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

Fitting Loss Distributions: Method of Moments and MLE

Updated 11 October 2026 · Fact-checked

Fitting a loss distribution means choosing parameter values so the model matches your claim data. Method of moments equates sample and model moments. Percentile matching equates sample and model percentiles. Maximum likelihood maximises the likelihood of the observed data, and handles truncated or censored claims by adjusting each term.

Understand Fitting Loss Distributions: Method of Moments and MLE

You have a set of claim amounts. You believe they come from a family such as exponential, gamma, Pareto or lognormal. The family has unknown parameters. Fitting means estimating those parameters from the data.

Method of moments (MoM) is the simplest idea. Write the model's theoretical moments as functions of the parameters. Set them equal to the sample moments. Solve. With one parameter you use the mean. With two parameters you use the first two moments, usually the mean and the variance, or E[X] and E[X²]. Remember the sample variance used in matching is often the one with divisor n unless the question says otherwise.

Percentile matching sets the model's chosen percentiles (for example the median and the 75th percentile) equal to the sample percentiles. It works well when the CDF has a closed form, such as the Pareto, Weibull or exponential. It is also useful when the moments do not exist.

Maximum likelihood estimation (MLE) picks the parameters that make the observed data most probable. For independent observations the likelihood is the product of the densities. You take logs, differentiate, set the derivative to zero and solve. MLE is usually preferred because its estimators are asymptotically unbiased, efficient and normally distributed. It also handles incomplete data naturally.

Incomplete data changes the likelihood terms. An exactly observed claim x contributes f(x). A claim censored at a policy limit u (you only know X > u) contributes S(u) = 1 − F(u). If data are truncated from below at a deductible d (you only see claims above d), each observed claim contributes f(x) ÷ S(d). Getting these contributions right is the main exam skill.

Key rules to remember

Method of moments: E[X^k; θ] = (1/n) Σ xᵢ^k, for k = 1, ..., number of parameters; Solve the equations for the parameters. For a two-parameter family use k = 1 and 2.
Percentile matching: F(πₚ; θ) = p, where πₚ is the sample 100p-th percentile; Use as many percentiles as parameters. State how you computed the sample percentile.
Log-likelihood, complete data: l(θ) = Σ ln f(xᵢ; θ); Solve dl/dθ = 0 and check it is a maximum.
Right-censored observation at u: contribution to L = S(u) = 1 − F(u); Applies when you only know the claim exceeds u, such as a policy limit.
Left-truncated observation at d: contribution to L = f(x) ÷ S(d), for x > d; Applies when claims below the deductible are not recorded.
Exponential MLE: λ̂ = n ÷ Σ xᵢ = 1 ÷ x̄ (rate λ); With censored data: λ̂ = (number of uncensored claims) ÷ (sum of all amounts, including censored).
Pareto moments (α, λ), f(x) = αλ^α ÷ (λ + x)^(α+1): E[X] = λ ÷ (α − 1) for α > 1; E[X²] = 2λ² ÷ ((α − 1)(α − 2)) for α > 2; Moments exist only above these values of α.
Gamma (α, λ) moments: E[X] = α ÷ λ; Var(X) = α ÷ λ²; MoM: λ̂ = x̄ ÷ s², α̂ = x̄² ÷ s².

How to solve Fitting Loss Distributions: Method of Moments and MLE questions

Use this method for any question that asks you to estimate parameters of a loss distribution.

1Identify the distribution and its parameterisation. Write down f(x), F(x) and the moments you need, using the form given in the question or tables.
2Identify the data type: complete, censored (policy limit), truncated (deductible), or grouped. Note which observations are exact.
3Decide the method asked for: moments, percentiles or MLE. If the question does not say, MLE is usually the better answer.
4For MoM or percentiles, write one equation per parameter and solve. For two moments, find the sample mean and the second sample moment or variance carefully.
5For MLE, write the likelihood with the correct term for each observation type. Take logs and simplify.
6Differentiate with respect to each parameter and set to zero. Solve. For Pareto and gamma, one parameter may need a numerical or reduced equation.
7Check the estimate is sensible: positive, inside the parameter range, and (for MoM) satisfying conditions such as α > 2.
8State the final estimates with units and, if asked, use them for a probability or expected claim.

Quickest way: Fast routes for common one- and two-parameter cases

When to use it: Use when time is short and the family is exponential, Pareto with known λ, or gamma with a two-moment fit.

Exponential: MLE equals MoM. The rate is 1 ÷ mean. With censoring, divide the number of uncensored claims by the total of all amounts.
Pareto with two parameters by MoM: use x̄ = λ ÷ (α − 1) and the second moment. Divide E[X²] by E[X]² to get 2(α − 1) ÷ (α − 2), solve for α, then find λ.
Gamma by MoM: α̂ = x̄² ÷ s² and λ̂ = x̄ ÷ s². Two lines.
Pareto with known λ by MLE: α̂ = n ÷ Σ ln((λ + xᵢ) ÷ λ).
For a truncated observation, divide by S(d). Often the S(d) terms simplify the exponential case, because memorylessness means you use (x − d).

