Integrated Data-to-Risk Case Workshop

From Incomplete Records to a Risk Decision

An aggregate-risk estimate is only as defensible as the path from data to model. In an integrated analysis, an actuary must identify what the records actually reveal, estimate parameters under an explicit model, apply the coverage terms in the correct order, and calculate risk measures using a stated quantile convention. A limit can make payments observable even when the underlying losses remain censored; confusing those two quantities can lead to unjustified conclusions about a different coverage.

This hypothetical case follows a portfolio of policy-years with a per-loss layer attaching at $100 and paying at most $200 per loss. The insurer’s payment is therefore $Y=\min((X-100)_+,200)$, where $X$ is the underlying loss and $(z)_+=\max(z,0)$. An observed payment of $200 tells us that $X\geq 300$, not that $X=300$. The payment itself, however, is known exactly.

Model and conventions: Treat each policy-year as an independent exposure. Let $q$ be the probability that an exposure produces a positive payment under the stated layer. Conditional on a positive payment, the payment is $100 with probability $\theta$ and $200 with probability $1-\theta$. For the aggregate model below, the portfolio has two such exposures, independently governed by the same fitted parameters. Use the left-quantile convention $\operatorname{VaR}_u(S)=\inf\{s:F_S(s)\geq u\}$, and define $\operatorname{TVaR}_u(S)=(1/(1-u))\int_u^1\operatorname{VaR}_r(S)\,dr$.

Step 1: Interpret the Censored Records

Suppose the hypothetical data contain 20 policy-years: 12 have zero payment, 4 have a $100 payment, and 4 have a $200 payment. The last four payments are capped by the layer limit. They establish that the corresponding underlying losses were at least $300, but they do not reveal how far above $300 those losses were.

The data support estimation of the probability of a positive payment and the conditional distribution of payments under this layer. They do not, by themselves, support an estimate of the underlying-loss distribution above $300. In particular, treating every capped loss as exactly $300 would add information that the records do not contain.

Worked Example: Estimate the Covered-Payment Model

Use the 20 policy-years above to estimate $q$ and $\theta$. Also calculate approximate standard errors for both estimates. State what the capped observations do and do not tell us about the underlying losses.

Method. Estimate $q$ by the fraction of exposures with positive payment. Estimate $\theta$ by the fraction of positive payments that equal $100. For a sample proportion $\hat p$ based on $n$ independent observations, use the estimated standard error $\sqrt{\hat p(1-\hat p)/n}$.

There are $4+4=8$ positive payments among 20 exposures, so \[ \hat q=\frac{8}{20}=0.40. \] There are four $100 payments among those eight positive payments, so \[ \hat\theta=\frac{4}{8}=0.50. \] The estimated standard error of $\hat q$ is \[ \sqrt{\frac{0.40(1-0.40)}{20}} =\sqrt{0.012} \approx 0.1095. \] The estimated standard error of $\hat\theta$, using the eight positive payments as the conditional sample, is \[ \sqrt{\frac{0.50(1-0.50)}{8}} =\sqrt{0.03125} \approx 0.1768. \]

Thus the fitted payment model assigns probability 0.60 to zero payment, 0.20 to a $100 payment, and 0.20 to a $200 payment for a policy-year. The relatively large standard errors reflect the modest sample sizes. The four capped records reveal that their underlying losses were at least $300, but they cannot identify the distribution of those losses above $300. The model is adequate for this capped payment layer as specified; it is not an estimate of uncapped severity.

Step 2: Build the Aggregate Distribution

Let $N$ be the number of positive payments in the two-exposure portfolio. Under the independent, common-probability assumptions, $N\sim\operatorname{Bin}(2,\hat q)$, and the aggregate payment $S$ is the sum of the two exposure payments. Conditional on a positive payment, the fitted payment is $100 or $200 with equal probability. The aggregate distribution can be built by conditioning on $N$: first find the probability of each claim count, then add the corresponding payments.

Result: For independent exposures with positive-payment probability $q$, the count of positive payments in $m$ exposures is binomial: $N\sim\operatorname{Bin}(m,q)$. If positive payments have a specified discrete distribution, the aggregate distribution is obtained by mixing the $n$-payment sums over $P(N=n)$. This compound binomial construction assumes independent exposures and the same payment model for each exposure.

Worked Example: Calculate the Two-Exposure Aggregate Distribution

Using $\hat q=0.40$ and equal conditional probabilities of $100 and $200, find the distribution of aggregate payment $S$ for two independent exposures.

Method. Condition on the number of positive payments. The count probabilities are \[ P(N=0)=0.60^2=0.36,\qquad P(N=1)=2(0.40)(0.60)=0.48,\qquad P(N=2)=0.40^2=0.16. \]

If $N=0$, then $S=0$, contributing probability 0.36. If $N=1$, the sole payment is $100 or $200 with conditional probability 0.50 each. This contributes probability $0.48(0.50)=0.24$ at each of $100 and $200.

