Lognormal And Pareto Tail Decay

Why Compare Severity Tails?

For a severity model, the survival probability $S_X(x)=P(X>x)$ gives the probability that one loss exceeds the threshold $x$. At a high threshold, this probability informs the assessment of unusually large losses. The lognormal and Pareto models can both represent right-skewed severity distributions, but their survival probabilities decay differently as $x$ grows.

The distinction is not always visible at a single threshold. A lognormal model can assign a higher exceedance probability than a particular Pareto model at moderate loss levels, yet eventually assign a lower probability at sufficiently large levels. The comparison therefore needs both the model parameters and the threshold.

Definition: If $\ln X\sim N(\mu,\sigma^2)$, where $\sigma>0$, then $X$ has a lognormal distribution with log-scale parameters $\mu$ and $\sigma$. For $x>0$, $$ S_X(x)=P(X>x)=\bar{\Phi}\left(\frac{\ln x-\mu}{\sigma}\right), $$ where $\bar{\Phi}(z)=1-\Phi(z)$ is the standard normal survival function. A Pareto Type I severity with scale $\theta>0$ and tail index $\alpha>0$ has support $x\geq\theta$ and $$ S_X(x)=\left(\frac{\theta}{x}\right)^\alpha,\qquad x\geq\theta. $$

These parameterizations are important. In the lognormal model, $\mu$ and $\sigma$ describe $\ln X$, not $X$ itself. In the Pareto model, $\theta$ is the minimum possible loss and $\alpha$ controls the rate of power-law decay. For $x<\theta$, the Pareto survival probability is 1 because every possible loss is at least $\theta$.

Calculating and Comparing Survival Probabilities

For a lognormal model, first standardize the logarithm of the threshold. For a Pareto Type I model, divide the scale by the threshold and raise the result to the tail index. A fair comparison uses the same threshold $x$ for both models. It is also useful to report the ratio $S_{\mathrm{LN}}(x)/S_{\mathrm{P}}(x)$: a ratio above 1 means the lognormal model assigns the greater exceedance probability at that threshold.

Worked Example: Comparing Tails at Two Loss Levels

Suppose a hypothetical loss has either a lognormal model with $\mu=0$, $\sigma=1$, or a Pareto Type I model with $\theta=1$, $\alpha=2$. Compare their survival probabilities at $x=4$ and $x=e^4$. The method is to calculate each survival probability under its own model, then compare them at each shared threshold.

At $x=4$, lognormal model. The standardized log-threshold is $$ z=\frac{\ln 4-0}{1}=1.386294. $$ Thus $S_{\mathrm{LN}}(4)=\bar{\Phi}(1.386294)\approx 0.08283$.

At $x=4$, Pareto model. $$ S_{\mathrm{P}}(4)=\left(\frac{1}{4}\right)^2=0.0625. $$ The ratio is $0.0828285/0.0625\approx 1.325$. At this threshold, the lognormal model assigns the greater probability of exceedance.

At $x=e^4$, lognormal model. Here $z=(\ln(e^4)-0)/1=4$, so $S_{\mathrm{LN}}(e^4)=\bar{\Phi}(4)\approx 0.00003167$.

At $x=e^4$, Pareto model. $$ S_{\mathrm{P}}(e^4)=\left(\frac{1}{e^4}\right)^2=e^{-8}\approx 0.00033546. $$ The ratio is $0.0000316712/0.000335463\approx 0.0944$. At this higher threshold the Pareto probability is larger. The lognormal probability rounds to 0.0000 at four decimal places, so its small unrounded value is also shown to avoid obscuring the comparison.

Check and common mistake. The Pareto probability at $x=4$ is $1/16$, and at $x=e^4$ it is $e^{-8}$, consistent with the same Pareto formula. A common mistake is to infer from the first comparison that the lognormal tail is always larger. The second comparison disproves that inference: the ordering can change with the threshold.

