Why Aggregate Recursion Outputs Need Independent Checks
A recursion can generate many aggregate probabilities quickly, but a long table is not evidence that the calculation is correct. A shifted severity index, a missing term, or a cutoff that is too low can produce plausible-looking values while distorting the distribution. A reliable check uses information that is independent of the individual recursion steps: total probability, model-implied moments, and direct calculations of selected probabilities.
This tutorial focuses on checking outputs from a discrete compound model. The same diagnostic principles apply when the frequency distribution is another member of the Panjer class, provided its corresponding recursion and boundary conditions are used. The checks do not replace the recursion; they test whether its results agree with the model that was specified.
Recursion and Its Indexing
For a compound Poisson model, assume \(N\sim\operatorname{Poisson}(\lambda)\), and assume independent, identically distributed, nonnegative integer severities with probabilities \(f_j\). The recursion starts at \(g_0=e^{-\lambda(1-f_0)}\) when zero severity is allowed. If severities are strictly positive, this reduces to \(g_0=e^{-\lambda}\). For \(k\geq 1\), the recursion is
Each term uses severity \(j\) and the already-computed aggregate probability at \(k-j\). Thus the severity index cannot exceed \(k\); severities larger than \(k\) cannot contribute to an aggregate of \(k\). This elementary index check catches many errors before any numerical comparison is needed.
The recursion follows from the probability generating function. If \(F(z)=\sum_{j\geq 0}f_jz^j\) is the severity generating function, then \(G(z)=\sum_{k\geq 0}g_kz^k=\exp(\lambda(F(z)-1))\). Differentiating gives \(G'(z)=\lambda F'(z)G(z)\). The coefficient of \(z^{k-1}\) on the left is \(kg_k\); on the right it is \(\lambda\sum_{j=1}^{k}j f_jg_{k-j}\). Equating coefficients yields the stated recursion. This derivation also explains both the factor \(j\) and the divisor \(k\); neither may be dropped.
Worked Checks: Direct Probabilities and Recursion
Worked Example: Checking a Low Aggregate Probability
A hypothetical portfolio has Poisson claim count with \(\lambda=0.8\). Each claim has severity 1 with probability 0.25 and severity 3 with probability 0.75. Use the recursion to find \(g_0,g_1,g_2\), then check \(g_2\) directly.
Method. Since severities are positive, \(g_0=e^{-0.8}\). For aggregate loss 1, only severity 1 can contribute:
For aggregate loss 2, the only possible severity index is \(j=1\), because \(f_2=0\). Therefore,
Direct check. Aggregate loss 2 requires exactly two claims, both of severity 1. Its probability is
The direct calculation agrees with the recursion. The common mistake avoided here is letting severity 3 contribute to aggregate loss 2, or including a nonexistent \(f_2\) term as if it were positive. The result is checked both structurally and numerically.
A direct calculation is especially useful at small aggregate values because there are usually only a few ways to form the total. At larger values, enumerate only when the number of combinations remains manageable. The purpose is not to duplicate the whole recursion by hand, but to independently test selected entries and their indices.
Worked Example: A Direct Check at a Larger Aggregate Value
Suppose the claim count is Poisson with \(\lambda=1\), and severity is 1 or 2, each with probability 0.5. Check the recursion output for aggregate loss 4 by enumerating possible claim counts and severity combinations.
With severity at least 1, aggregate loss 4 can arise from two claims both of size 2, three claims with two 1s and one 2, or four claims all of size 1. No other claim count can produce 4. Conditional on each count, the required severity pattern has probability \(0.5^n\) times its number of arrangements. Thus
The three bracketed terms are \(0.125\), \(0.0625\), and \(0.00260417\), respectively. Their sum is \(0.19010417\), so
For a recursion check, the needed earlier values are \(g_0=e^{-1}\), \(g_1=0.5e^{-1}\), \(g_2=0.625e^{-1}\), and \(g_3=(13/48)e^{-1}\). Substitution gives
The direct enumeration and recursion agree. A common error is to count the three-claim pattern as having one arrangement rather than \(\binom{3}{1}=3\); that would omit two possible placements of the severity-2 claim.
