Covariance of Ceded and Retained Losses

Why Ceded and Retained Losses Move Together

Reinsurance divides each covered loss between the insurer and the reinsurer. The resulting ceded and retained aggregate losses are not generally independent: they are built from allocations of the same claims, and both totals are affected by how many claims occur. To quantify their joint movement, separate these two sources of dependence: the relationship between ceded and retained payments on one claim, and variation in the claim count.

Throughout, consider a hypothetical portfolio over a specified period. Let \(N\) be the number of claims. For claim \(i\), let \(X_i\) be gross loss, \(C_i\) the amount ceded, and \(R_i\) the amount retained. Assume the treaty allocates every dollar of each gross loss between the two parties, with no unallocated payment, expense, or other cash flow. Then \(X_i=C_i+R_i\).

Definition: The ceded aggregate loss is \(C^A=\sum_{i=1}^{N}C_i\), the retained aggregate loss is \(R^A=\sum_{i=1}^{N}R_i\), and the gross aggregate loss is \(G=\sum_{i=1}^{N}X_i\). A sum over no claims is defined to be zero. Claim-level and aggregate losses are distinct random variables: \(C_i\) and \(R_i\) describe one claim, while \(C^A\) and \(R^A\) sum the allocations over the random count.

The key point is that aggregate covariance has two components. Even if the count were fixed, the ceded and retained amounts on each claim could be dependent. When the count is random, an unusually large count tends to raise both aggregate totals when their claim-level means are positive.

Covariance Formula for a Random Claim Count

Use the following assumptions: \(N\) is independent of the claim-level pairs \((C_i,R_i)\); those pairs are independent and identically distributed across claims; \(E[N]\) and \(\operatorname{Var}(N)\) are finite; and \(E[C_1]\), \(E[R_1]\), and \(E[|C_1R_1|]\) are finite. For nonnegative losses, the last condition is \(E[C_1R_1]<\infty\). It ensures the claim-level covariance exists. Define \(\mu_C=E[C_1]\), \(\mu_R=E[R_1]\), and \(\operatorname{Cov}(C_1,R_1)=E[C_1R_1]-\mu_C\mu_R\).

Theorem: Under these assumptions,

\operatorname{Cov}(C^A,R^A) =E[N]\operatorname{Cov}(C_1,R_1) +\operatorname{Var}(N)\mu_C\mu_R.

The first term is the claim-allocation contribution: with a fixed count, each claim pair contributes its own covariance. The second term is the frequency contribution: variation in \(N\) makes both aggregate totals rise or fall together according to their claim-level means.

Why the Formula Holds

Condition on the observed claim count. Given \(N=n\), there are \(n\) independent, identically distributed claim pairs. Thus the conditional means are \(n\mu_C\) and \(n\mu_R\), and the conditional covariance is \(n\operatorname{Cov}(C_1,R_1)\). The law of total covariance states that covariance equals the expected conditional covariance plus the covariance of the conditional means.

Condition on the count.
\(E[C^A\mid N]=N\mu_C\), \(E[R^A\mid N]=N\mu_R\), and \(\operatorname{Cov}(C^A,R^A\mid N)=N\operatorname{Cov}(C_1,R_1)\).

Average the conditional covariance.
\(E[\operatorname{Cov}(C^A,R^A\mid N)]=E[N]\operatorname{Cov}(C_1,R_1)\).

Find the covariance of the conditional means.
\(\operatorname{Cov}(E[C^A\mid N],E[R^A\mid N])=\operatorname{Cov}(N\mu_C,N\mu_R)=\operatorname{Var}(N)\mu_C\mu_R\).

Add the two contributions.
The law of total covariance gives the stated formula.

This derivation also shows why finite claim-level means alone are not enough. The product moment \(E[C_1R_1]\) must be finite to define the claim-level covariance and use the conditional covariance calculation. Nor is independence of \(N\) optional: if claim severity or allocation changes with the count, the conditional means and covariances need not have the forms used above.

Worked Applications

Worked Example: Quota Share with Poisson Frequency

A hypothetical insurer has a Poisson claim count with parameter \(\lambda=3\). Each gross claim is independently \(10\) or \(20\) dollars, each with probability \(0.5\). A quota-share treaty cedes \(40\%\) of each claim, so \(C_i=0.4X_i\) and \(R_i=0.6X_i\). Find the covariance of ceded and retained aggregate loss, and verify the gross-loss identity.

