Parametric TVaR Estimation From Fitted Loss Models

From Observed Losses to a Fitted TVaR

TVaR is a feature of a loss distribution, while a data set contains only a finite collection of observed losses. A parametric estimate connects the two: choose a distributional family, estimate its parameters from the observations, and substitute those estimates into the family’s quantile and tail-mean formulas. The resulting plug-in estimate is useful only to the extent that the chosen model and its assumptions are suitable for the losses being studied.

This tutorial focuses on that fitting-and-substitution process. It is aligned with the CAS Exam 9 (Risk Management for Actuaries) syllabus. The examples use complete observations and specified parametric families; they do not address estimation from censored data or simulation.

Definition: For a loss $X$, the $p$-quantile is $\operatorname{VaR}_p(X)=F_X^{-1}(p)$, using the generalized inverse $F_X^{-1}(p)=\inf\{x:F_X(x)\ge p\}$. For a continuous loss distribution with positive probability above this quantile, $\operatorname{TVaR}_p(X)=E[X\mid X>\operatorname{VaR}_p(X)]$. A parametric plug-in estimate replaces unknown model parameters in these quantities with estimates obtained from data.

What Is Being Estimated?

A parametric model specifies a family of distributions and a parameter vector. For example, a normal model has parameters $\mu$ and $\sigma^2$, while an exponential model can be written with rate $\lambda$. Fitting the model means using observations to estimate its unknown parameters. The fitted TVaR is then a model-based estimate of the tail mean, not the average of the largest observations in the sample.

A common fitting method is maximum likelihood. If $x_1,\ldots,x_n$ are independent observations from a density $f(x;\theta)$, their likelihood is $L(\theta)=\prod_{j=1}^n f(x_j;\theta)$. A maximum-likelihood estimate (MLE) is a parameter value that maximizes this likelihood, or equivalently its logarithm. The method relies on the assumed distribution and independence structure; it does not establish that those assumptions are true.

Modeling assumptions to state: Identify the loss unit and observation period; specify the distributional family and any fixed parameters; state whether observations are treated as independent and identically distributed; name the parameter-estimation method; and report the confidence level. A plug-in TVaR treats estimated parameters as fixed when applying the tail formula, so it does not by itself quantify sampling uncertainty in the estimate.

Normal Model: Fit, Then Calculate the Tail Mean

Suppose $X\sim N(\mu,\sigma^2)$, with $\sigma>0$. For independent, complete observations, the normal MLEs are $\hat{\mu}=\bar{x}$ and $\hat{\sigma}^2=\frac{1}{n}\sum_{j=1}^n(x_j-\bar{x})^2$. The variance estimate uses divisor $n$, not $n-1$; the latter gives the usual unbiased sample variance, but is not the normal maximum-likelihood estimate.

Let $z_p$ satisfy $\Phi(z_p)=p$, where $\Phi$ and $\phi$ are the standard normal distribution and density functions. For a continuous normal loss, the fitted formulas are

\widehat{\operatorname{VaR}}_p=\hat{\mu}+\hat{\sigma}z_p, \qquad \widehat{\operatorname{TVaR}}_p =\hat{\mu}+\hat{\sigma}\frac{\phi(z_p)}{1-p}.

The second formula is a tail mean, not a quantile. Its multiplier $\phi(z_p)/(1-p)$ is generally greater than $z_p$, so the fitted TVaR exceeds the fitted VaR.

Why the normal tail-mean formula holds: Write $X=\mu+\sigma Z$, where $Z$ is standard normal. The event $X>\mu+\sigma z_p$ is $Z>z_p$, which has probability $1-p$. Therefore $E[X\mid X>\mu+\sigma z_p]=\mu+\sigma E[Z\mid Z>z_p]$. Since $\phi'(z)=-z\phi(z)$, integration gives $\int_{z_p}^{\infty}z\phi(z)\,dz=\phi(z_p)$. Dividing this upper-tail integral by $1-p$ yields $E[Z\mid Z>z_p]=\phi(z_p)/(1-p)$, proving the formula.

