Stress-Testing Tail-Driven Rate Indications

Why Stress-Test Tail Assumptions?

A rate indication can change substantially when the assumed severity tail changes, even if the model’s general structure stays the same. The tail affects not only the expected loss, which drives the pure premium, but also high-confidence risk measures such as TVaR. A useful stress test therefore recalculates both expected losses and the risk load instead of adjusting only one component.

The calculations below use a hypothetical annual policy with at most one claim. This deliberately simple frequency-severity model makes the impact of the assumptions visible. It is not a claim that a Bernoulli frequency model or a Pareto severity model is appropriate for every portfolio; the model and its assumptions must be checked against the pricing context.

Model and assumptions: The annual claim count $N$ is Bernoulli with claim probability $q$, so $P(N=1)=q$ and $P(N=0)=1-q$. Given a claim, severity $X$ has a Pareto Type I distribution with minimum $x_m>0$ and shape $a>1$: $$ P(X>x)=\left(\frac{x_m}{x}\right)^a,\qquad x\ge x_m. $$ The annual aggregate loss is $L=NX$, interpreted as zero when $N=0$. Claim occurrence and severity are independent. The condition $a>1$ ensures finite mean severity.

For this model, the pure premium per exposure-year is $E[L]$. We will use a risk load equal to a stated fraction $\lambda$ of the difference between aggregate TVaR and expected aggregate loss. This is a pricing rule for the examples, not a universal formula: an insurer may use a different capital measure or loading rule.

Definitions: For confidence level $p$, $\operatorname{VaR}_p(L)$ is the $p$-quantile of annual aggregate loss. When the loss distribution is continuous at this quantile, $\operatorname{TVaR}_p(L)=E[L\mid L>\operatorname{VaR}_p(L)]$. Equivalently, for the model and quantiles used here, $$ \operatorname{TVaR}_p(L)=\operatorname{VaR}_p(L)+\frac{E[(L-\operatorname{VaR}_p(L))_+]}{1-p}, $$ where $z_+=\max(z,0)$. The latter expression also clarifies how the tail excess contributes to TVaR.

Pure Premium, VaR, and TVaR Under the Model

The Pareto mean is $E[X]=a x_m/(a-1)$. Since a claim occurs with probability $q$, the expected annual aggregate loss is

E[L]=qE[X]=q\frac{a x_m}{a-1}.

For any $t\ge x_m$, the probability that aggregate loss exceeds $t$ is the probability of a claim multiplied by the conditional severity tail:

P(L>t)=q\left(\frac{x_m}{t}\right)^a.

If $1-p<q$, the $p$-quantile is positive and lies in the Pareto tail. Solving $P(L>t)=1-p$ gives

\operatorname{VaR}_p(L)=x_m\left(\frac{q}{1-p}\right)^{1/a}.

This formula applies when the confidence level is above the probability of no claim, $p>1-q$. If $p\le 1-q$, the atom at zero means the VaR is zero; the positive-tail formula must not be used. In our examples, $p>1-q$.

Why the TVaR Formula Has This Form

Let $t=\operatorname{VaR}_p(L)$ be positive. The tail excess is zero when there is no claim or when the claim severity is at most $t$. For a Pareto severity, the expected excess conditional on a claim is

E[(X-t)_+] =\int_t^\infty P(X>x)\,dx =\int_t^\infty \left(\frac{x_m}{x}\right)^a dx =\frac{x_m^a t^{1-a}}{a-1}.

The integral is finite because $a>1$. Multiplying by $q$ accounts for the chance of a claim. The quantile equation implies $q x_m^a t^{-a}=1-p$, so

E[(L-t)_+] =q\frac{x_m^a t^{1-a}}{a-1} =\frac{(1-p)t}{a-1}.

Substituting this excess into the stop-loss expression for TVaR gives

\operatorname{TVaR}_p(L) =t+\frac{E[(L-t)_+]}{1-p} =t+\frac{t}{a-1} =\frac{a}{a-1}t.

Thus, for fixed $q$, $x_m$, and $p$, a smaller shape $a$ makes the tail heavier: it increases the VaR multiplier and the TVaR multiplier, and it also increases the mean when $a$ approaches 1. If $a\le 1$, the mean severity is infinite, so this finite-pure-premium calculation does not apply.