If $N=2$, each payment is independently $100 or $200 with probability 0.50. Both payments are $100 with probability $0.50^2=0.25$, giving an additional probability $0.16(0.25)=0.04$ at $200. One payment of each amount has probability $2(0.50)(0.50)=0.50$, giving probability $0.16(0.50)=0.08$ at $300. Both payments are $200 with probability $0.25$, giving probability $0.16(0.25)=0.04$ at $400.

Combining the contributions gives \[ \begin{array}{c|ccccc} s & 0 & 100 & 200 & 300 & 400\\ \hline P(S=s) & 0.36 & 0.24 & 0.28 & 0.08 & 0.04 \end{array} \] The probabilities sum to $0.36+0.24+0.28+0.08+0.04=1.00$, as required. This check catches omissions when the $N=1$ and $N=2$ cases both contribute to the same aggregate amount.

Step 3: Apply Discrete Quantile Conventions

For a discrete loss, the cumulative distribution function can jump over the target probability. Under the stated left-quantile convention, VaR is the smallest loss whose cumulative probability is at least the confidence level. Integrated-tail TVaR averages the quantiles over the upper probability interval. If the interval begins inside a probability mass at VaR, only the required fraction of that atom belongs to the tail average.

Derivation: Let $v=\operatorname{VaR}_u(S)$. By the left-quantile definition, the probability below $v$ is at most $u$, while the probability at or below $v$ is at least $u$. Thus the upper tail of probability $1-u$ contains all probability strictly above $v$, plus just enough probability at $v$ to make its total mass $1-u$. The integrated quantile $\int_u^1\operatorname{VaR}_r(S)\,dr$ is the total loss in that upper tail, including the required fractional mass at $v$. Dividing by $1-u$ gives the average loss in the tail. As $r$ increases across the upper probability interval, quantile values occur in ascending order of loss; equivalently, the same tail can be listed from largest loss downward in descending order.

Worked Example: Calculate VaR and TVaR from the Aggregate Distribution

For the aggregate distribution just calculated, find $\operatorname{VaR}_{0.90}(S)$ and integrated-tail $\operatorname{TVaR}_{0.90}(S)$.

Method. Accumulate the probability from the smallest loss upward to find the left quantile. Then average the upper 10% of outcomes, taking only the required fraction of the atom at the VaR if the upper tail starts within it.

The cumulative probabilities are \[ F_S(0)=0.36,\quad F_S(100)=0.60,\quad F_S(200)=0.88,\quad F_S(300)=0.96,\quad F_S(400)=1.00. \] Since $F_S(200)=0.88<0.90$ and $F_S(300)=0.96\geq0.90$, the left-quantile VaR is $\operatorname{VaR}_{0.90}(S)=\$300$.

The probability strictly above $300 is 0.04, all at $400. The upper 10% therefore contains all 0.04 probability at $400 and 0.06 probability at $300. Its total loss is \[ 0.04(400)+0.06(300)=16+18=34 \] in probability-weighted dollars. Dividing by the tail probability gives \[ \operatorname{TVaR}_{0.90}(S)=\frac{34}{0.10}=\$340. \]

As a check, the upper-tail weights $0.04+0.06=0.10$ equal the required tail probability. The method avoids treating the entire 0.08 atom at $300 as part of the upper 10%.

Coverage Changes and What the Data Can Support

A risk measure belongs to a specified loss definition and coverage. Here, the fitted payment outcomes are $0, $100, and $200 under an attachment of $100 and a maximum payment of $200. Changing the layer changes the payment variable. If the new payment depends on distinctions hidden by censoring, the existing records may not be enough to estimate it.

Worked Example: Compare a Lower Payment Limit

Suppose the insurer instead considers paying $\min(X,100)$ per loss, with no attachment and a $100 payment limit. Use the observed layer records to determine what can be concluded about the two-exposure aggregate under this alternative. State any limitation.

Method. Interpret each original payment category in terms of the underlying loss. A $100 original layer payment means $X=200$ under the specified attachment. An original payment of $200 means $X\geq300$. Both imply $X\geq100$, so either positive-payment category produces exactly $100 under the alternative coverage. A zero original payment only establishes $X\leq100$; at the boundary $X=100$, the alternative payment is $100, while for $X<100$ it is less. Therefore the records as stated do not identify the alternative payment for every zero-payment exposure.

To make a calculation, explicitly impose the additional assumption that every zero original payment corresponds to $X=0$, so its alternative payment is zero. Under that assumption, each exposure pays $100 with probability 0.40 and zero otherwise. For two exposures, aggregate payments are 0, $100, and $200 with probabilities $0.60^2=0.36$, $2(0.40)(0.60)=0.48$, and $0.40^2=0.16$, respectively. The cumulative probability at $100 is 0.84, so $\operatorname{VaR}_{0.90}(S)=\$200$. The top 10% lies entirely within the 0.16 probability at $200, giving $\operatorname{TVaR}_{0.90}(S)=\$200$.

