Risk Modelling and Survival Analysis · Compound distributions and their applications in risk modelling

Panjer Recursion and Numerical Methods for Aggregate Claims

Updated 11 October 2026 · Fact-checked

Aggregate claims S is the sum of a random number N of claim sizes. You find its distribution by convolution (exact but slow), Panjer recursion (exact and fast when N is Poisson, binomial or negative binomial and claims are discrete), or a normal approximation using E[S] and Var[S] (quick but rough).

Understand Recursive and Numerical Methods for Aggregate Claims

Aggregate claims S = X₁ + X₂ + … + X_N is the total paid by an insurer over a period. N is the number of claims and each X is a claim size. The aim is to find P(S = s) or P(S ≤ s), not just the mean and variance.

The convolution method works from first principles. Condition on N: P(S = s) = Σ P(N = n) × f^{*n}(s). Here f^{*n} is the distribution of the sum of n claims. You build f^{*2} from f, then f^{*3} from f^{*2} and f, and so on. It is exact, but the work grows fast. In an exam you use it only for small numbers of claims and a few claim sizes.

Panjer recursion removes the need for every convolution. It applies when N belongs to the (a, b, 0) class, so that P(N = k) = (a + b ÷ k) × P(N = k − 1) for k ≥ 1. Poisson, binomial, negative binomial and geometric all fit. Claim sizes must be discrete, on 0, 1, 2, … (or a continuous distribution discretised to a grid). You then get each g_s = P(S = s) from the earlier values g₀, g₁, …, g_{s−1}.

The normal approximation replaces S with a normal variable that has the same mean and variance. You need E[S] = E[N]E[X] and Var[S] = E[N]Var[X] + Var[N](E[X])². It is fast and works best when the expected number of claims is large. It ignores skewness, and S is usually right-skewed, so it tends to understate the upper tail. If S is discrete, apply a continuity correction.

The exam usually asks you to pick the right method, set it up correctly and then compute a few terms or a probability. Showing the formula, the parameter values and the working earns most of the marks.

Key rules to remember

(a, b, 0) class condition: P(N = k) = (a + b ÷ k) × P(N = k − 1), for k = 1, 2, 3, …; Use it to find a and b by dividing consecutive probabilities if the question does not give them.
Parameters a and b: Poisson(λ): a = 0, b = λ. Binomial(n, p), q = 1 − p: a = −p ÷ q, b = (n + 1)p ÷ q. Negative binomial (k, p), q = 1 − p, P(N = x) = C(x + k − 1, x) pᵏ qˣ: a = q, b = (k − 1)q.; Binomial has a negative. Geometric is the negative binomial with k = 1, so a = q, b = 0. Check the parametrisation against the Tables.
Panjer starting value: g₀ = P(S = 0) = G_N(f₀), where G_N is the PGF of N and f₀ = P(X = 0); If f₀ = 0, then g₀ = P(N = 0). For Poisson, g₀ = exp(λ(f₀ − 1)).
Panjer recursion: g_s = [1 ÷ (1 − a f₀)] × Σ (j = 1 to s) (a + b j ÷ s) f_j g_{s−j}, for s = 1, 2, 3, …; If f₀ = 0 the factor in front is 1. For compound Poisson this becomes g_s = (λ ÷ s) Σ j f_j g_{s−j}.
Convolution formula: P(S = s) = Σ (n ≥ 0) P(N = n) f^{*n}(s), with f^{*n}(s) = Σ_y f(y) f^{*(n−1)}(s − y); f^{*0}(0) = 1. For continuous claims, F_S(x) = Σ P(N = n) F^{*n}(x).
Mean and variance of S: E[S] = E[N]E[X]; Var[S] = E[N]Var[X] + Var[N](E[X])²; For compound Poisson with parameter λ, Var[S] = λE[X²].
Normal approximation: P(S ≤ s) ≈ Φ((s − E[S]) ÷ √Var[S]); For integer-valued S, use s + 0.5 for P(S ≤ s) and s − 0.5 for P(S ≥ s) (continuity correction).

How to solve Recursive and Numerical Methods for Aggregate Claims questions

Use this method for any question on finding or approximating the distribution of aggregate claims.

