Covariance Identities From First Principles

Why Covariance Identities Matter

When an insurer models two loss components, the total loss is their sum, but the components may move together. Covariance identities let an actuary calculate how that dependence affects an aggregate without rebuilding the entire distribution of the sum. The same identities also provide useful checks on calculations involving expected losses, fixed offsets, and aggregate variance.

The starting point is the definition of covariance as the expected product of deviations from the means. Expanding that definition gives a form that is often easier to calculate from a joint loss distribution.

Definition: For random variables $X$ and $Y$ with finite second moments, their covariance is $$ \operatorname{Cov}(X,Y)=E\big[(X-E[X])(Y-E[Y])\big]. $$ The finite-second-moment condition ensures the means, variances, and covariance used here are finite. The equivalent raw-moment identity is $$ \operatorname{Cov}(X,Y)=E[XY]-E[X]E[Y]. $$

The raw-moment form is especially convenient when the joint probabilities of paired losses are given. It requires the expected product $E[XY]$, which must be calculated from the joint outcomes; separate marginal means alone do not determine it.

Deriving the Basic Identities

First expand the definition. Since $E[X]$ and $E[Y]$ are constants, expectation is linear:

\begin{aligned} \operatorname{Cov}(X,Y) &=E\big[(X-E[X])(Y-E[Y])\big]\\ &=E[XY-XE[Y]-YE[X]+E[X]E[Y]]\\ &=E[XY]-E[X]E[Y]-E[Y]E[X]+E[X]E[Y]\\ &=E[XY]-E[X]E[Y]. \end{aligned}

This also explains why a joint distribution matters: the term $E[XY]$ depends on which values of $X$ and $Y$ occur together. For sums, the same definition and linearity give a further identity.

Center the sum.
For random variables $X$ and $Y$, $E[X+Y]=E[X]+E[Y]$, so $(X+Y)-E[X+Y]=(X-E[X])+(Y-E[Y])$.

Multiply by the centered third variable.
Multiply the centered sum by $Z-E[Z]$, then distribute across the two terms.

Take expectations term by term.
Linearity of expectation gives $\operatorname{Cov}(X+Y,Z)=\operatorname{Cov}(X,Z)+\operatorname{Cov}(Y,Z)$.

Applying this identity repeatedly gives the covariance of two sums. For four random variables with finite second moments,

\operatorname{Cov}(X+Y,U+V) =\operatorname{Cov}(X,U)+\operatorname{Cov}(X,V) +\operatorname{Cov}(Y,U)+\operatorname{Cov}(Y,V).

Each term represents one pairing between a component in the first sum and a component in the second. In particular, setting the two sums equal gives the variance identity for an aggregate:

\operatorname{Var}(X+Y) =\operatorname{Var}(X)+\operatorname{Var}(Y)+2\operatorname{Cov}(X,Y),

because $\operatorname{Cov}(X,X)=\operatorname{Var}(X)$, $\operatorname{Cov}(Y,Y)=\operatorname{Var}(Y)$, and $\operatorname{Cov}(X,Y)=\operatorname{Cov}(Y,X)$. These results do not require independence. If the variables are independent and have finite second moments, their covariance is zero, but independence is not an assumption in the identities themselves.

Theorem: If $c$ is a deterministic finite constant and the variables have finite second moments, then $$ E[X+c]=E[X]+c,\qquad \operatorname{Cov}(X+c,Y)=\operatorname{Cov}(X,Y),\qquad \operatorname{Cov}(c,Y)=0. $$ A fixed offset changes the mean but not covariance. The covariance identities for sums concern random-variable components; a deterministic constant contributes no covariance term.

Worked Calculations for Paired Losses

Worked Example: Covariance From a Joint Loss Distribution

A hypothetical insurer records two quarterly loss components, $X$ and $Y$, in thousands of dollars. Their joint distribution is:

Probability	$X$	$Y$
0.20	0	1
0.30	0	5
0.10	4	1
0.40	4	5

Calculate $\operatorname{Cov}(X,Y)$, then calculate $\operatorname{Cov}(X+Y,X)$. All losses are stated in thousands of dollars. Method: use the raw-moment identity, then bilinearity of covariance. This avoids treating the marginal distributions as if they provided the joint expected product.

