Why Check a Recursive Aggregate Distribution?
A Panjer recursion can produce a long table of aggregate probabilities, but a plausible-looking table is not necessarily correct. A missed term, an incorrect initial probability, or a loss-grid unit error can affect many later entries. Independent checks provide a way to detect such problems without repeating the recursion by the same method.
Let \(N\) be the number of claims, let \(X_1,X_2,\ldots\) be independent claim amounts with the same distribution as the severity \(X\), and define the aggregate loss by \(S=\sum_{r=1}^{N}X_r\), with \(S=0\) when \(N=0\). Write \(g_k=P(S=kh)\) for the recursive aggregate probabilities on a grid with spacing \(h\). The checks in this tutorial compare the probability table with total probability and with moments calculated independently from the frequency and severity distributions.
The aggregate moment checks require the frequency and severity moments. Let \(E[N]\) and \(\operatorname{Var}(N)\) be the frequency mean and variance, and let \(\mu_X=E[X]\) and \(\sigma_X^2=\operatorname{Var}(X)\) be the severity mean and variance. When claim amounts are independent of the count and are independent and identically distributed conditional on \(N\), the benchmark identities are:
A short justification follows from conditioning on \(N\). Given \(N=n\), the sum has mean \(n\mu_X\) and variance \(n\sigma_X^2\). Thus the law of total expectation gives \(E[S]=E[N\mu_X]\), and the law of total variance gives \(\operatorname{Var}(S)=E[N\sigma_X^2]+\operatorname{Var}(N\mu_X)\), which yields the identities above. These benchmarks are separate from the recursion: they use the moments of the input frequency and severity distributions.
Three Independent Checks
Use the checks in sequence. First establish whether the table is complete. Then check the sum of its probabilities. Finally, compare its moments with the frequency-severity benchmarks, taking care to use the same loss units throughout.
For a complete table, calculate \(\sum_k g_k\); it should equal 1. For a truncated table, this sum is the probability covered by the calculation, not a value that must equal 1. The omitted mass is \(1-\sum_k g_k\), if the displayed probabilities are otherwise correct.
For a complete table on a grid with spacing \(h\), calculate \(E[S]=\sum_k kh\,g_k\) and \(E[S^2]=\sum_k (kh)^2g_k\). Then calculate \(\operatorname{Var}(S)=E[S^2]-E[S]^2\).
Calculate the frequency and severity moments from their stated distributions, use the benchmark identities, and compare the results with the table. A discrepancy on a truncated grid may be expected; a discrepancy on a complete grid is a reason to investigate.
If the recursion uses integer indices for losses measured in units of \(h\), do not mistake the index \(k\) for a dollar amount unless \(h=1\) dollar. The probability-weighted mean in grid units is \(\sum k g_k\); in loss units it is \(h\sum k g_k\). The second moment must be scaled by \(h^2\).
Worked Checks
Worked Example: Auditing a Complete Compound Binomial Table
A hypothetical portfolio has \(N\sim\operatorname{Binomial}(2,0.5)\). Each claim independently has severity \(X=1\) or \(X=2\), each with probability \(0.5\). A recursion on a unit grid reports \(g_0=0.25\), \(g_1=0.25\), \(g_2=0.3125\), \(g_3=0.125\), and \(g_4=0.0625\). Check total probability, mean, and variance. The maximum possible aggregate loss is 4, so this table is complete.
Method and total probability. Sum the five entries:
The table passes the total-mass check.
Moment checks from the table. The loss values are 0, 1, 2, 3, and 4. Therefore,
Next,
Thus \(\operatorname{Var}(S)=3.625-(1.5)^2=3.625-2.25=1.375\).
Independent benchmarks. For the severity, \(\mu_X=(1+2)/2=1.5\), and \(E[X^2]=(1+4)/2=2.5\), so \(\sigma_X^2=2.5-(1.5)^2=0.25\). For \(N\sim\operatorname{Binomial}(2,0.5)\), \(E[N]=2(0.5)=1\) and \(\operatorname{Var}(N)=2(0.5)(0.5)=0.5\). Therefore,
The recursive table agrees with both benchmarks and sums to 1. The check also catches a common error: using \(E[X^2]\) in place of \(\operatorname{Var}(X)\) in the general variance identity would give \(1(2.5)+0.5(2.25)=3.625\), which is the aggregate second moment here, not its variance.
Worked Example: Recognizing a Truncated Poisson Recursion
A hypothetical aggregate model has \(N\sim\operatorname{Poisson}(1)\). Each claim is independently 1 or 2 loss units, with probability \(0.5\) for each amount. A recursion reports probabilities only for aggregate losses \(k=0,1,2,3,4\). Use the Poisson recursion to check the reported values, and explain why their sum and moments need not match the full-distribution benchmarks.
Method and recursive values. The initial probability is \(g_0=P(N=0)=e^{-1}=0.3678794\). With severity probabilities \(f_1=f_2=0.5\) and no zero-severity claims, the compound Poisson recursion is \(g_k=(1/k)\sum_{j=1}^k j f_jg_{k-j}\). Applying it successively gives:
The displayed mass is \(0.3678794+0.1839397+0.2299247+0.0996340+0.0699354=0.9513132\). This is not a failed total-probability check: the Poisson count is unbounded, and aggregate losses above 4 are possible. The omitted mass is approximately \(1-0.9513132=0.0486868\).
The partial first moment is \(1(0.1839397)+2(0.2299247)+3(0.0996340)+4(0.0699354)=1.2224\), to four decimal places. The partial second moment is \(1^2(0.1839397)+2^2(0.2299247)+3^2(0.0996340)+4^2(0.0699354)=3.1193\), to four decimal places. These are truncated sums, not the full aggregate moments.