1Read what is asked: exact probabilities for small values of s, a full distribution, or a probability in the tail. This decides the method.
2Write down the distribution of N and of X. Check whether X is discrete on 0, 1, 2, … and whether N is Poisson, binomial or negative binomial.
3If Panjer applies, find a and b (from the standard results or from the ratio P(N = k) ÷ P(N = k − 1)) and state f₀.
4Compute g₀ = G_N(f₀). Then apply g_s = [1 ÷ (1 − a f₀)] Σ (a + b j ÷ s) f_j g_{s−j} for s = 1, 2, … Keep a table of g_s and work to at least five or six decimal places.
5If only small totals are needed, convolution can be quicker. List the ways each total can arise for each n, and weight them by P(N = n).
6If a large expected number of claims is involved or an approximation is asked for, compute E[S] and Var[S]. Standardise and use Φ, with a continuity correction if S is integer-valued.
7Cumulate probabilities if the question asks for P(S ≤ s) or P(S > s). Check that the probabilities are sensible: non-negative, and the total is at most 1.
8State the answer with a short comment, for example that the normal approximation understates the right tail because S is skewed.

Quickest way: Table-first Panjer for compound Poisson

When to use it: Use when N is Poisson and claim sizes are on a small integer grid, and you need only the first few values of the distribution of S.

Write λ and f₁, f₂, … in a row at the top of your working.
Compute g₀ = e^(−λ(1 − f₀)). If f₀ = 0, this is e^(−λ).
Use g_s = (λ ÷ s) Σ j f_j g_{s−j}. Only terms with j ≤ s count, and any f_j that is zero drops out.
Add a quick check: for s = 1 or 2, list the combinations by hand and see if the answers match.
For tail probabilities, calculate 1 − Σ g_s rather than summing many terms.

Common mistakes in Recursive and Numerical Methods for Aggregate Claims

Using the wrong a and b for the binomial or negative binomial.
Students remember the Poisson case (a = 0, b = λ) and assume the others look similar, or mix up the parametrisations.
Fix: Derive a and b from P(N = k) ÷ P(N = k − 1) if unsure. Binomial has a negative a. For negative binomial check the Tables for which parameter is p and which is q.
Starting with g₀ = P(N = 0) when f₀ is not zero.
The shortcut g₀ = p₀ is true only when claims cannot be zero.
Fix: Always write g₀ = G_N(f₀). Also remember the 1 ÷ (1 − a f₀) factor in the recursion.
Getting the recursion indices wrong, for example using f_{s−j} g_j or summing from j = 0.
The convolution looks symmetric, but the weight (a + bj ÷ s) attaches to the claim size f_j.
Fix: Sum from j = 1 to s. Write f_j with weight (a + bj ÷ s) and g_{s−j} for the rest. Test the formula on g₁ first.
Using Var[S] = E[N]Var[X] only.
Students forget that the number of claims is also random.
Fix: Write Var[S] = E[N]Var[X] + Var[N](E[X])². For Poisson it simplifies to λE[X²], not λVar[X].
Omitting the continuity correction when S is discrete.
The normal approximation looks like a direct standardisation.
Fix: For integer S, use P(S ≤ s) ≈ Φ((s + 0.5 − μ) ÷ σ). Skip the correction when S is continuous.
Treating the normal approximation as accurate for the tail.
Students forget that S is skewed, especially with few expected claims or heavy-tailed claim sizes.
Fix: Say that the approximation is rough, and that it may understate high-percentile values. Mention a better method, such as a skewed distribution fitted on the first three moments or Panjer recursion.

Worked examples

Example 1

The number of claims N is Poisson with mean 2. Each claim is 1 with probability 0.5 and 2 with probability 0.5. Use Panjer recursion to find P(S = s) for s = 0, 1, 2, 3, and then P(S ≤ 3).

