Worked Problems in Aggregate Tail-Risk Calculation

Choosing a Method for Aggregate Tail Risk

A tail-risk calculation starts with the aggregate loss distribution, not with a formula chosen in isolation. The distribution may be given as a table, built from claim frequency and severity, or approximated by a continuous distribution. The correct value-at-risk (VaR) and tail value-at-risk (TVaR) calculation depends on which of these situations applies. A particular difficulty arises for discrete losses: the confidence-level probability may fall inside a point mass, so the tail must include only part of that mass.

All examples below concern hypothetical one-year aggregate losses. Loss amounts are undiscounted, and no mortality or interest assumptions are needed. When a model is used, its frequency, severity, and dependence assumptions are stated explicitly.

Definition: For an aggregate loss $X$ with distribution function $F_X(x)=P(X\le x)$, the value-at-risk at confidence level $p$, where $0<p<1$, is the lower quantile $$ \operatorname{VaR}_p(X)=\inf\{x:F_X(x)\ge p\}. $$ When $X$ has a finite mean, its tail value-at-risk is $$ \operatorname{TVaR}_p(X)=\frac{1}{1-p}\int_p^1 \operatorname{VaR}_q(X)\,dq. $$ This definition applies to both continuous and discrete distributions. For a continuous distribution, it agrees with $E[X\mid X>\operatorname{VaR}_p(X)]$. That conditional-expectation expression does not generally give the correct TVaR for a discrete distribution.

Discrete Distributions: Include the Right Part of the Quantile Mass

Suppose $v=\operatorname{VaR}_p(X)$ is a discrete loss value. By the lower-quantile definition, $F_X(v-)<p\le F_X(v)$, where $F_X(v-)=P(X<v)$. The upper tail with probability $1-p$ contains all losses greater than $v$, plus just enough probability at $v$ to make the tail probability equal to $1-p$.

\operatorname{TVaR}_p(X) = \frac{E[X\,\mathbf{1}_{\{X>v\}}]+v\bigl(F_X(v)-p\bigr)}{1-p}.

The term $F_X(v)-p$ is the amount of probability at $v$ that belongs to the upper tail under the quantile-integral definition. The formula works whether or not $p$ coincides with a cumulative probability. It also shows why simply averaging observations strictly greater than VaR can overstate TVaR: that calculation discards the probability mass at VaR that belongs to the upper tail.

Identify the distributional information.
Use a supplied table directly, derive the distribution from a frequency-severity model, or use a stated approximation only when its assumptions are appropriate.

Find the quantile.
Locate the first loss value $v$ for which $F_X(v)\ge p$. For a continuous distribution, solve $F_X(v)=p$.

Calculate the tail average.
For a discrete distribution, include all loss probability above $v$ and the required fraction at $v$. For a continuous distribution, integrate the quantile or use a valid conditional-tail formula.

Check the result.
TVaR should be at least VaR for an integrable loss distribution. For a discrete distribution, verify that the tail probability used in the numerator totals exactly $1-p$.

Worked Calculations

Worked Example: VaR and TVaR from a Tabulated Aggregate Distribution

A hypothetical one-year warranty portfolio has the following aggregate loss distribution, in dollars: $X=0, 50, 100, 200,$ or $500$, with respective probabilities $0.10, 0.20, 0.30, 0.25,$ and $0.15$. Calculate $\operatorname{VaR}_{0.80}(X)$ and $\operatorname{TVaR}_{0.80}(X)$. Use the tabulated distribution directly; no separate claim-frequency or severity model is assumed.

Method. Accumulate the tabulated probabilities to find the quantile, then use the discrete-tail formula. This avoids treating the probability at VaR as either wholly inside or wholly outside the tail.

The cumulative probabilities are

F_X(0)=0.10,\quad F_X(50)=0.30,\quad F_X(100)=0.60,\quad F_X(200)=0.85,\quad F_X(500)=1.00.

The first cumulative probability at least $0.80$ occurs at $200$, so $\operatorname{VaR}_{0.80}(X)=200$. Here $F_X(200)-0.80=0.85-0.80=0.05$ of the probability mass at $200$ is included in the upper tail. The probability strictly above $200$ is $0.15$, all at $500$. Therefore,

\operatorname{TVaR}_{0.80}(X) = \frac{500(0.15)+200(0.05)}{1-0.80} = \frac{75+10}{0.20} = 425.