Normalization and Moment Checks
For a complete aggregate distribution, \(\sum_{k\geq 0}g_k=1\). When only values through \(K\) have been calculated, the partial sum must be at most 1, and the residual \(1-\sum_{k=0}^{K}g_k\) is the probability of aggregate loss above \(K\). A partial sum that materially exceeds 1 signals an error such as duplicated mass, a faulty recursion term, or inconsistent inputs. A sum below 1 is not by itself evidence of an error; judge it against the cutoff and the expected tail.
For a compound Poisson model with finite severity moments, the aggregate mean and variance are
The variance uses the severity second raw moment, not just the severity variance. These model-implied quantities provide an independent check on a sufficiently complete recursive distribution. If the table is truncated, its partial probability-weighted moments will usually be below the full moments; compare them only after accounting for the omitted tail.
Worked Example: Interpreting a Truncated Table
Use the same model as the preceding example: \(N\sim\operatorname{Poisson}(1)\), with severities 1 and 2 each having probability 0.5. Calculate the probability mass and first two raw aggregate moments through \(K=4\), and compare them with the full model moments.
The recursion gives \(g_k=e^{-1}c_k\), where \(c_0=1\), \(c_1=0.5\), \(c_2=0.625\), \(c_3=13/48\), and \(c_4=73/384\). The sum of these coefficients is
Since \(e^{-1}\approx 0.367879\), the computed probability through 4 is \(2.5859375e^{-1}\approx 0.9513\). The remaining mass is approximately \(1-0.9513=0.0487\), which represents \(P(S>4)\). It should not be redistributed among the five displayed probabilities.
The partial first raw moment is
For the complete model, \(E[Y]=(1+2)/2=1.5\), so \(E[S]=1(1.5)=1.5\). The partial value 1.2224 is smaller because the omitted outcomes all exceed 4 and contribute positive loss. Similarly, \(E[Y^2]=(1+4)/2=2.5\), so \(\operatorname{Var}(S)=2.5\), and \(E[S^2]=2.5+1.5^2=4.75\). The partial second raw moment through 4 is
Both partial moments are below their full-model values, as expected. The check is informative because a table that claimed to include all mass through 4, or whose partial moment exceeded the model moment, would require investigation. The common mistake avoided is treating a truncated sum as a complete distribution and forcing it to total 1 by renormalizing.
Discretization Is Part of the Model
A recursion operates on the severity probabilities supplied to it. If an originally continuous severity is rounded to a grid, the resulting recursion describes that discretized model, not the original continuous model. The rounding convention can affect the mean, variance, and individual aggregate probabilities. A diagnostic comparison should therefore state the grid width and whether values are rounded down, up, or by another specified rule.
Worked Example: Sensitivity to the Severity Grid
A hypothetical portfolio has Poisson claim count with \(\lambda=2\). Before discretization, each claim severity is uniform on \((0,2)\), so its mean is 1 and its second raw moment is \(4/3\). Compare a grid width of 1 using lower-endpoint discretization with upper-endpoint discretization.
Under lower-endpoint discretization, a severity in \((0,1)\) is represented by 0 and one in \((1,2)\) by 1. Each value has probability 0.5. Hence \(E[Y]=0.5\), \(E[Y^2]=0.5\), and the aggregate mean and variance are \(2(0.5)=1\) and \(2(0.5)=1\). Zero-valued severities do not change aggregate loss, so the positive-loss claims have Poisson mean \(2(0.5)=1\). Therefore \(g_0=e^{-1}\approx0.3679\), and \(g_1=1e^{-1}\approx0.3679\).