Show the solution

N is Poisson, so a = 0 and b = λ = 2. Here f₀ = 0, f₁ = 0.5, f₂ = 0.5.
g₀ = e^(−2) = 0.135335.
The recursion is g_s = (2 ÷ s) Σ j f_j g_{s−j}.
g₁ = (2 ÷ 1)(1 × 0.5 × g₀) = g₀ = 0.135335.
g₂ = (2 ÷ 2)(1 × 0.5 × g₁ + 2 × 0.5 × g₀) = 0.067668 + 0.135335 = 0.203003.
g₃ = (2 ÷ 3)(1 × 0.5 × g₂ + 2 × 0.5 × g₁) = (2 ÷ 3)(0.101501 + 0.135335) = (2 ÷ 3)(0.236836) = 0.157891.
Check g₃ directly: N = 2 with claims (1, 2) or (2, 1) has probability 0.5 × 2e^(−2) = 0.135335. N = 3 with (1, 1, 1) has probability 0.125 × (8 ÷ 6)e^(−2) = 0.022556. The sum is 0.157891, which matches.
P(S ≤ 3) = 0.135335 + 0.135335 + 0.203003 + 0.157891 = 0.631564.

Answer: g₀ = 0.1353, g₁ = 0.1353, g₂ = 0.2030, g₃ = 0.1579, and P(S ≤ 3) ≈ 0.6316.

Example 2

Annual claims for a portfolio follow a compound Poisson distribution. The Poisson parameter is 50. Claim sizes have mean ₹20,000 and standard deviation ₹15,000. Use a normal approximation to estimate the probability that aggregate claims exceed ₹11,00,000.

Show the solution

E[X] = 20,000 and Var[X] = 15,000² = 22,50,00,000, so E[X²] = 22,50,00,000 + 40,00,00,000 = 62,50,00,000.
E[S] = λE[X] = 50 × 20,000 = ₹10,00,000.
For compound Poisson, Var[S] = λE[X²] = 50 × 62,50,00,000 = 3,125,00,00,000 = 3.125 × 10¹⁰.
Standard deviation = √(3.125 × 10¹⁰) = about ₹1,76,777.
S is continuous here, so no continuity correction is needed.
z = (11,00,000 − 10,00,000) ÷ 1,76,777 = 0.5657.
P(S > 11,00,000) ≈ 1 − Φ(0.5657). From tables, Φ(0.56) = 0.7123 and Φ(0.57) = 0.7157, so Φ(0.5657) ≈ 0.7145.
The probability is about 1 − 0.7145 = 0.2855.

Answer: About 0.285. This is rough, because S is right-skewed and the normal approximation may understate the true upper-tail probability.

Exam tips

Show a and b explicitly for the claim number distribution, and state f₀. Marks are often given for correct setup even if arithmetic slips.
Keep a clean table of g_s with at least five decimal places. Rounding early spoils later terms.
Check g₁ and g₂ by direct enumeration if time allows. It catches index errors in the recursion.
When asked to compare methods, discuss accuracy, workload and conditions. Say that Panjer needs discrete claims and an (a, b, 0) claim number, and that the normal approximation ignores skewness.
In the computer-based paper, write the recursion as a loop in R or Excel. State the grid used for discretising claim sizes and the length of the table.

Practice questions from Compound distributions and their applications in risk modelling

Recursive and Numerical Methods for Aggregate Claims: frequently asked questions

When can I use Panjer recursion?

Use it when the number of claims is Poisson, binomial, negative binomial or geometric, so that P(N = k) = (a + b ÷ k)P(N = k − 1). Claim sizes must be discrete on 0, 1, 2, … or discretised to such a grid. Claim sizes and claim numbers must be independent.

How do I get g₀ in Panjer recursion?

Use g₀ = G_N(f₀), the PGF of N evaluated at f₀ = P(X = 0). If claims cannot be zero, f₀ = 0 and g₀ = P(N = 0). For compound Poisson, g₀ = exp(λ(f₀ − 1)).

Is the convolution method or Panjer recursion better in the exam?

For a few small totals, convolution by direct enumeration can be quicker and also serves as a check. For a longer table of probabilities, Panjer recursion is faster and less error-prone. Choose by how many values of s are needed.

When is the normal approximation reliable for aggregate claims?

It works better when the expected number of claims is large and claim sizes are not very heavy-tailed. It is weaker with few claims or high skewness, because S is skewed and the normal is symmetric. Say this when you use it.