VaR and TVaR for Normal Losses

Why Normal-Loss VaR and TVaR Have Closed Forms

A normal model turns a confidence level into a loss threshold using a standard-normal quantile. The same model also gives a direct formula for the average loss above that threshold. These calculations are useful when a normal approximation is appropriate and the loss distribution is specified by its mean and standard deviation.

The key distinction is between a confidence level and its upper-tail probability. At confidence level $\alpha$, the loss exceeds its VaR with probability $1-\alpha$, not $\alpha$. TVaR averages the losses in that upper tail, so its formula uses both the standard-normal density at the VaR point and the tail probability.

Definition: Let $X$ be a loss with a continuous distribution, and let $0<\alpha<1$. Its Value at Risk at level $\alpha$ is the $\alpha$-quantile $q_\alpha$, satisfying $P(X\le q_\alpha)=\alpha$. Its Tail Value at Risk is the conditional mean of loss given that loss exceeds this threshold: $$ \operatorname{VaR}_\alpha(X)=q_\alpha,\qquad \operatorname{TVaR}_\alpha(X)=E[X\mid X>q_\alpha]. $$ For a continuous distribution, $P(X>q_\alpha)=1-\alpha$, so this conditional-tail definition applies directly.

VaR for a Normal Loss

Suppose $X\sim N(\mu,\sigma^2)$, where $\mu$ is the mean and $\sigma>0$ is the standard deviation. The notation uses $\sigma^2$ as the variance. Let $Z=(X-\mu)/\sigma$, so $Z\sim N(0,1)$. Write $\Phi$ for the standard-normal cumulative distribution function, $\phi$ for its density, and $z_\alpha=\Phi^{-1}(\alpha)$ for its $\alpha$-quantile.

Result: For $X\sim N(\mu,\sigma^2)$, $$ \operatorname{VaR}_\alpha(X)=\mu+\sigma z_\alpha. $$ The quantile $z_\alpha$ is determined by the confidence level, and the loss-scale conversion uses the standard deviation $\sigma$, not the variance $\sigma^2$.

To see why, standardize the loss. The event $X\le q_\alpha$ is equivalent to $(X-\mu)/\sigma\le(q_\alpha-\mu)/\sigma$. Its probability is $\alpha$ exactly when $(q_\alpha-\mu)/\sigma=z_\alpha$. Rearranging gives $q_\alpha=\mu+\sigma z_\alpha$.

Worked Example: VaR and TVaR at 97.5%

A hypothetical portfolio loss is $X\sim N(1000,200^2)$, with amounts in dollars. Find its 97.5% VaR and TVaR to the nearest cent. Use $z_{0.975}=1.96$ and $\phi(1.96)=0.0584409443$.

Method and tail probability. This is a continuous normal loss, so the VaR is its 97.5th percentile and the TVaR is the mean conditional on exceeding that percentile. Here $\mu=1000$, $\sigma=200$, $\alpha=0.975$, and $1-\alpha=0.025$.

VaR. Apply the normal quantile formula: $$ \operatorname{VaR}_{0.975}(X)=1000+200(1.96)=1000+392=\$1{,}392.00. $$

TVaR. The derivation below gives $\operatorname{TVaR}_\alpha(X)=\mu+\sigma\phi(z_\alpha)/(1-\alpha)$. Substituting the stated density value, $$ \operatorname{TVaR}_{0.975}(X) =1000+200\left(\frac{0.0584409443}{0.025}\right) =1000+467.5275544 =\$1{,}467.53. $$

The answer is checked against the threshold: TVaR, the mean of losses strictly above the 97.5th percentile, is greater than VaR, and $\$1{,}467.53>\$1{,}392.00$. The calculation avoids using the confidence level as the denominator; the relevant probability is the 2.5% upper tail.

Deriving TVaR from the Normal Density

For a continuous loss, conditioning on $X>q_\alpha$ means dividing the expected loss contributed by that upper tail by its probability $1-\alpha$. For a normal loss, standardization makes the tail expectation an integral involving the standard-normal density.

