Benchmarking Moment Approximations Against Exact Recursion

Why Benchmark an Approximation?

An aggregate-loss approximation is useful only insofar as it estimates the probability relevant to the decision. A method that is close near the mean may be less accurate in a tail, and two methods that match the same aggregate moments can still give noticeably different probabilities. A direct comparison with an exact recursion provides a benchmark: it shows the size and direction of the approximation error for a specified event.

This tutorial compares a normal approximation and a moment-matched gamma approximation with exact compound Poisson probabilities. The examples use hypothetical portfolios with discrete losses. Their aggregate losses lie on a lattice, so the normal approximation uses a continuity correction. The gamma approximation is continuous and is evaluated at the stated monetary threshold without a continuity correction. The comparisons are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Definition: For an event $A$, the signed probability error of an approximation is $P_{\text{approx}}(A)-P_{\text{exact}}(A)$. Its absolute probability error is $\left|P_{\text{approx}}(A)-P_{\text{exact}}(A)\right|$. A negative signed error means the approximation understates the event probability; a positive signed error means it overstates it.

The Exact Benchmark and the Approximation Rules

Let $N\sim\operatorname{Poisson}(\lambda)$ be the claim count, and let $X_1,X_2,\ldots$ be independent, identically distributed claim amounts, independent of $N$. The aggregate loss is $S=\sum_{j=1}^{N}X_j$, with $S=0$ when $N=0$. For a discrete severity distribution on positive integer loss units, write $f_j=P(X_1=j)$ and $p_k=P(S=k)$. The compound Poisson Panjer recursion starts with $p_0=e^{-\lambda}$ and gives $p_k$ for $k\geq1$.

p_k=\frac{\lambda}{k}\sum_{j=1}^{k}j f_jp_{k-j}.

The recursion provides the exact benchmark for the discretized model: sum its probabilities over the outcomes that satisfy the event. In the examples below, each severity is a fixed positive number of loss units, so the recursion simplifies to the Poisson probabilities for the claim count.

For a compound Poisson model, the aggregate mean and variance are $\mu_S=\lambda E[X_1]$ and $\sigma_S^2=\lambda E[X_1^2]$, respectively. The normal approximation uses $S\approx N(\mu_S,\sigma_S^2)$. If the aggregate lies on a lattice with spacing $h$, the continuity-corrected approximation to $P(S\leq c)$, where $c$ is a lattice point, evaluates the normal CDF at $c+h/2$. For $P(S\geq c)$, it evaluates the upper tail at $c-h/2$.

The moment-matched gamma approximation is a gamma distribution with the same mean and variance as $S$. For shape parameter $\alpha$ and scale parameter $\theta$, its mean is $\alpha\theta$, and its variance is $\alpha\theta^2$. Thus the matched parameters are $\alpha=\mu_S^2/\sigma_S^2$ and $\theta=\sigma_S^2/\mu_S$, when the mean and variance are positive. This approximation uses only those two moments; it does not preserve the exact lattice or any point mass, such as the mass at zero.

Why the Recursion and the Moment Matches Work

For a compound Poisson sum, the probability generating function of $S$ is $\exp\{\lambda(G_X(z)-1)\}$, where $G_X(z)=\sum_{j\geq1}f_jz^j$ is the severity probability generating function. Differentiating gives $G_S'(z)=\lambda G_X'(z)G_S(z)$. The coefficient of $z^{k-1}$ on the left is $kp_k$; on the right it is $\lambda\sum_{j=1}^{k}j f_jp_{k-j}$. Equating coefficients gives the stated recursion. The Poisson starting probability follows because $S=0$ exactly when there are no claims, given positive claim amounts.

The compound Poisson moment formulas follow from conditioning on the claim count. Given $N=n$, the sum has mean $nE[X_1]$ and variance $n\operatorname{Var}(X_1)$. Applying the laws of total expectation and total variance, with $E[N]=\operatorname{Var}(N)=\lambda$, yields

E[S]=\lambda E[X_1],\qquad \operatorname{Var}(S) =\lambda\operatorname{Var}(X_1)+\lambda(E[X_1])^2 =\lambda E[X_1^2].

Matching these two quantities to the gamma mean and variance equations gives $\alpha\theta=\mu_S$ and $\alpha\theta^2=\sigma_S^2$; solving gives the parameters above. Matching moments is a construction, not a guarantee that every threshold probability will be close.

