Parametric VaR from a Normal Loss Model

Introduction

When a loss model is specified by a probability distribution, a loss quantile can be calculated from the model rather than estimated by sorting a sample of observed losses. For a normal loss model, the calculation is direct: the VaR at a stated confidence level is determined by the model's mean, its standard deviation, and a standard-normal quantile. The calculation is simple, but interpreting the confidence level and using the correct tail are essential.

This tutorial derives and applies the normal-model formula. It assumes a loss amount $X$ for one specified period and a normal distribution with known model parameters. The calculations are aligned with the CAS Exam 9 (Risk Management for Actuaries) syllabus.

Definition: For a loss random variable $X$ and confidence level $p$, where $0<p<1$, define $\operatorname{VaR}_p(X)$ as the $p$-quantile of the loss distribution: $$ \operatorname{VaR}_p(X)=\inf\{x:F_X(x)\ge p\}. $$ For a continuous, strictly increasing distribution, this is the value $x$ satisfying $F_X(x)=p$. Thus, the probability of a loss exceeding this VaR is $1-p$.

Deriving VaR for a Normal Loss

Suppose the loss is modeled as $X\sim N(\mu,\sigma^2)$, where $\mu$ is the mean loss and $\sigma>0$ is the standard deviation. The second parameter in $N(\mu,\sigma^2)$ is the variance, not the standard deviation. Let $Z\sim N(0,1)$ be a standard-normal random variable, and let $\Phi$ be its cumulative distribution function. Define $z_p=\Phi^{-1}(p)$, the standard-normal quantile at probability $p$.

To derive the formula, standardize $X$. Since subtracting $\mu$ centers the loss and dividing by $\sigma$ gives it unit standard deviation, $(X-\mu)/\sigma$ has the standard-normal distribution. For any proposed quantile $x$,

F_X(x) =P(X\le x) =P\!\left(\frac{X-\mu}{\sigma}\le\frac{x-\mu}{\sigma}\right) =\Phi\!\left(\frac{x-\mu}{\sigma}\right).

The $p$-quantile is the value $x$ for which this cumulative probability equals $p$. Therefore, $\Phi((x-\mu)/\sigma)=p$, so $(x-\mu)/\sigma=\Phi^{-1}(p)=z_p$. Multiplying by $\sigma$ and adding $\mu$ gives the result.

Theorem: If $X\sim N(\mu,\sigma^2)$ with $\sigma>0$, then for $0<p<1$, $$ \operatorname{VaR}_p(X)=\mu+\sigma z_p, \qquad z_p=\Phi^{-1}(p). $$ Because the normal distribution is continuous, $P(X\le\operatorname{VaR}_p(X))=p$ and $P(X>\operatorname{VaR}_p(X))=1-p$.

The formula has a useful interpretation: $\mu$ sets the location of the loss distribution, while $\sigma z_p$ is the number of standard deviations between the mean and the selected quantile. At confidence levels above 50%, $z_p$ is positive, so the VaR exceeds the mean. At 50%, $z_p=0$, and the VaR equals the mean. At lower confidence levels, the quantile can fall below the mean.

Choosing the Standard-Normal Quantile

The confidence level $p$ is the cumulative probability to the left of the VaR. It is not the probability in the upper tail. For example, a 99% VaR uses the standard-normal quantile with 0.99 probability to its left; the probability of exceeding that VaR is 0.01.

Confidence level $p$	Upper-tail probability $1-p$	Approximate $z_p$
90%	10%	1.2816
95%	5%	1.6449
97.5%	2.5%	1.9600
99%	1%	2.3263
99.5%	0.5%	2.5758

These are one-sided quantiles. A two-sided normal interval may use a critical value associated with splitting probability across two tails, but that is not the quantile required for a one-sided loss VaR. Use the quantile for the stated cumulative probability $p$, not a critical value selected for a different purpose.

Worked Example: Calculating a 99% VaR in Thousands

A hypothetical insurer models the loss from one annual coverage period as $X\sim N(12,25)$, with $X$ measured in thousands of dollars. Calculate the 99% VaR and interpret its upper-tail probability. The model parameters are treated as fixed, and the normal model is assumed to apply to this loss measure.

Method. Apply $\operatorname{VaR}_p(X)=\mu+\sigma z_p$. The given variance is $25$, so the standard deviation is $\sigma=\sqrt{25}=5$ thousand dollars. The 99% confidence level requires $z_{0.99}=2.3263479$ (approximately).

Substitute the values, retaining the quantile precision in the calculation:

\operatorname{VaR}_{0.99}(X) =12+5(2.3263479) =12+11.6317395 =23.6317395\text{ thousand dollars}.

Rounded at the final step to three decimal places in thousands, the estimate is $23.632 thousand. Under the continuous normal model, $P(X\le 23.6317395)=0.99$, so the probability of exceeding this amount is $1-0.99=0.01$, or 1%.

Check. The 99% quantile must be above the mean because $z_{0.99}>0$; $23.632 thousand is above the $12 thousand mean. A result below the mean would signal a likely tail-selection or sign error. This method avoids confusing the variance, 25, with the standard deviation, 5.

Worked Applications and Quantile Traps

Worked Example: Using the Upper Tail for a 97.5% VaR

For a hypothetical annual claim-loss measure, assume $X\sim N(4.2,2.56)$, with amounts in millions of dollars. Calculate the 97.5% VaR and state the probability of exceeding it. The variance is in squared millions of dollars, the model parameters are fixed, and losses are assumed to follow the stated normal distribution.

