Simulation Estimates of Value-at-Risk and Tail Value-at-Risk

Why Estimate Tail Risk by Simulation?

For a complex aggregate-loss model, a direct calculation of its full probability distribution may be difficult or impractical. Simulation offers another route: generate many aggregate losses under the model, sort them, and use the upper part of the simulated sample to estimate value-at-risk and tail value-at-risk. The result depends on the simulated outcomes, so the estimates need not equal the model’s exact risk measures.

The key is to define precisely which simulated observations enter each estimate. A rank error can materially change a high-quantile estimate, and simply averaging every observation at or above the estimated value-at-risk can give the wrong tail average when there are ties.

Definition: Let $X$ be an aggregate loss with distribution function $F_X$. At confidence level $\alpha$, where $0<\alpha<1$, define value-at-risk by $$ \operatorname{VaR}_{\alpha}(X)=\inf\{x:F_X(x)\geq \alpha\}. $$ Define tail value-at-risk using the quantile integral $$ \operatorname{TVaR}_{\alpha}(X)=\frac{1}{1-\alpha}\int_{\alpha}^{1}\operatorname{VaR}_{p}(X)\,dp. $$ This definition applies to both continuous and discrete loss distributions. In a continuous model, when the relevant expectations exist, it agrees with the mean loss conditional on exceeding the $\alpha$-quantile. For a discrete distribution, the quantile-integral definition accounts for only the necessary fraction of any probability mass at the boundary.

Order Statistics from Simulated Losses

Suppose $X_1,\ldots,X_n$ are independent simulated outcomes from the aggregate-loss model of interest. Arrange them from smallest to largest and write the ordered values as $X_{(1)}\leq X_{(2)}\leq\cdots\leq X_{(n)}$. The empirical distribution assigns probability $1/n$ to each simulated outcome. Its quantile at level $\alpha$ is therefore the observation at rank $\lceil n\alpha\rceil$.

Simulation estimates: Under the empirical-quantile convention, $$ \widehat{\operatorname{VaR}}_{\alpha}=X_{(\lceil n\alpha\rceil)}. $$ For the corresponding empirical quantile-integral estimate of tail value-at-risk, set $r=n(1-\alpha)$, $k=\lfloor r\rfloor$, and $w=r-k$. Then $$ \widehat{\operatorname{TVaR}}_{\alpha} =\frac{\displaystyle\sum_{j=0}^{k-1}X_{(n-j)}+wX_{(n-k)}}{r}. $$ The sum is zero when $k=0$. Thus, the estimate includes the largest $k$ simulated losses in full and a fraction $w$ of the next loss. If $r$ is an integer, $w=0$ and the estimate is the average of the largest $r$ outcomes.

The value $r$ is the number of observations corresponding to the upper $1-\alpha$ fraction of the empirical distribution. It need not be an integer. When it is not, the fractional weight is not an arbitrary adjustment: it is the exact portion of the boundary step of the empirical quantile function included in the integral from $\alpha$ to 1.

Why the Tail-Average Formula Works

The empirical quantile function is constant over each probability interval of width $1/n$. Specifically, it equals $X_{(j)}$ over the interval from $(j-1)/n$ to $j/n$, apart from endpoint conventions that do not affect the integral. Integrating this step function from $\alpha$ to 1 adds the contributions of the upper order statistics. Each complete interval contributes its order statistic times $1/n$; if the integration starts partway through an interval, only the remaining fraction of that interval contributes.

Multiplying the integral by $n$ makes the total length of the integrated tail equal to $n(1-\alpha)=r$ observation-units. There are $k$ complete upper observations and a remaining fraction $w$, where $r=k+w$. The resulting sum is divided by $r$, giving the stated tail-average estimate. This also explains how the method handles ties: it weights positions in the ordered empirical distribution rather than including every tied value above a threshold.

Generate aggregate losses.
Simulate $n$ independent outcomes using the specified aggregate-loss model and retain the simulated values in consistent monetary units.

Sort and locate VaR.
Order the losses and take rank $\lceil n\alpha\rceil$, not rank $n\alpha$ rounded by an unrelated convention.

Average the upper tail.
Calculate $r=n(1-\alpha)$, then include the largest $k=\lfloor r\rfloor$ values and the fraction $w=r-k$ of the next value.

Compare with model values when available.
Use exact or otherwise known values as a benchmark, while recognizing that a finite simulation estimate can differ from that benchmark.

