Risk Measures from a Tabulated Aggregate Distribution

Reading a Tabulated Aggregate Distribution

When an aggregate loss is given as a finite probability table, the calculations for value-at-risk and tail value-at-risk depend on both the ordering of the loss amounts and the probability attached to each amount. The critical detail is that a confidence-level tail may contain only part of the probability at one loss value. Treating that entire probability as part of the tail can change the answer substantially.

Let $X$ be the aggregate loss, with possible values $x_1 < x_2 < \cdots < x_n$ and probabilities $p_i=P(X=x_i)$. A valid table has $p_i\geq 0$ and $\sum_{i=1}^{n}p_i=1$. For confidence level $\alpha$, where $0<\alpha<1$, value-at-risk is the smallest loss whose cumulative probability is at least $\alpha$. Tail value-at-risk averages the upper $1-\alpha$ fraction of the loss distribution, taking just enough of any probability mass at the boundary to make that tail probability exactly $1-\alpha$.

Definition: For a loss $X$ with distribution function $F_X$, define $\operatorname{VaR}_{\alpha}(X)=\inf\{x:F_X(x)\geq\alpha\}$. Define $\operatorname{TVaR}_{\alpha}(X)=\frac{1}{1-\alpha}\int_{\alpha}^{1}\operatorname{VaR}_{u}(X)\,du$. For a finite distribution, the integral averages the quantile values over the upper $1-\alpha$ fraction of probability, including only the needed fraction at the boundary value.

In an ordered table, write $F_X(x_i)=\sum_{j=1}^{i}p_j$. Then $\operatorname{VaR}_{\alpha}(X)=x_k$, where $k$ is the first index for which $F_X(x_k)\geq\alpha$. The mass strictly above this value is $1-F_X(x_k)$, and the part of the boundary mass needed in the tail is $F_X(x_k)-\alpha$. These two amounts add to $1-\alpha$.

\operatorname{TVaR}_{\alpha}(X) = \frac{ \displaystyle\sum_{i=k+1}^{n}x_i p_i +x_k\bigl(F_X(x_k)-\alpha\bigr) }{1-\alpha}

Why the Boundary-Mass Formula Works

The integral quantile definition gives a direct justification. For a discrete loss, the quantile function is constant over each interval of probability corresponding to a table entry. In particular, if $q=x_k=\operatorname{VaR}_{\alpha}(X)$, then the quantile equals $q$ from probability level $F_X(q-)$ up to $F_X(q)$, where $F_X(q-)=P(X<q)$. The integral from $\alpha$ to $F_X(q)$ therefore contributes $q$ times the interval length $F_X(q)-\alpha$. Every quantile level above $F_X(q)$ contributes its full outcome times its probability.

Order and check.
Arrange loss amounts from smallest to largest and confirm that the probabilities are nonnegative and sum to one.

Find the quantile.
Add probabilities cumulatively. The first loss for which the cumulative probability reaches or exceeds $\alpha$ is $\operatorname{VaR}_{\alpha}(X)$.

Assemble the tail.
Include every outcome strictly above the quantile, then include probability $F_X(q)-\alpha$ at $q$.

Average and verify.
Divide the probability-weighted tail loss by $1-\alpha$. Check that the included tail probabilities sum to $1-\alpha$.

If $F_X(q)=\alpha$ exactly, then $F_X(q)-\alpha=0$: none of the boundary mass is needed. The tail consists of all outcomes strictly above $q$, whose total probability is $1-\alpha$. If the cumulative probability jumps past $\alpha$, only the amount needed to fill the tail is included at $q$. This is why a discrete tail average is not generally the same as the conditional mean $E[X\mid X\geq q]$.

Worked Example: A Tail That Uses Part of the Quantile Mass

A hypothetical insurer models annual aggregate loss $X$, in thousands of dollars, with the following discrete distribution. Find $\operatorname{VaR}_{0.90}(X)$ and $\operatorname{TVaR}_{0.90}(X)$. This is a direct tabulated distribution; no interest rate or mortality assumption is used.

