Risk Measures Across Confidence Levels

Why the Confidence Level Matters

A risk measure is not a single description of a loss model unless its confidence level is specified. Raising the confidence level asks about a more extreme part of the loss distribution. Value at Risk (VaR) identifies a loss quantile; Tail Value at Risk (TVaR) averages quantiles in the upper tail. Both therefore depend on the selected level, and comparing them across levels is useful when a hypothetical insurer or risk manager is deciding how much of the tail to examine.

For example, a discrete loss model can have a range of confidence levels that all give the same VaR because the quantile remains at one probability mass. TVaR can still rise over that range: as the confidence level increases, its average is taken over a smaller, more extreme part of the tail. The definition of TVaR must account for probability atoms to capture this correctly.

Definition: Let $X$ be an integrable loss and let $0<p<1$. Its quantile at confidence level $p$ is $q_p=\inf\{x:F_X(x)\ge p\}$. With this quantile convention, $\operatorname{VaR}_p(X)=q_p$. Define TVaR by the average of the upper quantiles: $$ \operatorname{TVaR}_p(X)=\frac{1}{1-p}\int_p^1 q_u\,du. $$ The integral definition applies to both continuous and discrete loss distributions. In a discrete model, it includes only the fraction of a probability atom needed to make the tail probability $1-p$.

The confidence level $p$ is a probability, not a loss amount. For example, $p=0.95$ means the 95th percentile under the stated quantile convention. It does not mean that the loss is 95% of some amount. The risk measures are expressed in the same units as $X$, such as dollars.

Why Both Measures Are Nondecreasing

The quantile function is nondecreasing: if $u_1<u_2$, then $q_{u_1}\le q_{u_2}$. This immediately gives the result for VaR. Since VaR at confidence level $p$ is $q_p$, increasing $p$ cannot produce a lower quantile.

For TVaR, the relevant comparison is between two averages of the quantile function. Raising the confidence level removes the lower part of the tail interval and averages over only the more extreme quantiles. The argument works even when the quantile function has jumps, as it does for discrete losses.

Choose two levels.
Let $0<p_1<p_2<1$. The interval $[p_1,1]$ consists of $[p_1,p_2]$ followed by $[p_2,1]$.

Compare the interval averages.
Every quantile on the later interval is at least every quantile on the earlier interval, up to endpoints that do not affect the integral. Thus the average quantile on $[p_2,1]$ is at least the average on $[p_1,p_2]$.

Combine the averages.
The TVaR at $p_1$ is a weighted average of the averages on those two intervals, with weights proportional to their lengths. The average on $[p_2,1]$ is therefore at least the combined average.

Theorem: For an integrable loss and $0<p_1<p_2<1$, $$ \operatorname{VaR}_{p_1}(X)\le \operatorname{VaR}_{p_2}(X) \quad\text{and}\quad \operatorname{TVaR}_{p_1}(X)\le \operatorname{TVaR}_{p_2}(X). $$ The second inequality follows because the average quantile over $[p_1,1]$ combines a lower-interval average with the at-least-as-large average over $[p_2,1]$. This result does not require a continuous distribution.

For a continuous distribution with a strictly increasing quantile function, VaR generally rises strictly as $p$ rises. A discrete distribution can instead have a VaR plateau: several confidence levels may fall within the same jump of the cumulative distribution function. TVaR is also nondecreasing, but it may rise, stay level, or do both over successive ranges, depending on the losses in the tail.

Worked Example: Discrete Losses With a Fractional Tail Atom

A hypothetical equipment insurer models one claim loss $X$ as follows: $X=0$ with probability $0.60$, $X=100$ with probability $0.30$, and $X=400$ with probability $0.10$. Find VaR and TVaR at confidence levels $0.50$, $0.80$, $0.90$, and $0.95$, and compare the results. All amounts are in dollars.

Method. First locate each quantile from the cumulative probabilities. Then average the quantiles over the remaining probability interval using the TVaR integral. This avoids treating the whole probability atom at VaR as if it belonged in the upper tail.

The cumulative probabilities are $F_X(0)=0.60$, $F_X(100)=0.90$, and $F_X(400)=1.00$. Therefore the quantile function is 0 for $0\le u\le0.60$, 100 for $0.60<u\le0.90$, and 400 for $0.90<u\le1$. At a confidence level that equals a cumulative probability, the quantile convention gives the lower loss value at that boundary.

At $p=0.50$, $q_{0.50}=0$. The upper interval has length $1-0.50=0.50$, and its quantile integral is $0.10(0)+0.30(100)+0.10(400)=0+30+40=70$. Thus $\operatorname{TVaR}_{0.50}=70/0.50=140$.

