VaR from a Lognormal Loss Model

Why the Log Scale Matters

Loss amounts are positive and can be strongly right-skewed: many losses may be moderate, while a small number are much larger. A lognormal model represents this pattern by assuming that the logarithm of the loss is normally distributed. Its quantiles can therefore be calculated using a normal quantile, but the result must be transformed back to the original loss scale.

The key practical distinction is between parameters of the log loss and summaries of the loss in dollars. The parameter $\mu_L$ is the mean of log loss, not the mean loss. Likewise, $\sigma_L^2$ is the variance of log loss, not the variance of loss. Using either log-scale parameter as though it were a dollar-scale moment leads to incorrect results.

Definition: A positive loss $X$ has a lognormal distribution with log-scale parameters $\mu_L$ and $\sigma_L^2$ if $\ln X\sim N(\mu_L,\sigma_L^2)$, where $\sigma_L>0$. Thus, $\ln X$ is normally distributed with mean $\mu_L$ and variance $\sigma_L^2$, while $X$ itself is positive and right-skewed.

For a continuous loss distribution, $\operatorname{VaR}_{\alpha}(X)$ is the loss at cumulative probability $\alpha$, or equivalently the $\alpha$-quantile. The confidence level $\alpha$ is between zero and one. Write $z_\alpha$ for the standard normal quantile satisfying $P(Z\le z_\alpha)=\alpha$, where $Z\sim N(0,1)$.

Theorem: If $\ln X\sim N(\mu_L,\sigma_L^2)$, then for $0<\alpha<1$, $\operatorname{VaR}_{\alpha}(X)=\exp(\mu_L+\sigma_L z_\alpha)$.

Deriving the Lognormal VaR Formula

The transformation follows directly from the definition of a quantile. Since the logarithm is strictly increasing for positive values, the event that a loss is no greater than $x$ is equivalent to the event that its logarithm is no greater than $\ln x$. Standardizing that logarithm gives the normal cumulative distribution function.

Start with the lognormal model.
Let $W=\ln X$. The assumption is $W\sim N(\mu_L,\sigma_L^2)$.

Write the loss cumulative distribution function.
For $x>0$, $P(X\le x)=P(\ln X\le\ln x)=\Phi((\ln x-\mu_L)/\sigma_L)$, where $\Phi$ is the standard normal cumulative distribution function.

Set the cumulative probability equal to $\alpha$.
At the $\alpha$-quantile, $\Phi((\ln x-\mu_L)/\sigma_L)=\alpha$. Therefore, $(\ln x-\mu_L)/\sigma_L=z_\alpha$.

Return to the loss scale.
Rearranging gives $\ln x=\mu_L+\sigma_Lz_\alpha$. Exponentiating both sides gives $x=\exp(\mu_L+\sigma_Lz_\alpha)$, as required.

The formula requires a lognormal loss model with positive losses and the stated log-scale parameters. It does not say that every positive loss should be modeled as lognormal. The model is an assumption to be assessed for the application; if the loss distribution is not lognormal, this transformed normal quantile need not be an appropriate VaR estimate. Unlike a normal loss model, the lognormal model also cannot assign probability to a negative loss.

Worked Calculations

Worked Example: Calculating VaR From Log-Scale Parameters

A hypothetical insurer models the annual aggregate loss $X$, measured in dollars, as lognormal with $\ln X\sim N(\ln 80{,}000,0.20^2)$. Calculate $\operatorname{VaR}_{\alpha}(X)$ for $\alpha=\Phi(1)$, or approximately 0.8413. The model concerns losses only, so no interest rate or mortality assumption is used.

Method. Apply the lognormal quantile formula, using $z_{\Phi(1)}=1$. Here $\mu_L=\ln(80{,}000)$ and $\sigma_L=0.20$.

\[ \operatorname{VaR}_{\Phi(1)}(X) =\exp\bigl(\ln(80{,}000)+0.20(1)\bigr) =80{,}000\exp(0.20). \]

Since $\exp(0.20)\approx1.221402758$, the quantile is $80{,}000(1.221402758)\approx97{,}712.22$. Thus, $\operatorname{VaR}_{\alpha}(X)=\$97{,}712.22$, rounded to the nearest cent.

Check. The result exceeds the median, which is $\exp(\mu_L)=\$80{,}000$, as it should for a confidence level above 0.5. No interest or mortality assumption applies.

Worked Example: Recovering the Lognormal Parameters From Loss Moments

A hypothetical portfolio has positive annual losses with arithmetic mean $E[X]=\$100{,}000$ and variance $\operatorname{Var}(X)=30{,}000{,}000{,}000$ dollars squared. Assume $X$ is lognormal. Find its 50th-percentile VaR. This loss model has no interest or mortality assumption.

Method. First convert the arithmetic mean and variance to log-scale parameters. For a lognormal loss, the squared coefficient of variation is $\operatorname{Var}(X)/E[X]^2=\exp(\sigma_L^2)-1$. Here $E[X]^2=10{,}000{,}000{,}000$, so \[ \frac{\operatorname{Var}(X)}{E[X]^2} =\frac{30{,}000{,}000{,}000}{10{,}000{,}000{,}000}=3. \]

It follows that $\exp(\sigma_L^2)-1=3$, so $\exp(\sigma_L^2)=4$ and $\sigma_L^2=\ln 4$. The mean formula for a lognormal variable is $E[X]=\exp(\mu_L+\sigma_L^2/2)$. Therefore, \[ \mu_L=\ln(E[X])-\frac{\sigma_L^2}{2} =\ln(100{,}000)-\frac{\ln4}{2} =\ln(50{,}000). \]

At the 50th percentile, $z_{0.5}=0$. Thus, \[ \operatorname{VaR}_{0.5}(X)=\exp(\mu_L+\sigma_L z_{0.5}) =\exp(\ln(50{,}000))=\$50{,}000. \]

Check. The median is below the arithmetic mean, as expected for a right-skewed lognormal loss. Substituting the recovered parameters into the mean formula gives $\exp(\ln50{,}000+\ln4/2)=\exp(\ln100{,}000)=\$100{,}000$, confirming the conversion. No interest or mortality assumption applies.

