Risk Loadings Based on Variance and Tail Measures

Why Premium Principles Need a Risk Loading

A premium must cover more than an average loss if the insurer is to allow for uncertainty and adverse outcomes. A risk loading is the amount added to expected loss under a chosen premium principle. Different principles measure uncertainty in different ways: variance and standard deviation summarize dispersion around the mean, while value-at-risk and tail value-at-risk focus on a selected part of the loss distribution.

The resulting premiums are not automatically comparable just because they use the same loss model. The variance parameter has units, the standard-deviation parameter is dimensionless, and a VaR or TVaR premium depends on its confidence level. The chosen parameters and confidence level must therefore accompany any reported premium.

Definition: Let $X$ be a nonnegative loss with finite mean $\mu=E[X]$ and finite variance $\sigma^2=\operatorname{Var}(X)$. The variance premium and standard-deviation premium are $$ P_{\mathrm{var}}=E[X]+\alpha\operatorname{Var}(X), \qquad P_{\mathrm{sd}}=E[X]+\beta\sqrt{\operatorname{Var}(X)}, $$ where $\alpha\geq 0$ and $\beta\geq 0$. The variance parameter $\alpha$ has units inverse to the currency unit; $\beta$ is dimensionless.

For a confidence level $p$, with $0<p<1$, define the quantile by $ \operatorname{VaR}_p(X)=\inf\{x:F_X(x)\geq p\} $. The TVaR convention used here is the integral-quantile definition. It applies to both continuous and discrete losses.

Definition: The VaR premium is $P_{\mathrm{VaR}}=\operatorname{VaR}_p(X)$, and the TVaR premium is $P_{\mathrm{TVaR}}=\operatorname{TVaR}_p(X)$, where $$ \operatorname{TVaR}_p(X)=\frac{1}{1-p}\int_p^1 \operatorname{VaR}_u(X)\,du. $$ For a continuous loss distribution with no atom at $\operatorname{VaR}_p(X)$, this equals $E[X\mid X>\operatorname{VaR}_p(X)]$. For a discrete loss, the integral convention includes only the required fraction of probability at a quantile atom; it need not equal $E[X\mid X>\operatorname{VaR}_p(X)]$.

For any of these premium principles, the loading is $P-E[X]$. For the variance and standard-deviation principles, it is respectively $\alpha\operatorname{Var}(X)$ and $\beta\sqrt{\operatorname{Var}(X)}$. For the direct VaR and TVaR premiums, it is respectively $\operatorname{VaR}_p(X)-E[X]$ and $\operatorname{TVaR}_p(X)-E[X]$. A VaR loading can be negative: a selected quantile can be below the mean. A negative loading is a property of that choice of confidence level and loss distribution, not an arithmetic error.

Why the Tail Measure Uses an Upper-Tail Average

TVaR is more than the loss at a single percentile: it averages the quantiles from $p$ through the most extreme quantile. The integral definition also explains how to handle an atom at the VaR point. Informally, imagine drawing a percentile $U$ uniformly from $(0,1)$. The loss $\operatorname{VaR}_U(X)$ has the distribution of $X$; restricting $U$ to $(p,1)$ selects the upper $1-p$ fraction of the quantile scale. Its average is the integral in the definition.

In a continuous model, the upper $1-p$ fraction corresponds to the event $X>\operatorname{VaR}_p(X)$, so the integral reduces to the conditional tail expectation. For a discrete model, that event may contain more probability than $1-p$. The integral convention takes only the fraction needed to make up the specified tail probability, which is why using a conditional expectation beyond the VaR point can give the wrong answer.

Calculate the mean and dispersion.
Find $E[X]$, $\operatorname{Var}(X)$, and the standard deviation when the moment-based principles are needed.

Set each principle and parameter.
Record $\alpha$ with its currency units, $\beta$, and the confidence level $p$ before calculating premiums.

Find quantiles and tail averages.
Use the quantile definition for VaR and the integral-quantile convention for TVaR, especially when losses are discrete.

Report premiums and loadings.
Subtract the expected loss from each premium to interpret the risk loading and check whether it is positive.

Worked Comparisons

Worked Example: Four Premiums for a Discrete Loss

A hypothetical policy has loss $X$ equal to $\$0$, $\$100$, or $\$500$, with probabilities $0.40$, $0.40$, and $0.20$, respectively. Calculate the mean and variance, the variance premium for $\alpha=0.001$ per dollar, the standard-deviation premium for $\beta=0.25$, and the VaR and TVaR premiums at $p=0.80$. Report money to the cent.

Method. First calculate the moments directly from the probability distribution. Then use the quantile definition and the integral-quantile convention for the tail premiums.

Mean. The expected loss is $E[X]=0(0.40)+100(0.40)+500(0.20)=40+100=\$140$.

Second moment and variance. The second moment is $E[X^2]=0^2(0.40)+100^2(0.40)+500^2(0.20)=4{,}000+50{,}000=54{,}000$. Therefore, $\operatorname{Var}(X)=E[X^2]-E[X]^2=54{,}000-140^2=54{,}000-19{,}600=34{,}400$ dollars squared. The standard deviation is $\sqrt{34{,}400}=\$185.472\ldots$.

