Aggregate Loss Models and the Panjer Recursion

Let N be a non-negative integer-valued frequency with PMF p_k = P(N = k), and let X_1, X_2, ... be independent and identically distributed severities, independent of N. The aggregate loss is S = X_1 + ... + X_N, with S = 0 by convention when N = 0.

Aggregate models are the bridge from per-claim severity and per-period frequency to the portfolio quantity that drives premiums, reserves, and capital. ASTAM, GI 301, and CFE 101 each use S as a building block for pricing, ruin, and risk-measure computations.

Conditioning on N gives clean expressions for the mean, variance, and moment generating function of S in terms of the frequency and severity. These are the same identities Klugman Loss Models, 5th ed., Ch. 9 derives by tower property and conditional MGF.

When N is Poisson with mean λ, the variance simplifies to λ E[X^2] because Var(N) = E[N] = λ. That is the compound Poisson identity worth memorizing: variance of S equals the rate times the second raw moment of severity, not the second central moment.

A frequency PMF belongs to the (a, b, 0) class if there exist constants a and b such that the ratio of consecutive probabilities is linear in 1/k, starting from k = 1. Klugman shows there are exactly three nondegenerate members: Poisson, Binomial, and Negative Binomial.

Knowing the (a, b) values for each family is exam-critical because the Panjer recursion takes them as inputs. The Poisson sits at a = 0; Binomial sits at a < 0; Negative Binomial sits at 0 < a < 1.

When N is (a, b, 0) and severity X is integer-valued with PMF f(j) = P(X = j) for j = 0, 1, 2, ..., the PMF g(s) = P(S = s) satisfies a one-step recursion. The starting value depends on the frequency family.

The recursion is exact when the severity is genuinely integer-valued. Continuous severities (Pareto, lognormal, gamma) are first discretized onto a chosen span h, and the recursion is run on the discretized PMF. The Panjer output then converges to the true continuous aggregate distribution as h shrinks.

Continuous severity is mapped onto a lattice {0, h, 2h, 3h, ...} so the recursion can run. Two standard methods are the method of rounding (mass at jh equals F(jh + h/2) - F(jh - h/2)) and the method of local moment matching (mass at adjacent points chosen so the first one or two moments are preserved on each interval).

Method of rounding is the default on FAM and ASTAM because it is fastest. Local moment matching is preferred when the recursion needs to feed into a downstream calculation (stop-loss premiums, ruin probabilities) where preserving the mean matters more than the per-cell error.

The (a, b, 1) class allows p_0 to be modified (zero-truncated or zero-modified versions of Poisson, Binomial, and Negative Binomial), while keeping the linear ratio for k = 2, 3, .... The recursion stays nearly identical with only the starting value adjusted using the modified p_0.

Zero-truncated and zero-modified frequencies show up in actuarial work whenever the data-generating process censors zero counts (e.g., only policies with at least one claim are observed) or when an excess zero mass is added to model claim-free policies separately.

A claim count has p_0 = 0.5, p_1 = 0.3, p_2 = 0.135, p_3 = 0.0506. Compute consecutive ratios: p_1/p_0 = 0.6, p_2/p_1 = 0.45, p_3/p_2 = 0.375.

Solve a + b = 0.6 and a + b/2 = 0.45 from the (a, b, 0) defining equation. Subtraction gives b/2 = 0.15, so b = 0.3 and a = 0.3. Check at k = 3: a + b/3 = 0.3 + 0.1 = 0.4. Observed ratio is 0.375, close enough given rounding. With a = 0.3 and b = 0.3, the family is Negative Binomial with β/(1 + β) = 0.3 (so β = 3/7) and (r - 1)β/(1 + β) = 0.3 (so r = 2). The fitted family is NegBin(r = 2, β = 3/7).

Frequency is Poisson with λ = 2. Severity is integer-valued with f(1) = 0.5, f(2) = 0.3, f(3) = 0.2 (and f(0) = 0). Compute g(s) for s = 0, 1, 2, 3, 4, then P(S ≤ 3).

Starting value g(0) = exp(-λ(1 - f(0))) = exp(-2) ≈ 0.1353. With a = 0, b = 2, the recursion simplifies to g(s) = (2/s) Σ_{k=1}^{s} k f(k) g(s - k).

g(1) = (2/1)(1 · 0.5 · g(0)) = 1.0 · 0.1353 = 0.1353. g(2) = (2/2)(1 · 0.5 · g(1) + 2 · 0.3 · g(0)) = 0.5 · 0.1353 + 0.6 · 0.1353 = 0.1488. g(3) = (2/3)(1 · 0.5 · g(2) + 2 · 0.3 · g(1) + 3 · 0.2 · g(0)) = (2/3)(0.0744 + 0.0812 + 0.0812) = (2/3)(0.2367) = 0.1578. g(4) = (2/4)(1 · 0.5 · g(3) + 2 · 0.3 · g(2) + 3 · 0.2 · g(1) + 4 · 0 · g(0)) = 0.5 · (0.0789 + 0.0893 + 0.0812 + 0) = 0.1247.