First, the probabilities sum to $0.20+0.30+0.10+0.40=1.00$. The marginal means are

$$ E[X]=(0.20)(0)+(0.30)(0)+(0.10)(4)+(0.40)(4)=2, $$

and

$$ E[Y]=(0.20)(1)+(0.30)(5)+(0.10)(1)+(0.40)(5) =0.2+1.5+0.1+2.0=3.8. $$

The expected product is

$$ E[XY]=(0.20)(0)+(0.30)(0)+(0.10)(4)(1)+(0.40)(4)(5) =0+0+0.4+8=8.4. $$

Therefore,

\operatorname{Cov}(X,Y)=8.4-(2)(3.8)=8.4-7.6=0.8

in squared thousands of dollars. For the second covariance, apply the sum identity:

\operatorname{Cov}(X+Y,X) =\operatorname{Cov}(X,X)+\operatorname{Cov}(Y,X) =\operatorname{Var}(X)+\operatorname{Cov}(X,Y).

Calculate $\operatorname{Var}(X)$. Since $X$ is 4 with probability $0.50$ and 0 otherwise,

E[X^2]=(0.50)(4^2)+(0.50)(0^2)=8,\qquad \operatorname{Var}(X)=8-2^2=4.

Thus $\operatorname{Cov}(X+Y,X)=4+0.8=4.8$ squared thousands of dollars. As a check, $E[X+Y]=2+3.8=5.8$, while $E[(X+Y)X]=E[X^2]+E[XY]=8+8.4=16.4$. The raw-moment identity gives $16.4-(5.8)(2)=4.8$, the same result.

Worked Example: Variance of an Aggregate of Paired Losses

A hypothetical insurer has two annual loss components, $X$ and $Y$, measured in thousands of dollars. Their joint distribution is:

Probability	$X$	$Y$
0.50	2	1
0.30	4	3
0.20	1	5

Calculate the mean and variance of $S=X+Y$. Method: calculate the component moments and covariance, then use the aggregate variance identity. This avoids omitting the cross-covariance term.

The means are

$$ E[X]=(0.50)(2)+(0.30)(4)+(0.20)(1)=1+1.2+0.2=2.4, $$

$$ E[Y]=(0.50)(1)+(0.30)(3)+(0.20)(5)=0.5+0.9+1.0=2.4. $$

The second moments are

$$ E[X^2]=(0.50)(4)+(0.30)(16)+(0.20)(1)=2+4.8+0.2=7, $$

$$ E[Y^2]=(0.50)(1)+(0.30)(9)+(0.20)(25)=0.5+2.7+5=8.2. $$

Hence $\operatorname{Var}(X)=7-(2.4)^2=7-5.76=1.24$, and $\operatorname{Var}(Y)=8.2-(2.4)^2=8.2-5.76=2.44$. The expected product and covariance are

$$ E[XY]=(0.50)(2)(1)+(0.30)(4)(3)+(0.20)(1)(5)=1+3.6+1=5.6, $$

\operatorname{Cov}(X,Y)=5.6-(2.4)(2.4)=5.6-5.76=-0.16.

The mean aggregate loss is $E[S]=2.4+2.4=4.8$. Its variance is

\operatorname{Var}(S)=1.24+2.44+2(-0.16)=3.68-0.32=3.36.

Check directly from the aggregate outcomes: $S$ is 3, 7, and 6 with probabilities $0.50$, $0.30$, and $0.20$. Thus $E[S^2]=(0.50)(9)+(0.30)(49)+(0.20)(36)=4.5+14.7+7.2=26.4$, and $\operatorname{Var}(S)=26.4-(4.8)^2=26.4-23.04=3.36$. The two methods agree.

Worked Example: Deterministic Offsets and a Constant Aggregate

A hypothetical portfolio has paired losses $X$ and $Y$, in thousands of dollars, with probability $0.50$ on each outcome $(X,Y)=(3,1)$ and $(X,Y)=(1,3)$. Define reported quantities $A=X+5$ and $B=Y-2$, where 5 and 2 are fixed offsets in thousands of dollars. Find $E[A]$, $E[B]$, $\operatorname{Cov}(A,B)$, and $\operatorname{Var}(A+B)$. Method: use linearity for means and the constant-shift and aggregate-variance identities. This avoids treating a fixed offset as a random risk component.

The original means are $E[X]=(0.50)(3)+(0.50)(1)=2$ and $E[Y]=(0.50)(1)+(0.50)(3)=2$. Therefore,

E[A]=E[X]+5=7,\qquad E[B]=E[Y]-2=0.

The expected product of the original losses is $E[XY]=(0.50)(3)(1)+(0.50)(1)(3)=3$, so

\operatorname{Cov}(X,Y)=3-(2)(2)=-1.