Variance premium. The loading is $0.001(34{,}400)=\$34.40$, so $P_{\mathrm{var}}=140+34.40=\$174.40$.

Standard-deviation premium. The loading is $0.25\sqrt{34{,}400}=0.25(185.472\ldots)=\$46.368\ldots$, so $P_{\mathrm{sd}}=140+46.368\ldots=\$186.37$, rounded to cents.

VaR and TVaR premiums. The cumulative probabilities are $F_X(0)=0.40$, $F_X(100)=0.80$, and $F_X(500)=1$. Thus $\operatorname{VaR}_{0.80}(X)=\$100$. The upper 20% of the quantile scale consists entirely of the $\$500$ outcome, so $\operatorname{TVaR}_{0.80}(X)=\$500$.

Check. In the requested order, the premiums are $\$174.40$, $\$186.37$, $\$100.00$, and $\$500.00$. The VaR loading is $100-140=-\$40$, while the TVaR loading is $500-140=\$360$. The negative VaR loading is consistent with the selected quantile lying below the mean.

Common mistake avoided. Do not calculate the mean as $\$80$: the $\$500$ outcome contributes $500(0.20)=\$100$ to the mean. Also, at $p=0.80$, the VaR is $\$100$, but the upper 20% lies at $\$500$, making TVaR $\$500$, not $\$100$.

Worked Example: Tail Loadings for an Exponential Loss

A hypothetical loss is exponentially distributed with mean $\theta=\$1{,}000$, so its density is $f_X(x)=\frac{1}{1000}e^{-x/1000}$ for $x\geq0$. At $p=0.90$, calculate the variance, standard-deviation, VaR, and TVaR premiums using $\alpha=0.0002$ per dollar and $\beta=0.40$. Report money to the cent.

Moments. For this exponential model, $E[X]=\theta=\$1{,}000$ and $\operatorname{Var}(X)=\theta^2=1{,}000{,}000$ dollars squared. Thus the standard deviation is $\$1{,}000$. The variance loading is $0.0002(1{,}000{,}000)=\$200$, giving $P_{\mathrm{var}}=\$1{,}200.00$. The standard-deviation loading is $0.40(1{,}000)=\$400$, giving $P_{\mathrm{sd}}=\$1{,}400.00$.

VaR. The survival function is $S_X(x)=e^{-x/1000}$. Solving $S_X(q)=1-p=0.10$ gives $q=-1000\ln(0.10)=1000\ln(10)=\$2{,}302.585\ldots$. Therefore $P_{\mathrm{VaR}}=\$2{,}302.59$, with loading $\$1{,}302.59$.

TVaR derivation. For this continuous model, $\operatorname{TVaR}_{0.90}(X)=E[X\mid X>q]$. Integration by parts gives $$ E[X\mathbf{1}_{\{X>q\}}] =\int_q^\infty \frac{x}{1000}e^{-x/1000}\,dx =(q+1000)e^{-q/1000}. $$ Since $P(X>q)=e^{-q/1000}$, division by that probability yields $E[X\mid X>q]=q+1000$. Hence $P_{\mathrm{TVaR}}=\$3{,}302.585\ldots=\$3{,}302.59$, with loading $\$2{,}302.59$.

Check. The premiums are $\$1{,}200.00$, $\$1{,}400.00$, $\$2{,}302.59$, and $\$3{,}302.59$. The tail premiums exceed the mean here because both selected tail measures exceed $\$1{,}000$, and TVaR exceeds VaR because it averages losses beyond the VaR threshold.

Common mistake avoided. The exponential parameter here is the mean, $\theta=\$1{,}000$, not the rate $1/1000$. Using the rate as though it were the mean would produce incorrect quantiles, moments, and monetary units.

Worked Example: Equal Moments, Different Tail Rankings

Compare two hypothetical losses at $p=0.75$. Loss A takes values $\$0,\$100,\$200$ with probabilities $0.25,0.50,0.25$. Loss B takes values $\$0,\$150$ with probabilities $1/3,2/3$. Calculate the mean, variance, VaR and TVaR for both, then compare premiums for $\alpha=0.0001$ per dollar and $\beta=0.50$. Report money to the cent.

Loss A moments. $E[X_A]=0(0.25)+100(0.50)+200(0.25)=\$100$. Also, $E[X_A^2]=100^2(0.50)+200^2(0.25)=5{,}000+10{,}000=15{,}000$, so $\operatorname{Var}(X_A)=15{,}000-100^2=5{,}000$.

Loss B moments. $E[X_B]=0(1/3)+150(2/3)=\$100$. Also, $E[X_B^2]=150^2(2/3)=15{,}000$, so $\operatorname{Var}(X_B)=15{,}000-100^2=5{,}000$. The common standard deviation is $\sqrt{5{,}000}=\$70.7107\ldots$.

Moment-based premiums. Both variance premiums equal $100+0.0001(5{,}000)=\$100.50$. Both standard-deviation premiums equal $100+0.50(70.7107\ldots)=\$135.36$, rounded to cents.