Worked Example: Establishing a Base Indication

A hypothetical policy has annual claim probability $q=0.20$. Conditional on a claim, severity is Pareto with $x_m=\$1{,}000$ and $a=2$. At $p=0.95$, calculate the pure premium, VaR, TVaR, and indicated premium if the risk load is 15% of TVaR less expected aggregate loss.

Method. Calculate expected aggregate loss first, then the aggregate quantile and TVaR. Finally apply the specified load to the TVaR excess over the mean.

Pure premium. The mean severity is $$ E[X]=\frac{2(1{,}000)}{2-1}=\$2{,}000. $$ Therefore, $$ E[L]=0.20(\$2{,}000)=\$400. $$

VaR. The probability of no claim is $0.80$, so the 95th percentile is above the zero-loss atom. The required tail probability is $1-p=0.05$, and $$ \operatorname{VaR}_{0.95}(L) =1{,}000\left(\frac{0.20}{0.05}\right)^{1/2} =1{,}000\sqrt{4} =\$2{,}000. $$

TVaR. With $a=2$, $$ \operatorname{TVaR}_{0.95}(L) =\frac{2}{2-1}(\$2{,}000) =\$4{,}000. $$ The risk-load base is $\$4{,}000-\$400=\$3{,}600$. The load is $0.15(\$3{,}600)=\$540$, so the indicated premium is $\$400+\$540=\$940$ per exposure-year.

Check and common mistake. The model’s expected aggregate loss is $\$400$, while its TVaR is higher at $\$4{,}000$; the positive risk load is consistent with those figures. A common error is to apply 15% to TVaR itself, which would give a $\$600$ load. The stated rule applies 15% to TVaR less the mean, so the correct load is $\$540$.

Recalculating Under Tail and Scale Stresses

A controlled stress changes one assumption at a time while holding the others fixed. This helps distinguish a tail-shape effect from a change in overall severity level or claim probability. For each scenario, retain the same confidence level and loading rule; otherwise the resulting rate differences mix model changes with pricing-rule changes.

Worked Example: Stressing the Pareto Tail Shape

Keep the base example’s $q=0.20$, $x_m=\$1{,}000$, $p=0.95$, and 15% loading rule, but reduce the Pareto shape from $a=2$ to $a=1.5$. Recalculate the indication and compare it with the base.

Method. Recalculate both the mean and the tail measures using the stressed shape; changing $a$ affects all three.

The mean severity becomes $$ E[X]=\frac{1.5(\$1{,}000)}{1.5-1}=\$3{,}000, $$ so the pure premium is $0.20(\$3{,}000)=\$600$.

The aggregate VaR is $$ \operatorname{VaR}_{0.95}(L) =\$1{,}000\left(\frac{0.20}{0.05}\right)^{1/1.5} =\$1{,}000(4^{2/3}) \approx \$2{,}519.84. $$ Then $$ \operatorname{TVaR}_{0.95}(L) =\frac{1.5}{1.5-1}(\$2{,}519.84) =3(\$2{,}519.84) \approx \$7{,}559.53. $$ The risk-load base is $\$7{,}559.53-\$600=\$6{,}959.53$. The load is $0.15(\$6{,}959.53)=\$1{,}043.93$, giving an indicated premium of $\$600+\$1{,}043.93=\$1{,}643.93$.

Compared with the base indication of $\$940.00$, this is an increase of $\$703.93$. The change is not just a larger capital margin: the pure premium also rises from $\$400$ to $\$600$. This is why a tail-shape stress should recalculate expected loss rather than hold it fixed automatically.

Check and common mistake. Since $a=1.5>1$, the mean remains finite. Using the $a=2$ mean after changing the shape would incorrectly preserve a $\$400$ pure premium and understate the indicated rate.

Worked Example: Comparing Scale and Frequency Stresses

Starting from the base indication, compare two separate stresses at $p=0.95$ with a 15% loading rule: (A) increase $x_m$ to $\$1{,}250$, keeping $q=0.20$ and $a=2$; (B) increase $q$ to $0.25$, keeping $x_m=\$1{,}000$ and $a=2$. Identify which change produces the larger indicated-premium increase.