For the complete distribution, \(\mu_X=1.5\), \(E[X^2]=2.5\), and \(\sigma_X^2=2.5-1.5^2=0.25\). A Poisson random variable with parameter 1 has mean and variance both equal to 1. The benchmarks are therefore \(E[S]=1(1.5)=1.5\) and \(\operatorname{Var}(S)=1(0.25)+1(1.5)^2=2.5\). The full second moment is \(2.5+1.5^2=4.75\). The truncated sums fall below these values because the omitted outcomes have positive losses. Renormalizing the five displayed probabilities to sum to 1 would not restore the omitted outcomes or their contribution to the moments.
The check is to distinguish truncation from a recursion error: the probabilities shown are consistent with the recurrence, while the missing mass is explained by the cutoff. A common mistake is to insist that every finite prefix of an unbounded aggregate distribution must sum to 1.
Worked Example: Checking a Different Loss-Grid Unit
A hypothetical model has \(N\sim\operatorname{Binomial}(2,0.25)\), and each claim has severity 2 or 4 dollars with equal probability. The complete recursion is indexed in units of 2 dollars and gives \(g_0=0.5625\), \(g_1=0.1875\), \(g_2=0.203125\), \(g_3=0.03125\), and \(g_4=0.015625\). Check the distribution and its moments in dollars.
Method and total probability. The grid index \(k\) represents a dollar loss of \(2k\), not \(k\) dollars. The probability total is \(0.5625+0.1875+0.203125+0.03125+0.015625=1\).
Moment checks from the table. In dollars, the possible aggregate losses are 0, 2, 4, 6, and 8. Hence,
The dollar second moment is \(4(0.1875)+16(0.203125)+36(0.03125)+64(0.015625)=0.75+3.25+1.125+1=6.125\). The table variance is \(6.125-1.5^2=3.875\).
Independent benchmarks. The severity mean is \(\mu_X=(2+4)/2=3\) dollars, its second moment is \(E[X^2]=(4+16)/2=10\) dollars squared, and its variance is \(10-3^2=1\) dollar squared. The frequency mean is \(2(0.25)=0.5\), and its variance is \(2(0.25)(0.75)=0.375\). The benchmarks give \(E[S]=0.5(3)=1.5\) dollars and \(\operatorname{Var}(S)=0.5(1)+0.375(3^2)=0.5+3.375=3.875\) dollars squared. They match the table.
The common grid error is to calculate the mean as \(\sum k g_k=0.75\) and report it as dollars. That sum is a mean in grid units; multiplying by the spacing 2 gives the correct mean of 1.5 dollars. For the second moment, the required scale factor is \(2^2=4\).
Interpreting Discrepancies
A check is useful only when its assumptions match the calculation. Before treating a mismatch as an error, establish whether the probability table is complete, whether the severity and aggregate loss use the same grid, and whether the frequency and severity inputs are the ones actually used in the recursion. For a complete distribution with finite moments, all three checks should agree, subject only to any stated rounding.
| Observed result | Likely explanation to investigate |
|---|---|
| Probability sum below 1 | The grid may be truncated, or probability mass may have been omitted or miscalculated. |
| Probability sum above 1 | Look for duplicated entries, an incorrect initial probability, or an arithmetic error. |
| Mean differs by a constant factor | Check whether grid indices were converted to dollars using the grid spacing. |
| Mean matches but variance does not | Check the second moment, the subtraction of the squared mean, and whether the variance identity was applied correctly. |
| Small differences in rounded output | Compare using unrounded recursion values before deciding that the table fails. |
These checks are diagnostic rather than a proof that every individual recursion entry is correct. For example, two errors can partly offset in the total or in a moment. A strong audit checks total mass and both moments independently, inspects the initial condition and selected recursion steps, and confirms the units and support. This is particularly important for truncated calculations: their displayed probabilities can be internally correct even when their partial moments are not equal to full-distribution benchmarks.
Check Your Understanding
For each problem, decide whether the stated table is complete before interpreting its checks.
- A hypothetical model has \(N\sim\operatorname{Binomial}(3,0.5)\), with a fixed claim amount of 2 dollars. Find the aggregate mean and variance, and state whether the support is bounded. (Answer: mean 3 dollars; variance 3 dollars squared; maximum aggregate loss 6 dollars.)
- A compound Poisson model has frequency parameter 1 and a fixed severity of 1 loss unit. A recursion is shown only for aggregate losses 0, 1, and 2. Find the displayed probability total and the omitted probability. (Answer: displayed total \(2.5e^{-1}\approx0.9197\); omitted probability \(1-2.5e^{-1}\approx0.0803\).)
- A hypothetical model has \(N\sim\operatorname{Binomial}(1,0.4)\) and severity 1 or 3 dollars with equal probabilities. Find the aggregate mean, second moment, and variance. (Answer: mean 0.8 dollars; second moment 2 dollars squared; variance 1.36 dollars squared.)
- A complete recursion uses a grid spacing of 5 dollars and has \(\sum k g_k=0.6\) and \(\sum k^2g_k=1.2\). Find the aggregate mean and variance in dollars. (Answer: mean 3 dollars; second moment 30 dollars squared; variance 21 dollars squared.)
- A frequency distribution has mean 2 and variance 2. A claim severity has mean 4 dollars and variance 9 dollars squared. Find the aggregate mean and variance, assuming independent claim amounts and frequency. (Answer: mean 8 dollars; variance \(2(9)+2(4^2)=50\) dollars squared.)