Tail premiums. For A, the cumulative probability at $\$100$ is $0.75$, so $\operatorname{VaR}_{0.75}(X_A)=\$100$. Its upper 25% consists of the $\$200$ outcome, so $\operatorname{TVaR}_{0.75}(X_A)=\$200$. For B, the cumulative probability at $\$0$ is $1/3$, below $0.75$, and the cumulative probability at $\$150$ is 1. Therefore $\operatorname{VaR}_{0.75}(X_B)=\$150$. The upper 25% is drawn entirely from the $\$150$ outcome, so $\operatorname{TVaR}_{0.75}(X_B)=\$150$.

Check. The moment premiums are identical, but the VaR premiums are $\$100$ for A and $\$150$ for B; the TVaR premiums are $\$200$ for A and $\$150$ for B. Equal means and variances do not determine either tail ranking.

Common mistake avoided. Do not assume that matching mean and variance forces the same VaR or TVaR. Also, for A, the VaR loading is zero while the TVaR loading is $\$100$; the two tail measures summarize different parts of the distribution.

Interpreting and Comparing the Principles

The variance principle penalizes squared dispersion. If losses are measured in dollars, variance is measured in dollars squared, so $\alpha$ must convert that quantity back to dollars. The standard-deviation principle adds a multiple of a quantity already measured in dollars. For example, changing the currency unit from dollars to cents changes the numerical variance by a factor of $10{,}000$, but changes the standard deviation by a factor of 100. A variance loading parameter must be adjusted to the chosen currency unit if the premium is to remain economically equivalent.

VaR is a percentile threshold: at confidence level $p$, it marks the loss level at or above which the upper-tail probability is at most $1-p$, subject to possible probability mass at the threshold. TVaR averages the losses in that upper tail. Consequently, TVaR is at least VaR under the integral-quantile convention, but VaR can be below the mean. A direct VaR premium may therefore fail to provide a positive loading, as in the first worked example.

The four methods answer different pricing questions. Moment principles use global summaries of the distribution; tail principles emphasize outcomes at a chosen confidence level. Their numerical premiums are not a fair comparison of the methods unless the loading parameters and confidence level are specified. In particular, choosing a large $\alpha$ or $\beta$ can make a moment premium exceed a tail premium, and choosing a higher $p$ generally increases VaR and TVaR.

Key takeaway: State the loss model, the premium principle, its parameter or confidence level, and the resulting loading. Check units for the variance principle, use the integral-quantile convention for discrete TVaR, and do not assume that VaR must exceed expected loss.

Common Exam Traps

Using the wrong second moment. Calculate $E[X^2]$ by squaring each loss before weighting; then subtract $E[X]^2$ to obtain the variance.
Confusing variance and standard deviation. The variance premium uses $\alpha\operatorname{Var}(X)$; the standard-deviation premium uses $\beta\sqrt{\operatorname{Var}(X)}$.
Ignoring parameter units. A numerical value for $\alpha$ is incomplete unless its currency units are understood.
Using a strict conditional tail incorrectly for a discrete loss. If there is an atom at VaR, use the integral-quantile definition of TVaR rather than automatically conditioning on losses strictly greater than VaR.
Calling every risk loading positive. Direct VaR minus the mean can be negative at a lower confidence level or for a skewed distribution.

Check Your Understanding

For each problem, identify the requested premium principle and use the stated quantile convention. Report monetary answers to the nearest cent.

A hypothetical loss is $\$0$ or $\$200$, each with probability $0.50$. Find its mean and variance, then calculate the variance premium for $\alpha=0.002$ per dollar and the standard-deviation premium for $\beta=0.30$. (Answer: mean $\$100$, variance $10{,}000$, premiums $\$120.00$ and $\$130.00$)
A hypothetical loss is uniform on $[0,\$400]$. Find its mean, variance, VaR premium at $p=0.75$, and TVaR premium at $p=0.75$. (Answer: mean $\$200$, variance $13{,}333.33$, VaR $\$300.00$, TVaR $\$350.00$)
A hypothetical exponential loss has mean $\$500$. At $p=0.80$, calculate the VaR and TVaR premiums. (Answer: VaR $\$804.72$, TVaR $\$1{,}304.72$)
A hypothetical loss is $\$0,\$100,\$400$ with probabilities $0.20,0.50,0.30$. At $p=0.70$, calculate the VaR and TVaR premiums using the integral-quantile convention. (Answer: VaR $\$100.00$, TVaR $\$400.00$)
Loss C is $\$0$ or $\$200$, each with probability $0.50$. Loss D is $\$0$ with probability $0.80$ and $\$200$ with probability $0.20$. For each loss, calculate the mean and variance, then compare the variance premiums for $\alpha=0.001$ per dollar and the TVaR premiums at $p=0.80$. (Answer: C has mean $\$100$, variance $10{,}000$, variance premium $\$110.00$, and TVaR $\$200.00$; D has mean $\$40$, variance $6{,}400$, variance premium $\$46.40$, and TVaR $\$200.00$)

What You'll Learn