Write the conditional expectation.
For $q_\alpha=\mu+\sigma z_\alpha$, the upper-tail probability is $1-\alpha$, and $$ E[X\mid X>q_\alpha]=\frac{E[X\,\mathbf{1}_{\{X>q_\alpha\}}]}{1-\alpha}. $$

Standardize the tail integral.
Use $X=\mu+\sigma Z$, where $Z\sim N(0,1)$. The event $X>q_\alpha$ is $Z>z_\alpha$, so $$ E[X\,\mathbf{1}_{\{X>q_\alpha\}}] =\int_{z_\alpha}^{\infty}(\mu+\sigma z)\phi(z)\,dz. $$

Evaluate the two terms.
The integral of $\phi(z)$ above $z_\alpha$ is $1-\alpha$. Also, $\phi'(z)=-z\phi(z)$, so $$ \int_{z_\alpha}^{\infty}z\phi(z)\,dz =\left[-\phi(z)\right]_{z_\alpha}^{\infty} =\phi(z_\alpha), $$ because $\phi(z)$ tends to zero as $z$ tends to infinity.

Divide by the tail probability.
Thus the upper-tail contribution is $\mu(1-\alpha)+\sigma\phi(z_\alpha)$. Dividing by $1-\alpha$ gives $$ \operatorname{TVaR}_\alpha(X) =\mu+\sigma\frac{\phi(z_\alpha)}{1-\alpha}. $$

Result: If $X\sim N(\mu,\sigma^2)$, then $$ \operatorname{TVaR}_\alpha(X)=\mu+\sigma\frac{\phi(z_\alpha)}{1-\alpha}, \qquad z_\alpha=\Phi^{-1}(\alpha). $$ This formula follows from the conditional upper-tail mean for a continuous normal loss. It is not a formula for an arbitrary distribution, especially one with a probability atom at the quantile.

The density $\phi(z_\alpha)$ measures the standard-normal density at the standardized threshold; it is not itself the tail probability. The tail probability is $1-\alpha$. For reliable cent-level calculations, use unrounded quantiles and density values when available, and round only the final monetary result.

Worked Example: A 99% Tail Mean

A hypothetical reserve loss is $X\sim N(5000,1000^2)$, with amounts in dollars. Find its 99% VaR and TVaR to the nearest cent. Use $z_{0.99}=2.326347874$ and $\phi(z_{0.99})=0.0266521422$.

Method and tail probability. The model is continuous normal, with $\mu=5000$, $\sigma=1000$, $\alpha=0.99$, and upper-tail probability $1-\alpha=0.01$.

VaR. $$ \operatorname{VaR}_{0.99}(X) =5000+1000(2.326347874) =7326.347874 \approx \$7{,}326.35. $$

TVaR. Use the upper-tail density and the 1% tail probability: $$ \operatorname{TVaR}_{0.99}(X) =5000+1000\left(\frac{0.0266521422}{0.01}\right) =5000+2665.21422 \approx \$7{,}665.21. $$

The check is $\$7{,}665.21>\$7{,}326.35$, as expected for an average restricted to losses above the 99% VaR threshold. Using the unrounded density in the calculation matters: rounding an intermediate tail contribution too early can change the final cent.

Combining Independent Normal Losses

When loss components are independent and each is normal, their sum is normal. If $X_1\sim N(\mu_1,\sigma_1^2)$ and $X_2\sim N(\mu_2,\sigma_2^2)$ are independent, then $S=X_1+X_2$ has mean $\mu_1+\mu_2$ and variance $\sigma_1^2+\sigma_2^2$. Independence is an assumption here: without it, the variance of the sum also depends on covariance. Once the aggregate mean and standard deviation are found, the same normal VaR and TVaR formulas apply.

Worked Example: Risk Measures for Two Independent Losses

A hypothetical fund has two independent loss components, $X_1\sim N(120,30^2)$ and $X_2\sim N(80,40^2)$, in dollars. Find the 95% VaR and TVaR of $S=X_1+X_2$, to the nearest cent. Use $z_{0.95}=1.644853627$ and $\phi(z_{0.95})=0.1031356404$.

