Deriving TVaR For Normal Losses

Why Normal-Loss TVaR Has a Closed Form

Value at Risk identifies a loss quantile, while Tail Value at Risk (TVaR) summarizes the average loss in the upper tail beyond that quantile. For a normal loss, this tail average can be calculated directly from the normal density. The formula is useful when a normal model is a reasonable approximation and the mean, standard deviation, and confidence level are specified.

Let $X$ be a loss with $X\sim N(\mu,\sigma^2)$, where $\mu$ is the mean and $\sigma>0$ is the standard deviation. Let $p$ be the confidence level, with $0<p<1$. Write $\Phi$ for the standard normal cumulative distribution function, $\phi$ for its density, and $z_p=\Phi^{-1}(p)$ for its $p$-quantile.

Definition: For a continuous loss distribution, $\operatorname{TVaR}_p(X)$ is the expected loss conditional on exceeding its $p$-quantile. Equivalently, it is the average of the upper quantiles: $\operatorname{TVaR}_p(X)=E[X\mid X>\operatorname{VaR}_p(X)]=\frac{1}{1-p}\int_p^1\operatorname{VaR}_q(X)\,dq$. The normal distribution is continuous, so the conditional-expectation definition applies directly. The probability of exceeding $\operatorname{VaR}_p(X)$ is $1-p$.

For a normal loss, standardization gives $\operatorname{VaR}_p(X)=\mu+\sigma z_p$. This is the cutoff separating the upper tail from the rest of the distribution. The desired TVaR is not this cutoff: it is the mean of losses above it.

Deriving the Normal TVaR Formula

Set $Z=(X-\mu)/\sigma$, so $Z\sim N(0,1)$ and $X=\mu+\sigma Z$. The event $X>\operatorname{VaR}_p(X)$ is equivalent to $Z>z_p$. Its probability is $1-\Phi(z_p)=1-p$. Thus the conditional mean can be computed by integrating $X$ over that upper tail and dividing by its probability.

Write the conditional expectation.
$E[X\mid X>\operatorname{VaR}_p(X)]=\frac{1}{1-p}\int_{z_p}^{\infty}(\mu+\sigma z)\phi(z)\,dz$.

Separate the constant and variable terms.
The integral of $\mu\phi(z)$ from $z_p$ to infinity is $\mu[1-\Phi(z_p)]=\mu(1-p)$.

Evaluate the standard normal first-moment integral.
Since $\phi'(z)=-z\phi(z)$, integrating $z\phi(z)=-\phi'(z)$ gives $\int_{z_p}^{\infty}z\phi(z)\,dz=\phi(z_p)$. Here $\phi(z)$ tends to zero as $z$ tends to infinity.

Divide by the tail probability.
The numerator is $\mu(1-p)+\sigma\phi(z_p)$. Dividing by $1-p$ yields the result.

\operatorname{TVaR}_p(X)=\mu+\sigma\frac{\phi(z_p)}{1-p}, \qquad z_p=\Phi^{-1}(p).

This derivation depends on the normal model and on $\sigma>0$. The denominator is the probability of the upper tail, not the confidence level itself. The result can also be read as the mean $\mu$ plus a positive tail adjustment. At a fixed confidence level, that adjustment is proportional to $\sigma$.

Key distinction: $\operatorname{VaR}_p(X)=\mu+\sigma z_p$, whereas $\operatorname{TVaR}_p(X)=\mu+\sigma\phi(z_p)/(1-p)$. A TVaR calculation needs both the quantile $z_p$ and the density value $\phi(z_p)$.

Applying the Formula

A practical calculation has four steps: identify $\mu$ and $\sigma$, find $z_p$, evaluate $\phi(z_p)/(1-p)$, and substitute. The second parameter in $N(\mu,\sigma^2)$ is the variance; take its square root before using the formula as the standard deviation.

Worked Example: A 95% TVaR for a Normal Annual Loss

A hypothetical insurer models an annual portfolio loss $X$ in dollars as $X\sim N(1{,}000^2,200^2)$ is not the intended parameterization; instead, the model is $X\sim N(1000,200^2)$. Calculate the 95% VaR and TVaR. Use full precision in intermediate calculations and round money to the nearest cent.

Method. Use the normal quantile for VaR and the derived density-over-tail-probability factor for TVaR. Here $\mu=\$1{,}000$, $\sigma=\$200$, $p=0.95$, and $1-p=0.05$.

From the standard normal distribution, $z_{0.95}=1.6448536$ and $\phi(z_{0.95})\approx0.1031356$. Therefore,

\operatorname{VaR}_{0.95}(X)=1000+200(1.6448536)=1328.97072\approx \$1{,}328.97.

For TVaR, the density-to-tail factor is $0.1031356/0.05=2.062712$. Substituting,

\operatorname{TVaR}_{0.95}(X)=1000+200(2.062712) =1412.5424\approx \$1{,}412.54.

The result is greater than the VaR cutoff, as an average of losses above that cutoff must be. Common mistake avoided: dividing the density by $p=0.95$ would use the confidence level rather than the upper-tail probability. The correct denominator is $1-p=0.05$.

