How Severity Model Choice Changes VaR and TVaR

Why Severity Model Choice Matters

Two fitted severity distributions can describe ordinary losses similarly while assigning very different probabilities to very large losses. Their expected severities may be close—or even equal—yet their estimates of high-quantile risk can diverge. For an insurer assessing a large individual claim, this difference can matter more than a small difference in the fitted mean.

This tutorial compares VaR and TVaR under alternative severity families. The calculations treat each candidate model and its fitted parameters as fixed. They therefore show the effect of distributional model choice, not parameter uncertainty. A severity model describes the amount of an individual loss $X$; it does not, by itself, describe the number of losses or their aggregate.

Definition: For a loss $X$ with distribution function $F_X$, the VaR at probability level $p$ is the quantile $$ \operatorname{VaR}_p(X)=\inf\{x:F_X(x)\ge p\}. $$ For a continuous distribution with a unique $p$-quantile $q_p$, the TVaR at level $p$ is the mean loss in the upper tail: $$ \operatorname{TVaR}_p(X)=E[X\mid X>q_p]. $$ Here $0<p<1$. The conditional-mean definition applies directly to continuous severities; for distributions with atoms, tail expectation requires a convention for allocating probability at the quantile.

VaR locates a threshold: for a continuous model, a proportion $p$ of losses are at or below it. TVaR describes the average loss among those exceeding that threshold. Thus, VaR and TVaR answer related but distinct questions, and a model comparison should not substitute one for the other.

Useful Formulas for Two Severity Families

For an exponential severity with mean $\theta>0$, the survival function is $S_X(x)=e^{-x/\theta}$ for $x\ge 0$. For a Type I Pareto severity with scale $\theta>0$ and shape $\alpha>0$, the survival function is $S_X(x)=(\theta/x)^\alpha$ for $x\ge\theta$. Its mean exists when $\alpha>1$, and its TVaR is finite under the same condition.

Result: For an exponential severity with mean $\theta$, and a Type I Pareto severity with scale $\theta$ and shape $\alpha>1$, the VaR and TVaR are $$ \begin{aligned} \operatorname{VaR}_p^{\mathrm{Exp}}&=-\theta\ln(1-p), & \operatorname{TVaR}_p^{\mathrm{Exp}}&=\operatorname{VaR}_p^{\mathrm{Exp}}+\theta,\\ \operatorname{VaR}_p^{\mathrm{Par}}&=\theta(1-p)^{-1/\alpha}, & \operatorname{TVaR}_p^{\mathrm{Par}}&=\frac{\alpha}{\alpha-1}\operatorname{VaR}_p^{\mathrm{Par}}. \end{aligned} $$ The exponential mean is $\theta$. The Pareto mean is $\alpha\theta/(\alpha-1)$.

These formulas follow by solving the survival equation $S_X(q_p)=1-p$ for the quantile, then calculating the mean above that quantile. For a continuous nonnegative loss with $S_X(q_p)=1-p$, the conditional-mean definition gives

E[X\mid X>q_p] =q_p+\frac{\int_{q_p}^{\infty}S_X(t)\,dt}{S_X(q_p)}.

To see why, write $X=q_p+(X-q_p)$ conditional on $X>q_p$. The conditional expected excess is the integral of its conditional survival probabilities, and $P(X>t\mid X>q_p)=S_X(t)/S_X(q_p)$ for $t\ge q_p$. For the exponential model, this ratio is $e^{-(t-q_p)/\theta}$, whose integral from $q_p$ to infinity is $\theta$. For the Pareto model, it is $(q_p/t)^\alpha$, whose integral is $q_p/(\alpha-1)$ when $\alpha>1$. Substitution yields the stated TVaR formulas.

Worked Comparisons

Worked Example: Same Mean, Different Models at the 99th Percentile

A hypothetical insurer fits two models to individual claim severities, measured in dollars. Model E is exponential with mean $\theta=\$1{,}000$. Model P is Type I Pareto with scale $\theta=\$500$ and shape $\alpha=2$. Both models have mean $\$1{,}000$. Calculate each model's 99% VaR and TVaR, then compare the estimates.

Method and assumptions. Treat the fitted distributions as continuous models with fixed parameters, and set $p=0.99$, so $1-p=0.01$. For Model E, $\operatorname{VaR}_{0.99}=-\theta\ln(0.01)$ and $\operatorname{TVaR}_{0.99}=\operatorname{VaR}_{0.99}+\theta$.