Worked Example: A Central Probability for a One-Unit Claim Model

A hypothetical portfolio has $N\sim\operatorname{Poisson}(4)$. Each claim costs exactly $1,000, independently of the count. Estimate the probability that aggregate claims are at most $4,000 using exact recursion, the continuity-corrected normal approximation, and the moment-matched gamma approximation. Report probabilities to four decimals and compare absolute errors.

Method and exact benchmark. Let one loss unit be $1,000. Then $S/1000=N$, so the event is $N\leq4$. In the Panjer recursion, $f_1=1$ and all other $f_j=0$; hence $p_k=4p_{k-1}/k$, starting from $p_0=e^{-4}$. The successive probabilities are $p_1=4e^{-4}$, $p_2=8e^{-4}$, $p_3=(32/3)e^{-4}$, and $p_4=(32/3)e^{-4}$. Therefore

P(S\leq4000)=e^{-4}\left(1+4+8+\frac{32}{3}+\frac{32}{3}\right) =0.6288.

Normal approximation. Here $E[X_1]=1000$ and $E[X_1^2]=1{,}000{,}000$. Thus $\mu_S=4{,}000$ and $\sigma_S^2=4{,}000{,}000$, so $\sigma_S=2{,}000$. The lattice spacing is $1,000. For the event $S\leq4{,}000$, apply the continuity correction at $4,500:

P(S\leq4000)\approx \Phi\left(\frac{4500-4000}{2000}\right) =\Phi(0.25)=0.5987.

Moment-matched gamma approximation. In thousands of dollars, the aggregate mean is $4$ and variance is $4$. Matching the gamma mean and variance gives $\alpha=4^2/4=4$ and $\theta=4/4=1$. A gamma variable with integer shape 4 and scale 1 has CDF $1-e^{-x}(1+x+x^2/2+x^3/6)$. Therefore

P(S\leq4000)\approx 1-e^{-4}\left(1+4+8+\frac{64}{6}\right)=0.5665.

The normal absolute error is $|0.5987-0.6288|=0.0301$; the gamma absolute error is $|0.5665-0.6288|=0.0623$. For this central event, the continuity-corrected normal approximation is closer. This comparison is specific to this model and event; it does not establish a universal ranking.

Worked Example: Comparing Upper-Tail Probabilities

A hypothetical account again has a Poisson claim count with mean 4 and claims of exactly $1,000. Compare the exact probability of aggregate claims of at least $8,000 with the normal and moment-matched gamma estimates. Report probabilities and absolute errors to four decimals.

Exact recursion. The event is $N\geq8$, so sum the exact recursion through 7 and subtract from 1. Using $p_k=4p_{k-1}/k$,

P(N\geq8)=1-e^{-4}\left(1+4+8+\frac{32}{3} +\frac{32}{3}+\frac{128}{15}+\frac{256}{45}+\frac{1024}{315}\right) =0.0511.

Normal approximation. The mean and standard deviation remain $4,000 and $2,000. The event $S\geq8{,}000$ starts at a lattice point, so the continuity-corrected upper-tail boundary is $7,500:

P(S\geq8000)\approx 1-\Phi\left(\frac{7500-4000}{2000}\right) =1-\Phi(1.75)=0.0401.

Gamma approximation. The matched gamma parameters remain shape 4 and scale 1 in thousands of dollars. Its survival probability at 8 is

P(S\geq8000)\approx e^{-8}\left(1+8+\frac{8^2}{2}+\frac{8^3}{6}\right) =0.0424.

The normal absolute error is $|0.0401-0.0511|=0.0110$ using the rounded probabilities; using unrounded values it is approximately $0.0111$. The gamma absolute error is approximately $0.0088$. Here the gamma estimate is closer, although it still understates the exact tail. Reported errors should ideally be calculated from unrounded probabilities and rounded only at the end.

Worked Example: A Different Loss Unit and a Central Threshold

A hypothetical portfolio has $N\sim\operatorname{Poisson}(3)$, and every claim costs exactly $2,000. Estimate $P(S\leq$6,000)$ by each method and compare absolute errors. Give probabilities to four decimals.

Exact recursion. In units of $2,000, $S/2000=N$; the event is $N\leq3$. The Poisson recursion gives $p_k=3p_{k-1}/k$ and $p_0=e^{-3}$, so

P(S\leq6000)=P(N\leq3) =e^{-3}\left(1+3+\frac{9}{2}+\frac{9}{2}\right) =0.6472.

