How Coverage Changes Aggregate Risk Measures

Why Coverage Changes Risk Measures

Risk measures usually describe the loss for a specified party. A ground-up loss is the amount before insurance coverage is applied; an insurer’s payment, or a policyholder’s retained loss, can have a different distribution. A deductible can create an atom at zero, a payment limit can create an atom at its maximum, and coinsurance can rescale payments. Those changes can move both VaR and TVaR.

The order of operations matters. Applying a deductible and limit to each claim and then adding the payments is generally different from applying one deductible and limit to the total aggregate loss. The first step in a calculation is therefore to identify exactly which loss the coverage applies to.

Definition: Let $X\geq 0$ be a ground-up loss. For deductible $d\geq 0$, payment limit $u\geq 0$, and coinsurance fraction $c$ with $0\leq c\leq 1$, define the covered payment per loss by $$ Y^L=c\min\big((X-d)_+,u\big), \qquad (z)_+=\max(z,0). $$ Here $u$ is the maximum amount to which coinsurance is applied, so the maximum payment is $cu$. With no finite payment limit, take $u=\infty$. This convention applies the deductible first, then the payment limit, then coinsurance.

For a portfolio with claim losses $X_1,\ldots,X_N$, per-claim coverage produces the aggregate payment $T=\sum_{j=1}^{N}c\min((X_j-d)_+,u)$. An aggregate layer instead applies coverage to the total $S=\sum_{j=1}^{N}X_j$, for example $T_{\mathrm{agg}}=c\min((S-D)_+,U)$, with aggregate attachment $D$ and payment limit $U$. These are different random variables in general.

For any random loss $Z$, use the lower-quantile convention $\operatorname{VaR}_p(Z)=\inf\{z:F_Z(z)\geq p\}$, where $0<p<1$. Define TVaR by the average of upper quantiles: $\operatorname{TVaR}_p(Z)=(1-p)^{-1}\int_p^1\operatorname{VaR}_r(Z)\,dr$. This definition handles discrete and mixed distributions, including probability mass at the quantile.

How to Recalculate VaR and TVaR

The dependable approach is to determine the distribution after coverage, then calculate its quantile and upper-tail average. A useful identity makes the TVaR calculation practical even when the distribution has atoms.

Theorem: For a loss $Z$ with finite expectation and $q=\operatorname{VaR}_p(Z)$, $$ \operatorname{TVaR}_p(Z)=q+\frac{E[(Z-q)_+]}{1-p}. $$ The excess term accounts for the part of the upper tail above the quantile. If there is probability mass at $q$, only the fraction needed to make up the upper $1-p$ probability is included in the tail average.

To justify the identity, write $F_Z(q-)=P(Z<q)$ and $F_Z(q)=P(Z\leq q)$. By the quantile convention, $F_Z(q-)<p\leq F_Z(q)$. In the quantile integral from $p$ to 1, the portion from $p$ to $F_Z(q)$ contributes $(F_Z(q)-p)q$. Quantiles above $q$ contribute $E[Z\,1_{\{Z>q\}}]$. Thus the integral is

(F_Z(q)-p)q+E[Z\,1_{\{Z>q\}}] =(1-p)q+E[(Z-q)_+].

Dividing by $1-p$ gives the result. This also explains why simply averaging observations strictly above $q$ can be wrong: when $q$ is an atom, the upper $1-p$ portion may include only part of that atom.

Theorem: If $g$ is strictly increasing and continuous and $Z=g(X)$, then $$ \operatorname{VaR}_p(Z)=g\big(\operatorname{VaR}_p(X)\big). $$ The continuity condition matters for the stated infimum-quantile convention. Coverage functions commonly have flat portions and create atoms, so a safe general procedure is to construct the transformed distribution and find its quantile directly.

TVaR does not generally obey the analogous shortcut. Even when a coverage function is monotone, the payment above the covered-loss quantile can be clipped or rescaled, and the probability mass at that quantile may change. Calculate TVaR from the covered distribution or use its expected-excess identity.

Worked Examples

Worked Example: A Deductible, Coinsurance, and a Payment Limit

A hypothetical single claim $X$ has the following distribution: $P(X=0)=0.20$, $P(X=100)=0.30$, $P(X=300)=0.30$, and $P(X=600)=0.20$, with amounts in dollars. Coverage has deductible $d=\$100$, payment limit $u=\$250$, and coinsurance fraction $c=0.80$. Find ground-up and covered-payment VaR and TVaR at $p=0.80$.