Worked Simulation Estimates

Worked Example: An Integer-Sized Tail Under an Exponential Model

A hypothetical insurer models one-year aggregate loss $X$ as exponential with mean $\theta=\$1{,}000$, so $F_X(x)=1-e^{-x/1000}$ for $x\geq0$. Eight independent simulated losses, in dollars, are $400,550,700,850,1{,}000,1{,}350,1{,}800,2{,}600$. Estimate $\operatorname{VaR}_{0.75}$ and $\operatorname{TVaR}_{0.75}$, then compare them with the exact model values.

Method: sort and use the empirical quantile and tail average. The values are already sorted, $n=8$, and $\alpha=0.75$. The VaR rank is $$ \lceil n\alpha\rceil=\lceil8(0.75)\rceil=\lceil6\rceil=6. $$ Thus, $\widehat{\operatorname{VaR}}_{0.75}=X_{(6)}=\$1{,}350$.

The tail size is $r=8(1-0.75)=2$, so $k=2$ and $w=0$. The tail estimate is the mean of the two largest observations: $$ \widehat{\operatorname{TVaR}}_{0.75} =\frac{\$2{,}600+\$1{,}800}{2} =\$2{,}200. $$

For an exponential loss with mean $\theta$, $F_X(x)=1-e^{-x/\theta}$. Solving $F_X(x)=\alpha$ gives $\operatorname{VaR}_{\alpha}(X)=-\theta\ln(1-\alpha)$. The exponential memoryless property gives $\operatorname{TVaR}_{\alpha}(X)=\operatorname{VaR}_{\alpha}(X)+\theta$. Therefore, at $\alpha=0.75$, the exact values are $$ \operatorname{VaR}_{0.75}(X)=-\$1{,}000\ln(0.25)=\$1{,}386.29, \qquad \operatorname{TVaR}_{0.75}(X)=\$1{,}386.29+\$1{,}000=\$2{,}386.29. $$ The simulation estimates are below these exact values in this particular sample. This is a sample comparison, not a claim that simulation must systematically underestimate. The calculations avoid the mistake of averaging all eight losses when estimating tail value-at-risk.

Worked Example: A Fractional Tail Under an Exponential Model

For a second hypothetical aggregate-loss model, let $X$ be exponential with mean $\theta=\$1{,}200$. A simulation produces these ten ordered losses, in dollars: $250,400,500,650,800,950,1{,}100,1{,}400,1{,}900,3{,}000$. Estimate both risk measures at $\alpha=0.75$, and compare with the exact values.

Method: use the empirical quantile, then fractionally weight the boundary loss. Here $n\alpha=10(0.75)=7.5$, so the VaR rank is $\lceil7.5\rceil=8$, and $\widehat{\operatorname{VaR}}_{0.75}=\$1{,}400$.

The tail size is $r=10(0.25)=2.5$, giving $k=2$ and $w=0.5$. Include the two largest losses in full and half of the next largest: $$ \widehat{\operatorname{TVaR}}_{0.75} =\frac{\$3{,}000+\$1{,}900+0.5(\$1{,}400)}{2.5} =\frac{\$5{,}600}{2.5} =\$2{,}240. $$

The exact exponential-model values are $$ \operatorname{VaR}_{0.75}(X)=-\$1{,}200\ln(0.25)=\$1{,}663.55, \qquad \operatorname{TVaR}_{0.75}(X)=\$1{,}663.55+\$1{,}200=\$2{,}863.55. $$ Amounts are rounded to the nearest cent. Both simulated estimates are below the exact values for these ten outcomes. The common mistake is to round the tail size $2.5$ to 3 and average the three largest losses; that gives $(\$3{,}000+\$1{,}900+\$1{,}400)/3=\$2{,}100$, which does not equal the quantile-integral estimate. The boundary observation must receive weight 0.5, not weight 1.

Worked Example: Ties in a Discrete Aggregate-Loss Model

A hypothetical portfolio has aggregate loss $X$ with probabilities $P(X=\$0)=0.5$, $P(X=\$100)=0.3$, and $P(X=\$300)=0.2$. Eight simulated outcomes are $\$0,\$0,\$0,\$100,\$100,\$100,\$300,\$300$. Estimate VaR and TVaR at $\alpha=0.75$, and compare with the exact model values.

