Normal-Model Value-at-Risk and Tail Value-at-Risk

Why Normal-Model Tail Measures Have Closed Forms

A normal loss model is often used as a tractable approximation when an aggregate loss reflects many sources of variation. Its mean locates the distribution, and its standard deviation measures the scale of variation. Once those parameters and a confidence level are specified, both value-at-risk and tail value-at-risk can be calculated using the standard normal distribution.

The calculations depend on using the correct tail. At confidence level $\alpha$, the upper-tail probability is $1-\alpha$. The normal-model TVaR formula divides by this tail probability, not by $\alpha$. A second essential check is that the standard normal quantile is the upper quantile $z_\alpha$, not the lower quantile $z_{1-\alpha}$.

Definition: Let loss $X$ have a normal distribution $X\sim N(\mu,\sigma^2)$, where $\sigma>0$. For $0<\alpha<1$, let $z_\alpha$ be the standard normal quantile satisfying $\Phi(z_\alpha)=\alpha$, where $\Phi$ is the standard normal cumulative distribution function, and let $\phi$ be its density. With the continuous-loss quantile convention, $\operatorname{VaR}_\alpha(X)$ is the $\alpha$-quantile of $X$, and $\operatorname{TVaR}_\alpha(X)$ is the expected loss conditional on exceeding that quantile. For a normal loss, these are $$ \operatorname{VaR}_\alpha(X)=\mu+\sigma z_\alpha, \qquad \operatorname{TVaR}_\alpha(X)=\mu+\sigma\frac{\phi(z_\alpha)}{1-\alpha}. $$

Here $\mu$ and $\sigma$ must use the same monetary unit as $X$. The second argument in $N(\mu,\sigma^2)$ is the variance, so the standard deviation used in the formulas is its square root. The formulas assume a continuous normal loss distribution with positive standard deviation. In particular, the conditional-tail definition of TVaR is unambiguous because a continuous distribution has zero probability of being exactly at its quantile.

Deriving the TVaR Formula

Standardize the loss by writing $Z=(X-\mu)/\sigma$. Then $Z\sim N(0,1)$, and $X>\mu+\sigma z_\alpha$ is equivalent to $Z>z_\alpha$. Since the standard normal distribution is continuous, $P(Z>z_\alpha)=1-\alpha$. The expected value of $Z$ in this upper tail is therefore

E[Z\mid Z>z_\alpha] =\frac{\int_{z_\alpha}^{\infty}t\phi(t)\,dt}{1-\alpha}.

The standard normal density satisfies $\phi'(t)=-t\phi(t)$. Thus $t\phi(t)=-\phi'(t)$, and integrating gives

\int_{z_\alpha}^{\infty}t\phi(t)\,dt =-\int_{z_\alpha}^{\infty}\phi'(t)\,dt =-\bigl[\phi(t)\bigr]_{z_\alpha}^{\infty} =\phi(z_\alpha),

because $\phi(t)$ tends to zero as $t$ tends to infinity. Rescaling from $Z$ to $X=\mu+\sigma Z$ gives the formula:

E[X\mid X>\operatorname{VaR}_\alpha(X)] =\mu+\sigma E[Z\mid Z>z_\alpha] =\mu+\sigma\frac{\phi(z_\alpha)}{1-\alpha}.

This derivation also explains why TVaR exceeds VaR for a non-degenerate normal loss: the conditional mean of observations above the quantile lies above that quantile. TVaR is not obtained by simply adding a fixed multiple of the standard deviation that works at every confidence level; both $z_\alpha$ and the density-to-tail ratio depend on $\alpha$.

Calculating VaR and TVaR

Worked Example: A Hypothetical Insurance Portfolio at 95%

A hypothetical insurer models its annual portfolio loss as $X\sim N(40{,}000,10{,}000^2)$, with amounts in dollars. Calculate the 95% VaR and TVaR, and state how far the TVaR is above the mean. Use $z_{0.95}=1.644854$ and $\phi(z_{0.95})=0.103136$, each rounded to six decimal places.

Method and parameters. The model has mean $\mu=\$40{,}000$, standard deviation $\sigma=\$10{,}000$, and confidence level $\alpha=0.95$. Its upper-tail probability is $1-\alpha=0.05$.

VaR. Substitute the 95% standard normal quantile:

\operatorname{VaR}_{0.95}(X) =40{,}000+10{,}000(1.644854) =40{,}000+16{,}448.54 =\$56{,}448.54.

