-sign loadings the covariance contribution is negative, and even for same-sign loadings the final correlation must be recomputed from all three pooled moments.

Key takeaway: Pooled dependence combines within-regime covariance with covariance among regime means. A shared shock contributes covariance according to the product of its loadings, while also increasing each affected variance. Calculate the pooled moments before calculating the pooled correlation.

Check Your Understanding

For each problem, calculate pooled moments before standardizing to a correlation where required. Report correlations to four decimal places.

A hypothetical two-regime model has equally likely regimes. In regime 1, the bivariate normal variables \(X,Y\) have means \((4,5)\), variances \((1,4)\), and covariance \(0.5\). In regime 2, they have means \((8,9)\), the same variances, and covariance \(1.5\). Find the pooled correlation. (Answer: \(0.7906\))
Let \(X=A+Z\) and \(Y=B-Z\), where \(A,B,Z\) are mutually independent normal variables with mean zero and variances \(3,5,2\). Find the correlation of \(X,Y\). (Answer: \(-0.3381\))
Regime 1 has probability \(0.25\), and regime 2 has probability \(0.75\). Conditional on regime 1, the means of \(X,Y\) are \((0,0)\); conditional on regime 2, they are \((4,2)\). In both regimes, the conditional variances are \(4\) for \(X\) and \(1\) for \(Y\), and the conditional covariance is \(1\). Find the pooled correlation. (Answer: \(0.7143\))
Let \(H\) be Bernoulli with \(P(H=1)=0.20\). It is independent of mutually independent, mean-zero normal variables \(A,B\) with variances \(2,8\). Define \(X=A+H\) and \(Y=B+4H\). Find the pooled correlation. (Answer: \(0.1340\))
Suppose two regimes have identical conditional means, identical conditional variances, and conditional covariances \(2\) and \(-1\). The first regime has probability \(0.40\). Find the pooled covariance and explain whether a between-regime mean term contributes. (Answer: pooled covariance \(0.20(2)+0.60(-1)=0.20\); the between-regime mean term is zero because the conditional means are identical.)

What You'll Learn