Parametric VaR and TVaR from Aggregate Models

From a Fitted Aggregate Model to Tail Risk

A fitted aggregate-loss model turns assumptions about claim frequency and severity into a distribution for total loss. VaR identifies a threshold at a chosen confidence level; TVaR measures the average of the worst tail beyond that level. For a discrete aggregate model, however, the threshold itself can have positive probability. Correctly allocating only the needed fraction of that probability is essential to a consistent TVaR calculation.

The methods below apply when model parameters are treated as fixed at their fitted values. For a compound model, assume claim count $N$ is independent of the claim amounts $Y_1,Y_2,\ldots$, and that the claim amounts are identically distributed. The aggregate loss is $S=\sum_{j=1}^N Y_j$, with $S=0$ when $N=0$. These assumptions define the model; if the frequency and severity are dependent, or the claim amounts are not identically distributed, the compound formulas used here need not apply.

Definition: For a loss $S$ with distribution function $F_S(s)=P(S\leq s)$, the lower $\alpha$-quantile, or $\operatorname{VaR}_{\alpha}(S)$, is $$ q_\alpha=\inf\{s:F_S(s)\geq\alpha\}. $$ For $0<\alpha<1$, define TVaR by the upper-tail quantile average $$ \operatorname{TVaR}_{\alpha}(S)=\frac{1}{1-\alpha}\int_{\alpha}^{1}\operatorname{VaR}_{u}(S)\,du. $$ This convention includes only as much probability at the VaR threshold as is needed to make the upper tail have probability $1-\alpha$.

TVaR When the Aggregate Distribution Has Atoms

Let $q=q_\alpha$. Under the stated lower-quantile definition, $F_S(q-)\leq\alpha\leq F_S(q)$, where $F_S(q-)=P(S<q)$. The portion of the probability at $q$ included in the upper tail is $F_S(q)-\alpha$. The resulting formula is

\operatorname{TVaR}_{\alpha}(S) =\frac{E[S\mathbf{1}_{\{S>q\}}]+q\big(F_S(q)-\alpha\big)}{1-\alpha}.

The first term totals losses strictly greater than $q$. The second adds the required fraction of the probability at $q$. If the distribution is continuous at $q$, then $F_S(q)=\alpha$, and the boundary contribution is zero; in that case TVaR is the conditional mean above the quantile, when that conditional expectation is defined.

Find the quantile.
Use the fitted aggregate distribution to locate the smallest $q$ whose cumulative probability is at least $\alpha$.

Separate the strict tail and the boundary.
Calculate $E[S\mathbf{1}_{\{S>q\}}]$ and $F_S(q)-\alpha$. Do not use all of $P(S=q)$ unless that is exactly the needed boundary probability.

Normalize by the tail probability.
Add the strict-tail moment to $q$ times the boundary probability, then divide by $1-\alpha$.

The formula follows directly from the quantile-average definition. For $u$ between $F_S(q-)$ and $F_S(q)$, the quantile is $q$; this interval has length $P(S=q)$. The portion of that interval lying above $\alpha$ has length $F_S(q)-\alpha$, so its contribution to the integral is $q(F_S(q)-\alpha)$. For $u>F_S(q)$, the quantiles correspond to losses strictly greater than $q$. Their integral is $E[S\mathbf{1}_{\{S>q\}}]$, the upper-tail quantile representation of the strict-tail first moment. Adding the two pieces and dividing by the integration interval length $1-\alpha$ gives the stated formula. In particular, a discrete distribution does not generally have $\operatorname{TVaR}_{\alpha}(S)=E[S\mid S\geq q]$: that conditional mean includes the entire boundary mass, not just the required fraction.

Worked Examples

Worked Example: Compound Poisson Loss With a Fixed Claim Amount

A hypothetical portfolio has annual claim count $N\sim\operatorname{Poisson}(2)$. Every claim is exactly $1,000, and claim counts follow the stated Poisson model. Find the 90% VaR and TVaR of annual aggregate loss. Give money to the nearest cent.

Method. Since every claim is $1,000, write aggregate loss in thousands of dollars as $S/1000=N$. For a Poisson random variable with mean 2, $P(N=k)=e^{-2}2^k/k!$. The cumulative probabilities needed to locate the 90th percentile are

F_S(3)=P(N\leq3)\approx0.8571235,\qquad F_S(4)=P(N\leq4)\approx0.9473470.

