Tail-Risk Loads Using Exceedance Measures

Why Use an Exceedance Measure?

Expected layer loss is the mean amount a layer is expected to pay over the chosen time period. It is central to pricing, but it does not describe how severe losses can be in the upper tail. Two layers with the same expected loss can have different upper-tail outcomes, and a reinsurer or insurer may require compensation for holding capital against those outcomes.

Percentile measures and tail-value-at-risk (TVaR) summarize different aspects of the loss distribution. A percentile identifies a threshold. TVaR averages losses in the upper tail, with a precise adjustment when the threshold itself has probability mass. To use either measure in a price, first distinguish the tail measure from the tail capital above expected loss and from any additional compensation charged for that capital.

Definition: Let $L$ be a nonnegative loss over a specified period, with distribution function $F_L(x)=P(L\le x)$ and finite mean $E[L]$. For $0<p<1$, the $p$-percentile, or value-at-risk at level $p$, is $$ \operatorname{VaR}_p(L)=\inf\{x:F_L(x)\ge p\}. $$ Define tail-value-at-risk at level $p$ by averaging the upper quantiles: $$ \operatorname{TVaR}_p(L)=\frac{1}{1-p}\int_p^1 \operatorname{VaR}_q(L)\,dq. $$ The tail capital under either measure is the measure less expected loss: $\operatorname{VaR}_p(L)-E[L]$ or $\operatorname{TVaR}_p(L)-E[L]$, respectively. These definitions assume the relevant expectations are finite.

Here $L$ may be an annual aggregate layer loss or another loss variable being priced. The time period and loss basis must be consistent: for example, do not compare an annual expected loss with a percentile calculated for a single occurrence. A percentile is a threshold, not an average. TVaR is an average of upper-tail outcomes and is generally at least as large as the percentile at the same level.

Deriving the Tail-Average Formula

The quantile definition gives a useful formula that works for both continuous and discrete loss distributions. It is especially important for discrete distributions, where the percentile threshold can occur with positive probability.

Set the threshold.
Let $v=\operatorname{VaR}_p(L)$. By the generalized-inverse definition, $F_L(v)\ge p$, while the probability strictly below $v$ is no greater than $p$.

Identify the quantiles equal to the threshold.
For quantile levels from $p$ through $F_L(v)$, the quantile equals $v$. This contributes $v(F_L(v)-p)$ to the integral of upper quantiles.

Add the strict exceedances.
The quantiles above $F_L(v)$ correspond to losses strictly greater than $v$. Their integrated contribution is $E[L\,\mathbf{1}_{\{L>v\}}]$.

Divide by the tail probability.
The upper tail has probability $1-p$. Dividing the two contributions by $1-p$ gives the equivalent formula below.

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

The term $F_L(v)-p$ is the amount of probability mass at the threshold that must be included to make the selected tail have probability exactly $1-p$. If the distribution is continuous at $v$ and $F_L(v)=p$, there is no threshold mass to add. In that case, when the conditional mean exists, TVaR equals $E[L\mid L>v]$. With an atom at $v$, the formula instead includes the needed fraction of that mass.

Worked Calculations

Worked Example: Exponential Annual Layer Loss

A hypothetical catastrophe layer has annual aggregate loss $L$ measured in thousands of dollars. Assume $L$ has an exponential distribution with mean $40$, so its density is $f_L(x)=\frac{1}{40}e^{-x/40}$ for $x\ge0$. Find the 95th percentile, 95% TVaR, and TVaR tail capital. Also calculate a risk load equal to 6% of that tail capital. Ignore expenses, investment income, and taxes.

Method. For an exponential loss with mean $40$, the survival function is $P(L>x)=e^{-x/40}$. Set the survival probability at the 95th percentile to $1-0.95=0.05$:

e^{-v/40}=0.05,\qquad v=-40\ln(0.05)=40\ln(20)=119.8293.

Thus the 95th percentile is $119.8293$ thousand dollars. The exponential distribution is continuous, and its memoryless property gives $E[L-v\mid L>v]=40$. Therefore $E[L\mid L>v]=v+40$, which is also the 95% TVaR:

\operatorname{TVaR}_{0.95}(L)=119.8293+40=159.8293.

Expected loss is $40$ thousand dollars. The TVaR tail capital is $159.8293-40=119.8293$ thousand dollars. The 6% capital-compensation load is $0.06(119.8293)=7.189758$ thousand dollars, and the indicated amount before other provisions is $40+7.189758=47.189758$ thousand dollars. Rounding final dollar amounts to the nearest dollar, tail capital is $119,829, the load is $7,190, and the indicated amount is $47,190. This example uses the continuous-distribution case; the next example shows why strict exceedances alone do not always give TVaR.

Worked Example: A Discrete Loss With an Atom at the Percentile

A hypothetical annual layer loss, in thousands of dollars, is $0$ with probability $0.60$, $20$ with probability $0.25$, and $100$ with probability $0.15$. Calculate its 80th percentile, 80% TVaR, expected loss, and TVaR tail capital. If the capital-compensation rate is 5%, find the additional load.

Method. The cumulative probabilities are $F_L(0)=0.60$, $F_L(20)=0.85$, and $F_L(100)=1$. Since $0.60<0.80\le0.85$, the 80th percentile is $v=20$. The upper tail must have probability $1-0.80=0.20$. It contains all $0.15$ probability at $100$ and another $0.05$ probability at the threshold $20$. Hence

\operatorname{TVaR}_{0.80}(L) =\frac{(100)(0.15)+(20)(0.05)}{0.20} =\frac{15+1}{0.20} =80.