Loss $x_i$ (thousands of dollars)	Probability $p_i$	Cumulative probability
0	0.7000	0.7000
20	0.1500	0.8500
50	0.1000	0.9500
100	0.0500	1.0000

Method. Use the cumulative-probability definition for VaR, then construct exactly the upper $1-0.90=0.10$ probability for TVaR. The probabilities sum to $0.7000+0.1500+0.1000+0.0500=1.0000$.

The cumulative probability first reaches $0.90$ at loss 50, since $0.8500<0.90\leq0.9500$. Thus $\operatorname{VaR}_{0.90}(X)=50$ thousand dollars. Above 50, the table has probability $0.0500$, all at loss 100. The tail needs probability $0.1000$, so the required probability at loss 50 is $0.1000-0.0500=0.0500$. Equivalently, the boundary amount is $F_X(50)-0.90=0.9500-0.9000=0.0500$.

\operatorname{TVaR}_{0.90}(X) = \frac{(100)(0.0500)+(50)(0.0500)}{0.1000} = \frac{5.000+2.500}{0.1000} = 75

Therefore, $\operatorname{TVaR}_{0.90}(X)=75$ thousand dollars, or $75,000. The included tail probabilities are $0.0500+0.0500=0.1000$, as required. Common mistake avoided: using the entire probability $0.1000$ at loss 50 would include a tail larger than 10%; only half of that boundary mass belongs in this tail calculation.

Worked Tables: Exact and Fractional Boundaries

An efficient solution makes the tail probability explicit before calculating its weighted loss. This prevents a frequent denominator error: dividing by the probability of all outcomes at or above VaR, rather than by the required tail probability $1-\alpha$. The next examples use the same procedure in two different boundary situations.

Worked Example: A High Confidence Level with a Large Boundary Jump

A hypothetical equipment pool has aggregate loss $X$, measured in thousands of dollars, with this discrete model. Calculate $\operatorname{VaR}_{0.95}(X)$ and $\operatorname{TVaR}_{0.95}(X)$. No interest or mortality assumption applies; the table specifies the complete loss distribution.

Loss $x_i$	Probability $p_i$	Cumulative probability
0	0.8000	0.8000
40	0.1200	0.9200
80	0.0600	0.9800
200	0.0200	1.0000

Method. Locate the first cumulative probability at least $0.95$. The cumulative probability at 40 is $0.9200$, and at 80 it is $0.9800$, so $\operatorname{VaR}_{0.95}(X)=80$ thousand dollars. The probabilities sum to $1.0000$.

The required tail probability is $1-0.95=0.0500$. Losses strictly above 80 have probability $0.0200$, with weighted loss $(200)(0.0200)=4.000$ thousand-dollar probability units. The remaining boundary probability is $0.0500-0.0200=0.0300$, also obtained as $0.9800-0.9500=0.0300$. Its weighted loss is $(80)(0.0300)=2.400$.

\operatorname{TVaR}_{0.95}(X) = \frac{4.000+2.400}{0.0500} = 128 \text{ thousand dollars}

The answer is $128,000. The included probability is $0.0200+0.0300=0.0500$, and the weighted total divided by that probability is 128. Common mistake avoided: including all $0.0600$ probability at 80 would use $0.0800$ tail probability in total, not the specified $0.0500$.

Worked Example: When the Cumulative Probability Equals the Confidence Level

A hypothetical portfolio has aggregate loss $X$, in thousands of dollars, with the distribution below. Calculate $\operatorname{VaR}_{0.90}(X)$ and $\operatorname{TVaR}_{0.90}(X)$, and explain whether any mass at the VaR value enters the tail. The model is the given finite distribution; interest and mortality assumptions are not relevant.

Loss $x_i$	Probability $p_i$	Cumulative probability
0	0.4000	0.4000
10	0.3000	0.7000
30	0.2000	0.9000
60	0.0800	0.9800
150	0.0200	1.0000

Method. The cumulative probability first reaches $0.90$ at loss 30, so $\operatorname{VaR}_{0.90}(X)=30$ thousand dollars. The probabilities check to $0.4000+0.3000+0.2000+0.0800+0.0200=1.0000$.