Method: use the ordered sample and the quantile-integral tail convention. With $n=8$, the empirical VaR rank is $\lceil8(0.75)\rceil=6$, so $\widehat{\operatorname{VaR}}_{0.75}=\$100$. The empirical tail size is $r=8(0.25)=2$; the largest two outcomes are both $\$300$. Hence $\widehat{\operatorname{TVaR}}_{0.75}=(\$300+\$300)/2=\$300$.

For the exact distribution, the cumulative probabilities at $\$0,\$100,\$300$ are 0.5, 0.8, and 1.0. The smallest loss whose cumulative probability is at least 0.75 is $\$100$, so $\operatorname{VaR}_{0.75}(X)=\$100$. The upper 25% of probability consists exactly of the 20% mass at $\$300$ and 5% of the mass at $\$100$. Thus $$ \operatorname{TVaR}_{0.75}(X) =\frac{0.20(\$300)+0.05(\$100)}{0.25} =\frac{\$65}{0.25} =\$260. $$ The simulation tail estimate is $\$300$, compared with the exact $\$260$. Averaging every simulated loss at least as large as the VaR would include three $\$100$ outcomes and two $\$300$ outcomes, giving $(3(\$100)+2(\$300))/5=\$180$. That is not the quantile-integral estimate: it includes more than the required upper-quarter mass. This example also shows why the conditional mean above a threshold is not interchangeable with the quantile-integral definition at an atom.

Interpreting and Checking the Estimates

A simulation estimate is a summary of the generated outcomes, not an exact replacement for the model’s distribution. When a known model value is available, comparing it with the estimate is a useful check on the calculation and a way to see how the particular simulated sample behaves. A difference by itself does not establish an error; first verify the model, confidence level, ordering, and tail weights.

A disciplined calculation makes the procedure auditable. Record the confidence level and number of simulated outcomes, display the relevant order-statistic rank, and show which observations contribute to the tail average. In a fractional-tail calculation, state the boundary weight explicitly. In a sample with ties, follow the chosen quantile-integral convention rather than treating every observation equal to a threshold as automatically belonging to the tail.

Check the direction. VaR at a high confidence level should come from the upper portion of the ordered losses, not the lower portion.
Check the rank. Under the stated empirical convention, the rank is $\lceil n\alpha\rceil$.
Check the tail size. The denominator for the tail average is $r=n(1-\alpha)$, including when $r$ is fractional.
Check the units. If the simulated losses are in thousands of dollars, keep the calculations in thousands and convert only when reporting.
Check ties deliberately. For a discrete distribution, distinguish the quantile-integral tail average from the average of all outcomes at or above estimated VaR.

Key takeaway: Sort the simulated aggregate losses, estimate VaR with the empirical quantile order statistic, and estimate TVaR by averaging exactly the upper $n(1-\alpha)$ observation-units, using a fractional boundary weight when needed. Compare with known model values only after applying the same risk-measure convention.

Check Your Understanding

For each sample, use the empirical quantile and quantile-integral tail-average conventions stated above. Amounts are in dollars unless otherwise specified.

A simulation of $n=12$ aggregate losses gives the ordered values $100,150,180,220,260,300,350,400,500,650,900,1{,}200$. Find the estimated VaR and TVaR at $\alpha=0.75$. (Answer: VaR $=\$500$; TVaR $=\$916.67$)
Eight ordered simulated losses are $40,60,75,90,110,160,240,500$. Find both estimates at $\alpha=0.875$. (Answer: VaR $=\$240$; TVaR $=\$500$)
A model specifies exponential aggregate loss with mean $\$800$. Calculate its exact VaR and TVaR at $\alpha=0.90$, using the exponential-model formulas. (Answer: VaR $=\$1{,}842.07$; TVaR $=\$2{,}642.07$)
Six ordered simulated losses are $10,20,30,50,80,140$. Estimate VaR and TVaR at $\alpha=0.75$, showing the fractional boundary weight. (Answer: VaR $=\$80$; TVaR $=\$120$)
A discrete aggregate-loss model has $P(X=\$0)=0.4$, $P(X=\$50)=0.4$, and $P(X=\$200)=0.2$. A simulation of eight outcomes contains four zeros, two losses of $\$50$, and two losses of $\$200$. Find the sample estimates and the exact model values at $\alpha=0.75$. (Answer: sample VaR $=\$50$, sample TVaR $=\$200$; exact VaR $=\$50$, exact TVaR $=\$170$)

What You'll Learn