Method and claim moments. The count satisfies \(E[N]=\operatorname{Var}(N)=3\), and it is independent of the iid severities. First calculate the severity moments:

E[X_i]=0.5(10)+0.5(20)=15,\qquad E[X_i^2]=0.5(10^2)+0.5(20^2)=250, $$ $$ \operatorname{Var}(X_i)=250-15^2=25.

Therefore \(\mu_C=0.4(15)=6\), \(\mu_R=0.6(15)=9\), and, because both allocations are constant multiples of \(X_i\),

\operatorname{Cov}(C_i,R_i) =(0.4)(0.6)\operatorname{Var}(X_i) =0.24(25)=6.

Aggregate covariance. Apply the random-count formula:

\operatorname{Cov}(C^A,R^A) =3(6)+3(6)(9) =18+162 =180\text{ dollars}^2.

Check. For every claim, \(C_i+R_i=0.4X_i+0.6X_i=X_i\). Summing gives \(C^A+R^A=G\) for every possible claim count and every set of severities. The expected gross aggregate loss is \(E[G]=E[N]E[X_i]=3(15)=45\) dollars; the expected ceded and retained totals are \(3(6)=18\) and \(3(9)=27\), which sum to \(45\). No rounding is needed.

Common mistake avoided: using only \(E[N]\operatorname{Cov}(C_i,R_i)=18\) omits the \(162\text{ dollars}^2\) frequency contribution. The correct total is \(180\text{ dollars}^2\).

Worked Example: Fixed Count and a Per-Claim Layer

A hypothetical portfolio has exactly two claims. Each gross claim is \(8\) or \(18\) dollars with probability \(0.5\), independently. A per-claim excess-of-loss treaty attaches at \(10\) dollars with no upper limit: \(C_i=(X_i-10)_+\), while the insurer retains \(R_i=\min(X_i,10)\). Find the covariance of aggregate ceded and retained loss.

Method. Since \(N=2\) is fixed, \(\operatorname{Var}(N)=0\), so aggregate covariance is twice the claim-level covariance. First list the allocations:

X_i=8:\ (C_i,R_i)=(0,8),\qquad X_i=18:\ (C_i,R_i)=(8,10).

The claim-level means are \(\mu_C=0.5(0)+0.5(8)=4\) and \(\mu_R=0.5(8)+0.5(10)=9\). The product moment is \(E[C_iR_i]=0.5(0\cdot8)+0.5(8\cdot10)=40\). Hence

\operatorname{Cov}(C_i,R_i)=40-(4)(9)=4\text{ dollars}^2, \qquad \operatorname{Cov}(C^A,R^A)=2(4)=8\text{ dollars}^2.

Check. For the two possible severities, the allocations sum to gross loss: \(0+8=8\) and \(8+10=18\). Therefore the aggregate identity holds for any pair of claims. The result is exact; no rounding is needed.

Common mistake avoided: treating the claim-level covariance as the aggregate covariance would give \(4\text{ dollars}^2\). There are exactly two independent claim pairs, so their covariances add to \(8\text{ dollars}^2\).

Worked Example: Random Frequency with a Quota Share

A hypothetical portfolio has claim count \(N=0,2,4\) with probabilities \(0.25,0.50,0.25\), respectively. Each gross claim is independently \(10\) or \(30\) dollars with equal probability. A \(50\%\) quota share divides each claim equally between ceded and retained loss. Find the aggregate covariance and check the expected gross total.

Step 1: Count moments. The count mean and second moment are

E[N]=0(0.25)+2(0.50)+4(0.25)=2, $$ $$ E[N^2]=0^2(0.25)+2^2(0.50)+4^2(0.25)=6, \qquad \operatorname{Var}(N)=6-2^2=2.

Step 2: Claim moments. The gross severity has \(E[X_i]=0.5(10)+0.5(30)=20\) and \(E[X_i^2]=0.5(100)+0.5(900)=500\), so \(\operatorname{Var}(X_i)=500-20^2=100\). Thus \(\mu_C=\mu_R=10\), and \(\operatorname{Cov}(C_i,R_i)=0.5(0.5)(100)=25\).

Step 3: Aggregate covariance.

\operatorname{Cov}(C^A,R^A) =2(25)+2(10)(10) =50+200 =250\text{ dollars}^2.

