A Worked VaR and TVaR Calculation Workshop

Why Tail Calculations Need Care

VaR identifies a loss threshold at a chosen confidence level; TVaR summarizes the losses in the tail beyond that level. Calculating both requires attention to how probability accumulates at the threshold. For a continuous loss distribution, the probability at one exact value is zero. For a finite or mixed distribution, there may be a positive probability at the VaR itself, and part of that probability can belong in the tail average.

The method below applies to a loss random variable $X$ with a finite mean and a confidence level $p$ satisfying $0<p<1$. Losses are treated as positive amounts, and results are stated in the same units as $X$. The quantile convention is the generalized inverse of the cumulative distribution function.

Definition: Let $F_X(x)=P(X\leq x)$. The Value-at-Risk at confidence level $p$ is $$ \operatorname{VaR}_p(X)=\inf\{x:F_X(x)\geq p\}. $$ The Tail Value-at-Risk is the mean of the upper $1-p$ fraction of the loss distribution, defined for distributions with finite mean by $$ \operatorname{TVaR}_p(X)=\frac{1}{1-p}\int_p^1 \operatorname{VaR}_u(X)\,du. $$

A Tail Formula That Includes Probability at VaR

Write $q=\operatorname{VaR}_p(X)$. The defining quantile convention gives $F_X(q-)\leq p\leq F_X(q)$, where $F_X(q-)=P(X<q)$. All probability strictly above $q$ belongs in the upper tail. If $P(X=q)>0$, only enough of the probability at $q$ is added to make the tail probability exactly $1-p$.

This gives a useful calculation formula:

\operatorname{TVaR}_p(X) = \frac{E[X\,\mathbf{1}_{\{X>q\}}]+(F_X(q)-p)q}{1-p}.

Here $\mathbf{1}_{\{X>q\}}$ is 1 when $X>q$ and 0 otherwise. The amount $F_X(q)-p$ is the probability from the atom at $q$ needed for the tail average. If there is no atom at $q$, then $F_X(q)=p$ for a continuous distribution and this term is zero.

Locate the quantile.
Find the first $q$ for which $F_X(q)\geq p$, and check the probability immediately below $q$ when the distribution has an atom.

Identify the upper-tail probability.
The tail used for TVaR must have total probability $1-p$. Include all losses above $q$, then the required fraction of probability at $q$.

Average the selected tail.
Calculate the probability-weighted loss in that tail and divide by $1-p$. Check that the probabilities included total $1-p$.

Why the Tail Formula Holds

The quantile-integral definition averages the quantiles from level $p$ to level 1. These quantiles represent the upper $1-p$ fraction of the distribution, ordered from the threshold upward. Every outcome above $q$ is therefore included. When a loss atom occurs at $q$, the quantile remains equal to $q$ across the probability interval occupied by that atom. Only the portion of that interval above level $p$ is included, and its probability is $F_X(q)-p$.

Consequently, the total probability selected is $P(X>q)+F_X(q)-p$. Since $P(X>q)=1-F_X(q)$, this equals $1-p$. The corresponding total loss is $E[X\mathbf{1}_{\{X>q\}}]+q(F_X(q)-p)$; dividing by $1-p$ gives the stated TVaR formula. This argument also explains why, for a continuous distribution, TVaR can be calculated as the conditional mean above $q$: the probability at $q$ is zero, so the selected upper tail is simply $X>q$.

Worked Example: A Finite-Support Loss with a VaR Atom

A hypothetical equipment insurer models one policy-year loss $X$ in dollars as follows: $X=0,100,300,$ or $700$, with probabilities $0.40,0.30,0.20,$ and $0.10$, respectively. Calculate VaR and TVaR at $p=0.80$. The probabilities sum to $0.40+0.30+0.20+0.10=1.00$, so this is a valid distribution.

Method. Accumulate the probabilities to locate the generalized-inverse quantile, then use the atom-adjusted upper-tail formula.

The cumulative probabilities are $F_X(0)=0.40$, $F_X(100)=0.70$, $F_X(300)=0.90$, and $F_X(700)=1.00$. Since $F_X(100)=0.70<0.80$ and $F_X(300)=0.90\geq0.80$,

\operatorname{VaR}_{0.80}(X)=300.

The required upper-tail probability is $1-0.80=0.20$. Losses strictly above 300 contribute probability $0.10$, all at 700. The amount of probability at 300 to include is $F_X(300)-p=0.90-0.80=0.10$. The selected tail probability is therefore $0.10+0.10=0.20$, as required.

\operatorname{TVaR}_{0.80}(X) = \frac{(700)(0.10)+(300)(0.10)}{0.20} = \frac{70+30}{0.20} = 500.

Check. The selected tail contains probability $0.20$ and weighted loss $100$, so its average is $100/0.20=500$. The result lies between the selected losses 300 and 700. A common mistake is to calculate $E[X\mid X>300]=700$; that conditions on probability $0.10$, not the required upper-tail probability $0.20$, and omits the necessary part of the atom at 300.

Continuous Losses: Conditional Tail Means

For a continuous distribution with a strictly increasing CDF around $q$, $F_X(q)=p$ and $P(X=q)=0$. The upper-tail formula then reduces to

\operatorname{TVaR}_p(X)=E[X\mid X>q] =\frac{E[X\mathbf{1}_{\{X>q\}}]}{1-p}.

The conditional-mean form is convenient when the conditional distribution above the quantile is easy to identify. It should not be used without checking continuity at the threshold: when there is positive probability at $q$, conditioning only on $X>q$ may select too little probability.