Since $F_S(3)<0.90$ and $F_S(4)\geq0.90$, the quantile is $q=4$ thousand dollars. Thus $\operatorname{VaR}_{0.90}(S)=\$4,000.00$.

For a Poisson variable, $E[N\mathbf{1}_{\{N>4\}}]=2P(N\geq4)$. This identity follows by shifting the Poisson probability index in the sum of $kP(N=k)$ over $k\geq5$. Therefore, in thousands of dollars,

E[N\mathbf{1}_{\{N>4\}}] =2\big(1-F_S(3)\big) \approx 0.2857530790.

The probability at the boundary to include in the upper 10% is $F_S(4)-0.90\approx0.0473469827$. Apply the atom formula:

\operatorname{TVaR}_{0.90}(S) =\frac{0.2857530790+4(0.0473469827)}{0.10} \approx4.7514101\text{ thousand dollars}.

Rounded to the nearest cent, $\operatorname{TVaR}_{0.90}(S)=\$4,751.41$. The strict-tail moment and boundary term are in thousands of dollars before division. The result exceeds the VaR, as an upper-tail average should. The common mistake avoided here is truncating the cumulative probabilities or using the full probability at $N=4$, rather than the fraction needed to complete the upper 10% tail.

Worked Example: Recursion, a Discrete Severity Model, and a Boundary Atom

A hypothetical insurer models claim count as $N\sim\operatorname{Poisson}(1)$, independent of claim amounts. Each claim is $1,000 or $2,000 with probability $1/2$ each. Use Panjer recursion to find the 90% VaR and TVaR. Report money to the nearest cent.

Method and units. Measure aggregate loss in thousands of dollars, so each claim amount $Y$ is 1 or 2, each with probability $f_1=f_2=0.5$. Let $p_k=P(S=k)$ in these units. For a compound Poisson model with parameter $\lambda$, the recursion is $p_0=e^{-\lambda}$ and $p_k=(\lambda/k)\sum_{j=1}^{k}j f_jp_{k-j}$. Here $\lambda=1$, so $p_0=e^{-1}\approx0.3678794412$.

For $k=1$, only $j=1$ contributes; for $k=2,3,4$, substitute the nonzero severity probabilities:

\begin{aligned} p_1&=0.5p_0=0.1839397206,\\ p_2&=\frac{0.5p_1+1p_0}{2}=0.2299246508,\\ p_3&=\frac{0.5p_2+1p_1}{3}=0.0996340153,\\ p_4&=\frac{0.5p_3+1p_2}{4}=0.0699354146. \end{aligned}

Summing the probabilities gives $F_S(3)=p_0+p_1+p_2+p_3\approx0.8813778279$ and $F_S(4)=F_S(3)+p_4\approx0.9513132425$. Thus the 90% quantile is 4 thousand dollars, or $\operatorname{VaR}_{0.90}(S)=\$4,000.00$. The boundary probability needed is $F_S(4)-0.90\approx0.0513132425$.

The compound Poisson mean is $E[S]=\lambda E[Y]=1(1.5)=1.5$ thousand dollars. The first moment from outcomes at or below 4 is

1p_1+2p_2+3p_3+4p_4 \approx0.1839397206+0.4598493015+0.2989020460+0.2797416585 =1.2224327266.

Subtracting this from the total mean gives $E[S\mathbf{1}_{\{S>4\}}]\approx1.5-1.2224327266=0.2775672734$ thousand dollars. Therefore

\operatorname{TVaR}_{0.90}(S) =\frac{0.2775672734+4(0.0513132425)}{0.10} \approx4.8282024\text{ thousand dollars}.

The result is $\$4,828.20$ to the nearest cent. It is greater than the VaR, and the numerator accounts for the strict tail plus only the boundary mass required by the quantile convention. A common error is to include all of $p_4$ in the tail; that would describe a different tail probability and would not be the specified 90% TVaR.

Worked Example: Normal Approximation to a Compound Aggregate Model

A hypothetical portfolio has $N\sim\operatorname{Poisson}(2)$ claims per year, independent of claim amounts. Each claim is exactly $1,000. Use a normal approximation to estimate 95% VaR and TVaR. Use $z_{0.95}=1.644854$ and $\phi(z_{0.95})=0.103136$, where $\phi$ is the standard normal density. Report money to the nearest cent.

Method. For a compound Poisson aggregate, $E[S]=\lambda E[Y]$ and $\operatorname{Var}(S)=\lambda E[Y^2]$, provided the count is Poisson and independent of identically distributed severities. Thus

\mu_S=2(1000)=\$2,000,\qquad \sigma_S^2=2(1000^2)=2,000,000, \qquad \sigma_S=\sqrt{2,000,000}\approx\$1,414.2136.