Worked Example: Normal MLE From Four Losses

A hypothetical portfolio has four independent, complete annual losses, in thousands of dollars: 98, 100, 102, and 104. Assume these observations come from a normal distribution and estimate its parameters by maximum likelihood. Calculate the fitted 95% VaR and TVaR. Use $z_{0.95}=1.6449$ and $\phi(z_{0.95})=0.103135$; use these displayed approximations throughout.

Method and fitted parameters. The sample mean is $\bar{x}=(98+100+102+104)/4=404/4=101$. The squared deviations from 101 are $9,1,1,9$, with sum $20$. Thus the MLE variance is $20/4=5$ and $\hat{\sigma}=\sqrt{5}\approx2.236068$, in thousands of dollars.

Fitted VaR. Substitute the fitted mean, fitted standard deviation, and stated quantile: $101+\sqrt{5}(1.6449)=101+3.678108\ldots=104.6781$ thousand dollars, rounded to four decimals.

Fitted TVaR. The normal tail multiplier is $0.103135/(1-0.95)=0.103135/0.05=2.0627$. Therefore, $101+\sqrt{5}(2.0627)=101+4.612337\ldots=105.6123$ thousand dollars, rounded to four decimals.

Check and common mistake. The fitted tail mean, 105.6123, exceeds the fitted 95% quantile, 104.6781, as expected. A common error is to report $101+\sqrt{5}(1.6449)$ as TVaR; that is the fitted VaR because it uses the quantile multiplier rather than the tail-mean multiplier.

A normal model is mathematically convenient, but it assigns positive probability to negative values. For loss data, that feature should be considered rather than ignored. A normal fit may be a reasonable approximation in some settings, but if negative losses are implausible or the upper tail is poorly represented, the family may be unsuitable for the intended use.

Exponential Model: Rate and Memoryless Tail

For a nonnegative loss $X$ with exponential rate $\lambda>0$, the survival function is $S_X(x)=e^{-\lambda x}$ for $x\ge0$, and $E[X]=1/\lambda$. For independent complete observations, the likelihood is proportional to $\lambda^n e^{-\lambda\sum x_j}$; maximizing it gives $\hat{\lambda}=n/\sum x_j=1/\bar{x}$.

The $p$-quantile solves $e^{-\lambda q}=1-p$, so $q=-\ln(1-p)/\lambda$. The exponential distribution has the memoryless property: conditional on exceeding $q$, the amount in excess of $q$ has the same exponential distribution as $X$. Thus $E[X\mid X>q]=q+1/\lambda$, and the plug-in formula is

\widehat{\operatorname{TVaR}}_p =\frac{-\ln(1-p)+1}{\hat{\lambda}} =\bar{x}\bigl[-\ln(1-p)+1\bigr].

Worked Example: Exponential Fit to Complete Losses

A hypothetical insurer records three independent, complete claim amounts of 1, 3, and 8 thousand dollars. Assume an exponential model with an unknown rate and estimate it by maximum likelihood. Find the fitted 90% TVaR.

Fit the rate. The total loss is $1+3+8=12$ thousand dollars, so the mean is $12/3=4$ thousand dollars. The MLE is $\hat{\lambda}=3/12=0.25$ per thousand dollars. Equivalently, the fitted mean is $1/0.25=4$ thousand dollars.

Calculate the fitted quantile. Since $-\ln(0.10)=2.302585$, the fitted 90% VaR is $2.302585/0.25=9.210340$ thousand dollars.

Calculate the fitted TVaR. Add the exponential mean excess $1/\hat{\lambda}=4$: $9.210340+4=13.210340$ thousand dollars. The fitted 90% TVaR is therefore 13.2103 thousand dollars, rounded to four decimals.

Check and common mistake. The result exceeds the fitted quantile by exactly the fitted mean, as the exponential memoryless property requires. A rate-unit error can produce an incorrect answer: because losses were entered in thousands of dollars, the fitted rate is per thousand dollars, and the calculated risk amounts are also in thousands of dollars.

A Fixed Threshold Is Not Necessarily an Estimated Parameter

Some analyses model losses only above a known threshold. If $X=c+Z$, where $c$ is a fixed threshold and $Z$ is exponential with rate $\lambda$, then the observations used to fit the excess model are $x_j-c$. For a fixed, specified $c$, the MLE is $\hat{\lambda}=n/\sum_j(x_j-c)$. The tail quantile and tail mean for $X$ are each the corresponding quantity for $Z$, plus $c$. This calculation assumes the threshold is fixed in advance; estimating a threshold is a different fitting problem.