Worked Example: A Uniform Loss Distribution

Suppose a hypothetical covered loss $X$, in dollars, is uniformly distributed from 200 to 1,200. Calculate VaR and TVaR at $p=0.90$. Under this model, the density is $1/1000$ on $[200,1200]$, and the CDF on that interval is $F_X(x)=(x-200)/1000$.

Method. Solve $F_X(q)=0.90$, then average the uniform distribution over the remaining upper-tail interval.

Set $(q-200)/1000=0.90$. Thus $q-200=900$, so $q=1{,}100$ dollars. The CDF check is $F_X(1{,}100)=900/1000=0.90$, giving $\operatorname{VaR}_{0.90}(X)=\$1{,}100$.

The upper 10% of this uniform distribution is the interval from 1,100 to 1,200. Its conditional mean is the midpoint:

\operatorname{TVaR}_{0.90}(X) = E[X\mid X>1{,}100] = \frac{1{,}100+1{,}200}{2} = \$1{,}150.

Check. The interval from 1,100 to 1,200 has probability $100/1000=0.10$, exactly $1-p$, and its average loss is \$1,150. The result exceeds the VaR threshold, as expected for the average of losses in the upper tail.

Mixed Distributions: A Point Mass and a Continuous Component

A mixed distribution combines discrete probability at one or more values with a continuous component. The CDF can jump at an atom and increase continuously between jumps. For such a model, locate VaR by checking both the value just below a candidate threshold and the CDF at the threshold; then use the atom-adjusted formula rather than automatically taking a strict conditional tail mean.

Worked Example: A Mixed Loss with an Atom at VaR

A hypothetical contract has loss $X=100$ dollars with probability $0.30$. With probability $0.70$, its loss is uniformly distributed from 0 to 1,000 dollars. Calculate VaR and TVaR at $p=0.20$. The mixture probabilities sum to $0.30+0.70=1.00$.

Method. Find the jump in the mixture CDF at 100, then combine the continuous losses strictly above 100 with the required portion of the atom.

Immediately below 100, only the uniform component contributes, so $F_X(100-)=0.70(100/1000)=0.07$. At 100, the point mass is added: $F_X(100)=0.07+0.30=0.37$. Since $0.07<0.20\leq0.37$,

\operatorname{VaR}_{0.20}(X)=100.

The required tail probability is $1-0.20=0.80$. The probability of a loss strictly above 100 is $0.70(900/1000)=0.63$. The remaining probability needed at the threshold is $F_X(100)-p=0.37-0.20=0.17$. The tail probability check is $0.63+0.17=0.80$.

Conditional on coming from the uniform component and exceeding 100, the loss is uniform from 100 to 1,000, with mean $(100+1{,}000)/2=550$. The weighted loss above 100 is therefore $0.63(550)=346.50$. The selected mass at 100 contributes $0.17(100)=17.00$.

\operatorname{TVaR}_{0.20}(X) = \frac{346.50+17.00}{0.80} = \frac{363.50}{0.80} = \$454.38.

Check. The calculation averages exactly 80% of the probability, including 63% strictly above 100 and 17% at 100. Conditioning only on $X>100$ would average a tail with probability 0.63, rather than the required 0.80, and would give a different quantity.

Common Calculation Traps

Using the wrong quantile convention. VaR is the first value where the CDF reaches or exceeds $p$, not necessarily a value satisfying $F_X(q)=p$. In a jump, the CDF can cross $p$ without equaling it.
Ignoring the atom at the threshold. For a discrete or mixed loss, $X>q$ may have probability less than $1-p$. Include probability $F_X(q)-p$ at $q$ to complete the tail.
Dividing by the wrong probability. TVaR at level $p$ averages a tail of probability $1-p$, not $p$. Check the included probabilities before dividing.
Relying on a continuous-tail formula in a mixed model. The expression $E[X\mid X>q]$ applies directly when there is no mass at $q$. Otherwise use the atom-adjusted calculation.

Key takeaway: First verify the quantile using the CDF. Then construct a tail with probability exactly $1-p$. For a continuous distribution this is usually the region above VaR; for a distribution with an atom at VaR, part of the atom may also be needed.

Check Your Understanding

For each problem, calculate VaR and TVaR at the stated confidence level. Loss amounts are in dollars unless otherwise stated.

A loss takes values 0, 50, 120, and 400 with probabilities 0.20, 0.35, 0.25, and 0.20. Find VaR and TVaR at $p=0.60$. (Answer: VaR = $120; TVaR = $260.)
A loss takes values 0, 200, 500, and 1,000 with probabilities 0.50, 0.20, 0.20, and 0.10. Find VaR and TVaR at $p=0.75$. (Answer: VaR = $500; TVaR = $700.)
A loss is uniform from $200 to $1,200. Find VaR and TVaR at $p=0.90$. (Answer: VaR = $1,100; TVaR = $1,150.)
A loss is $300 with probability 0.40; otherwise it is uniform from $0 to $1,000. Find VaR and TVaR at $p=0.30$. (Answer: VaR = $300; TVaR = $510.)
A loss has an exponential distribution with rate $0.01$ per dollar, so $S_X(x)=e^{-0.01x}$ for $x\geq0$. Find VaR and TVaR at $p=0.90$, rounding to the nearest cent. (Answer: VaR = $230.26; TVaR = $330.26.)

What You'll Learn