This result is conditional on the stated extra assumption about zero payments. Without information distinguishing underlying losses below or at $100, the alternative aggregate distribution is not identified by these records alone. The calculation avoids silently assuming that a zero payment means a zero underlying loss.

Sampling Uncertainty and Reporting

The fitted aggregate distribution treats $\hat q$ and $\hat\theta$ as if they were known. They are estimates from finite data. The standard errors calculated earlier show that both have substantial sampling variability, and uncertainty in these parameters can change the aggregate probabilities and, in turn, VaR and TVaR. In discrete models, a small parameter change can move a cumulative probability across the confidence level, changing VaR by an entire payment increment.

For a quick approximate uncertainty summary, a Wald interval for a proportion estimate $\hat p$ is $\hat p\pm1.96\sqrt{\hat p(1-\hat p)/n}$. For the positive-payment probability, this gives $0.40\pm1.96(0.1095)$, or approximately $[0.185,0.615]$. This approximation is imprecise for small samples and is not a confidence interval for VaR itself. To assess uncertainty in a risk measure, an analyst would need to propagate parameter uncertainty through the aggregate model, for example by repeatedly simulating parameter values and recalculating the risk measure. Any such result should state its method and assumptions.

A useful report separates what is directly observed from what is model-based: the exposure counts and capped payments are observed; $q$, $\theta$, and the aggregate risk measures are estimated or calculated under stated assumptions. It should also identify the payment layer, the exposure period, independence assumptions, the left-quantile convention, and whether uncertainty in fitted parameters is included. These details make the conclusion interpretable and prevent a capped-payment estimate from being mistaken for an uncapped-loss estimate.

Common Exam Traps

Equating a capped payment with the underlying loss. An insurer payment of $200 under this layer indicates $X\geq300$, not $X=300$. Do not infer a tail above the cap without additional data or assumptions.
Mixing conditional and unconditional probabilities. The estimated probability of a $100 payment per exposure is $0.40(0.50)=0.20$, not 0.50. The latter is conditional on a positive payment.
Adding count probabilities without conditioning on payment sizes. Knowing $P(N=2)$ does not put all that probability at one aggregate amount; the two payment amounts must also be combined.
Using the wrong mass at the VaR in TVaR. Integrated-tail TVaR includes only the fraction of an atom needed to fill the upper tail, not automatically the entire atom.
Changing coverage without checking identifiability. A new attachment or limit can require distinctions that the censored records do not reveal. State any additional assumption before calculating.

Key takeaway: An integrated data-to-risk analysis must preserve the distinction between observed payments and latent losses, carry the fitted model through the aggregate calculation, and state the quantile convention and uncertainty treatment. The numerical risk measure is meaningful only for the data, coverage, and assumptions that produced it.

Check Your Understanding

Use the stated conventions and show the conditioning or tail-mass calculation needed for each result.

A hypothetical set of 30 exposures has 21 zero payments, 6 payments of $50, and 3 payments at a layer cap of $150. Estimate the probability of a positive covered payment and the conditional probability of a $50 payment. What does a capped payment establish about the underlying loss? (Answer: $\hat q=0.3000$; conditional probability $=0.6667$; the underlying loss is at least the payment cap plus the attachment, but is not known exactly.)
Let $N\sim\operatorname{Bin}(2,0.50)$, and suppose every positive payment is exactly $100. Find the aggregate distribution, $\operatorname{VaR}_{0.75}(S)$, and $\operatorname{TVaR}_{0.75}(S)$. (Answer: $P(S=0)=0.25,\ P(S=100)=0.50,\ P(S=200)=0.25;\ \operatorname{VaR}_{0.75}(S)=\$100;\ \operatorname{TVaR}_{0.75}(S)=\$200$.)
A discrete aggregate loss has probabilities 0.50 at $0, 0.30 at $100, and 0.20 at $500. Find $\operatorname{VaR}_{0.80}(S)$ and $\operatorname{TVaR}_{0.80}(S)$ under the left-quantile and integrated-tail conventions. (Answer: $\operatorname{VaR}_{0.80}(S)=\$100;\ \operatorname{TVaR}_{0.80}(S)=\$500$.)
Let $N\sim\operatorname{Bin}(2,0.25)$. Conditional on a positive payment, payment is $100 or $200 with equal probability, independently across exposures. Find the aggregate distribution and the 90% VaR and TVaR. (Answer: probabilities at $0,100,200,300,400$ are $0.5625,0.1875,0.203125,0.03125,0.015625$; $\operatorname{VaR}_{0.90}(S)=\$200;\ \operatorname{TVaR}_{0.90}(S)=\$262.50$.)
For $n=100$ independent observations with 25 positive payments, calculate the estimated standard error of the positive-payment probability and its approximate 95% Wald interval. (Answer: $\hat q=0.25$; standard error $=\sqrt{0.25(0.75)/100}\approx0.0433$; interval $[0.1651,0.3349]$.)

What You'll Learn