At $p=0.80$, $q_{0.80}=100$. Only $0.10$ of the probability mass at 100 remains above $p=0.80$, followed by all $0.10$ at 400. The tail integral is $0.10(100)+0.10(400)=10+40=50$, and the tail length is $0.20$. Hence $\operatorname{TVaR}_{0.80}=50/0.20=250$.

At $p=0.90$, $q_{0.90}=100$, because $F_X(100)=0.90$. The interval $(0.90,1]$ contains only the loss of 400, with probability $0.10$. Thus $\operatorname{TVaR}_{0.90}=0.10(400)/0.10=400$.

At $p=0.95$, $q_{0.95}=400$. The remaining tail is still entirely at 400, so $\operatorname{TVaR}_{0.95}=0.05(400)/0.05=400$.

The results are $(\operatorname{VaR},\operatorname{TVaR})=(0,140)$, $(100,250)$, $(100,400)$, and $(400,400)$, respectively. Both measures are nondecreasing. In particular, VaR is constant from 0.80 to 0.90, while TVaR increases because that smaller tail contains a larger share of the loss at 400.

Check. At $p=0.80$, the tail probability is $0.20$, and the fractional tail losses sum to 50 dollars. Dividing gives 250 dollars, as required.

Continuous Losses Across Several Levels

For a continuous uniform loss on $[a,b]$, where $a<b$, the quantile at level $u$ is $q_u=a+(b-a)u$. Integrating this line over the upper quantile interval gives $$ \operatorname{TVaR}_p(X) =\frac{1}{1-p}\int_p^1 \bigl(a+(b-a)u\bigr)\,du =\frac{q_p+b}{2}. $$ This is the average of the quantile at the lower end of the tail and the upper endpoint. It relies on the uniform model; it is not a general formula for continuous losses.

Worked Example: A Uniform Loss at Four Confidence Levels

A hypothetical covered repair cost $X$ is uniformly distributed from 200 dollars to 1,400 dollars. Calculate VaR and TVaR at $p=0.25$, $0.50$, $0.75$, and $0.90$, then find the increases between successive levels.

Model and method. Here $X\sim\operatorname{Uniform}(200,1400)$, so the width is $1400-200=1200$. Its quantile is $q_p=200+1200p$, and its TVaR is $(q_p+1400)/2$. The loss is continuous, so there is no probability atom to adjust.

At $p=0.25$, $q_{0.25}=200+1200(0.25)=200+300=500$. Then $\operatorname{TVaR}_{0.25}=(500+1400)/2=950$.

At $p=0.50$, $q_{0.50}=200+1200(0.50)=200+600=800$. Then $\operatorname{TVaR}_{0.50}=(800+1400)/2=1100$.

At $p=0.75$, $q_{0.75}=200+1200(0.75)=200+900=1100$. Then $\operatorname{TVaR}_{0.75}=(1100+1400)/2=1250$.

At $p=0.90$, $q_{0.90}=200+1200(0.90)=200+1080=1280$. Then $\operatorname{TVaR}_{0.90}=(1280+1400)/2=1340$.

From 0.25 to 0.50, VaR increases by $800-500=300$ dollars and TVaR by $1100-950=150$ dollars. From 0.50 to 0.75, the increases are $1100-800=300$ and $1250-1100=150$. From 0.75 to 0.90, they are $1280-1100=180$ and $1340-1250=90$. The values rise at each step, as expected. The smaller final changes reflect the narrower confidence-level interval, not a reversal in the direction of either risk measure.

Check. The 90th-percentile loss is 1,280 dollars, which lies between 200 and 1,400. The mean quantile over the final 10% of the uniform distribution is halfway from 1,280 to 1,400, or 1,340 dollars.

Reading a Discrete Risk Profile

A sequence of risk measures across confidence levels is sometimes called a risk profile. In a discrete model, its shape can reveal how probability is concentrated among tail losses. A flat VaR segment means that the selected quantile has not moved to the next loss value; it does not mean that every loss in the corresponding tail is the same.

Worked Example: Comparing Changes Across a VaR Plateau

For a hypothetical shipment policy, a claim loss $X$ is 0 dollars with probability $0.70$, 300 dollars with probability $0.20$, and 900 dollars with probability $0.10$. Calculate VaR and TVaR at $p=0.60$, $0.70$, $0.85$, $0.90$, and $0.95$. Identify where VaR is flat and where TVaR changes.

Method. The cumulative probabilities are 0.70 at 0 dollars, 0.90 at 300 dollars, and 1.00 at 900 dollars. Thus the quantile intervals have loss values 0, 300, and 900, with respective probability lengths 0.70, 0.20, and 0.10.

At $p=0.60$, VaR is 0. The upper interval contains $0.10$ at 0, $0.20$ at 300, and $0.10$ at 900. Its integral is $0.10(0)+0.20(300)+0.10(900)=0+60+90=150$. Therefore TVaR is $150/0.40=375$ dollars.

