Deriving the Panjer Aggregate Recursion

Why the Aggregate Recursion Works

A compound aggregate model combines a random number of claims with a random payment on each claim. Direct convolution can calculate the probability of each aggregate amount, but the calculation grows as the target amount increases. The Panjer recursion replaces repeated convolution with a sequential calculation: once the probability at zero is known, each new aggregate probability is obtained from earlier ones and the severity probabilities.

The recursion is useful only when its assumptions match the model. In particular, the claim count must be independent of the claim severities, the severities must be identically distributed on a common integer lattice, and the frequency probabilities must satisfy the required recursion. It does not directly calculate probabilities for an unmodified continuous severity distribution.

Definition: Let $N$ be a nonnegative integer-valued claim count with probabilities $p_n=P(N=n)$. Let $X_1,X_2,\ldots$ be independent and identically distributed nonnegative integer-valued claim payments, independent of $N$, with $f_j=P(X_i=j)$. The aggregate payment is $S=\sum_{i=1}^{N}X_i$, where the empty sum is zero, and its probabilities are $g_k=P(S=k)$.

For the standard $(a,b,0)$ frequency class, assume that the frequency probabilities satisfy $n p_n=(a n+b)p_{n-1}$ for every integer $n\geq 1$. The parameters $a$ and $b$ describe the frequency recursion; they are not severity parameters. Examples include Poisson and binomial frequencies. The recursion derived below also allows $f_0>0$, meaning an individual claim can have a zero payment.

Deriving the Compound Recursion

The derivation uses probability generating functions (pgfs). For a nonnegative integer-valued random variable $Z$, its pgf is $E[z^Z]$, represented by the power series whose coefficient of $z^k$ is $P(Z=k)$. Write $P_N(z)=\sum_{n\geq0}p_nz^n$ for the frequency pgf and $F(z)=\sum_{j\geq0}f_jz^j$ for the severity pgf. Independence gives the aggregate pgf $G(z)=P_N(F(z))=\sum_{k\geq0}g_kz^k$.

First apply the frequency recursion to the derivative of $P_N$. The coefficient of $z^{n-1}$ in $P_N'(z)$ is $n p_n$. Therefore,

P_N'(z)=\sum_{n\geq1}(an+b)p_{n-1}z^{n-1} =a\bigl(zP_N'(z)+P_N(z)\bigr)+bP_N(z) =azP_N'(z)+(a+b)P_N(z).

Next differentiate $G(z)=P_N(F(z))$. The chain rule gives $G'(z)=P_N'(F(z))F'(z)$. Substituting the frequency identity with $F(z)$ in place of $z$, and using $P_N'(F(z))F'(z)=G'(z)$, gives

$$ (1-aF(z))G'(z)=(a+b)G(z)F'(z). $$

To extract the probability at aggregate amount $k\geq1$, compare coefficients of $z^{k-1}$. On the left, the coefficient from $G'$ is $kg_k$, while that from $F G'$ is $\sum_{j=0}^{k-1}f_j(k-j)g_{k-j}$. On the right, the coefficient from $G F'$ is $\sum_{j=1}^{k}j f_jg_{k-j}$. Separating the $j=0$ term on the left and collecting terms yields the recursion.

Theorem: Under the independence, identical-severity, nonnegative integer-lattice, and $(a,b,0)$ frequency assumptions above, let $g_0=P_N(f_0)=\sum_{n\geq0}p_n f_0^n$. If $1-af_0\ne0$, then for each $k\geq1$, $$ g_k=\frac{1}{1-af_0}\sum_{j=1}^{k}\left(a+\frac{bj}{k}\right)f_jg_{k-j}. $$

In the coefficient calculation, the terms involving $f_jg_{k-j}$ combine as $a(k-j)+(a+b)j=ak+bj$; division by $k$ produces $a+bj/k$. This explains the weight in the formula. The factor $1-af_0$ comes from the $j=0$ term, so it must not be dropped when severity has positive probability at zero and $a\ne0$.

The starting value matters as much as the recurrence. If $f_0=0$, then $g_0=p_0$: the aggregate is zero only when there are no claims. If $f_0>0$, the aggregate can also be zero when claims occur but all have zero payment. In that case, $g_0$ is the frequency pgf evaluated at $f_0$, not simply $p_0$.

