Risk Modelling and Survival Analysis · Loss distributions, with and without risk sharing

Loss Distributions and Their Properties for Claim Sizes

Updated 11 October 2026 · Fact-checked

A loss distribution is a continuous, positive probability model for the size of a single claim. The main ones are exponential, gamma, lognormal, Pareto, Weibull and Burr. To answer a question, write down the density or survival function, find the moments, then judge the tail. Pareto, lognormal and Burr have heavy tails.

Understand Loss Distributions and Their Properties

A loss distribution models the amount of one claim, X, given that a claim occurs. X is positive and continuous, and usually right-skewed: most claims are small, a few are very large. The model you pick decides how much probability sits in the large claims, and that drives premiums, reserves and reinsurance cost.

The simplest model is the exponential. It has one parameter and a constant hazard rate. It is light-tailed and rarely fits real claims well, but it is the benchmark. The gamma adds a shape parameter and gives a more flexible body with a similar light tail. The Weibull has survival function exp(−c x^γ). Its hazard rate rises if γ > 1 and falls if γ < 1, so with γ < 1 it has a tail heavier than the exponential.

The lognormal, Pareto and Burr are the usual heavy-tailed choices. If ln X is normal, X is lognormal. The Pareto has a power-law tail, so large claims are far more likely than under the exponential. The Burr adds a third parameter and includes the Pareto as a special case when γ = 1. A practical test: light-tailed distributions have a moment generating function that exists for some t > 0. Heavy-tailed ones do not. For heavy-tailed models, higher moments can be infinite.

Choosing a model uses evidence. Check the shape of the data, the skewness, and how the empirical mean excess or hazard behaves. Also check whether the model's moments exist. Motor own-damage claims might suit a lognormal or gamma. Liability or large property claims often need a Pareto or Burr. Once a family is chosen, you fit its parameters, for example by moments or maximum likelihood.

Key rules to remember

Exponential(λ): f(x) = λe^(−λx); S(x) = e^(−λx); E[X] = 1/λ; Var(X) = 1/λ²; Constant hazard rate λ. MGF = λ ÷ (λ − t) for t < λ.
Gamma(α, λ): f(x) = λ^α x^(α−1) e^(−λx) ÷ Γ(α); E[X] = α/λ; Var(X) = α/λ²; MGF = (λ ÷ (λ − t))^α for t < λ. Light-tailed. α = 1 gives the exponential.
Lognormal(μ, σ²): E[X^k] = exp(kμ + k²σ²/2); E[X] = exp(μ + σ²/2); Var(X) = exp(2μ + σ²) × (exp(σ²) − 1); μ and σ² are the mean and variance of ln X, not of X. The median is e^μ. The MGF does not exist for t > 0.
Pareto(α, λ): S(x) = (λ ÷ (λ + x))^α; f(x) = αλ^α ÷ (λ + x)^(α+1); E[X] = λ ÷ (α − 1) for α > 1; Var(X) = αλ² ÷ ((α − 1)²(α − 2)) for α > 2; E[X^k] = k! λ^k ÷ ((α−1)(α−2)...(α−k)) only when α > k. This is the two-parameter form on the support x > 0.
Weibull(c, γ): S(x) = exp(−c x^γ); f(x) = cγ x^(γ−1) exp(−c x^γ); E[X^k] = Γ(1 + k/γ) ÷ c^(k/γ); Hazard rate is cγ x^(γ−1). It increases for γ > 1, is constant for γ = 1 and decreases for γ < 1.
Burr(α, λ, γ): S(x) = (λ ÷ (λ + x^γ))^α; f(x) = αγ λ^α x^(γ−1) ÷ (λ + x^γ)^(α+1); E[X^k] = λ^(k/γ) Γ(k/γ + 1) Γ(α − k/γ) ÷ Γ(α); The k-th moment exists only for −γ < k < αγ. With γ = 1 it reduces to the Pareto.
Light vs heavy tail test: Light-tailed: M_X(t) exists for some t > 0. Heavy-tailed: it does not.; Exponential, gamma and Weibull with γ ≥ 1 are light-tailed. Lognormal, Pareto and Burr are heavy-tailed. Weibull with γ < 1 is also usually treated as heavy-tailed.

How to solve Loss Distributions and Their Properties questions

Use this order for any question on loss distributions. It works for calculation questions and for discussion questions on model choice.

1Identify the distribution and its parameterisation. Check the Tables for the form given, and note the support and the parameter meanings.
2Write down the density f(x) or survival function S(x) you need before substituting numbers.
3For probabilities of exceeding a value, use S(x) directly. For the lognormal, standardise ln x and use normal tables.
4For moments, use the standard E[X^k] formula. First check the condition for the moment to exist, for example α > k for the Pareto.
5Compute variance as E[X²] − (E[X])², and standard deviation or coefficient of variation if asked.
6Comment on the tail: does the MGF exist, is the hazard rate increasing, constant or decreasing, and is the mean above the median.
7State the result with units in rupees where relevant, and say what the answer means for the insurer.
8For model choice, link the data features (skew, large outliers, hazard behaviour) to the family, and state the assumptions.

Quickest way: Match tail and moments to the family in under a minute

When to use it: Use this for MCQs and for the first part of a written question that asks which distribution suits a description or what is a property of it.

