TVaR For Compound Frequency-Severity Losses

From Claim Counts to Aggregate Losses

A risk measure applied to one claim does not automatically describe the risk of a full year of claims. Annual losses depend both on how many claims occur and on how large those claims are. TVaR for a compound frequency-severity model therefore starts with the distribution of the aggregate loss, not just the distribution of an individual claim.

Definition: Let $N$ be the number of claims in a fixed period, and let $X_1,X_2,\ldots$ be the claim severities. The aggregate loss is $S=\sum_{j=1}^{N}X_j$, with $S=0$ when $N=0$. In a standard compound model, $N$ is independent of the severities, and the severities are independent and identically distributed. Aggregate TVaR at confidence level $p$ is calculated from the distribution of $S$, not from the distribution of a single $X_j$.

For a discrete aggregate distribution, define $\operatorname{VaR}_p(S)=\inf\{s:F_S(s)\ge p\}$. TVaR is the average of the upper $1-p$ fraction of the distribution, including only the required fraction of probability at the VaR threshold when that threshold has positive probability. This convention is also called quantile-based TVaR; it avoids treating all outcomes tied at the threshold as if they were necessarily in the tail.

\operatorname{TVaR}_p(S) = \frac{1}{1-p}\int_p^1 \operatorname{VaR}_q(S)\,dq = \operatorname{VaR}_p(S) + \frac{E[(S-\operatorname{VaR}_p(S))_+]}{1-p},

Here $z_+=\max(z,0)$. The second form is useful when the aggregate distribution is known: calculate the VaR threshold, then calculate the expected amount by which aggregate loss exceeds it. The formula applies to discrete and continuous losses under the quantile-based definition above. In a discrete distribution, simply calculating $E[S\mid S>\operatorname{VaR}_p(S)]$ can give the wrong answer because it omits the necessary fraction of probability at the threshold.

Constructing the Compound Distribution

If claim severities are discrete, the distribution of $S$ can be built by conditioning on the claim count. For $n$ claims, the aggregate has the $n$-fold convolution of the severity distribution. For $n=0$, it has all its probability at zero.

P(S=s)=\sum_{n=0}^{\infty}P(N=n)\,P(X_1+\cdots+X_n=s),

The term for $n=0$ is $P(N=0)$ when $s=0$, and zero otherwise. In practical calculations, the count distribution may have a finite range, or a numerical method may approximate an infinite sum. The conditioning formula follows by partitioning the event $S=s$ according to the mutually exclusive possibilities $N=0,1,2,\ldots$: each conditional probability is weighted by the probability of its count.

Worked Example: Exact TVaR From a Small Compound Distribution

A hypothetical annual portfolio has $N\sim\operatorname{Binomial}(2,0.5)$. Conditional on the count, each claim independently has severity of either $10,000 or $30,000, each with probability 0.5. Assume the count is independent of severities. Find aggregate VaR and TVaR at $p=0.80$, reporting amounts in thousands of dollars.

Method. First combine the possible counts and claim severities into the distribution of $S$. Then locate the 80th percentile and average the upper 20% of aggregate outcomes.

The count probabilities are $P(N=0)=0.25$, $P(N=1)=0.50$, and $P(N=2)=0.25$. With no claims, $S=0$ with probability 0.25. With one claim, $S=10$ and $S=30$ each receive probability $0.50(0.50)=0.25$. With two claims, the sums $20,40,60$ have conditional probabilities $0.25,0.50,0.25$, so their unconditional probabilities are $0.0625,0.125,0.0625$.

Combining the outcomes and ordering them gives:

\begin{array}{c|rrrrrr} s & 0 & 10 & 20 & 30 & 40 & 60\\ \hline P(S=s) & 0.25 & 0.25 & 0.0625 & 0.25 & 0.125 & 0.0625\\ F_S(s) & 0.25 & 0.50 & 0.5625 & 0.8125 & 0.9375 & 1.0000 \end{array}

Since the cumulative probability is 0.5625 at 20 and 0.8125 at 30, $\operatorname{VaR}_{0.80}(S)=30$. The upper 20% consists of all probability at 40 and 60, totaling $0.125+0.0625=0.1875$, plus 0.0125 probability at the threshold 30. The weighted tail amount is $40(0.125)+60(0.0625)+30(0.0125)=5+3.75+0.375=9.125$ thousand dollars. Thus TVaR is $9.125/0.20=45.625$ thousand dollars.