Step 1: Identify the standard deviation. Since the variance is $2.56$, $\sigma=\sqrt{2.56}=1.6$ million dollars.

Step 2: Select the correct quantile. For a 97.5% VaR, use $z_{0.975}=1.959964$, approximately. This puts 0.975 probability to the left and 0.025 to the right.

Step 3: Calculate the quantile.

\operatorname{VaR}_{0.975}(X) =4.2+1.6(1.959964) =4.2+3.135942 =7.335942\text{ million dollars}.

Rounded at the final step to three decimal places in millions, the VaR is $7.336 million. The modeled probability of a loss greater than this amount is $1-0.975=0.025$, or 2.5%.

Common mistake and correction. A wrong calculation might use $z_{0.025}=-1.959964$, giving $4.2+1.6(-1.959964)=1.064058$ million dollars. That is the 2.5th percentile, not the 97.5th percentile: it leaves only 2.5% of the model distribution to its left. The requested VaR has 97.5% to its left, so the correct quantile is $z_{0.975}=1.959964$ and the answer is $7.336 million.

Worked Example: Comparing Two Confidence Levels

A hypothetical risk manager models a one-year loss as $X\sim N(2.8,0.64)$, with amounts in millions of dollars. Calculate the 90% and 99.5% VaRs and the difference between them. Assume the same loss definition and normal model at both confidence levels, with fixed parameters.

Step 1: Find the standard deviation. The stated variance is $0.64$ million dollars squared, so $\sigma=\sqrt{0.64}=0.8$ million dollars.

Step 2: Calculate the 90% VaR. Use $z_{0.90}=1.2815516$:

\operatorname{VaR}_{0.90}(X) =2.8+0.8(1.2815516) =2.8+1.0252413 =3.8252413\text{ million dollars}.

Step 3: Calculate the 99.5% VaR. Use $z_{0.995}=2.5758293$:

\operatorname{VaR}_{0.995}(X) =2.8+0.8(2.5758293) =2.8+2.0606634 =4.8606634\text{ million dollars}.

Step 4: Compare and check. Rounded at the final step to three decimal places in millions, the estimates are $3.825 million and $4.861 million. Their difference, calculated before rounding, is $4.8606634-3.8252413=1.0354221$, or $1.035 million to three decimal places. The higher confidence level has the higher VaR because $z_{0.995}>z_{0.90}$. This comparison uses the same mean and standard deviation for both estimates.

Assumptions, Interpretation, and Common Errors

The normal VaR formula is exact when the loss variable has the stated normal distribution and its parameters are the $\mu$ and $\sigma$ used in the calculation. If these parameters are estimates, the result is a plug-in estimate of model VaR; uncertainty in those estimates is not included in the formula. The model must also describe the same loss measure and time period for which VaR is requested. A mean or standard deviation measured in different units or over a different period cannot be substituted without a justified adjustment.

Normality has substantive consequences. A normal distribution is unbounded in both directions, so it assigns positive probability to negative losses as well as arbitrarily large positive losses. For a given confidence level, the formula still calculates the quantile of that normal model, but the result is only as suitable as the modeling assumption. In particular, if the modeled quantity cannot be negative, a normal model may assign probability to values that do not match the quantity's possible outcomes. The formula does not itself correct that feature.

VaR is a quantile, not a statement that losses cannot exceed the reported amount. For a continuous normal model, a $p$-level VaR is exceeded with probability $1-p$. It does not specify how large losses are when they exceed VaR. Keep the interpretation tied to the modeled distribution and period.

The following checks help catch common calculation errors:

Read the distribution parameters correctly. In $N(\mu,\sigma^2)$, take the square root of the second parameter to obtain $\sigma$ before multiplying by a normal quantile.
Use the stated confidence level as a left-tail cumulative probability. A 99% VaR uses $z_{0.99}$, not $z_{0.01}$.
Keep the units consistent. If $\mu$ and $\sigma$ are stated in thousands of dollars, the resulting VaR is also in thousands of dollars.
Check the direction. For a confidence level above 50%, $z_p>0$, so the normal-model VaR must exceed $\mu$.
Round only at the end. Retaining adequate precision in $z_p$ and intermediate products avoids unnecessary rounding differences.

Key takeaway: For $X\sim N(\mu,\sigma^2)$, calculate $\operatorname{VaR}_p(X)$ as $\mu+\sigma\Phi^{-1}(p)$. Use the left-cumulative probability $p$, distinguish variance from standard deviation, and interpret the exceedance probability as $1-p$.

Check Your Understanding

For each problem, apply the normal-model formula and retain precision until the final reported answer. Each loss and parameter is hypothetical.

A one-period loss is modeled as $X\sim N(6,2.25)$, with amounts in millions of dollars. Find the 95% VaR. (Answer: $8.467 million)
A hypothetical reserve loss is $X\sim N(15,16)$, with amounts in thousands of dollars. Find the 99% VaR and state the modeled exceedance probability. (Answer: $24.305 thousand; 1%)
A loss is modeled as $X\sim N(3.1,0.36)$, in millions of dollars. Calculate its 97.5% VaR, using the standard deviation rather than the variance. (Answer: $4.276 million)
For $X\sim N(10,9)$, calculate the 90% and 99% VaRs in the same units as $X$, then give their difference. (Answer: 13.845 and 16.979; difference 3.134 units)
A normal loss model has mean $12 million. Its 95% VaR is $17.4 million. Infer the model standard deviation, then estimate the 99% VaR. Use $z_{0.95}=1.6448536$ and $z_{0.99}=2.3263479$. (Answer: standard deviation $3.283 million; 99% VaR $19.637 million)

What You'll Learn