Expected loss is

$$ E[L]=(0)(0.60)+(20)(0.25)+(100)(0.15)=5+15=20. $$

Therefore TVaR tail capital is $80-20=60$ thousand dollars, and the 5% capital-compensation load is $0.05(60)=3$ thousand dollars. The expected loss plus this load is $20+3=23$ thousand dollars. As a check, the TVaR calculation averages exactly $0.20$ probability of outcomes: the numerator $16$ thousand-dollar-probability units divided by $0.20$ gives $80$ thousand dollars.

Common mistake and correction. Using $E[L\mid L>20]=100$ as TVaR is wrong: it averages only the strict exceedances and omits the $0.05$ probability at $20$ needed to fill the 20% tail. The correct TVaR is $80$. When the strict-exceedance conditional mean is defined, it cannot be less than TVaR: TVaR averages the strict exceedances together with any needed threshold mass, whose value is $v$. If there are no strict exceedances, the conditional mean given $L>v$ is undefined.

Worked Example: Applying an Occurrence Layer to a Uniform Loss

For one hypothetical occurrence, let ground-up loss $X$, measured in thousands of dollars, be uniform on $[0,200]$. A layer pays $50$ thousand dollars excess of a $50$ thousand-dollar attachment, so its payment is $L=\min((X-50)^+,50)$, where $a^+=\max(a,0)$. Find the 40th percentile, 40% TVaR, expected layer payment, and TVaR tail capital. Use an 8% capital-compensation rate to calculate the additional load.

Method. If $X\le50$, then $L=0$, with probability $50/200=0.25$. If $50<X<100$, then $L=X-50$, giving density $1/200$ for $0<L<50$. If $X\ge100$, the layer pays its $50$ thousand-dollar limit, with probability $100/200=0.50$. Thus, for $0\le y<50$, $F_L(y)=0.25+y/200$, followed by an atom of $0.50$ at $50$.

Solve $F_L(v)=0.40$ within the continuous part:

0.25+\frac{v}{200}=0.40,\qquad v=200(0.15)=30.

The 40th percentile is $30$ thousand dollars. The 40% upper tail has probability $0.60$. It contains the $0.50$ probability at $50$ and the $0.10$ probability in the continuous layer-loss interval from $30$ to $50$, since $(50-30)/200=0.10$. The tail-loss numerator is

(50)(0.50)+\int_{30}^{50}\frac{y}{200}\,dy =25+\frac{50^2-30^2}{400} =25+4 =29.

Therefore $\operatorname{TVaR}_{0.40}(L)=29/0.60=48.3333$ thousand dollars. Expected layer payment consists of the continuous payments and the limit payments:

E[L]=\int_0^{50}\frac{y}{200}\,dy+(50)(0.50) =\frac{50^2}{400}+25 =6.25+25 =31.25.

TVaR tail capital is $48.3333-31.25=17.0833$ thousand dollars. The 8% load is $0.08(17.0833)=1.3667$ thousand dollars, giving an expected loss plus capital-compensation indication of $31.25+1.3667=32.6167$ thousand dollars. Rounded to the nearest dollar, expected loss is $31,250, tail capital is $17,083, the load is $1,367, and the indication is $32,617. The check $F_L(30)=0.25+30/200=0.40$ confirms the percentile, and the tail numerator divided by its probability gives the stated TVaR.

Turning a Tail Measure Into a Risk Load

A tail measure is not itself a unique pricing rule. A common capital-based convention is to calculate tail capital as $\operatorname{TVaR}_p(L)-E[L]$, then multiply it by a specified cost-of-capital rate. Under that convention, the amount added to expected loss is $c(\operatorname{TVaR}_p(L)-E[L])$, where $c$ is the selected rate. A different convention might specify a percentile-based capital amount or a target premium directly. State the convention and confidence level before calculating; do not call the tail measure, tail capital, and dollar load by the same name.

The choice of percentile or tail-average measure also matters. VaR reports a threshold and says nothing about how large losses beyond it may be. TVaR responds to the severity of those outcomes by averaging the selected upper tail. Discrete layer payments often have atoms—for example, zero payments below an attachment or capped payments at a limit—so the atom-aware formula is essential. Even where the loss model is continuous, confirm that the modeled loss variable and time basis match the layer being priced.

Key takeaway: Calculate expected loss separately from the tail measure. For TVaR, average exactly the upper probability $1-p$, including only the required fraction of any mass at the percentile threshold. Then subtract expected loss if the pricing convention defines capital as excess over the mean.

Check Your Understanding

For each problem, use the stated tail-capital convention: tail capital equals the requested tail measure minus expected loss. All amounts are in thousands of dollars unless stated otherwise.

A hypothetical annual loss is exponential with mean $25$. Find its 90th percentile and 90% TVaR. (Answer: $\operatorname{VaR}_{0.90}=57.5646$, $\operatorname{TVaR}_{0.90}=82.5646$.)
A layer's annual loss is $0$ with probability $0.50$, $10$ with probability $0.30$, and $40$ with probability $0.20$. Find its 75th percentile, 75% TVaR, expected loss, and TVaR tail capital. (Answer: percentile $10$, TVaR $34$, expected loss $11$, tail capital $23$.)
Let annual loss $L$ be uniform on $[0,100]$. Find its 80th percentile, 80% TVaR, expected loss, and TVaR tail capital. (Answer: percentile $80$, TVaR $90$, expected loss $50$, tail capital $40$.)
A hypothetical annual layer loss is $0$ with probability $0.70$, $30$ with probability $0.20$, and $90$ with probability $0.10$. At level $p=0.80$, find the percentile, TVaR, expected loss, and tail capital; then calculate the additional load at a 4% cost-of-capital rate. (Answer: percentile $30$, TVaR $60$, expected loss $15$, tail capital $45$, additional load $1.8$.)

What You'll Learn