Worked Applications

Worked Example: Poisson Frequency and Two Positive Payment Sizes

A hypothetical portfolio has $N\sim\operatorname{Poisson}(1)$ claims. Conditional on a claim, its payment is 1 with probability $2/3$ and 2 with probability $1/3$. Claims are independent of the count and of one another. Calculate $g_0,g_1,g_2,$ and $g_3$.

Method and assumptions. Here $a=0$, $b=1$, $f_1=2/3$, $f_2=1/3$, and all other $f_j$ are zero. Since severity is strictly positive, $g_0=p_0=e^{-1}$. The recursion becomes $g_k=\sum_{j=1}^k(j/k)f_jg_{k-j}$.

For $k=1$, only $j=1$ contributes: $g_1=(1)(2/3)e^{-1}=(2/3)e^{-1}$. For $k=2$, the $j=1$ term is $(1/2)(2/3)(2/3)e^{-1}=(2/9)e^{-1}$, and the $j=2$ term is $(2/2)(1/3)e^{-1}=(1/3)e^{-1}$. Thus $g_2=(5/9)e^{-1}$.

For $k=3$, the $j=1$ term is $(1/3)(2/3)(5/9)e^{-1}=(10/81)e^{-1}$, and the $j=2$ term is $(2/3)(1/3)(2/3)e^{-1}=(12/81)e^{-1}$. There is no $j=3$ term because $f_3=0$. Therefore $g_3=(22/81)e^{-1}$.

Using $e^{-1}=0.367879\ldots$, the probabilities, rounded only at the final step to four decimal places, are $g_0=0.3679$, $g_1=0.2453$, $g_2=0.2044$, and $g_3=0.0999$. As a check, each calculation uses only earlier aggregate probabilities and the stated severity masses. A common mistake this avoids is using the claim count probability $p_k$ as the aggregate probability $g_k$; the two distributions are different because each claim can pay 1 or 2.

Worked Example: Binomial Frequency With a Zero-Payment Atom

A hypothetical contract has $N\sim\operatorname{Binomial}(2,0.5)$ reported claims. Each claim produces payment 0 with probability 0.2 and payment 1 with probability 0.8, independently of all other claims. Find the aggregate probabilities $g_0,g_1,g_2$.

Method and assumptions. For a binomial frequency with $m=2$ and claim probability $q=0.5$, the frequency recursion has $a=-q/(1-q)=-1$ and $b=(m+1)q/(1-q)=3$. The severity masses are $f_0=0.2$, $f_1=0.8$, and $f_2=0$. First calculate $g_0=P_N(0.2)=0.25+0.5(0.2)+0.25(0.2)^2=0.25+0.10+0.01=0.36$.

The recursion denominator is $1-af_0=1-(-1)(0.2)=1.2$. For $k=1$, the sole term has weight $a+b=2$, so $g_1=(2)(0.8)(0.36)/1.2=0.576/1.2=0.48$. For $k=2$, the $j=1$ weight is $-1+3/2=0.5$; the $j=2$ term is zero because $f_2=0$. Hence $g_2=(0.5)(0.8)(0.48)/1.2=0.192/1.2=0.16$.

The results sum to $0.36+0.48+0.16=1$, as required. They also check directly: aggregate payment 1 arises from one claim paying 1 and the other paying 0, or from a single reported claim paying 1, giving $0.5(0.8)+0.25(2)(0.8)(0.2)=0.48$. A common mistake this avoids is omitting the denominator: doing so would give $g_1=0.576$, inconsistent with the model.

Worked Example: Negative Binomial Frequency

Suppose a hypothetical portfolio has a negative binomial claim count with parameters $r=2$ and $q=0.5$, using $P(N=n)=\binom{n+r-1}{n}(1-q)^r q^n$ for $n=0,1,\ldots$. Claim payments are 1 or 2, each with probability $0.5$, independently of the count and one another. Calculate $g_0,g_1,g_2,$ and $g_3$.

Method and assumptions. The frequency probability ratio is $p_n/p_{n-1}=q(n+r-1)/n=0.5+0.5/n$, so $a=0.5$ and $b=0.5$. Because $f_0=0$, $g_0=p_0=(1-q)^r=0.25$, and the denominator in the recursion is 1.

