Common Errors in Aggregate Recursion and Approximation

Why Aggregate Calculations Need Checks

Aggregate-loss calculations can look plausible while containing a serious error. A severity probability may be assigned to the wrong grid point; a recursion may use an initialization valid for a different frequency class; or a finite table may be treated as though it contained all probability. Normal approximations have their own traps: the variance may be parameterized incorrectly, or the continuity correction may move the cutoff in the wrong direction.

This tutorial focuses on diagnosing those errors. Let $N$ be the number of claims, let $X_1,X_2,\ldots$ be independent, identically distributed claim severities independent of $N$, and define aggregate loss by $S=\sum_{r=1}^{N}X_r$, with $S=0$ when $N=0$. For a lattice calculation, take a grid width $h>0$ and write $f_j=P(X=jh)$ and $g_k=P(S=kh)$. The probabilities $f_j$ and $g_k$ are indexed by grid units, not by dollar amounts.

Definition: A severity distribution is represented on a lattice of width $h$ when its possible values are $0,h,2h,\ldots$. Its probability at $jh$ is $f_j$. The compound aggregate probability at $kh$ is $g_k$. If severity is strictly positive, $f_0=0$, and $g_0=P(N=0)$.

The lattice may be an exact representation of the model, or it may result from discretizing a severity distribution. In either case, the choice of $h$ must be consistent: a loss of $2h$ contributes to severity index 2, not index $2h$. In particular, if $h$ is measured in thousands of dollars, the indices remain whole-number grid positions.

Panjer Recursion: Check the Frequency Class

For a frequency distribution in the Panjer $(a,b,0)$ class, the probabilities $p_n=P(N=n)$ satisfy $p_n=(a+b/n)p_{n-1}$ for $n\geq1$. Assuming the claim severities are independent and identically distributed on the stated lattice and independent of $N$, the compound probabilities satisfy the following recursion. The displayed form allows a zero-severity probability $f_0$; the denominator must be nonzero.

Theorem: For an $(a,b,0)$ frequency distribution and lattice severity probabilities $f_j$, let $g_0=P(N=0)$ when severities are strictly positive. For $k\geq1$, $$ g_k=\frac{1}{1-af_0}\sum_{j=1}^{k}\left(a+\frac{bj}{k}\right)f_jg_{k-j}. $$ When $f_0=0$, the denominator is 1. The formula is not, by itself, the general recursion for every frequency in the $(a,b,1)$ class.

A short derivation explains both the index $j/k$ and the denominator. Let $F(z)=\sum_{j\geq0}f_jz^j$ and $G(z)=\sum_{k\geq0}g_kz^k$ be the severity and aggregate probability generating functions. For an $(a,b,0)$ frequency, its probability generating function $P_N(z)$ satisfies $(1-az)P_N'(z)=(a+b)P_N(z)$, as follows by multiplying $p_n=(a+b/n)p_{n-1}$ by $n z^{n-1}$ and summing. Since $G(z)=P_N(F(z))$, differentiation gives $(1-aF(z))G'(z)=(a+b)G(z)F'(z)$. Equating coefficients of $z^{k-1}$, separating the $f_0g_k$ term on the left, and dividing by $k$ yields the stated recursion. Thus $f_j$ pairs with $g_{k-j}$: a claim at grid index $j$ leaves aggregate index $k-j$.

A frequency distribution in the $(a,b,1)$ class instead satisfies the Panjer relation only for $n\geq2$; $p_0$ and $p_1$ need not obey the relation that would extend the $(a,b,0)$ formula down to $n=1$. The appropriate recursion includes a correction for that first probability.

Theorem: For an $(a,b,1)$ frequency distribution, with $p_0=P(N=0)$, $p_1=P(N=1)$, and lattice severity probabilities $f_j$, the modified recursion is $$ g_k=\frac{1}{1-af_0}\left[\bigl(p_1-(a+b)p_0\bigr)f_k+\sum_{j=1}^{k}\left(a+\frac{bj}{k}\right)f_jg_{k-j}\right],\qquad k\geq1. $$ When $p_1=(a+b)p_0$, the correction term vanishes. Do not use the $(a,b,0)$ form for an $(a,b,1)$ frequency unless this condition holds.

Worked Examples: Indexing, Initialization, and Probability Mass

Worked Example: A Truncated Poisson Recursion

A hypothetical portfolio has $N\sim\operatorname{Poisson}(0.8)$ claims. Each claim is exactly $1000$ dollars. Calculate the aggregate probabilities at $0$, $1000$, and $2000$ dollars, and determine how much probability is omitted if the table stops at $2000$.