Method and coverage transformation. Apply the deductible, then the limit, then coinsurance: $Y^L=0.80\min((X-100)_+,250)$. At the four possible ground-up losses, the payments are respectively $\$0,\$0,\$160,\$200$. Thus $P(Y^L=0)=0.50$, $P(Y^L=160)=0.30$, and $P(Y^L=200)=0.20$.

Ground-up VaR. The cumulative probabilities for $X=0,100,300,600$ are $0.20,0.50,0.80,1.00$. Therefore $\operatorname{VaR}_{0.80}(X)=\$300$.

Covered VaR. The covered cumulative probability reaches $0.50$ at $\$0$ and $0.80$ at $\$160$. Therefore $\operatorname{VaR}_{0.80}(Y^L)=\$160$.

Ground-up TVaR. Using $q=\$300$, the only loss above $q$ is $\$600$, with probability $0.20$. Hence $\operatorname{TVaR}_{0.80}(X)=300+[0.20(600-300)]/0.20=\$600$.

Covered TVaR. With $q=\$160$, the payment above $q$ is $\$40$ with probability $0.20$. Therefore $\operatorname{TVaR}_{0.80}(Y^L)=160+[0.20(200-160)]/0.20=\$200$. The top 20% of covered outcomes are exactly the outcomes paying $\$200$, which checks the result.

Common mistake avoided. Transforming ground-up TVaR as if it were a loss quantile would give $0.80\min((600-100),250)=\$200$ here by coincidence. That shortcut is not justified: the TVaR calculation must use the covered distribution and its upper-tail probability.

Worked Example: Summing Payments Under Per-Claim Coverage

A hypothetical portfolio has exactly two independent claims. Each claim is $\$100$ with probability $0.50$ and zero otherwise. Each claim is subject to a $\$40$ deductible and 50% coinsurance, with no payment limit. Find ground-up aggregate and covered aggregate VaR and TVaR at $p=0.75$.

Method and aggregate distributions. The aggregate ground-up loss $S=X_1+X_2$ is $\$0,\$100,\$200$. Independence gives probabilities $0.25,0.50,0.25$, respectively: neither claim occurs with probability $0.5^2=0.25$, exactly one occurs with probability $2(0.5)(0.5)=0.50$, and both occur with probability $0.25$.

Each claim payment is $Y_j^L=0.50(X_j-40)_+$. A zero claim pays $\$0$, and a $\$100$ claim pays $0.50(100-40)=\$30$. Consequently the covered aggregate $T=Y_1^L+Y_2^L$ takes values $\$0,\$30,\$60$, with probabilities $0.25,0.50,0.25$.

VaR calculations. For $S$, the cumulative probability is $0.25$ at zero and $0.75$ at $\$100$, so $\operatorname{VaR}_{0.75}(S)=\$100$. For $T$, the cumulative probability is $0.25$ at zero and $0.75$ at $\$30$, so $\operatorname{VaR}_{0.75}(T)=\$30$.

TVaR calculations. For $S$, use $q=\$100$. The excess above $q$ is $\$100$ with probability $0.25$, giving $100+[0.25(100)]/0.25=\$200$. For $T$, use $q=\$30$. Its excess above $q$ is $\$30$ with probability $0.25$, giving $30+[0.25(30)]/0.25=\$60$. In each case the upper quarter consists of the largest aggregate outcome, confirming both calculations.

Common mistake avoided. This coverage applies separately to each claim before the payments are added. It would be a different contract to apply one deductible to $S$. Here, one claim produces a $\$30$ payment; two claims produce $\$60$.

Worked Example: An Aggregate Layer and a Quantile Atom

A hypothetical aggregate ground-up loss $S$ equals $\$0$ with probability $0.40$, $\$100$ with probability $0.35$, and $\$300$ with probability $0.25$. An aggregate layer attaches at $D=\$80$, pays 100% of covered loss, and has payment limit $U=\$120$. Find ground-up and layer-payment VaR and TVaR at $p=0.60$.

Method and transformed distribution. The layer payment is $T_{\mathrm{agg}}=\min((S-80)_+,120)$. For aggregate losses $\$0,\$100,\$300$, the payments are $\$0,\$20,\$120$. Their probabilities remain $0.40,0.35,0.25$. The layer has an atom at its maximum payment because every ground-up loss of at least $\$200$ would produce the same payment; in this distribution, the $\$300$ outcome contributes that atom.