For Model E, $-\ln(0.01)=4.605170186$. Therefore,

\operatorname{VaR}_{0.99}^{\mathrm{E}} =1{,}000(4.605170186) =\$4{,}605.17, \qquad \operatorname{TVaR}_{0.99}^{\mathrm{E}} =4{,}605.170186+1{,}000 =\$5{,}605.17.

For Model P, the mean is $2(500)/(2-1)=\$1{,}000$, confirming the equal-mean condition. Its quantile multiplier is $(0.01)^{-1/2}=10$, and its TVaR multiplier is $2/(2-1)=2$. Thus,

\operatorname{VaR}_{0.99}^{\mathrm{P}} =500(10) =\$5{,}000.00, \qquad \operatorname{TVaR}_{0.99}^{\mathrm{P}} =2(5{,}000) =\$10{,}000.00.

The Pareto estimate is $\$394.83$ higher for VaR and $\$4{,}394.83$ higher for TVaR. The models have the same mean, but the Pareto model assigns more weight to very large losses, producing a much larger average above the 99% quantile. Common mistake avoided: assuming equal means imply equal high-tail risk.

Worked Example: Changing Pareto Shape While Holding the Mean Fixed

A hypothetical analyst compares two Type I Pareto models for individual losses in dollars. Model A has scale $\theta=500$ and shape $\alpha=2$. Model B has shape $\alpha=3$; choose its scale so that its mean also equals $\$1{,}000$. Find both models' 99% VaR and TVaR.

Method and assumptions. Both models are continuous Type I Pareto distributions with shape greater than 1, so their means and TVaRs are finite. Use $E[X]=\alpha\theta/(\alpha-1)$ to find Model B's scale before calculating its tail measures.

Model A has mean $2(500)/(2-1)=\$1{,}000$. For Model B, solve $3\theta/(3-1)=1{,}000$, giving $\theta=2{,}000/3=\$666.6667$. At $p=0.99$, $(1-p)^{-1}=100$. The quantile multipliers are $100^{1/2}=10$ for Model A and $100^{1/3}=4.641588834$ for Model B.

\begin{aligned} \operatorname{VaR}_{0.99}^{\mathrm{A}} &=500(10)=\$5{,}000.00, & \operatorname{TVaR}_{0.99}^{\mathrm{A}} &=2(5{,}000)=\$10{,}000.00,\\ \operatorname{VaR}_{0.99}^{\mathrm{B}} &=\frac{2{,}000}{3}(4.641588834)=\$3{,}094.39, & \operatorname{TVaR}_{0.99}^{\mathrm{B}} &=\frac{3}{2}(3{,}094.392556)=\$4{,}641.59. \end{aligned}

Model B has both a lower 99% VaR and a lower 99% TVaR, despite having the same mean. Increasing the Pareto shape while adjusting the scale to preserve the mean makes this model's upper tail lighter at the selected probability level. Common mistake avoided: changing the shape but forgetting to adjust the scale, which would no longer hold the mean constant.

Worked Example: Why the Probability Level Can Change the Comparison

Use the same hypothetical Model E, exponential with mean $\$1{,}000$, and Model P, Type I Pareto with scale $\$500$ and shape $2$. Calculate VaR and TVaR at 95% and 99.5%, and determine whether the ranking of the VaR estimates is the same at both levels.

Method and assumptions. Use the exponential and Pareto formulas above, with fixed parameters and probabilities $p=0.95$ and $p=0.995$. Retain full precision in intermediate calculations and round dollar amounts to cents at the end.

At $p=0.95$, $-\ln(0.05)=2.995732274$, while $(0.05)^{-1/2}=\sqrt{20}=4.472135955$. Therefore,

\begin{aligned} \operatorname{VaR}_{0.95}^{\mathrm{E}} &=1{,}000(2.995732274)=\$2{,}995.73, & \operatorname{TVaR}_{0.95}^{\mathrm{E}} &=2{,}995.732274+1{,}000=\$3{,}995.73,\\ \operatorname{VaR}_{0.95}^{\mathrm{P}} &=500(4.472135955)=\$2{,}236.07, & \operatorname{TVaR}_{0.95}^{\mathrm{P}} &=2\bigl(500(4.472135955)\bigr)=\$4{,}472.14. \end{aligned}