At $p=0.70$, the quantile is still 0 because the cumulative probability at 0 is 0.70. The upper interval now contains $0.20$ at 300 and $0.10$ at 900. Its integral is $60+90=150$, so TVaR is $150/0.30=500$ dollars.

At $p=0.85$, VaR is 300. The upper interval contains $0.05$ of the probability mass at 300 and all $0.10$ at 900. Its integral is $0.05(300)+0.10(900)=15+90=105$. Thus TVaR is $105/0.15=700$ dollars.

At $p=0.90$, VaR is 300 under the stated quantile convention, since the cumulative probability at 300 equals 0.90. The remaining interval contains only the 900-dollar loss, so TVaR is $0.10(900)/0.10=900$ dollars.

At $p=0.95$, VaR is 900, and the upper interval still contains only that loss. TVaR is $0.05(900)/0.05=900$ dollars.

The ordered pairs are $(0,375)$, $(0,500)$, $(300,700)$, $(300,900)$, and $(900,900)$, with VaR listed first. VaR is flat from 0.60 to 0.70 and from 0.85 to 0.90. TVaR rises over both intervals because removing lower quantiles leaves a more severe average tail. It is flat from 0.90 to 0.95 because the entire remaining tail has the same loss value.

Check. At 0.85, only 0.15 of probability remains. The relevant losses contribute 15 and 90 dollars to the quantile integral, for a total of 105; $105/0.15=700$, between the VaR of 300 and the maximum loss of 900.

Common Mistake: Treating TVaR as the Loss at VaR

A frequent error is to report VaR as TVaR, or to compute TVaR as the conditional mean of losses strictly greater than VaR without checking whether VaR is an atom. In the shipment example at $p=0.85$, this shortcut would use only losses strictly greater than 300 and report 900 dollars. That is not the quantile-integral TVaR: the upper 15% includes 5% of the quantile mass at 300 as well as the 10% at 900, giving 700 dollars.

The correction is to use the fraction of the VaR atom needed to fill the tail probability $1-p$. If $q_p$ is a discrete loss and the cumulative probability below it is $F_X(q_p-)$, the amount of probability at $q_p$ included in the upper quantile interval is $F_X(q_p)-p$, when $p$ lies within that atom. Combine that fraction with all probability above $q_p$, then divide the resulting loss-weighted total by $1-p$. The quantile-integral definition handles the same calculation without a separate atom formula.

The opposite shortcut can also mislead: multiplying the full probability of the VaR atom by its loss overcounts the upper tail whenever only part of that atom belongs to the remaining probability $1-p$. Always check that the probability included in the calculation sums to exactly $1-p$.

Key takeaway: Increasing the confidence level cannot decrease VaR or TVaR. VaR may remain unchanged across a discrete probability atom, while TVaR reflects how the average upper tail changes. For discrete losses, use fractional tail mass; for continuous losses, integrate the quantile function over the tail interval.

Check Your Understanding

For each problem, state VaR and TVaR in the requested order. Amounts are in dollars unless otherwise noted.

A hypothetical loss is uniform on $[0,800]$. Find VaR and TVaR at confidence levels 0.60 and 0.90. (Answer: At 0.60, VaR = 480 and TVaR = 640; at 0.90, VaR = 720 and TVaR = 760.)
A discrete loss is 0 with probability 0.50, 200 with probability 0.30, and 700 with probability 0.20. Find the VaR and TVaR pairs at levels 0.40, 0.80, and 0.90. (Answer: $(0,333.33)$, $(200,700)$, and $(700,700)$, respectively.)
A hypothetical loss is 50 with probability 0.20, 150 with probability 0.50, and 450 with probability 0.30. Find VaR and TVaR at levels 0.20, 0.50, 0.70, and 0.90, and explain whether both measures are nondecreasing. (Answer: The pairs are $(50,262.50)$, $(150,330)$, $(150,450)$, and $(450,450)$; both measures are nondecreasing.)
Suppose a continuous exponential loss has mean 100 dollars. Using $\operatorname{VaR}_p=-100\ln(1-p)$ and $\operatorname{TVaR}_p=\operatorname{VaR}_p+100$, find both measures at levels 0.90 and 0.99. Round to the nearest cent. (Answer: At 0.90, VaR = $230.26 and TVaR = $330.26; at 0.99, VaR = $460.52 and TVaR = $560.52.)
A discrete loss is 0 with probability 0.60, 100 with probability 0.30, and 400 with probability 0.10. At level 0.80, explain why using only losses strictly greater than VaR gives the wrong TVaR, then give the correct TVaR. (Answer: VaR is 100, and the upper 20% includes 10% at 100 and 10% at 400; TVaR is $ (10+40)/0.20=250$ dollars, not 400 dollars.)

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