TVaR. First calculate the density-to-tail ratio, then multiply by the standard deviation:

\frac{\phi(z_{0.95})}{1-0.95} =\frac{0.103136}{0.05} =2.06272,

\operatorname{TVaR}_{0.95}(X) =40{,}000+10{,}000(2.06272) =40{,}000+20{,}627.20 =\$60{,}627.20.

The TVaR is $\$20{,}627.20$ above the mean, and it exceeds the VaR by $\$4{,}178.66$. The calculations use the displayed six-decimal density and quantile; rounding the final monetary results to cents gives the amounts shown. Using unrounded normal values instead can change the final cents slightly.

Check and common mistake avoided. The TVaR is above the 95% VaR, as expected. Dividing the density by $0.95$, rather than by the upper-tail probability $0.05$, would give a tail increment of only $10{,}000(0.103136/0.95)=\$1{,}085.64$, an implausibly small increment for the mean of the worst 5% of losses. The denominator must be $1-\alpha$.

Worked Example: A High-Confidence Capital Estimate

For a hypothetical specialty insurer, annual loss is modeled as $X\sim N(120{,}000,30{,}000^2)$, in dollars. Calculate VaR and TVaR at 99%, using $z_{0.99}=2.326348$ and $\phi(z_{0.99})=0.026652$ rounded to six decimal places. Round final amounts to cents.

Method. The mean is $\$120{,}000$, the standard deviation is $\$30{,}000$, and the upper-tail probability is $1-0.99=0.01$.

VaR calculation:

\operatorname{VaR}_{0.99}(X) =120{,}000+30{,}000(2.326348) =120{,}000+69{,}790.44 =\$189{,}790.44.

TVaR calculation: The tail factor is $0.026652/0.01=2.6652$, so

\operatorname{TVaR}_{0.99}(X) =120{,}000+30{,}000(2.6652) =120{,}000+79{,}956.00 =\$199{,}956.00.

The 99% TVaR is $\$10{,}165.56$ above the 99% VaR and $\$79{,}956.00$ above the mean. This is consistent with the interpretation that TVaR averages the losses in the upper 1% of the model. The computation uses the displayed rounded quantile and density; rounding the resulting monetary amounts to cents introduces no further change here.

Check and common mistake avoided. The quantile is positive because this is an upper-tail confidence level above 50%. Substituting the 1% quantile, $z_{0.01}=-2.326348$, would calculate a lower-tail value, not the 99% VaR.

Applying the Formulas to a Normal Aggregate

A linear combination of jointly normal losses is normally distributed. If $X_1$ and $X_2$ are jointly normal, then $X_1+X_2$ has mean $E[X_1]+E[X_2]$ and variance $\operatorname{Var}(X_1)+\operatorname{Var}(X_2)+2\operatorname{Cov}(X_1,X_2)$. The covariance term matters unless the losses are independent (in which case it is zero). Once the aggregate mean and standard deviation are found, the normal-model formulas apply to that aggregate.

Worked Example: Correlated Normal Loss Components

A hypothetical risk portfolio has two jointly normal annual loss components, measured in thousands of dollars. Their models are $X_1\sim N(80,20^2)$ and $X_2\sim N(50,15^2)$, and their correlation is $0.4$. Calculate the aggregate $X=X_1+X_2$ VaR and TVaR at 97.5%. Use $z_{0.975}=1.959964$ and $\phi(z_{0.975})=0.058445$, rounded to six decimal places. Give final answers to the nearest $0.1$ thousand dollars.

Method and aggregate parameters. The aggregate mean is

$$ E[X]=80+50=130 $$

in thousands of dollars. The covariance is the correlation times the product of standard deviations:

\operatorname{Cov}(X_1,X_2) =0.4(20)(15)=120

in squared thousands of dollars. Therefore, the aggregate variance and standard deviation are

\operatorname{Var}(X) =20^2+15^2+2(120) =400+225+240 =865, \qquad \sigma_X=\sqrt{865}=29.410882.

Because the components are jointly normal, $X\sim N(130,865)$, with amounts in thousands of dollars. The upper-tail probability at 97.5% is $0.025$.

VaR calculation:

\operatorname{VaR}_{0.975}(X) =130+29.410882(1.959964) \approx130+57.644270 =187.644270.

TVaR calculation: The density-to-tail ratio is $0.058445/0.025=2.3378$. Thus

\operatorname{TVaR}_{0.975}(X) =130+29.410882(2.3378) \approx130+68.756760 =198.756760.