Method and assumptions. The component losses are independent and normal, so their sum is normal. Add the means and add the variances; do not add standard deviations.

Aggregate parameters. $$ E[S]=120+80=200, \qquad \operatorname{Var}(S)=30^2+40^2=900+1600=2500. $$ Therefore the aggregate standard deviation is $\sqrt{2500}=50$, and $S\sim N(200,50^2)$.

VaR. $$ \operatorname{VaR}_{0.95}(S) =200+50(1.644853627) =282.24268135 \approx \$282.24. $$

TVaR. The upper-tail probability is $1-0.95=0.05$: $$ \operatorname{TVaR}_{0.95}(S) =200+50\left(\frac{0.1031356404}{0.05}\right) =200+103.1356404 \approx \$303.14. $$

The calculation is consistent with the 5% upper-tail interpretation: TVaR exceeds the 95% threshold, since $\$303.14>\$282.24$. The main error avoided is treating the variance sum, 2500, as the standard deviation; the correct scale parameter is its square root, 50.

Common Exam Traps and Interpretation

A normal model is fully specified by its mean and variance, but the risk formulas use the standard deviation as the multiplier of $z_\alpha$ and $\phi(z_\alpha)$. Confirm whether a problem supplies a standard deviation or a variance before substituting. Also check the confidence level carefully: a 99% calculation has upper-tail probability 0.01, while a 95% calculation has upper-tail probability 0.05.

For example, in the first worked example, a common wrong calculation is to divide the density contribution by 0.975: $$ 1000+200\left(\frac{0.0584409443}{0.975}\right) \approx \$1{,}011.99. $$ That denominator is the confidence level, not the probability of exceeding VaR. The correction is to divide by $1-0.975=0.025$, which gives TVaR of $\$1{,}467.53$. A second frequent error is to use $200^2$ in place of $200$; the scale in both normal formulas is the standard deviation, not the variance.

A normal distribution has support over all real numbers, so it assigns some probability to negative losses. Whether that is a reasonable approximation depends on the application. The formulas remain mathematically valid for the specified normal model, but the model's suitability should not be assumed merely because the calculations are convenient. These formulas and techniques are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, identify the standard deviation and the upper-tail probability before applying the formulas. Round dollar answers to the nearest cent; use the supplied quantile and density values without intermediate rounding.

A hypothetical loss has distribution $X\sim N(600,50^2)$. At the 90% confidence level, use $z_{0.90}=1.281551566$ and $\phi(z_{0.90})=0.175498332$. Find VaR and TVaR. (Answer: VaR = $664.08; TVaR = $687.75.)
A hypothetical claim loss has distribution $X\sim N(1200,300^2)$. At the 95% confidence level, use $z_{0.95}=1.644853627$ and $\phi(z_{0.95})=0.1031356404$. Find VaR and TVaR. (Answer: VaR = $1,693.46; TVaR = $1,818.81.)
A hypothetical loss has distribution $X\sim N(400,25^2)$. At the 99% confidence level, use $z_{0.99}=2.326347874$ and $\phi(z_{0.99})=0.0266521422$. Find VaR and TVaR. (Answer: VaR = $458.16; TVaR = $466.63.)
Two independent hypothetical losses satisfy $X_1\sim N(50,10^2)$ and $X_2\sim N(70,24^2)$. For their sum, find the 90% VaR and TVaR using $z_{0.90}=1.281551566$ and $\phi(z_{0.90})=0.175498332$. (Answer: VaR = $153.32; TVaR = $165.63.)
A hypothetical loss $X$ has distribution $N(300,50^2)$, and a modified loss is $Z=1.2X+75$. Find the 95% VaR and TVaR of $Z$, using $z_{0.95}=1.644853627$ and $\phi(z_{0.95})=0.1031356404$. (Answer: VaR = $533.69; TVaR = $558.76.)

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