Adding or subtracting deterministic constants leaves covariance unchanged; consequently $\operatorname{Cov}(A,B)=-1$. Verify directly: $(A,B)$ is $(8,-1)$ or $(6,1)$, each with probability $0.50$. Thus $E[AB]=(0.50)(-8)+(0.50)(6)=-1$, and $E[A]E[B]=(7)(0)=0$, giving covariance $-1$.

For the aggregate, $A+B=X+Y+3$. In both outcomes $X+Y=4$, so $A+B=7$ with probability 1 and its variance is zero. The variance identity confirms this: $X$ and $Y$ each have variance $(0.50)(3^2)+(0.50)(1^2)-2^2=5-4=1$. Since constants do not change variance, $\operatorname{Var}(A)=\operatorname{Var}(B)=1$. Therefore,

\operatorname{Var}(A+B)=1+1+2(-1)=0,

which agrees with the constant aggregate obtained directly.

Common Exam Trap: Dropping the Cross-Term

A frequent error is to calculate the variance of a sum as the sum of the component variances, without checking whether the covariance term is zero. That shortcut is justified only when the covariance is zero. Independence is one sufficient condition, but it is not needed for the general identity.

Common mistake: For the second worked example, writing $\operatorname{Var}(X+Y)=1.24+2.44=3.68$ omits the cross-term. The correct calculation is $$ \operatorname{Var}(X+Y)=1.24+2.44+2(-0.16)=3.36. $$ The incorrect value also fails the direct check $E[S^2]-(E[S])^2=26.4-23.04=3.36$.

A related mistake is to assume that adding a fixed amount to a loss changes its covariance with another loss. A constant changes the mean, but its covariance with any random variable is zero. Distinguish a deterministic offset from an additional random loss component.

Using the Identities Efficiently

For paired-loss calculations, a reliable sequence is to establish the means, obtain $E[XY]$ from the joint outcomes, and then apply the raw-moment identity. For an aggregate, expand the covariance or variance using the sum identities rather than assuming a dependence structure that the data do not establish.

Use $E[X+Y]=E[X]+E[Y]$ to obtain an aggregate mean.
Use $\operatorname{Cov}(X,Y)=E[XY]-E[X]E[Y]$ when joint outcomes are available.
Use $\operatorname{Cov}(X+Y,Z)=\operatorname{Cov}(X,Z)+\operatorname{Cov}(Y,Z)$ to expand a covariance with a sum.
Use $\operatorname{Var}(X+Y)=\operatorname{Var}(X)+\operatorname{Var}(Y)+2\operatorname{Cov}(X,Y)$ for a two-component aggregate.
Use a direct calculation from aggregate outcomes as a check when the joint distribution is small enough to enumerate.

These identities are useful because they isolate the contribution of each component and its pairwise covariance. They simplify aggregate calculations while keeping the dependence information visible.

Check Your Understanding

Work each problem from the stated information, then compare with the final answer.

A paired-loss model, in thousands of dollars, has outcomes $(X,Y)=(1,2)$ with probability $0.40$ and $(3,4)$ with probability $0.60$. Calculate $E[X]$, $E[Y]$, and $\operatorname{Cov}(X,Y)$. (Answer: $E[X]=2.2$, $E[Y]=3.2$, and $\operatorname{Cov}(X,Y)=0.96$, in squared thousands of dollars.)
Suppose $E[U]=5$, $E[V]=2$, and $E[UV]=13$. Calculate $\operatorname{Cov}(U+7,V-4)$. (Answer: $13-(5)(2)=3$.)
For random losses $X,Y,Z$, suppose $\operatorname{Cov}(X,Z)=0.5$ and $\operatorname{Cov}(Y,Z)=-0.2$. Calculate $\operatorname{Cov}(X+Y,Z)$. (Answer: $0.5-0.2=0.3$.)
An aggregate has two components with variances 6 and 9 and covariance $-2$. Calculate the aggregate variance. (Answer: $6+9+2(-2)=11$.)
For paired losses, $E[X]=3$, $E[Y]=4$, $E[X^2]=14$, $E[Y^2]=20$, and $E[XY]=10$. Calculate $\operatorname{Cov}(X,Y)$ and $\operatorname{Var}(X+Y)$. (Answer: $\operatorname{Cov}(X,Y)=10-(3)(4)=-2$; $\operatorname{Var}(X+Y)=(14-9)+(20-16)+2(-2)=5$.)

What You'll Learn