The requested results are VaR $=\$187{,}600$ and TVaR $=\$198{,}800$, after rounding to the nearest $0.1$ thousand dollars. TVaR exceeds VaR by about $11.112490$ thousand dollars. The aggregate variance includes $2\operatorname{Cov}(X_1,X_2)$; omitting that term would incorrectly treat the components as independent.

Check and common mistake avoided. The positive correlation increases the aggregate variance relative to an independent-components calculation. The standard deviation is $\sqrt{865}$, not $865$, because the normal formulas require a standard deviation.

Interpretation, Assumptions, and Exam Traps

For a normal model, VaR marks a percentile of the loss distribution, while TVaR is the mean loss conditional on being beyond that percentile. The difference is visible in the formulas: VaR uses the standardized quantile $z_\alpha$, whereas TVaR uses the upper-tail density ratio $\phi(z_\alpha)/(1-\alpha)$. At a fixed mean and standard deviation, increasing $\alpha$ raises both measures because it moves attention farther into the upper tail.

Use the tail probability correctly. At confidence $\alpha$, the TVaR denominator is $1-\alpha$, not $\alpha$.
Use the upper quantile. A 95% VaR uses $z_{0.95}$, which is positive, not $z_{0.05}$.
Keep variance and standard deviation distinct. If a model is stated as $N(\mu,\sigma^2)$, use $\sigma$, not $\sigma^2$, in the formulas.
Aggregate before applying the formulas. For correlated jointly normal components, include the covariance terms in the aggregate variance.
Check whether the model is plausible. A normal distribution assigns positive probability to negative losses. If negative modeled losses are material or the actual loss distribution is strongly skewed, the normal model may be a poor representation; the closed-form results are then only as appropriate as the model assumption.

These formulas are aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus. In a calculation, state the distribution, its parameters, the confidence level and the corresponding upper-tail probability before substituting values. That sequence makes unit errors, incorrect tail choices and omitted dependence terms easier to detect.

Key takeaway: For $X\sim N(\mu,\sigma^2)$, calculate $\operatorname{VaR}_\alpha(X)=\mu+\sigma z_\alpha$, then calculate $\operatorname{TVaR}_\alpha(X)=\mu+\sigma\phi(z_\alpha)/(1-\alpha)$. The denominator is the probability in the upper tail.

Check Your Understanding

For each problem, identify the distribution parameters and confidence level before applying the formulas. Round monetary answers as indicated.

A hypothetical warranty portfolio has annual loss $X\sim N(25{,}000,5{,}000^2)$, in dollars. At 90%, use $z_{0.90}=1.281552$ and $\phi(z_{0.90})=0.175498$, each to six decimal places. Find VaR and TVaR to cents. (Answer: VaR = $31,407.76; TVaR = $33,774.90.)
A hypothetical loss, measured in thousands of dollars, is $X\sim N(200,40^2)$. At 97.5%, use $z_{0.975}=1.959964$ and $\phi(z_{0.975})=0.058445$. Find both measures to three decimal places in thousands. (Answer: VaR = 278.399 thousand; TVaR = 293.512 thousand.)
A hypothetical insurer models annual loss as $X\sim N(75{,}000,12{,}000^2)$. At 95%, use $z_{0.95}=1.644854$ and $\phi(z_{0.95})=0.103136$. Find VaR and TVaR to cents, and calculate how far TVaR is above VaR. (Answer: VaR = $94,738.25; TVaR = $99,752.64; difference = $5,014.39.)
Two independent hypothetical loss components are normal, with $X_1\sim N(30,8^2)$ and $X_2\sim N(20,6^2)$, in thousands of dollars. Find the sum's VaR and TVaR at 95%, using $z_{0.95}=1.644854$ and $\phi(z_{0.95})=0.103136$. Give answers to three decimals in thousands. (Answer: VaR = 66.449 thousand; TVaR = 70.627 thousand.)
A hypothetical loss is $X\sim N(100,25^2)$, in thousands of dollars. Calculate VaR and TVaR at both 95% and 99%, using $z_{0.95}=1.644854$, $\phi(z_{0.95})=0.103136$, $z_{0.99}=2.326348$, and $\phi(z_{0.99})=0.026652$. Give answers to three decimals in thousands. (Answer: At 95%, VaR = 141.121 and TVaR = 151.568 thousand; at 99%, VaR = 158.159 and TVaR = 166.630 thousand.)

What You'll Learn