As a check, the excess over the VaR threshold has expectation $10(0.125)+30(0.0625)=1.25+1.875=3.125$ thousand dollars. The excess formula gives $30+3.125/0.20=45.625$, matching the direct tail average. Common mistake: using only outcomes strictly greater than 30 gives $(40(0.125)+60(0.0625))/0.1875=46.6667$ thousand dollars. That conditional average excludes the threshold mass needed to make the tail probability exactly 0.20, so it is not quantile-based TVaR here.

Estimating TVaR From Simulated Aggregate Losses

When a compound distribution is too large or complex to construct directly, simulation can produce aggregate-loss observations. Each simulation trial should represent one full period: generate a claim count, generate that many severities, and sum them. The result is a sample of aggregate losses, on which the empirical quantile and empirical tail average can be calculated.

Worked Example: Monte Carlo Estimate From Annual Losses

A hypothetical model has annual frequency $N\sim\operatorname{Poisson}(1)$, independent across years. Every claim has a fixed severity of $10,000. A simulation of ten years produces claim counts $0,0,1,1,2,2,3,3,4,6$. Estimate aggregate VaR and TVaR at $p=0.80$, in thousands of dollars, using the empirical distribution with equal weight $1/10$ on each simulated year.

Method. Convert each count to an annual aggregate loss by multiplying by 10, then sort the ten aggregate losses. The empirical VaR is the smallest value whose cumulative sample proportion is at least 0.80.

The ordered aggregate losses are $0,0,10,10,20,20,30,30,40,60$. The first eight observations are at most 30, so the empirical cumulative proportion at 30 is $8/10=0.80$; hence the empirical VaR is 30 thousand dollars. The empirical upper 20% contains the two largest observations, 40 and 60. Its average is $(40+60)/2=50$ thousand dollars, so estimated TVaR is 50 thousand dollars.

Check using the excess formula: the average positive excess above 30 is $((40-30)+(60-30))/10=4$ thousand dollars. Therefore the estimate is $30+4/0.20=50$ thousand dollars. Common mistake: averaging the two largest individual claims is not an option here; the simulated observations are annual aggregate losses, and their count and severity have already been combined. This small simulation illustrates the calculation, not the precision that a larger run would provide.

Aggregate Tail Risk Is Not Claim-Level Tail Risk

The tail of an individual severity describes unusually large claims. The tail of aggregate loss also reflects claim count, sums of multiple claims, and the probability of no claims. Even when the same severity model is used, aggregate VaR and TVaR need not equal the corresponding claim-level measures or a simple multiple of them.

Worked Example: Allowing for Probability at the VaR Threshold

A hypothetical annual portfolio has $N\sim\operatorname{Binomial}(2,0.5)$. Independent claim severities are $10,000 with probability 0.75 and $50,000 with probability 0.25. Calculate aggregate TVaR at $p=0.80$, in thousands of dollars, and compare it with the TVaR of one individual claim.

Method. Calculate the aggregate distribution by count, find its 80th-percentile threshold, and take the appropriate upper-tail fraction. The count probabilities are 0.25, 0.50, and 0.25 for $N=0,1,2$, respectively.

For $N=0$, $S=0$ with probability 0.25. For $N=1$, the unconditional probabilities at 10 and 50 are $0.50(0.75)=0.375$ and $0.50(0.25)=0.125$. For $N=2$, the conditional sums 20, 60, and 100 have probabilities $0.75^2=0.5625$, $2(0.75)(0.25)=0.375$, and $0.25^2=0.0625$. Multiplying by $P(N=2)=0.25$ gives aggregate probabilities 0.140625, 0.09375, and 0.015625.