Check. Given \(N\), the conditional covariance is \(25N\), whose expectation is \(25(2)=50\). The covariance of the conditional means \(10N\) and \(10N\) is \(100\operatorname{Var}(N)=200\). These add to \(250\), confirming the formula. Also \(C_i+R_i=X_i\) claim by claim, so \(C^A+R^A=G\) exactly. The expected gross total is \(E[G]=2(20)=40\) dollars, equal to the sum of the expected ceded and retained totals, \(20+20\). No rounding is needed.

Common mistake avoided: setting \(N\) equal to its mean and treating it as fixed would retain only the \(50\text{ dollars}^2\) claim-allocation term. The count variability contributes another \(200\text{ dollars}^2\).

The Gross-Loss Identity and Its Limits

The identity between gross, ceded, and retained aggregate losses is not a covariance result; it is an accounting identity. If each claim is fully allocated, then \(X_i=C_i+R_i\). For every outcome,

G=\sum_{i=1}^{N}X_i =\sum_{i=1}^{N}(C_i+R_i) =C^A+R^A.

It remains true when \(N=0\), because all three sums are zero. It also implies \(E[G]=E[C^A]+E[R^A]\) whenever the expectations are finite. This is a useful arithmetic check, but it does not imply that ceded and retained aggregate losses are independent or that their covariance is zero.

The formula for covariance has narrower conditions than the accounting identity. The identity only requires the claim-level allocations to sum to gross loss. The covariance formula additionally relies on independent, identically distributed claim pairs, independence between the count and those pairs, suitable finite moments, and a finite claim-level product moment. For nonnegative losses, if \(E[C_1R_1]\) is infinite, the covariance formula cannot be used as a finite covariance calculation. If the count depends on claim severity, or the claim-level pairs are dependent across claims, derive the covariance for that model rather than applying this formula unchanged.

Common Exam Traps

Dropping the frequency term. A random count contributes \(\operatorname{Var}(N)\mu_C\mu_R\); it does not disappear merely because allocations are made claim by claim.
Confusing claim and aggregate covariance. With fixed \(N=n\), aggregate covariance is \(n\operatorname{Cov}(C_1,R_1)\), not just the covariance for one claim.
Assuming the two aggregates are independent. The same count affects both, and claim-level allocations may also be dependent.
Using insufficient moment assumptions. Finite \(E[C_1]\) and \(E[R_1]\) do not guarantee a finite covariance. Check that \(E[|C_1R_1|]\) is finite.
Checking only expected totals. Matching expected ceded plus retained loss to expected gross loss is a useful check, but the stronger identity holds outcome by outcome when every claim is fully allocated.

Key takeaway: Under the stated random-sum assumptions, aggregate covariance equals the expected number of claims times claim-level covariance, plus the claim-count variance times the product of claim-level means. Separately, complete claim-level allocation guarantees \(C^A+R^A=G\) for every outcome.

Check Your Understanding

For each problem, find the requested covariance or verify the allocation identity. Treat all dollar amounts as hypothetical, and assume the count is independent of iid claim-level losses unless the problem says otherwise.

A portfolio has exactly four claims. Each gross claim is \(6\) or \(14\) dollars with equal probability, and a quota share cedes \(25\%\). Find the covariance of ceded and retained aggregate loss. (Answer: \(12\text{ dollars}^2\))
The claim count is Poisson with mean \(5\). Each claim is \(0\) or \(20\) dollars with equal probability, and \(30\%\) is ceded under quota share. Find the aggregate covariance. (Answer: \(210\text{ dollars}^2\))
The count is \(0,1,3\) with probabilities \(0.50,0.25,0.25\). Claims are \(4\) or \(12\) dollars with equal probability, and each is split equally. Find the aggregate covariance. (Answer: \(28\text{ dollars}^2\))
There are exactly three claims, each independently \(12\) or \(22\) dollars with equal probability. A per-claim layer cedes \(\min((X_i-10)_+,5)\), and the remainder is retained. Find the covariance of aggregate ceded and retained loss. (Answer: \(15.75\text{ dollars}^2\))
Suppose a treaty allocates every claim as \(X_i=C_i+R_i\). State the relationship among \(G,C^A,R^A\), including when \(N=0\). (Answer: \(G=C^A+R^A\) for every outcome, including \(N=0\), when all three sums are zero.)

What You'll Learn