Why Compare Recursion With Approximations?
An aggregate-loss approximation is most useful when its accuracy is understood in context. A normal approximation can be quick to calculate, while a translated gamma approximation can reflect skewness through a third-moment match. Panjer recursion, when its assumptions apply and the grid is manageable, gives an exact probability calculation for the specified discrete model. Comparing all three for the same portfolio shows both the value and the limitations of each method.
The calculations below use one hypothetical portfolio throughout. Each claim has a fixed severity of $1,000, and the number of claims is Poisson with mean 1. Aggregate losses are measured in thousands of dollars, so the possible totals are the nonnegative integers. This simple model makes it possible to check every probability while illustrating how errors vary by threshold. The comparison method is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.
Exact Probabilities by Panjer Recursion
For a compound Poisson model with strictly positive integer severities, the aggregate probability at zero is \(p_0=e^{-\lambda}\). For integers \(k\geq1\), Panjer recursion gives
The restriction to strictly positive severities matters here: it gives \(p_0=e^{-\lambda}\). A model that puts probability on zero severity requires the corresponding zero-severity mass to be included in the aggregate probability generating function; the formula for \(p_0\) then changes.
A short derivation explains the recursion. The probability generating function of the aggregate is \(P(z)=\exp\{\lambda(F(z)-1)\}\), where \(F(z)=\sum_{j\geq1}f_jz^j\) is the severity probability generating function. Differentiating gives \(P'(z)=\lambda F'(z)P(z)\). Equating coefficients of \(z^{k-1}\) on both sides gives \(kp_k=\lambda\sum_{j=1}^k j f_jp_{k-j}\), which is the stated recursion. The derivation relies on the compound Poisson frequency model and the integer severity grid.
Worked Example: Build the Exact Tail Probabilities
For the hypothetical portfolio, calculate \(P(S\geq k)\) for \(k=1,2,3,4,5\) using Panjer recursion. Each claim is exactly $1,000 and \(N\sim\operatorname{Poisson}(1)\). Give probabilities to four decimal places.
Method. Since \(f_1=1\) and \(f_j=0\) for \(j\neq1\), the recursion reduces to \(p_k=p_{k-1}/k\). Start with \(p_0=e^{-1}=0.3678794\). Then
For each integer threshold, sum the probabilities below it and subtract from 1. For example, \(P(S\geq3)=1-(p_0+p_1+p_2)=1-(0.3678794+0.3678794+0.1839397)=0.0803015\), which is \(0.0803\) to four decimals. The resulting tails are \(0.6321,0.2642,0.0803,0.0190,0.0037\) for \(k=1,2,3,4,5\), respectively.
Check. These values decrease as the threshold rises, as upper-tail probabilities must. The recursion also gives \(p_0+p_1+p_2+p_3+p_4+p_5\approx0.9994\), leaving a small positive probability above 5; the probabilities have not been mistakenly normalized after truncation. A common mistake is to report \(p_k\) as \(P(S\geq k)\). The tail includes every point probability from \(k\) upward, not only the mass at \(k\).
Moment Approximations for the Same Model
For a compound Poisson aggregate, the first three cumulants are \(\kappa_r=\lambda E[X_1^r]\) for \(r=1,2,3\). Here \(\lambda=1\) and \(X_1=1\), so \(\kappa_1=\kappa_2=\kappa_3=1\). Thus \(E[S]=1\), \(\operatorname{Var}(S)=1\), and the third cumulant is 1. The normal approximation uses \(Y_N\sim N(1,1)\), where the second parameter is the variance.
The normal distribution is continuous, while \(S\) takes integer values. For the event \(S\geq k\), a unit-grid continuity correction places the continuous cutoff halfway between \(k-1\) and \(k\): use \(P(Y_N\geq k-0.5)\). This is an approximation, not an identity.
For the translated gamma approximation, write \(Y_G=c+G\), where \(G\) has a gamma distribution with shape \(\alpha\) and scale \(\theta\). Its mean, variance, and third cumulant are \(c+\alpha\theta\), \(\alpha\theta^2\), and \(2\alpha\theta^3\). Matching the three aggregate cumulants gives
The translation is part of the fit: using an unshifted gamma with shape 1 and scale 1 would match the mean and variance here, but not the third cumulant. For a gamma distribution with integer shape 4, the survival probability at \(y\geq0\) is \(e^{-y}(1+y+y^2/2+y^3/6)\). Applying the same unit-grid continuity correction to \(S\geq k\), the gamma argument is \(((k-0.5)-c)/\theta=2k+1\). Although this fitted translated gamma can assign probability to negative values, it remains a moment approximation, not a model of the aggregate’s exact support.
Worked Example: Compare All Three Methods Across Thresholds
For the same hypothetical portfolio, compare the exact tail probabilities with the continuity-corrected normal and translated gamma estimates at thresholds \(k=1,2,3,4,5\). Show the absolute error of each approximation, using the exact tail as the benchmark.