Scenario A: severity scale. The mean severity is $2(\$1{,}250)=\$2{,}500$, so the pure premium is $0.20(\$2{,}500)=\$500$. The VaR is $$ \$1{,}250\left(\frac{0.20}{0.05}\right)^{1/2} =\$1{,}250(2) =\$2{,}500. $$ Thus TVaR is $2(\$2{,}500)=\$5{,}000$. The load is $0.15(\$5{,}000-\$500)=\$675$, and the indication is $\$500+\$675=\$1{,}175$. Relative to the base, the increase is $\$235$.

Scenario B: claim probability. The mean severity remains $\$2{,}000$, so the pure premium is $0.25(\$2{,}000)=\$500$. The VaR is $$ \$1{,}000\left(\frac{0.25}{0.05}\right)^{1/2} =\$1{,}000\sqrt{5} \approx \$2{,}236.07. $$ TVaR is $2(\$2{,}236.07)=\$4{,}472.14$. The load is $0.15(\$4{,}472.14-\$500)=\$595.82$, and the indication is $\$500+\$595.82=\$1{,}095.82$. Relative to the base, the increase is $\$155.82$.

Conclusion and check. Under these assumptions, the scale stress increases the indication more than the claim-probability stress: $\$235>\$155.82$. Both stresses raise the pure premium to $\$500$, but their TVaRs differ. The comparison is conditional on the chosen model and loading rule; it is not a universal ranking of scale and frequency risk.

Common Exam Trap: The Aggregate Percentile Is Not the Severity Percentile

The positive-loss distribution has an atom $1-q$ at zero. When $p>1-q$, the aggregate tail probability $1-p$ must be divided by $q$ to obtain the conditional severity tail probability at the aggregate VaR. In the base case, $1-p=0.05$ and $q=0.20$, so the conditional severity exceedance probability is $0.05/0.20=0.25$. The relevant severity percentile is therefore 75%, not 95%.

A wrong calculation would use the 95th percentile of severity: $$ \$1{,}000\left(\frac{1}{0.05}\right)^{1/2} \approx \$4{,}472.14. $$ That is not the aggregate 95th percentile. The correct aggregate VaR is $\$2{,}000$, obtained from $P(L>t)=qP(X>t)=0.05$. Substituting the wrong threshold would also produce an incorrect TVaR and risk load.

For a practical stress test, document the baseline assumptions, change one assumption at a time where possible, and recalculate the pure premium, VaR, TVaR, load, and final indication consistently. Check both that the quantile lies above the zero-loss atom and that the stressed severity shape still gives a finite mean.

Check Your Understanding

Use the Bernoulli-frequency and Pareto severity model described above. Round money to the nearest cent. Each problem is hypothetical.

A policy has $q=0.10$, $x_m=\$500$, and $a=2$. At $p=0.95$, calculate the pure premium and TVaR. (Answer: pure premium $\$100.00$; TVaR $\$1{,}414.21$)
For $q=0.25$, $x_m=\$800$, and $a=2$, calculate the 95% aggregate VaR and TVaR. (Answer: VaR $\$1{,}788.85$; TVaR $\$3{,}577.71$)
Let $q=0.20$, $x_m=\$1{,}000$, $a=3$, and $p=0.95$. With a load of 10% of TVaR less expected aggregate loss, calculate the indicated premium. (Answer: pure premium $\$300.00$; TVaR $\$2{,}381.10$; indicated premium $\$508.11$)
Let $q=0.15$, $x_m=\$1{,}200$, $a=2$, and $p=0.90$. If the load is 20% of TVaR less expected aggregate loss, calculate the indicated premium. (Answer: pure premium $\$360.00$; TVaR $\$2{,}939.39$; indicated premium $\$875.88$)
For $q=0.20$, $x_m=\$1{,}000$, $p=0.95$, and a 10% loading rule, compare the indication at $a=2$ with the indication at $a=1.25$. (Answer: $a=2$ indication $\$760.00$; $a=1.25$ indication $\$2{,}415.72$; increase $\$1{,}655.72$)

What You'll Learn