Thus the ordered aggregate distribution has probabilities $P(S=0)=0.25$, $P(S=10)=0.375$, $P(S=20)=0.140625$, $P(S=50)=0.125$, $P(S=60)=0.09375$, and $P(S=100)=0.015625$. The cumulative probability through 20 is 0.765625, and through 50 it is 0.890625. Therefore $\operatorname{VaR}_{0.80}(S)=50$.

The upper 20% consists of all probability above 50, namely 0.09375 at 60 and 0.015625 at 100, plus 0.090625 at 50. These probabilities total $0.09375+0.015625+0.090625=0.20$. The tail amount is $60(0.09375)+100(0.015625)+50(0.090625)=5.625+1.5625+4.53125=11.71875$ thousand dollars. Aggregate TVaR is $11.71875/0.20=58.59375$ thousand dollars.

For an individual claim, the 80th-percentile VaR is 50 thousand dollars because the severity CDF at 10 is only 0.75. The upper 20% of individual claims all have severity 50, so individual-claim TVaR is 50 thousand dollars. It is not the aggregate TVaR: the aggregate model combines both the count and the severity, producing a different loss distribution. Common mistake: substituting an individual-claim TVaR for aggregate TVaR ignores the possibility of multiple claims in a period.

Practical Checks

Before finalizing a compound-loss TVaR calculation, check that the count-severity assumptions match the model, that the aggregate probabilities sum to one, and that the tail probability used is $1-p$. For discrete losses, verify that the tail includes exactly $1-p$ probability, using a fractional amount at the VaR threshold if necessary.

Specify the period and model.
Identify the claim count distribution, severity distribution, and any independence assumptions. Keep the period consistent across all losses.

Obtain aggregate losses.
Use count conditioning and severity convolutions for an exact distribution, or simulate a count and its severities for each period.

Locate the aggregate VaR.
Use the aggregate CDF or ordered aggregate simulation results, not individual-claim ranks.

Average the correct tail.
Include the full probability above VaR and only the necessary fraction at VaR. Check the resulting tail probability is $1-p$.

These calculations are aligned with the CAS Exam 9 (Risk Management for Actuaries) syllabus because they connect frequency-severity modeling to tail-risk measurement. The central discipline is to define the loss variable being measured: for annual capital or portfolio risk, that variable is generally the aggregate loss for the period.

Check Your Understanding

For each problem, work from the aggregate-loss distribution or simulated aggregate observations. Amounts are in thousands of dollars unless stated otherwise.

A hypothetical portfolio has $N\sim\operatorname{Bernoulli}(0.4)$, and each claim is exactly 25. Find aggregate VaR and TVaR at $p=0.75$. (Answer: VaR $=25$; TVaR $=25$ thousand dollars.)
A hypothetical portfolio has $N\sim\operatorname{Bernoulli}(0.5)$. A single claim is 10 or 30, each with probability 0.5, independently of $N$. Construct the aggregate outcomes and find VaR and TVaR at $p=0.75$. (Answer: VaR $=10$; TVaR $=30$ thousand dollars.)
A hypothetical portfolio has $N\sim\operatorname{Binomial}(2,0.5)$ and fixed claim severity 12. Find the aggregate VaR and TVaR at $p=0.75$, accounting for the probability at the threshold. (Answer: VaR $=12$; TVaR $=24$ thousand dollars.)
Eight simulated aggregate annual losses, already ordered, are $0,5,10,10,15,20,30,50$. Using equal empirical weights and $p=0.75$, find empirical VaR and TVaR. (Answer: VaR $=20$; TVaR $=40$ thousand dollars.)
An aggregate loss has probabilities $P(S=0)=0.20$, $P(S=10)=0.30$, $P(S=20)=0.30$, and $P(S=50)=0.20$. Find VaR and TVaR at $p=0.75$, using the quantile-based definition. (Answer: VaR $=20$; TVaR $=44$ thousand dollars.)

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