Normal approximation. The claim mean is $2,000 and its second moment is $4{,}000{,}000$ dollars squared. Thus $\mu_S=$6,000$, $\sigma_S^2=12{,}000{,}000$ dollars squared, and \(\sigma_S=\sqrt{12{,}000{,}000}\approx$3,464.1. With lattice spacing $2,000, the corrected boundary for \(S\leq$6,000 is $7,000:

P(S\leq6000)\approx \Phi\left(\frac{7000-6000}{\sqrt{12{,}000{,}000}}\right) =\Phi(0.2887)\approx0.6136.

Moment-matched gamma approximation. In thousands of dollars the mean is 6 and variance is 12. Therefore $\alpha=6^2/12=3$ and $\theta=12/6=2$. At $6,000, the gamma argument divided by the scale is $6/2=3$. The shape-3 CDF is $1-e^{-y}(1+y+y^2/2)$, giving

P(S\leq6000)\approx1-e^{-3}\left(1+3+\frac{9}{2}\right)=0.5768.

The normal absolute error is approximately $|0.6136-0.6472|=0.0336$; the gamma absolute error is approximately $|0.5768-0.6472|=0.0704$. Both approximations understate this exact central probability, and the normal estimate is closer for this example.

Reading the Comparison Correctly

The three examples show why a benchmark should identify the event as carefully as it identifies the distribution. In the first model, the gamma approximation has a smaller error for the upper tail, but a larger error for the central cumulative probability. The third example changes the claim amount and frequency while keeping a deterministic severity, and the relative performance changes again. An approximation is not simply “accurate” or “inaccurate” without specifying a threshold and an error measure.

For any comparison, use the same model parameters and the same event in all methods. Calculate the exact recursion probability first, then evaluate each approximation at the correct threshold. Report signed error if the direction of bias matters, or absolute error if comparing closeness. For a very small exact probability, relative error can also be informative, but it can become large or unstable when the benchmark probability is near zero.

Specify the event.
Translate the wording into an exact inequality, such as $S\leq c$ or $S\geq c$, and retain the strict or inclusive boundary as stated.

Establish the benchmark.
Use exact recursion probabilities and sum the appropriate lattice outcomes. Check that the event includes the intended boundary mass.

Apply each approximation consistently.
Use aggregate mean and variance for the normal model, apply continuity correction for a lattice event, and evaluate the matched gamma CDF or survival function at the original threshold.

Measure and interpret error.
Subtract the exact probability from the approximation to find signed error; take its absolute value for closeness. Do not infer general performance from one threshold.

A frequent exam trap is omitting the continuity correction or applying it in the wrong direction. For example, approximating $P(S\leq$4,000)$ in the first example at $4,000 gives $\Phi(0)=0.5000$, not the corrected estimate $0.5987$. The event contains the entire lattice mass at $4,000, so the normal boundary belongs halfway between $4,000 and the next possible value, at $4,500. For an upper-tail event beginning at $8,000, the corresponding boundary is halfway below it, at $7,500. The gamma approximation is continuous, so the same half-unit adjustment is not applied to it.

Key takeaway: Exact recursion is the reference probability for the specified discrete model. Compare normal and moment-matched estimates against that same event, use the lattice continuity correction for the normal method, and evaluate error from unrounded probabilities.

Check Your Understanding

Use exact recursion as the benchmark where requested. Round probabilities to four decimals.

A hypothetical portfolio has $N\sim\operatorname{Poisson}(2)$, with every claim costing $500. Find the exact probability that aggregate claims are at most $1,000. (Answer: $5e^{-2}=0.6767$)
A hypothetical portfolio has $N\sim\operatorname{Poisson}(4)$, with each claim costing $1,000. Find the exact probability of at least six claims. (Answer: $1-e^{-4}(1+4+8+32/3+32/3+128/15)=0.2149$)
A hypothetical portfolio has $N\sim\operatorname{Poisson}(3)$, with each claim costing $2,000. Find the exact probability aggregate claims are at most $4,000. (Answer: $e^{-3}(1+3+9/2)=0.4232$)
For a Poisson(4) count with $1,000 fixed claim amounts, approximate the probability of at most three claims by a continuity-corrected normal distribution. (Answer: $\Phi(-0.25)=0.4013$)
For a Poisson(3) count with $2,000 fixed claim amounts, match a gamma distribution to aggregate mean and variance, then approximate the probability aggregate claims exceed $10,000. (Answer: shape $3$, scale $2$ thousand dollars; $e^{-5}(1+5+25/2)=0.1247$)

What You'll Learn