The upper tail consists of probability $0.15$ at $500$ and probability $0.05$ at $200$, totaling $0.20$. Its average loss is $425$, which exceeds the VaR of $200$, as expected.

Worked Example: Constructing an Aggregate Distribution from Frequency and Severity

A hypothetical one-year device-repair plan has claim count $N\sim\operatorname{Bernoulli}(0.40)$: there is either no claim or one claim. Conditional on a claim, its severity $Y$ is $100$ dollars with probability $0.75$ and $300$ dollars with probability $0.25$. The count and severity mechanism is as specified, and there are no other payments. Calculate $\operatorname{VaR}_{0.80}(X)$ and $\operatorname{TVaR}_{0.80}(X)$ for aggregate loss $X$.

Method. First combine the frequency and severity probabilities to obtain the aggregate distribution, then use the discrete-tail formula. The common mistake to avoid is calculating $E[X\mid X>\operatorname{VaR}]$ as TVaR without checking for a point mass at VaR.

The probability of no claim, and hence zero aggregate loss, is $P(X=0)=1-0.40=0.60$. For one claim of $100$, the aggregate probability is $0.40(0.75)=0.30$. For one claim of $300$, it is $0.40(0.25)=0.10$. Thus,

P(X=0)=0.60,\qquad P(X=100)=0.30,\qquad P(X=300)=0.10.

The cumulative probability at $0$ is $0.60$, and at $100$ it is $0.60+0.30=0.90$. Therefore $\operatorname{VaR}_{0.80}(X)=100$. The probability strictly above $100$ is $0.10$, and the probability at $100$ included in the tail is $F_X(100)-0.80=0.90-0.80=0.10$. Hence,

\operatorname{TVaR}_{0.80}(X) = \frac{300(0.10)+100(0.10)}{0.20} = \frac{30+10}{0.20} = 200.

A strict conditional calculation would give $E[X\mid X>100]=300$, because the only loss strictly greater than $100$ is $300$. That is not TVaR here: it ignores the $0.10$ probability at $100$ needed to fill the upper tail of size $0.20$. The correct tail average is $200$.

Worked Example: Continuous Aggregate Loss Using a Normal Approximation

For a hypothetical portfolio, suppose the one-year aggregate loss is approximated by $X\sim N(\mu,\sigma^2)$, with $\mu=\$1{,}200{,}000$ and $\sigma=\$300{,}000$. This is an approximation to the aggregate distribution, not an assertion that actual losses can be negative. Calculate VaR and TVaR at confidence level $0.95$. Report amounts in thousands of dollars to three decimal places.

Method. Apply the continuous normal quantile and tail-mean formulas. The approximation is appropriate only insofar as a normal model is adequate for the aggregate loss in question; it should not be imposed automatically on a strongly skewed or heavy-tailed aggregate distribution.

Let $Z=(X-\mu)/\sigma$, so $Z\sim N(0,1)$. The standard normal 95th percentile is $z_{0.95}=1.644854$. Therefore,

\operatorname{VaR}_{0.95}(X) = \mu+\sigma z_{0.95} = 1{,}200{,}000+300{,}000(1.644854) = 1{,}693{,}456.2.

For a standard normal variable, the upper-tail conditional mean above $z_p$ is $\phi(z_p)/(1-p)$, where $\phi$ is the standard normal density. At $z_{0.95}=1.644854$, $\phi(z_{0.95})\approx0.103136$. Thus,

\operatorname{TVaR}_{0.95}(X) = \mu+\sigma\frac{\phi(z_{0.95})}{0.05} = 1{,}200{,}000+300{,}000\left(\frac{0.103136}{0.05}\right) \approx 1{,}818{,}816.

In thousands of dollars, the results rounded to three decimal places are VaR $=1{,}693.456$ and TVaR $=1{,}818.816$. TVaR exceeds VaR, consistent with the fact that it averages losses in the upper five percent of the continuous distribution.

Why the Discrete Formula Works

The quantile-integral definition gives a direct justification. For $q$ between $F_X(v-)$ and $F_X(v)$, the lower quantile $\operatorname{VaR}_q(X)$ equals $v$. Since $F_X(v-)<p\le F_X(v)$, integrating from $p$ through $F_X(v)$ contributes $v(F_X(v)-p)$. Quantiles above $F_X(v)$ account for the outcomes strictly greater than $v$; their total contribution to the quantile integral is $E[X\mathbf{1}_{\{X>v\}}]$. Dividing the resulting tail integral by its length $1-p$ yields

\operatorname{TVaR}_p(X) = \frac{v(F_X(v)-p)+E[X\mathbf{1}_{\{X>v\}}]}{1-p}.