Why the Pareto Tail Eventually Dominates

The change in ordering has a mathematical explanation. Write $t=\ln x$. For a lognormal model, the standardized threshold is $z=(t-\mu)/\sigma$, which grows linearly in $t$. The normal survival probability at large positive $z$ is approximately $\phi(z)/z$, where $\phi(z)=e^{-z^2/2}/\sqrt{2\pi}$. In contrast, a Pareto survival probability is a power of $x$, so its logarithm decreases linearly in $\ln x$.

Write the ratio.
For $x\geq\theta$, divide the lognormal survival probability by the Pareto survival probability: $$ R(x)=\frac{\bar{\Phi}(z)}{(\theta/x)^\alpha},\qquad z=\frac{\ln x-\mu}{\sigma}. $$

Use the large-$z$ normal-tail approximation.
As $z$ grows, $\bar{\Phi}(z)\sim \phi(z)/z$. Taking logarithms gives $$ \ln R(x)\sim -\frac{z^2}{2}-\ln z-\frac{1}{2}\ln(2\pi)+\alpha\ln(x/\theta). $$

Compare the rates of growth.
Since $z=(\ln x-\mu)/\sigma$, the term $-z^2/2$ decreases quadratically in $\ln x$, while $\alpha\ln(x/\theta)$ increases only linearly in $\ln x$. Therefore $\ln R(x)\to-\infty$, and hence $R(x)\to 0$.

The approximation used here follows from the standard normal tail’s large-$z$ behavior; it can also be motivated by integration by parts in the integral of the normal density from $z$ to infinity. The key consequence is the limit: for fixed valid lognormal and Pareto parameters,

\lim_{x\to\infty}\frac{S_{\mathrm{LN}}(x)}{S_{\mathrm{P}}(x)}=0.

Thus, although a lognormal survival probability can exceed a particular Pareto survival probability at some finite thresholds, it is eventually smaller than that Pareto survival probability. This is an asymptotic statement, not a rule for every threshold. It does not identify the threshold at which the ordering changes, and there may be no crossing in a particular practical range of losses.

Worked Example: An Extreme Threshold with Different Parameters

Consider hypothetical models with $\ln X\sim N(1,0.75^2)$ for the lognormal severity and a Pareto Type I severity with $\theta=1$, $\alpha=1.5$. Compare their probabilities of exceeding $x=e^4$. The method is to standardize the lognormal threshold and evaluate the Pareto power separately.

Lognormal calculation. The lognormal parameters are $\mu=1$ and $\sigma=0.75$. Therefore $$ z=\frac{\ln(e^4)-1}{0.75}=\frac{3}{0.75}=4. $$ It follows that $S_{\mathrm{LN}}(e^4)=\bar{\Phi}(4)\approx 0.00003167$.

Pareto calculation. The threshold exceeds $\theta=1$, so the Pareto survival formula applies: $$ S_{\mathrm{P}}(e^4)=\left(\frac{1}{e^4}\right)^{1.5}=e^{-6}\approx 0.00247875. $$

Comparison and check. The ratio is $0.0000316712/0.00247875\approx 0.01278$. The lognormal exceedance probability is about 1.278% of the Pareto exceedance probability at this threshold. Both results have the expected direction for this high threshold: the Pareto tail is larger. The lognormal probability is displayed in scientific form because rounding it to four decimal places would produce 0.0000. Do not calculate the Pareto exponent as $-4$: raising $e^{-4}$ to $1.5$ gives $e^{-6}$.

Interpreting the Comparison

The phrase “heavier tail” needs a comparison target. In this tutorial, the asymptotic comparison is between survival probabilities: for a fixed Pareto Type I model and a fixed lognormal model, the Pareto survival probability eventually exceeds the lognormal survival probability. This does not mean that every Pareto model has a larger survival probability than every lognormal model at all loss levels. Changing $\theta$, $\alpha$, $\mu$, or $\sigma$ can change finite-threshold comparisons.

For a practical comparison, state all parameters, choose the loss threshold of interest, and calculate both survival probabilities. If the purpose is to assess extremely large losses, also consider whether the threshold is far enough into the tail for asymptotic behavior to matter. A threshold described as “large” in dollars may not be extreme relative to a model’s scale or log-scale parameters.