Worked Example: A 99% TVaR in a Large-Loss Portfolio

A hypothetical commercial portfolio has annual loss $X\sim N(1{,}500{,}000,300{,}000^2)$, with amounts in dollars. Find its 99% VaR and TVaR, rounding to the nearest dollar.

Method. At $p=0.99$, the upper-tail probability is $1-p=0.01$. Use $z_{0.99}=2.3263479$ and $\phi(z_{0.99})\approx0.0266521$. The parameters are $\mu=\$1{,}500{,}000$ and $\sigma=\$300{,}000$.

First calculate the quantile:

\operatorname{VaR}_{0.99}(X) =1{,}500{,}000+300{,}000(2.3263479) =2{,}197{,}904.37 \approx \$2{,}197{,}904.

Next, the tail factor is $0.0266521/0.01=2.66521$. Thus,

\operatorname{TVaR}_{0.99}(X) =1{,}500{,}000+300{,}000(2.66521) =2{,}299{,}563 \approx \$2{,}299{,}563.

The TVaR exceeds the 99% VaR, consistent with its interpretation as an average in the upper 1% of losses. Common mistake avoided: using $0.99$ as the denominator would produce a tail adjustment far too small. The tail probability is $0.01$, even though the confidence level is $0.99$.

Worked Example: TVaR When the Model Is Stated in Thousands

A hypothetical risk assessment models annual loss in thousands of dollars as $X\sim N(50,10^2)$. Calculate the 90% VaR and TVaR in thousands of dollars, rounding to two decimal places.

Method. Interpret the second parameter as the variance: $\mu=50$ and $\sigma=\sqrt{10^2}=10$, both in thousands of dollars. At $p=0.90$, $1-p=0.10$, $z_{0.90}=1.2815516$, and $\phi(z_{0.90})\approx0.1754983$.

The VaR is

\operatorname{VaR}_{0.90}(X)=50+10(1.2815516)=62.815516\approx 62.82

thousand dollars. For TVaR, $0.1754983/0.10=1.754983$, so

\operatorname{TVaR}_{0.90}(X)=50+10(1.754983)=67.54983\approx 67.55

thousand dollars. This corresponds to $\$67{,}549.83$. Common mistake avoided: using $10^2=100$ as the standard deviation would confuse variance with standard deviation and inflate the tail adjustment by a factor of ten.

Interpretation and Model Limitations

For a fixed confidence level, the formula separates the location of the loss distribution from its spread: $\mu$ shifts TVaR, and $\sigma$ scales the tail adjustment. A higher confidence level generally means a smaller upper-tail probability and a farther-out quantile, so the average of that more extreme tail is higher. This makes the confidence level an essential part of any reported TVaR.

A normal distribution has support over the entire real line, including negative values. If $X$ represents a loss, negative values can be interpreted as gains under a signed net-loss convention, but they may be implausible for some nonnegative gross-loss measures. The closed-form formula is mathematically correct for the normal model; whether the normal model is suitable for a particular loss measure is a separate modeling judgment. Do not silently truncate or otherwise modify the distribution and still use this formula: changing the model changes the tail calculation.

This normal-loss calculation is aligned with the CAS Exam 9 (Risk Management for Actuaries) syllabus. On an exam or in practice, show the confidence level, the corresponding $z_p$, the upper-tail probability, and the density value used. Those details make it possible to detect the most frequent errors: confusing variance with standard deviation, using $p$ instead of $1-p$, and reporting VaR when the question asks for TVaR.

Check Your Understanding

For each hypothetical normal loss, identify the standard deviation before applying the formula. Answers are rounded as indicated.

A one-year loss has distribution $X\sim N(400,25^2)$, in dollars. Find its 95% VaR and TVaR. Use $z_{0.95}=1.6449$ and $\phi(z_{0.95})/(0.05)=2.0627$. (Answer: VaR = $\$441.12$; TVaR = $\$451.57$, to the nearest cent.)
A hypothetical portfolio loss has distribution $X\sim N(800{,}000,120{,}000^2)$. Find the 97.5% TVaR. Use $z_{0.975}=1.9600$ and $\phi(z_{0.975})/(0.025)=2.3378$. (Answer: approximately $\$1{,}080{,}536$, to the nearest dollar.)
A loss model is $X\sim N(3{,}000,500^2)$, in dollars. Calculate its 90% VaR and TVaR using $z_{0.90}=1.2816$ and $\phi(z_{0.90})/(0.10)=1.7550$. (Answer: VaR = $\$3{,}640.80$; TVaR = $\$3{,}877.50$.)
A model is given as $X\sim N(12{,}000,4{,}000{,}000)$, with both parameters in squared-dollar convention as $N(\mu,\sigma^2)$. Find the 99% TVaR using $z_{0.99}=2.3263$ and $\phi(z_{0.99})/(0.01)=2.6652$. (Answer: $\$17{,}330.40$.)
A normal loss has mean $\$500$, and its 95% TVaR is $\$900$. Estimate its standard deviation using $\phi(z_{0.95})/(0.05)=2.0627$. (Answer: $\sigma=400/2.0627\approx \$193.92$.)

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