VaR calculations. For $S$, the cumulative probability is $0.40$ at zero and $0.75$ at $\$100$, so $\operatorname{VaR}_{0.60}(S)=\$100$. For $T_{\mathrm{agg}}$, the cumulative probability is $0.40$ at zero and $0.75$ at $\$20$, so $\operatorname{VaR}_{0.60}(T_{\mathrm{agg}})=\$20$.

TVaR calculations. For $S$, the excess above $\$100$ is $\$200$ with probability $0.25$. Thus $\operatorname{TVaR}_{0.60}(S)=100+[0.25(200)]/0.40=\$225$. For the layer payment, the excess above $\$20$ is $\$100$ with probability $0.25$. Thus $\operatorname{TVaR}_{0.60}(T_{\mathrm{agg}})=20+[0.25(100)]/0.40=\$82.50$.

As a check, the upper 40% of layer outcomes comprises the entire 25% at $\$120$ and 15% at $\$20$. Their average is $[0.25(120)+0.15(20)]/0.40=\$82.50$.

Common mistake and correction. Applying the layer function to ground-up TVaR would give $\min((225-80)_+,120)=\$120$, which is not the layer TVaR. TVaR averages outcomes in the upper tail; the correct calculation includes both the $\$120$ outcomes and the required fraction of the $\$20$ outcomes.

Interpretation and Exam Traps

When $0\leq c\leq 1$, $d\geq0$, and the payment limit is applied as defined above, each covered payment is no greater than its ground-up loss. For per-claim coverage, summing this inequality over claims gives $T\leq S$ outcome by outcome. An aggregate layer with nonnegative attachment and a payment fraction no greater than one likewise pays no more than the aggregate loss. Thus the covered-loss distribution cannot have larger quantiles than the corresponding ground-up loss. Its TVaR is also no larger, because the quantiles at every probability level are no larger.

This comparison depends on the specified coverage and loss basis. It does not mean that every risk measure can be obtained by applying the coverage formula to the same ground-up risk measure. A deductible or cap changes which outcomes occupy the tail, while coinsurance and limits can change the size of each tail outcome.

Do not confuse per-claim and aggregate terms. Transform each $X_j$ before summing for per-claim coverage. Sum the losses first for an aggregate layer.
Check the stated limit convention. In this tutorial, $u$ caps the amount after the deductible and before coinsurance; the maximum payment is $cu$. A policy limit defined on a different basis needs a different formula.
Respect atoms in VaR. Find the first value whose cumulative probability reaches $p$; do not interpolate across a probability jump unless a different convention is explicitly specified.
Respect atoms in TVaR. Use the quantile-integral definition or the expected-excess formula. Averaging only outcomes strictly above VaR can omit the part of a quantile atom needed in the upper tail.

Key takeaway: Write the coverage transformation precisely, construct the covered-loss distribution, and then calculate its VaR and TVaR. A quantile transformation is available under appropriate conditions, but TVaR generally requires a fresh upper-tail calculation.

Check Your Understanding

For each problem, distinguish the loss to which coverage applies and use the lower-quantile VaR convention.

A single loss $X$ is uniform on $[0,500]$ dollars. A policy has deductible $\$50$, payment limit $\$200$ after the deductible, and coinsurance $c=0.80$. Find ground-up and covered-payment VaR at $p=0.90$, and covered-payment TVaR at that level. (Answer: ground-up VaR $\$450$; covered VaR $\$160$; covered TVaR $\$160$.)
An aggregate loss $S$ is $\$0,\$50,\$150,\$250$ with probabilities $0.20,0.30,0.30,0.20$. An aggregate layer has attachment $\$40$ and payment limit $\$100$. Find the layer-payment VaR and TVaR at $p=0.70$. (Answer: VaR $\$100$; TVaR $\$100$.)
A ground-up aggregate loss has VaR at $p=0.95$ of $\$800{,}000$ and TVaR at that level of $\$1{,}100{,}000$. Coverage is pure 70% coinsurance, with no deductible or limit. Find both covered risk measures. (Answer: VaR $\$560{,}000$; TVaR $\$770{,}000$.)
A single loss $X$ is $\$0,\$100,\$400$ with probabilities $0.50,0.30,0.20$. Coverage has deductible $\$50$, payment limit $\$200$, and no coinsurance. Find ground-up and covered-payment VaR and TVaR at $p=0.60$. (Answer: ground-up VaR $\$100$, ground-up TVaR $\$250$; covered VaR $\$50$, covered TVaR $\$125$.)

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