Under upper-endpoint discretization, severities in \((0,1)\) are represented by 1 and those in \((1,2)\) by 2, again each with probability 0.5. Now \(E[Y]=1.5\) and \(E[Y^2]=2.5\), giving aggregate mean \(2(1.5)=3\) and variance \(2(2.5)=5\). With strictly positive severity, \(g_0=e^{-2}\approx0.1353\) and \(g_1=2(0.5)e^{-2}=e^{-2}\approx0.1353\).
For comparison, the undiscretized compound model has aggregate mean \(2(1)=2\) and variance \(2(4/3)=8/3\). The lower and upper grid rules place the aggregate means at 1 and 3, respectively, on opposite sides of the continuous-model mean. Their probabilities at zero also differ substantially. The check is to compare the recursion output with the discretized inputs actually used, and then assess the grid sensitivity; it is not valid to expect either discretized table to reproduce the continuous model exactly.
Common Errors and a Reliable Audit Sequence
When a computed output fails a check, locate the error systematically rather than adjusting values until the table looks plausible. The following sequence separates implementation problems from expected truncation or discretization effects.
Record the frequency distribution and parameter, the severity support and probabilities, and the aggregate-loss grid. Check that the frequency parameter refers to the same period and scale as the severity amounts.
Verify \(g_0\), then confirm that each term for \(g_k\) uses \(f_j\) and \(g_{k-j}\) only for feasible indices. If severity is positive, \(g_0=e^{-\lambda}\); if zero severity is possible, use \(e^{-\lambda(1-f_0)}\).
Compute the partial sum and compare it with 1, allowing for loss above the cutoff. Compare probability-weighted moments with model-implied moments only when the omitted tail is negligible or otherwise accounted for.
For a small aggregate value or a value with few possible claim combinations, calculate its probability directly. Agreement tests the recursion's indices, coefficients, and severity inputs.
Extend the maximum aggregate value to test for omitted mass. If severity was discretized, compare plausible grid widths or rounding rules and report their effect rather than concealing it by renormalization.
Typical warning signs include a partial probability sum greater than 1, a nonzero \(g_k\) below the minimum possible aggregate amount, a direct calculation that disagrees with the recursion, or moments that conflict with the specified model after the tail has been captured. By contrast, a sum below 1 at a modest cutoff and moments below the full-model values can be entirely consistent with omitted high losses.
Check Your Understanding
For each problem, identify what is being checked and retain any exponential factor in exact answers where convenient.
- A compound Poisson model has \(\lambda=0.6\), with severity 1 or 2 each having probability 0.5. Find \(g_0\), \(g_1\), and \(g_2\) using the recursion. (Answer: \(g_0=e^{-0.6}\), \(g_1=0.3e^{-0.6}\), \(g_2=0.345e^{-0.6}\approx0.1893\).)
- A compound Poisson model has \(\lambda=1.2\), with severity 1 of probability 0.25 and severity 3 of probability 0.75. Find \(g_2\) and verify which claim pattern can produce it. (Answer: Two claims both of severity 1; \(g_2=0.045e^{-1.2}\approx0.0136\).)
- Claims have Poisson frequency with \(\lambda=2\) and fixed severity 2. Give the aggregate mean, variance, and probability of zero aggregate loss. (Answer: Mean \(4\), variance \(8\), and \(P(S=0)=e^{-2}\approx0.1353\).)
- For \(\lambda=1\) and severities 1 and 2 each with probability 0.5, the recursion is calculated only through \(k=2\). Find the partial probability sum and the omitted mass above 2. (Answer: \((1+0.5+0.625)e^{-1}=2.125e^{-1}\approx0.7817\); omitted mass is approximately \(0.2183\).)
- A Poisson model has \(\lambda=2\) and original severity uniform on \((0,2)\). Compare aggregate means under grid-width-1 lower-endpoint and upper-endpoint discretization. (Answer: Lower-endpoint mean \(1\); upper-endpoint mean \(3\); undiscretized mean \(2\).)