Method. Use the recursion results above for the exact column. For the normal estimate, calculate \(1-\Phi(k-1.5)\), since the mean and standard deviation are both 1 and the corrected cutoff is \(k-0.5\). For the translated gamma estimate, substitute \(y=2k+1\) into the shape-4 survival formula.
At \(k=2\), the corrected normal cutoff is \(1.5\), so its standardized value is \((1.5-1)/1=0.5\), and the estimate is \(P(Z\geq0.5)=0.3085\). The translated gamma argument is \(y=5\), giving \(e^{-5}(1+5+25/2+125/6)=0.2650\). The exact tail is \(0.2642\), so the absolute errors are \(0.0443\) and \(0.0008\), respectively.
At \(k=4\), the corrected normal cutoff is \(3.5\), with standardized value \(2.5\). Its upper tail is \(0.0062\). The translated gamma argument is \(y=9\), giving \(e^{-9}(1+9+81/2+729/6)=e^{-9}(172)=0.0212\). Compared with the exact tail \(0.0190\), the absolute errors are \(0.0128\) for normal and \(0.0022\) for translated gamma.
The full comparison, rounded to four decimal places, is:
| Threshold \(k\) | Exact recursion | Normal estimate | Normal absolute error | Translated gamma estimate | Gamma absolute error |
|---|---|---|---|---|---|
| 1 | 0.6321 | 0.6915 | 0.0593 | 0.6472 | 0.0151 |
| 2 | 0.2642 | 0.3085 | 0.0443 | 0.2650 | 0.0008 |
| 3 | 0.0803 | 0.0668 | 0.0135 | 0.0818 | 0.0015 |
| 4 | 0.0190 | 0.0062 | 0.0128 | 0.0212 | 0.0022 |
| 5 | 0.0037 | 0.0002 | 0.0034 | 0.0049 | 0.0013 |
For example, the exact tail at \(k=5\), before rounding, is approximately \(0.0036598\), and the normal tail is approximately \(0.0002326\); their difference rounds to \(0.0034\). Small differences between displayed rounded values and displayed errors can occur because errors are calculated before rounding. A common mistake is to standardize the uncorrected threshold \(k\), or to use \(k+0.5\) for an upper tail. For an integer event \(S\geq k\) on a unit grid, this comparison uses \(k-0.5\).
Reading the Comparison Carefully
The table shows why a single statement such as “the gamma approximation is better” needs a specified threshold. For these five thresholds, the translated gamma estimate has the smaller absolute error at each one. Its advantage is especially clear at \(k=2\), where it is close to the recursively calculated tail. That observation is specific to this model and these thresholds; it is not a general guarantee about translated gamma approximations.
The normal estimate is too high at \(k=1\) and \(k=2\), then too low from \(k=3\) onward in this comparison. The translated gamma estimates are slightly high at each displayed threshold. These directional patterns do not establish a universal bias. The key exam skill is to calculate a benchmark under the stated model, apply each approximation with the correct parameters and cutoff, and compare the resulting errors rather than assuming one method must dominate.
Exact recursion is also conditional on having a suitable discrete severity model. The recursion above uses integer-valued severities and a compound Poisson frequency. A continuous severity cannot be inserted directly into this discrete recurrence: it must first be represented on a specified grid. That discretization would introduce an additional approximation, separate from the normal or translated gamma approximation compared here.
State the frequency distribution, severity probabilities, and loss unit before calculating aggregate probabilities.
Use the model’s probability recursion to obtain exact point probabilities and sum them to the requested tail.
Match the required moments, then use a continuity correction consistent with the aggregate loss grid.
Calculate absolute errors from unrounded probabilities and avoid extending the result beyond the thresholds examined.
Check Your Understanding
Use the hypothetical portfolio defined above unless a question states otherwise. Round probability answers to four decimal places.
- Use the exact recursion benchmark to find \(P(S\geq2)\). (Answer: 0.2642)
- For \(P(S\geq3)\), what is the continuity-corrected normal cutoff, its standardized value, and the resulting upper-tail estimate? (Answer: cutoff \(2.5\), standardized value \(1.5\), estimate \(0.0668\))
- For \(P(S\geq1)\), evaluate the translated gamma estimate using its fitted parameters and the continuity correction. (Answer: \(e^{-3}(1+3+3^2/2+3^3/6)\approx0.6472\))
- Use the recursion to find the point probability \(P(S=2)\), and distinguish it from the tail \(P(S\geq2)\). (Answer: \(p_2=e^{-1}/2\approx0.1839\))
- At \(k=4\), which approximation has the smaller absolute error against recursion, and what are the two errors? (Answer: translated gamma; gamma error \(0.0022\), normal error \(0.0128\))