Common mistakes in Fitting Loss Distributions: Method of Moments and MLE

Using f(x) for a censored claim instead of S(u).
Students treat every recorded number as an exact claim.
Fix: Ask for each data point: do I know the exact value, or only that it is above a limit? Use f(x) for exact and S(u) for censored.
Forgetting to divide by S(d) for claims truncated at a deductible.
The data look like a normal sample, so the conditioning is missed.
Fix: If small claims are not observed, every observed claim is conditional on X > d. Use f(x) ÷ S(d).
Using the wrong variance divisor in method of moments.
Students automatically use n − 1 from statistics practice.
Fix: In MoM you match the second moment about the mean using the sample moments, with divisor n, unless the question states otherwise. State your choice clearly.
Applying Pareto moment formulas when α ≤ 2.
The solved α is not checked against the condition for the moment to exist.
Fix: Check α > 2 for the variance. If it fails, say MoM is not valid and use percentile matching or MLE.
Mixing up rate and scale for exponential and gamma.
Different sources write f(x) = λe^(−λx) or (1/θ)e^(−x/θ).
Fix: Write the density first and keep to that form. Then mean is 1/λ for rate, θ for scale.
Not checking that the stationary point is a maximum, or ignoring the parameter range.
Time pressure after a long derivation.
Fix: Check the second derivative is negative, or note the log-likelihood is concave, and confirm the estimate lies in the allowed range.

Worked examples

Example 1

Five claims (₹ thousand) are 2, 3, 5, 6 and 9. A gamma distribution with parameters α and λ (mean α ÷ λ, variance α ÷ λ²) is fitted by method of moments, using the sample variance with divisor n. Find α̂ and λ̂.

Show the solution

Sample mean: (2 + 3 + 5 + 6 + 9) ÷ 5 = 25 ÷ 5 = 5.
Sum of squares: 4 + 9 + 25 + 36 + 81 = 155. Second moment = 155 ÷ 5 = 31.
Variance with divisor n: 31 − 5² = 31 − 25 = 6.
λ̂ = mean ÷ variance = 5 ÷ 6 = 0.8333.
α̂ = mean² ÷ variance = 25 ÷ 6 = 4.1667. Check: α̂ ÷ λ̂ = 4.1667 ÷ 0.8333 = 5.

Answer: α̂ = 25/6 ≈ 4.167 and λ̂ = 5/6 ≈ 0.833 (per ₹ thousand).

Example 2

An insurer records 6 claims under a policy with a limit of ₹10 lakh. Four claims are exact: ₹2, ₹3, ₹4 and ₹7 lakh. Two claims are censored at the limit of ₹10 lakh. Assuming claims are exponential with rate λ, find the MLE of λ and the estimated probability that a claim exceeds ₹5 lakh.

Show the solution

Exact claims contribute f(x) = λe^(−λx). Censored claims contribute S(10) = e^(−10λ).
Likelihood: L = λ⁴ e^(−λ(2 + 3 + 4 + 7)) × e^(−λ × 10 × 2) = λ⁴ e^(−λ(16 + 20)) = λ⁴ e^(−36λ).
Log-likelihood: l = 4 ln λ − 36λ.
Differentiate: dl/dλ = 4/λ − 36 = 0, so λ̂ = 4 ÷ 36 = 1/9 = 0.1111 per lakh.
Second derivative is −4/λ², which is negative, so this is a maximum.
P(X > 5) = e^(−5λ̂) = e^(−5/9) = e^(−0.5556) ≈ 0.574.

Answer: λ̂ = 1/9 ≈ 0.111 per lakh (mean about ₹9 lakh). Estimated P(X > ₹5 lakh) ≈ 0.574.

Exam tips

Write the likelihood contribution for each data type before doing any algebra. Marks are often given for the correct contributions even if the algebra slips.
Always state the parameterisation you use for Pareto, gamma and exponential, and keep it consistent. The Formulae and Tables book may use a different form from your notes.
In Paper B (R), know how to write the negative log-likelihood as a function and minimise it with optim or nlm. Give starting values and check convergence.
For MCQs, test quickly whether the data are censored or truncated. Many wrong options come from using the complete-data formula.
Say why you prefer MLE or MoM when asked to comment: MLE is generally more efficient and handles incomplete data, while MoM is simpler and gives starting values.

Practice questions from Loss distributions, with and without risk sharing

Fitting Loss Distributions: Method of Moments and MLE: frequently asked questions

What is the difference between method of moments and maximum likelihood?

Method of moments sets the model's moments equal to the sample moments. Maximum likelihood chooses the parameters that make the observed data most probable. MLE usually has better large-sample properties and handles censoring and truncation directly. MoM is easier to compute.

How do I fit a Pareto distribution by method of moments?

Write E[X] = λ ÷ (α − 1) and E[X²] = 2λ² ÷ ((α − 1)(α − 2)). Match them to the sample first and second moments. Solve for α first by dividing, then find λ. Check that α > 2.

How does censoring change the likelihood?

A claim censored at a limit u contributes S(u) instead of f(x). You only know it exceeded u. Exact claims still contribute f(x).

Does the MLE for the gamma distribution have a closed form?

Not for both parameters together. The equation for the shape parameter involves the digamma function, so you solve numerically. If the shape is known, the rate has a simple closed form. In exams you are usually asked for method of moments for the gamma.

When should I use percentile matching?

Use it when the question asks for it, or when the CDF is simple and moments are hard or do not exist. It is common for the Pareto, Weibull and lognormal.