Worked Example: Comparing Probabilities at a Scaled Threshold

A hypothetical portfolio is modeled either by a lognormal severity with $\mu=\ln(1000)$, $\sigma=0.5$, or by a Pareto Type I severity with $\theta=1000$, $\alpha=2$. Compare the probabilities of a loss exceeding $1000e^2$ dollars. The method is to use the common threshold in both formulas, keeping its scale relative to each model explicit.

Threshold. The threshold is $1000e^2\approx \$7{,}389.06$. It exceeds the Pareto scale $1000$, so the Pareto survival formula is valid.

Lognormal probability. $$ z=\frac{\ln(1000e^2)-\ln(1000)}{0.5} =\frac{2}{0.5}=4. $$ Thus $S_{\mathrm{LN}}(1000e^2)=\bar{\Phi}(4)\approx 0.00003167$.

Pareto probability. $$ S_{\mathrm{P}}(1000e^2) =\left(\frac{1000}{1000e^2}\right)^2 =e^{-4}\approx 0.01831564. $$

Comparison and check. The ratio is $0.0000316712/0.01831564\approx 0.001729$. The Pareto probability is about $1/0.001729\approx 578.4$ times the lognormal probability at this threshold. The scale cancels correctly from the Pareto expression because the threshold is $e^2$ times $\theta$. A common mistake is to use $x=1000e^2$ in the lognormal formula without first subtracting $\mu=\ln(1000)$; that would ignore the lognormal model’s dollar scale.

Common Exam Traps

Mixing original and log scales. For the lognormal model, standardize $\ln x$, not $x$: use $(\ln x-\mu)/\sigma$.
Comparing different thresholds. A comparison of survival probabilities is meaningful only when both models are evaluated at the same $x$.
Ignoring Pareto support. The expression $(\theta/x)^\alpha$ applies for $x\geq\theta$; below the scale, the survival probability is 1.
Confusing a finite-threshold result with an asymptotic result. A tail’s eventual ordering does not guarantee that ordering at a particular moderate threshold.
Rounding away a small probability. A value such as $0.00003167$ becomes 0.0000 to four decimal places. Preserve a scientific-form value when needed to make a tail comparison informative.

Key takeaway: Calculate the lognormal tail as a normal survival probability and the Pareto Type I tail as a power of the scale-to-threshold ratio. At finite loss levels either model may have the larger survival probability, but as $x\to\infty$, the lognormal-to-Pareto survival ratio tends to zero for fixed parameters. This comparison is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Check Your Understanding

For each question, compare probabilities at the stated threshold or identify the limiting tail relationship.

A lognormal severity has $\mu=0$, $\sigma=1$, and a Pareto Type I severity has $\theta=1$, $\alpha=2$. At $x=2$, calculate both survival probabilities and their ratio, lognormal divided by Pareto. (Answer: $S_{\mathrm{LN}}(2)\approx0.2441$, $S_{\mathrm{P}}(2)=0.2500$, ratio $\approx0.9764$.)
A lognormal severity has $\mu=\ln(500)$, $\sigma=0.4$. Find the probability that a loss exceeds $500e^{0.8}$. (Answer: $z=2$, so the probability is $\bar{\Phi}(2)\approx0.0228$.)
A Pareto Type I severity has $\theta=3$, $\alpha=1.5$. Find the probability that a loss exceeds 12, and state why the Pareto formula is valid. (Answer: $12\geq3$, so $S_{\mathrm{P}}(12)=(3/12)^{1.5}=0.1250$.)
A lognormal severity has $\mu=2$, $\sigma=0.5$. Calculate the probability of exceeding $e^4$, retaining a useful value for this small tail. (Answer: $z=(4-2)/0.5=4$, so the probability is approximately $0.00003167$, or $3.167\times10^{-5}$.)
For any fixed lognormal parameters $\mu\in\mathbb{R}$, $\sigma>0$, and Pareto Type I parameters $\theta>0$, $\alpha>0$, what is the limit of $S_{\mathrm{LN}}(x)/S_{\mathrm{P}}(x)$ as $x\to\infty$, and what does it imply? (Answer: The limit is 0; the Pareto survival probability eventually exceeds the lognormal survival probability.)

What You'll Learn