Ask first: is the tail exponential-like, power-like or log-normal-like? Power-like means Pareto or Burr.
Check for a constant hazard: exponential only. Rising hazard: Weibull with γ > 1. Falling hazard: Pareto or Weibull with γ < 1.
For a Pareto or Burr, look at α. If the question asks about a mean or variance, test α > 1 or α > 2 (or the Burr condition) before calculating.
For a lognormal, remember mean = exp(μ + σ²/2) and median = exp(μ). The mean is always above the median.
For exceedance probabilities, use S(x) straight away rather than integrating the density.

Common mistakes in Loss Distributions and Their Properties

Using the wrong parameterisation, for example treating the Pareto as having support x > λ.
Different books define the Pareto, Weibull and Burr differently, and students remember a form from another text.
Fix: Use the form in the IAI Tables. Check the survival function in the Tables before you start. State the form you are using.
Treating lognormal μ and σ² as the mean and variance of X.
The names sound like mean and variance, and they are for the underlying normal variable.
Fix: Remember that μ and σ² belong to ln X. Convert with E[X] = exp(μ + σ²/2) and the variance formula.
Quoting a Pareto mean or variance without checking α.
Students substitute into λ ÷ (α − 1) automatically.
Fix: State the condition first: α > 1 for the mean and α > 2 for the variance. If it fails, say the moment is infinite.
Saying a distribution is heavy-tailed because it has a large mean or variance.
Tail weight is confused with scale.
Fix: Tail weight is about how fast S(x) falls at large x. Use the MGF test or compare survival functions, not the size of the mean.
Mixing up the Weibull shape effect on the hazard rate.
The hazard cγ x^(γ−1) is not memorised.
Fix: Remember γ > 1 means rising hazard, γ < 1 means falling hazard and γ = 1 gives the exponential.
Forgetting to standardise ln x for lognormal probabilities.
Students apply normal tables to x itself.
Fix: Use P(X > x) = 1 − Φ((ln x − μ) ÷ σ).

Worked examples

Example 1

Claim sizes follow a Pareto distribution with α = 3 and λ = ₹20,000. Find the mean, the standard deviation, and the probability that a claim exceeds ₹30,000.

Show the solution

The mean exists because α = 3 > 1. E[X] = λ ÷ (α − 1) = 20,000 ÷ 2 = ₹10,000.
The variance exists because α = 3 > 2. Var(X) = αλ² ÷ ((α − 1)²(α − 2)) = 3 × 400,000,000 ÷ (4 × 1) = 300,000,000.
Standard deviation = √300,000,000 ≈ ₹17,321.
For the probability, S(x) = (λ ÷ (λ + x))^α. With x = 30,000: S = (20,000 ÷ 50,000)³ = 0.4³ = 0.064.

Answer: Mean ₹10,000, standard deviation about ₹17,321 and P(X > ₹30,000) = 0.064. The standard deviation is well above the mean, which is typical of a heavy tail.

Example 2

Claim sizes are lognormal with μ = 8 and σ = 1 (amounts in rupees). Find the mean and median, and the probability that a claim exceeds ₹5,000. Comment on the result.

Show the solution

Mean = exp(μ + σ²/2) = exp(8 + 0.5) = e^8.5 ≈ 2,980.96 × 1.64872 ≈ ₹4,915.
Median = e^μ = e^8 ≈ ₹2,981.
For the probability, ln 5,000 ≈ 8.5172. Standardise: z = (8.5172 − 8) ÷ 1 = 0.5172.
P(X > 5,000) = 1 − Φ(0.5172). Φ(0.51) ≈ 0.6950 and Φ(0.52) ≈ 0.6985, so Φ(0.5172) ≈ 0.6975.
So P(X > 5,000) ≈ 1 − 0.6975 = 0.3025.
The mean is well above the median because the distribution is right-skewed. A few large claims pull the mean up.

Answer: Mean ≈ ₹4,915, median ≈ ₹2,981 and P(X > ₹5,000) ≈ 0.30. The mean exceeds the median, which shows right skew.

Exam tips

Write the survival function first. Most tail and exceedance questions are one line once S(x) is written.
Always state the condition for a moment to exist when the question uses a Pareto or Burr. Examiners give marks for it.
In discussion questions, link each choice of distribution to a data feature: skewness, extreme outliers, hazard behaviour. Do not just list the distributions.
In MCQs, check whether the parameterisation in the question matches the Tables before you calculate.
For computer-based work, show the formula and the parameter values you used before quoting any numerical result.

Practice questions from Loss distributions, with and without risk sharing

Loss Distributions and Their Properties in other exams

The same ground in other exams, if you are preparing for more than one or want another angle on it.

Loss Distributions and Their Properties: frequently asked questions

What is the difference between light-tailed and heavy-tailed distributions?

A light-tailed distribution has a moment generating function that exists for some t > 0, so its tail falls at least as fast as an exponential. A heavy-tailed one does not, and large claims are much more likely. In actuarial work, exponential and gamma are light-tailed, while Pareto, lognormal and Burr are heavy-tailed.

How do I choose a loss distribution for claim amounts?

Look at the data: its shape, skewness and the behaviour of the largest claims. Use exponential or gamma for lighter tails, and lognormal, Pareto or Burr when there are very large claims. Then fit the parameters and test whether the model describes the data well.

When does a Pareto distribution have no mean or variance?

For the two-parameter Pareto, the mean is infinite if α ≤ 1 and the variance is infinite if α ≤ 2. In general the k-th moment exists only if α > k. Check this condition before using any moment formula.

Why is the lognormal mean larger than its median?

The median is e^μ, while the mean is e^(μ + σ²/2). Since σ² > 0, the mean is always larger. This reflects the long right tail.