Approximate $S$ by a normal random variable with this mean and variance. The normal 95th percentile is $\mu_S+\sigma_S z_{0.95}$:

\operatorname{VaR}_{0.95}(S) \approx 2000+1414.2136(1.644854) \approx \$4,326.17.

For a continuous normal distribution, the upper-tail conditional mean is $\mu_S+\sigma_S\phi(z_{0.95})/(1-0.95)$. Hence

\operatorname{TVaR}_{0.95}(S) \approx2000+1414.2136\frac{0.103136}{0.05} \approx \$4,917.13.

Both values are approximations, not exact quantiles or tail averages of the original discrete compound Poisson model. The calculation is internally ordered: the estimated TVaR exceeds the estimated VaR. A common error is using the claim count parameter as the aggregate mean in dollars; the mean must include the claim amount, and the variance uses $E[Y^2]$, not just the claim count variance.

Choosing and Checking a Calculation Method

For a discrete severity distribution on a regular monetary grid, recursion can produce aggregate probabilities; cumulative sums then locate VaR, while tail probabilities or first moments supply the TVaR numerator. For small models, direct convolution or conditioning may be practical as well. If the aggregate distribution is available only through numerical calculation, sum the strict-tail contributions and add the boundary term explicitly. A normal approximation can be useful when it is appropriate to the model and scale, but its continuous TVaR formula does not correct for atoms in the original distribution.

Simulation provides another estimate when an analytic or recursive distribution is inconvenient. Generate many aggregate losses from the fitted frequency and severity model, sort them, and estimate the quantile from the ordered losses. For TVaR, use a tail average consistent with the chosen quantile convention. If simulated losses tie at the estimated quantile, averaging every observation at or above that threshold may include too much boundary mass. Fractional weighting, or an equivalent empirical quantile-integral calculation, preserves the convention. Simulation error is a separate issue from this boundary calculation.

Check the units. If recursion uses thousands of dollars, both the strict-tail moment and the boundary contribution are in thousands until converted.
Check the quantile. Under the lower-quantile convention, $F_S(q-)\leq\alpha\leq F_S(q)$; a value with $F_S(q)<\alpha$ cannot be the VaR.
Check the tail amount. The strict-tail moment and the quantile boundary term must describe a total probability of $1-\alpha$.
Check the ordering. For a nonnegative loss with finite upper-tail mean, TVaR should not be below the corresponding VaR.

These calculations are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus. In exam-style problems, stating the quantile convention and showing the boundary contribution makes it possible to distinguish a correct tail calculation from a numerical answer that uses a different definition.

Check Your Understanding

For each problem, identify the quantile convention and keep track of the units before reporting the result.

A hypothetical portfolio has $N\sim\operatorname{Poisson}(1)$, with every claim exactly $250. At confidence level 80%, find VaR and TVaR. Round money to the nearest cent. (Answer: VaR $=\$500.00$; TVaR $\approx\$629.55$)
A fitted aggregate model has $F_S(6)=0.91$, $F_S(7)=0.97$, and $E[S\mathbf{1}_{\{S>7\}}]=0.24$, with losses in thousands of dollars. At confidence level 95%, find VaR and TVaR. (Answer: VaR $=\$7,000.00$; TVaR $=\$7,600.00$)
A compound Poisson model has $N\sim\operatorname{Poisson}(2)$, independent of fixed $1,000 claims. Use a normal approximation and $z_{0.95}=1.644854$, $\phi(z_{0.95})=0.103136$ to estimate 95% VaR and TVaR. Round money to the nearest cent. (Answer: approximate VaR $=\$4,326.17$; approximate TVaR $=\$4,917.13$)
At confidence level 90%, a discrete fitted model has $q=\operatorname{VaR}_{0.90}=12$, $F_S(12)=0.94$, and $E[S\mathbf{1}_{\{S>12\}}]=0.78$. Find TVaR in the same loss units. (Answer: $\operatorname{TVaR}_{0.90}=12.6$)
A hypothetical compound Poisson model has $N\sim\operatorname{Poisson}(1)$ and every claim is exactly $500. Use the lower-quantile convention to find the 90% VaR and TVaR. Round money to the nearest cent. (Answer: VaR $=\$1,000.00$; TVaR $\approx\$1,518.19$)

What You'll Learn