The tail probability is $0.1000$. The probability strictly above 30 is $0.0800+0.0200=0.1000$, exactly the required amount. The boundary amount is $F_X(30)-0.90=0.9000-0.9000=0$, so no probability at 30 enters the tail. The weighted tail loss is $(60)(0.0800)+(150)(0.0200)=4.800+3.000=7.800$.

\operatorname{TVaR}_{0.90}(X) = \frac{7.800}{0.1000} = 78 \text{ thousand dollars}

Thus, $\operatorname{TVaR}_{0.90}(X)=78$ thousand dollars, or $78,000. The probability included is $0.0800+0.0200=0.1000$. Common mistake avoided: adding probability at 30 merely because 30 is the VaR would make the tail larger than 10%; an exact cumulative match means the boundary contribution is zero.

Common Interpretation and Calculation Traps

The key distinction is between a quantile threshold and a tail average. VaR identifies a loss amount; it does not say that every occurrence of that amount must be included in the tail used for TVaR. The integral quantile definition resolves this even when the distribution has atoms, that is, positive probability at individual loss values.

For example, in the first worked example, conditioning on $X\geq50$ would include probability $0.1000$ at 50 and $0.0500$ at 100. The conditional mean would be

E[X\mid X\geq50] = \frac{(50)(0.1000)+(100)(0.0500)}{0.1500} = \frac{10.000}{0.1500} = 66.666\ldots \text{ thousand dollars}.

That is not the 90% TVaR of 75 thousand dollars. The conditional mean uses 15% probability, whereas TVaR at 90% averages exactly the upper 10%. In general, $E[X\mid X\geq\operatorname{VaR}_{\alpha}(X)]$ is not the discrete-TV aR calculation when the quantile has positive probability. The prescribed fractional-boundary calculation is essential.

A reliable final check is to verify three items: the table probabilities sum to one; the quantile is the first value whose cumulative probability reaches $\alpha$; and the probabilities included in the tail sum to $1-\alpha$. Also retain the table’s units throughout. If losses are tabulated in thousands of dollars, a result of 75 is $75,000, not $75.

Key takeaway: Find the first cumulative probability at least $\alpha$ for VaR. For TVaR, include all loss mass strictly above that quantile and only $F_X(\operatorname{VaR}_{\alpha})-\alpha$ probability at the quantile, then divide the weighted tail loss by $1-\alpha$.

Check Your Understanding

For each finite distribution, probabilities are listed in increasing order of loss. Calculate the requested VaR and TVaR, checking that the included tail probability is exactly $1-\alpha$.

A hypothetical aggregate loss, in thousands of dollars, is 0 with probability 0.6000, 10 with probability 0.2500, and 40 with probability 0.1500. Find $\operatorname{VaR}_{0.80}$ and $\operatorname{TVaR}_{0.80}$. (Answer: VaR = 10 thousand dollars; TVaR = 32.5 thousand dollars.)
A hypothetical loss is 0 with probability 0.5000, 20 with probability 0.3000, and 70 with probability 0.2000. Find $\operatorname{VaR}_{0.90}$ and $\operatorname{TVaR}_{0.90}$. (Answer: VaR = 70; TVaR = 70, in the same units as the losses.)
A hypothetical aggregate loss, in thousands of dollars, is 0 with probability 0.7000, 25 with probability 0.1500, 100 with probability 0.1000, and 250 with probability 0.0500. Find $\operatorname{VaR}_{0.85}$ and $\operatorname{TVaR}_{0.85}$. (Answer: VaR = 25 thousand dollars; TVaR = 150 thousand dollars.)
A hypothetical loss is 5 with probability 0.2000, 15 with probability 0.4000, 50 with probability 0.3000, and 120 with probability 0.1000. Find $\operatorname{VaR}_{0.75}$ and $\operatorname{TVaR}_{0.75}$. (Answer: VaR = 50; TVaR = 78, in the same units as the losses.)
A hypothetical aggregate loss, in thousands of dollars, is 0 with probability 0.8200, 30 with probability 0.1000, 90 with probability 0.0600, and 300 with probability 0.0200. Find $\operatorname{VaR}_{0.95}$ and $\operatorname{TVaR}_{0.95}$. (Answer: VaR = 90 thousand dollars; TVaR = 174 thousand dollars.)

What You'll Learn