Method and assumptions. Use a grid width $h=\$1000$, so the severity is at index 1: $f_1=1$, with $f_0=0$ and all other $f_j=0$. A Poisson frequency has $a=0$, $b=\lambda=0.8$, and $p_0=e^{-0.8}$. The $(a,b,0)$ recursion therefore gives $g_k=(0.8/k)g_{k-1}$.

Initialization. Since a positive claim cannot sum to zero, $g_0=P(N=0)=e^{-0.8}=0.449328964\ldots$.

At index 1. $g_1=(0.8/1)g_0=0.8(0.449328964\ldots)=0.359463171\ldots$.

At index 2. $g_2=(0.8/2)g_1=0.4(0.359463171\ldots)=0.143785268\ldots$.

The displayed masses sum to $0.449328964\ldots+0.359463171\ldots+0.143785268\ldots=0.952577404\ldots$. Thus the omitted probability above $2000$ dollars is $1-0.952577404\ldots=0.047422596\ldots$. To six decimal places, the sum is $0.952577$ and the omitted probability is $0.047423$. As a check, these are the Poisson probabilities for $N=0,1,2$ and $N\geq3$, respectively. The common mistake is to renormalize the three displayed values to sum to 1; they describe an incomplete table, not a conditional distribution.

Worked Example: Severity Values Are Not Grid Indices

A hypothetical claim count is Poisson with mean 1. Each claim is either $1000$ dollars or $2000$ dollars, each with probability $0.5$. Use a $1000$-dollar grid to calculate $g_1$ and $g_2$.

Method and assumptions. Here $h=\$1000$, so $1000=h$ has index 1 and $2000=2h$ has index 2. Thus $f_1=f_2=0.5$, $f_0=0$, $a=0$, $b=1$, and $g_0=e^{-1}=0.367879441\ldots$.

For $k=1$, only the $j=1$ term contributes: $$ g_1=\left(0+\frac{1}{1}\right)f_1g_0 =0.5(0.367879441\ldots) =0.183939721\ldots. $$

For $k=2$, both severity indices contribute: $$ g_2=\left(0+\frac{1}{2}\right)f_1g_1+ \left(0+\frac{2}{2}\right)f_2g_0 =0.5(0.5)(0.183939721\ldots)+1(0.5)(0.367879441\ldots) =0.229924651\ldots. $$

The direct check at $2000$ dollars is one claim of $2000$, with probability $e^{-1}(0.5)$, or two claims both of $1000$, with probability $(e^{-1}/2)(0.5^2)$. Their sum is $0.5e^{-1}+0.125e^{-1}=0.625e^{-1}=0.229924651\ldots$, agreeing with the recursion. Treating the $2000$-dollar loss as index 2000 would put it on the wrong lattice; the index is the loss divided by the grid width.

Worked Example: The First-Claim Correction in the (a,b,1) Class

A hypothetical portfolio has exactly one claim: $P(N=1)=1$ and $P(N=0)=0$. Its severity is $1000$ dollars with probability $0.4$ and $2000$ dollars with probability $0.6$. Calculate the aggregate distribution on a $1000$-dollar grid.

Method and assumptions. This frequency has $a=b=0$ in the relation for $n\geq2$, so it is an $(a,b,1)$ example. Set $p_0=0$, $p_1=1$, $f_0=0$, $f_1=0.4$, and $f_2=0.6$. Then $g_0=p_0=0$, and the modified recursion has correction $p_1-(a+b)p_0=1$.

At $k=1$, the summation terms are zero because $a=b=0$, so $g_1=[1f_1]/1=0.4$. At $k=2$, $g_2=[1f_2]/1=0.6$. All other $g_k$ are zero, and the total probability is $0.4+0.6=1$, as required. Applying the $(a,b,0)$ recursion would instead give $g_1=g_2=0$, because it would omit the first-claim correction. That result fails the total-probability check and contradicts $S=X_1$ when $N=1$.

Aggregate Moments: Use the Second Moment of Severity

A normal approximation requires the mean and variance of the aggregate, not merely the mean and variance of one claim. For independent, identically distributed severities independent of the frequency, conditioning on $N$ gives $$ E[S]=E[N]E[X],\qquad \operatorname{Var}(S)=E[N]\operatorname{Var}(X)+\operatorname{Var}(N)(E[X])^2. $$ For a Poisson frequency with mean $\lambda$, $\operatorname{Var}(N)=E[N]=\lambda$, so $$ E[S]=\lambda E[X],\qquad \operatorname{Var}(S)=\lambda\left(\operatorname{Var}(X)+(E[X])^2\right)=\lambda E[X^2]. $$

The final equality follows from $E[X^2]=\operatorname{Var}(X)+(E[X])^2$. The term $\lambda E[X^2]$ matters: using only $\lambda\operatorname{Var}(X)$ drops the variation caused by the random number of claims. These formulas assume a Poisson frequency and independent, identically distributed severities independent of $N$; they should not be transferred unchanged to an arbitrary frequency model.