Cumulative: P(S ≤ 3) = g(0) + g(1) + g(2) + g(3) = 0.1353 + 0.1353 + 0.1488 + 0.1578 = 0.5772. The remaining mass 0.4228 sits at S = 4, 5, 6, ..., decaying geometrically as the recursion is iterated further.

Use Example 2 to verify the moment formulas. Severity has E[X] = 1 · 0.5 + 2 · 0.3 + 3 · 0.2 = 1.7 and E[X^2] = 1 · 0.5 + 4 · 0.3 + 9 · 0.2 = 3.5. By the compound Poisson moment formulas, E[S] = λ E[X] = 2 · 1.7 = 3.4 and Var(S) = λ E[X^2] = 2 · 3.5 = 7.0.

An exam graders’ trick: candidates who run the recursion correctly but forget to recover the moments lose easy partial credit. Always sanity-check g(s) by computing Σ s g(s) and comparing to the closed-form mean. Here Σ s g(s) over s = 0, ..., 8 reproduces 3.4 to within rounding.

Same model as Example 2, but each claim is reduced by a per-claim deductible d = 1 (with the loss to the insurer truncated at zero). The new severity is Y = (X - 1)_+ with PMF f_Y(0) = 0.5, f_Y(1) = 0.3, f_Y(2) = 0.2.

Apply Panjer to (Y, N): g_Y(0) = exp(-2(1 - 0.5)) = exp(-1) ≈ 0.3679. g_Y(1) = (2/1)(1 · 0.3 · 0.3679)/(1 - 0) = 0.2207. g_Y(2) = (2/2)(1 · 0.3 · 0.2207 + 2 · 0.2 · 0.3679) = 0.0662 + 0.1471 = 0.2133.

Cumulative through 2: 0.3679 + 0.2207 + 0.2133 = 0.8019. The deductible has compressed the aggregate distribution: most of the new mass piles up at 0 because each individual claim now has 50% probability of being absorbed by the deductible. This is exactly how ASTAM tests coverage modifications inside an aggregate frame.

Trap 1: Using the variance formula with Var(X) instead of E[X^2] for compound Poisson. Var(S) = λ E[X^2], not λ Var(X). The mistake adds back a missing λ · E[X]^2 = (mean of S) · E[X], which can be a large miss when severity has a sizable mean.

Trap 2: Running Panjer with f(0) > 0 but using the simplified recursion (denominator 1) intended for f(0) = 0. The full recursion has denominator (1 - a f(0)); for compound Poisson a = 0 so this simplifies, but for Binomial and Negative Binomial it does not.

Trap 3: Discretizing a heavy-tailed severity (Pareto, lognormal) onto a span h that is too coarse. The aggregate distribution’s right tail then collapses onto too few mass points and stop-loss premiums are biased downward. Klugman recommends starting with h equal to a small fraction of E[X] and halving it until the answer stabilizes.

Panjer is the right tool when (i) the frequency is (a, b, 0) or (a, b, 1) and (ii) the support of S can be truncated at a manageable upper bound. For a portfolio with E[N] in the tens and severity that can be discretized on a span of a few hundred lattice points, Panjer is faster and more accurate than Monte Carlo.

Monte Carlo wins when the frequency leaves the (a, b, 0)/(a, b, 1) class (e.g., mixture frequencies, frequency with random parameters), when severity has multi-dimensional structure (deductibles, limits, coinsurance interacting with policy-level features), or when the lattice would have to be enormous. Most production aggregate-loss code combines the two: Panjer for closed-form components and simulation for the rest.

The aggregate loss S is the input to ruin theory (Cramer-Lundberg surplus model), to risk-measure-based capital (VaR, TVaR), and to stop-loss reinsurance pricing (E[(S - d)_+]). Each of these downstream uses needs the full distribution of S, not just its mean and variance, which is why the Panjer recursion is the standard ASTAM and CFE 101 computational tool.

Klugman, Loss Models, 5th ed., Ch. 9 develops compound moments, the (a, b, 0) class in Ch. 6, and the recursion proper in Ch. 9; Hardy, Quantitative Enterprise Risk Management, 2nd ed., Ch. 4-5 connects aggregate distributions to risk measures and ruin.