For $k=1$, the weight is $a+b=1$, giving $g_1=(1)(0.5)(0.25)=0.125$. For $k=2$, the $j=1$ weight is $0.5+0.5/2=0.75$, and the $j=2$ weight is 1. Thus $g_2=(0.75)(0.5)(0.125)+(1)(0.5)(0.25)=0.046875+0.125=0.171875$.

For $k=3$, the $j=1$ weight is $0.5+0.5/3=2/3$, and the $j=2$ weight is $0.5+1/3=5/6$. The $j=3$ term is zero. Therefore $g_3=(2/3)(0.5)(0.171875)+(5/6)(0.5)(0.125)=0.0572917+0.0520833=0.109375$.

To four decimal places, $g_0=0.2500$, $g_1=0.1250$, $g_2=0.1719$, and $g_3=0.1094$. As a check, conditioning on $N$ gives $g_2=0.125+0.1875(0.25)=0.171875$, from one payment of 2 or two payments of 1. This method avoids treating the negative binomial frequency as Poisson: its $a$ term is nonzero and changes the weights at every aggregate amount.

Boundary Conditions and Practical Checks

The theorem uses an $(a,b,0)$ frequency recursion, which holds for every $n\geq1$. Some frequency models are instead specified by the $(a,b,1)$ condition: $n p_n=(an+b)p_{n-1}$ for $n\geq2$, with $p_0$ and $p_1$ given separately. In that case, the same aggregate formula needs a boundary correction. Define $C=p_1-(a+b)p_0$. The derivative identity for the frequency pgf includes this extra constant, and coefficient extraction adds $C f_k$ to the numerator:

g_k=\frac{1}{1-af_0}\left[ \sum_{j=1}^{k}\left(a+\frac{bj}{k}\right)f_jg_{k-j}+C f_k \right],\qquad k\geq1.

Thus one must identify whether the frequency recursion starts at $n=1$ or $n=2$ before applying a compound recursion. Using the $(a,b,0)$ formula unchanged for an $(a,b,1)$ model can give incorrect aggregate probabilities even when the recurrence appears similar.

A practical calculation should establish the lattice unit, specify $f_0$, calculate $g_0$ from the frequency pgf, and then proceed in increasing order of $k$. The support must be expressed in the chosen unit: if payments are multiples of 100 dollars, for example, the recursion index counts units of 100 dollars, not individual dollars. Finally, probabilities should be nonnegative and the calculated probabilities should approach a total of 1 as the aggregate range is extended. A partial sum below 1 is not itself an error, since larger aggregate amounts may remain.

Key takeaway: The aggregate recursion follows from composing the frequency and severity pgfs, then equating coefficients. Its correct use depends on independence, a shared nonnegative integer lattice, the frequency boundary condition, and the appropriate starting value $g_0=P_N(f_0)$.

Check Your Understanding

For each problem, identify the frequency recursion and severity support before applying the aggregate recursion.

A hypothetical portfolio has $N\sim\operatorname{Poisson}(2)$; a payment is 1 with probability 0.75 and 2 with probability 0.25. Find $g_0,g_1,$ and $g_2$. (Answer: $g_0=0.1353,\ g_1=0.2030,\ g_2=0.2199$, to four decimal places.)
Let $N$ have a geometric distribution on $0,1,\ldots$, with $P(N=n)=0.6(0.4)^n$. Each claim pays exactly 1. Find $P(S=3)$. (Answer: $0.0384$.)
Let $N\sim\operatorname{Binomial}(3,0.25)$, and let an independent claim payment be 0 with probability 0.1, 1 with probability 0.6, and 2 with probability 0.3. Find $g_0$ and $g_1$. (Answer: $g_0=0.4655,\ g_1=0.2703$, to four decimal places.)
Let $N$ have a negative binomial distribution with $r=2,q=0.5$, using the parameterization in the worked example. If each claim pays 1 or 2 with equal probability, find $g_3$. (Answer: $0.1094$, to four decimal places.)
A frequency model satisfies the $(a,b,1)$ recursion and has $p_0=0.3$, $p_1=0.25$. Severity has $f_0=0.2$, $f_1=0.5$, $f_2=0.3$, and $a=0.4,b=0.2$. What boundary-correction constant $C$ must be used? (Answer: $C=p_1-(a+b)p_0=0.25-(0.6)(0.3)=0.07$.)

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