The same reasoning explains the continuous formula. If $X$ is continuous with distribution function $F_X$, then $\operatorname{VaR}_q(X)=F_X^{-1}(q)$. In the quantile integral, substitute $q=F_X(x)$, so $dq=f_X(x)\,dx$. The range $q\in[p,1]$ corresponds to $x\ge v=\operatorname{VaR}_p(X)$. Consequently,

\int_p^1 \operatorname{VaR}_q(X)\,dq = \int_v^\infty x f_X(x)\,dx = E[X\mathbf{1}_{\{X>v\}}],

and division by $1-p=P(X>v)$ gives $E[X\mid X>v]$. This derivation relies on continuity at the quantile; with an atom at $v$, strict exceedance does not have probability $1-p$, which is why the discrete adjustment is necessary.

Common Exam Traps and Method Checks

Using the wrong quantile convention. With the lower-quantile definition, choose the smallest $x$ for which $F_X(x)\ge p$. Do not interpolate between table values unless the model specifies a continuous distribution that supports interpolation.
Dropping the atom at VaR. For a discrete distribution, the tail generally contains part of the probability at VaR. Use $F_X(v)-p$, not zero and not the entire probability $P(X=v)$.
Confusing confidence level with tail probability. At confidence level $p=0.95$, the upper tail has probability $1-p=0.05$. Dividing the tail expectation by $p$ would use the wrong portion of the distribution.
Applying a continuous formula to a discrete model. The identity $\operatorname{TVaR}_p(X)=E[X\mid X>\operatorname{VaR}_p(X)]$ applies when the distribution is continuous at the quantile. Check for atoms before using it.
Choosing a distribution before examining the information. Use a table when aggregate probabilities are supplied; build a distribution when the frequency and severity structure is given and manageable; use a normal approximation only when it is explicitly supported by the model or problem assumptions.

Key takeaway: VaR identifies a quantile, while TVaR averages the upper tail. For discrete aggregate losses, that tail includes all losses above VaR and only the required fraction of the probability mass at VaR. Verify the tail probability and the chosen distributional method before finalizing either measure.

Check Your Understanding

For each problem, use the stated model and the lower-quantile convention. Amounts are hypothetical one-year losses.

An aggregate loss is $0$, $100$, or $400$ dollars with probabilities $0.20$, $0.50$, and $0.30$, respectively. Find $\operatorname{VaR}_{0.60}$ and $\operatorname{TVaR}_{0.60}$. (Answer: VaR $=\$100$; TVaR $=\$325$.)
A loss is $0$ with probability $0.50$, $2$ with probability $0.25$, and $10$ with probability $0.25$. Calculate $\operatorname{VaR}_{0.75}$ and $\operatorname{TVaR}_{0.75}$, and state why the strict conditional expectation above VaR is not the discrete-tail calculation. (Answer: VaR $=2$; TVaR $=10$; the upper tail is exactly the probability mass at $10$, so strict exceedance happens to give 10 here, while the mass at the quantile contributes zero.)
A hypothetical one-year plan has $N\sim\operatorname{Bernoulli}(0.20)$ and pays exactly $\$500$ if a claim occurs, with no payment otherwise. Find $\operatorname{VaR}_{0.90}$ and $\operatorname{TVaR}_{0.90}$. (Answer: VaR $=\$500$; TVaR $=\$500$.)
An aggregate loss is modeled as $X\sim N(\$800{,}000,(\$100{,}000)^2)$. Using $z_{0.975}=1.96$ and $\phi(1.96)\approx0.05844$, calculate VaR and TVaR at $0.975$, to the nearest dollar. (Answer: VaR $=\$996{,}000$; TVaR $\approx\$1{,}033{,}760$.)
An aggregate loss takes values $0$, $100$, and $300$ dollars with probabilities $0.70$, $0.20$, and $0.10$, respectively. Calculate $\operatorname{VaR}_{0.85}$ and $\operatorname{TVaR}_{0.85}$. (Answer: VaR $=\$100$; TVaR $=\$233.33$.)

What You'll Learn