At $p=0.995$, $-\ln(0.005)=5.298317367$, while $(0.005)^{-1/2}=\sqrt{200}=14.142135624$. Hence,

\begin{aligned} \operatorname{VaR}_{0.995}^{\mathrm{E}} &=1{,}000(5.298317367)=\$5{,}298.32, & \operatorname{TVaR}_{0.995}^{\mathrm{E}} &=5{,}298.317367+1{,}000=\$6{,}298.32,\\ \operatorname{VaR}_{0.995}^{\mathrm{P}} &=500(14.142135624)=\$7{,}071.07, & \operatorname{TVaR}_{0.995}^{\mathrm{P}} &=2\bigl(500(14.142135624)\bigr)=\$14{,}142.14. \end{aligned}

At 95%, the Pareto VaR is lower than the exponential VaR, but at 99.5% it is higher. The TVaR is higher under the Pareto model at both levels. Thus a statement such as “one model is more conservative” needs a specified risk measure and probability level. Common mistake avoided: inferring the ranking at an extreme percentile from the ranking at a less extreme percentile.

Interpreting Model Comparisons

The examples isolate a key point: the mean summarizes the center of a severity distribution, whereas high-level VaR and TVaR depend strongly on its tail. In the Pareto family, the shape parameter controls tail decay: smaller $\alpha$ gives a heavier tail. The scale must also be considered, because changing it changes the overall severity level and, generally, the mean.

A useful comparison process is to calculate each candidate's mean, then calculate the requested tail measures at the same probability level and in the same units. Do not infer tail risk from the mean alone. Also check whether the model supports the requested quantity: for a Type I Pareto with $\alpha\le 1$, the mean and conditional tail mean are infinite, so a finite TVaR formula cannot be applied.

State the fitted model.
Identify the severity family, its parameterization, and the fitted parameter values. Check that the formula's conditions hold.

Fix the comparison target.
Use the same probability level, loss definition, and currency units for all candidate models.

Calculate and interpret both measures.
Find VaR from the quantile equation and TVaR from a conditional-tail formula. Compare the results without treating them as interchangeable.

A common computational trap is to divide VaR by $1-p$ to obtain TVaR. This is not a general formula. For example, for the Pareto model with scale $\$500$, shape $2$, and $p=0.99$, VaR is $\$5{,}000$. The incorrect calculation $5{,}000/(1-0.99)=\$500{,}000$ is not the conditional mean of losses above the quantile. The correct calculation is $2(5{,}000)=\$10{,}000$, using the Pareto TVaR formula. Always use a formula justified by the fitted distribution or evaluate the appropriate tail expectation.

Key takeaway: Similar fitted means do not guarantee similar VaR or TVaR. State the severity family and parameterization, calculate tail measures at the required probability level, and verify that the model's tail moments exist.

Check Your Understanding

Assume all severities below are continuous and all fitted parameters are fixed. Calculate each requested quantity, retaining full precision until the final dollar amount.

A hypothetical exponential severity has mean $\$800$. Find its 95% VaR and TVaR. (Answer: VaR = $2,396.59; TVaR = $3,196.59)
A hypothetical Type I Pareto severity has scale $\$400$ and shape $2$. Find its 96% VaR and TVaR, and state whether the model has a finite mean. (Answer: VaR = $2,000.00; TVaR = $4,000.00; the mean is finite)
A hypothetical Type I Pareto severity has mean $\$1{,}500$, shape $3$, and probability level $p=0.99$. First find its scale, then calculate VaR and TVaR. (Answer: Scale = $1,000.00; VaR = $4,641.59; TVaR = $6,962.38)
Compare an exponential severity with mean $\$1{,}200$ and a Type I Pareto severity with scale $\$600$, shape $2$, at $p=0.99$. Which has the larger TVaR, and by how much? (Answer: Pareto TVaR is larger by $5,273.80; exponential TVaR = $6,726.20 and Pareto TVaR = $12,000.00)
A hypothetical Type I Pareto severity has scale $\$500$ and shape $1.5$. At $p=0.99$, calculate its mean, VaR, and TVaR. (Answer: Mean = $1,500.00; VaR = $10,772.17; TVaR = $32,316.52)

What You'll Learn