Worked Example: Finding the Confidence Level at a Given Loss Amount

A hypothetical reinsurer models a claim amount $X$, in dollars, as lognormal with $\ln X\sim N(\ln 120{,}000,(\ln2)^2)$. Find the confidence level $\alpha$ for which $\operatorname{VaR}_{\alpha}(X)=\$480{,}000$, and state the corresponding upper-tail probability. No interest rate or mortality assumption applies.

Method. Take logarithms of the proposed quantile and standardize. From the quantile formula, \[ z_\alpha=\frac{\ln(480{,}000)-\ln(120{,}000)}{\ln2} =\frac{\ln(480{,}000/120{,}000)}{\ln2} =\frac{\ln4}{\ln2}=2. \]

Therefore $\alpha=\Phi(2)\approx0.9772$, to four decimal places. The upper-tail probability is $1-\alpha\approx1-0.9772=0.0228$, or about 2.28%.

Check. Substituting $z_\alpha=2$ back into the VaR formula gives $\exp(\ln120{,}000+2\ln2)=120{,}000(4)=\$480{,}000$. This confirms both the quantile and its confidence level.

Log-Scale Parameters Are Not Loss Moments

A parameterization is often given as $\ln X\sim N(\mu_L,\sigma_L^2)$, but calculations may instead begin with the arithmetic mean and variance of $X$. The two pairs of quantities are related, but they are not interchangeable. For a lognormal loss,

E[X]=\exp\left(\mu_L+\frac{\sigma_L^2}{2}\right), \qquad \operatorname{Var}(X)=\left(\exp(\sigma_L^2)-1\right)\exp(2\mu_L+\sigma_L^2).

Conversely, if the arithmetic mean $m=E[X]>0$ and variance $s^2=\operatorname{Var}(X)$ are known, then

\sigma_L^2=\ln\left(1+\frac{s^2}{m^2}\right), \qquad \mu_L=\ln(m)-\frac{\sigma_L^2}{2}.

These relationships follow from the normal moment-generating formula: if $W\sim N(\mu_L,\sigma_L^2)$, then $E[\exp(tW)]=\exp(t\mu_L+t^2\sigma_L^2/2)$. Setting $t=1$ gives the lognormal mean, and setting $t=2$ gives $E[X^2]$. Subtracting $E[X]^2$ from $E[X^2]$ yields the displayed variance. Solving the mean and variance equations for $\mu_L$ and $\sigma_L^2$ gives the conversion formulas.

The median of a lognormal loss is $\exp(\mu_L)$, since $z_{0.5}=0$. The arithmetic mean is $\exp(\mu_L+\sigma_L^2/2)$, which is greater than the median whenever $\sigma_L>0$. A high-confidence VaR is further into the upper tail: for $\alpha>0.5$, $z_\alpha>0$, so its exponent exceeds $\mu_L$.

Common Mistake and Practical Checks

A common error is to treat the median parameter $\exp(\mu_L)$ as though it were the arithmetic mean, or to insert a dollar-scale mean directly in place of $\mu_L$. Consider the first worked example. Its median is \$80,000 and $\sigma_L=0.20$, but its arithmetic mean is $80{,}000\exp(0.20^2/2)=\$81{,}616.11$. An incorrect calculation that uses this mean as the lognormal median gives $\$81{,}616.11\exp(0.20)\approx\$99{,}686.14$. The correct calculation uses the actual median: $\$80{,}000\exp(0.20)=\$97{,}712.22$. The incorrect result is too high because it applies the lognormal adjustment twice.

Before finalizing a calculation, check that the chosen $z_\alpha$ corresponds to the requested cumulative confidence level, that $\sigma_L$ is the square root of the log variance, and that exponentiation has returned the answer to the original loss scale. The upper-tail probability associated with a continuous-model VaR is $1-\alpha$; it is not $\alpha$.

Key takeaway: For a lognormal loss, calculate the normal quantile on the log scale as $\mu_L+\sigma_Lz_\alpha$, then exponentiate. If the input is an arithmetic mean and variance, convert them to log-scale parameters first.

Check Your Understanding

For each problem, assume the stated lognormal model is appropriate. Losses are measured in dollars; no interest or mortality assumption applies.

A loss has $\ln X\sim N(\ln40{,}000,0.10^2)$. Find its 50th-percentile VaR. (Answer: \$40,000.00)
A loss has $\ln X\sim N(\ln30{,}000,0.25^2)$. Find its VaR at $\alpha=\Phi(1)$. (Answer: \$38,520.76)
A loss has median \$90,000 and log-scale standard deviation $0.30$. Find its VaR at $\alpha=\Phi(2)$. (Answer: \$163,990.69)
A lognormal loss has arithmetic mean \$60,000 and variance $3{,}600{,}000{,}000$ dollars squared. Find its 50th-percentile VaR. (Answer: \$42,426.41)
A lognormal loss has median \$75,000 and log-scale standard deviation $\ln2$. A proposed VaR is \$300,000. Find its confidence level and upper-tail probability. (Answer: confidence level $\Phi(2)\approx0.9772$; upper-tail probability $\approx0.0228$)

What You'll Learn