Worked Example: Exponential Excesses Above a Fixed Threshold

A hypothetical severity model treats losses above a fixed threshold of 2 thousand dollars as $X=2+Z$, where $Z$ is exponential. Three complete losses are 4, 6, and 8 thousand dollars. Estimate the exponential rate from the excesses and calculate the fitted 80% TVaR for the original loss $X$.

Fit the excess distribution. The excesses are $4-2=2$, $6-2=4$, and $8-2=6$ thousand dollars. Their sum is $12$, their mean is $12/3=4$, and the rate MLE is $\hat{\lambda}=3/12=0.25$ per thousand dollars.

Calculate the excess tail mean. Since $-\ln(1-0.80)=-\ln(0.20)=1.609438$, the fitted 80% VaR for $Z$ is $1.609438/0.25=6.437752$ thousand dollars. Its fitted TVaR is $6.437752+1/0.25=10.437752$ thousand dollars.

Restore the threshold. Because $X=2+Z$, add 2 thousand dollars to the excess TVaR. The fitted TVaR for $X$ is $10.437752+2=12.437752$, or 12.4378 thousand dollars rounded to four decimals.

Check and common mistake. The result is the excess tail mean plus the fixed threshold, not the excess tail mean alone. The threshold is added once because it is a fixed location shift in every modeled loss; it is not included when estimating the exponential rate from excesses.

Practical Reporting and Limits of a Plug-In Estimate

A useful report makes the calculation reproducible. State the fitted family, parameter estimates, fitting method, confidence level, and units. Also state material assumptions, including whether the data are treated as independent and identically distributed and whether the observations are complete. In the examples, the numbers are hypothetical and the model assumptions are imposed for illustration.

The result is a plug-in estimate: uncertainty in the estimated parameters is not included in the displayed TVaR. Two samples from the same underlying process can produce different fitted parameters and hence different fitted TVaRs. The calculation also depends on the chosen distributional family, especially in the tail. A precise numerical result does not remove that dependence.

Parameter estimates must match the model used in the tail formula. For example, the exponential rate MLE from complete observations is $n/\sum x_j$, not $1/\sum x_j$; the latter has the wrong units and omits the number of observations. Similarly, a normal MLE variance uses divisor $n$, and a fixed-threshold exponential fit uses excesses rather than the original losses. These distinctions can materially change the resulting tail estimate.

Key takeaway: Parametric TVaR estimation has three linked steps: specify a model and its assumptions, estimate its parameters from data using a stated method, and substitute those estimates into the model’s quantile and tail-mean formulas. Check the parameterization, units, and tail multiplier before reporting the result.

Check Your Understanding

For each problem, treat observations as independent and complete where a fitting assumption is needed. Answers are rounded as stated.

Losses in thousands of dollars are 7, 9, and 11. Fit a normal model by maximum likelihood and calculate the 90% TVaR. Use $z_{0.90}=1.2816$ and $\phi(z_{0.90})=0.1755$. (Answer: approximately 11.87 thousand dollars.)
Exponential losses in thousands of dollars are 2, 5, and 5. Fit the rate by maximum likelihood and calculate the 95% TVaR. Use $-\ln(0.05)=2.995732$. (Answer: 15.9829 thousand dollars.)
A fixed threshold is 3 thousand dollars, and losses are 5, 7, and 9 thousand dollars. Fit an exponential model to the excesses and calculate the 50% TVaR for the original loss. Use $\ln 2=0.693147$. (Answer: 9.7726 thousand dollars.)
A normal model is fitted by maximum likelihood with $\hat{\mu}=50$ and $\hat{\sigma}=10$, in thousands of dollars. Calculate the 97.5% TVaR using $z_{0.975}=1.96$ and $\phi(1.96)=0.05844$. (Answer: 73.3760 thousand dollars.)
Two complete exponential losses are $1,000 and $3,000. Fit the rate by maximum likelihood and calculate the 80% TVaR in dollars. Use $-\ln(0.20)=1.609438$. (Answer: $5,218.88.)

What You'll Learn