Worked Example: Moment Parameterization and Continuity Correction

A hypothetical aggregate has $N\sim\operatorname{Poisson}(2)$, with each claim equal to 1 or 3 lattice units, each with probability $0.5$. Approximate $P(S\geq7)$ by a normal distribution with a continuity correction.

Method and assumptions. The claim mean is $E[X]=(1)(0.5)+(3)(0.5)=2$. Its second moment is $E[X^2]=(1^2)(0.5)+(3^2)(0.5)=5$, so $\operatorname{Var}(X)=5-2^2=1$. Since the count is Poisson with $\lambda=2$, the aggregate mean is $2(2)=4$ and the aggregate variance is $2(5)=10$. Approximate $S$ by $Y\sim N(4,10)$, where 10 is the variance and $\sqrt{10}$ is the standard deviation.

The aggregate lies on the integer lattice. The event $S\geq7$ includes the mass at 7, so its normal boundary is halfway between 6 and 7, namely 6.5. Therefore $$ P(S\geq7)\approx P(Y\geq6.5) =1-\Phi\left(\frac{6.5-4}{\sqrt{10}}\right) =1-\Phi(0.7906) \approx 0.2146. $$ The probability is reported to four decimal places. As a parameter check, $\operatorname{Var}(S)=\lambda E[X^2]=10$, not $\lambda\operatorname{Var}(X)=2$. As a boundary check, using 7 instead of 6.5 would give $1-\Phi(0.9487)\approx0.1714$, an inappropriate cutoff for this inclusive lattice event. The normal approximation is an approximation, not an exact aggregate probability.

Practical Checks Before Accepting a Result

A calculation is easier to audit when its conventions are stated before the arithmetic. Record the grid width and translate every possible severity into an integer index. State the frequency class and its parameters, then verify the initial value $g_0$. For positive severities, $g_0=P(N=0)$; when zero severities are possible, the compound probability at zero can also include claims whose severities are all zero.

When a recursion is evaluated only through index $K$, the sum $\sum_{k=0}^{K}g_k$ is generally less than 1. The difference $1-\sum_{k=0}^{K}g_k$ is the omitted upper-tail probability if the table includes all indices from 0 through $K$. It is not a numerical defect simply because the sum is below 1. Renormalizing the entries changes their interpretation to probabilities conditional on $S\leq Kh$.

For a normal approximation, write the distribution as $N(\mu,\sigma^2)$, identify $\sigma=\sqrt{\sigma^2}$, and check that the units of the cutoff match those of the aggregate. For a lattice event $S\geq mh$, where $m$ is an integer, the usual continuity-corrected boundary is $(m-\tfrac12)h$. For $S\leq mh$, it is $(m+\tfrac12)h$. If the threshold is not itself a lattice point, first identify the first lattice value included in the event. Continuity correction addresses the discrete-to-continuous boundary; it does not repair an incorrect mean, variance, or severity grid.

Key takeaway: Before trusting an aggregate probability, verify the severity indices, frequency-class formula, initial mass, and probability-table range. Before trusting a normal tail, verify the aggregate moments and move the cutoff by half a lattice step in the direction implied by the event.

Check Your Understanding

For each problem, state the calculation or check that controls the answer.

A Poisson frequency has mean $0.5$, and each claim equals one grid unit. Find $g_0$, $g_1$, $g_2$, and the probability omitted when the table stops at index 2. (Answer: $g_0=0.606531$, $g_1=0.303265$, $g_2=0.075816$, omitted probability $=0.014388$, to six decimals.)
A severity can be $3000$ or $6000$ dollars, and the loss grid width is $3000$ dollars. What indices should be used for these severities? (Answer: indices 1 and 2, respectively.)
A lattice aggregate has approximate normal mean $4h$ and standard deviation $2h$. Use a continuity correction to approximate $P(S\geq5h)$. (Answer: use boundary $4.5h$, giving $1-\Phi(0.25)\approx0.4013$.)
A Poisson count has mean 3, and claim severity has mean 2 and variance 4. Find the aggregate mean and variance. (Answer: mean $6$, variance $24$, since $E[X^2]=4+2^2=8$.)
A recursion has produced probabilities only for indices 0 through $K$, and their sum is $0.93$. What is the omitted probability if these are all consecutive lower indices, and should the entries be renormalized for an unconditional aggregate distribution? (Answer: omitted probability $0.07$; do not renormalize for the unconditional distribution.)

What You'll Learn