Compound Binomial Aggregate Probabilities by Recursion

Why a Fixed Number of Claim Opportunities Matters

A compound aggregate model combines a random number of claim payments with a severity distribution. In some settings, the number of claims is generated by a fixed set of possible claim opportunities: each of $m$ exposures either produces one claim or produces none. The resulting claim count cannot exceed $m$. That upper bound distinguishes the binomial model from a count model that allows arbitrarily many claims.

When claim amounts are integer-valued, the Panjer recursion calculates aggregate probabilities sequentially. For a binomial count, its parameters include a negative value of $a$. The recursion remains valid, but that sign and the finite claim-count limit are important checks on the calculation.

Definition: Let $N\sim\operatorname{Bin}(m,\pi)$, where $m$ is a fixed positive integer and $0<\pi<1$. Thus, $N$ counts successes in $m$ independent Bernoulli trials, each with success probability $\pi$. Let $X_1,X_2,\ldots$ be independent and identically distributed nonnegative integer-valued claim severities, independent of $N$. The aggregate loss is $S=\sum_{r=1}^{N}X_r$, with $S=0$ when $N=0$. Write $f_j=P(X_r=j)$ and $g_k=P(S=k)$.

For the basic recursion, claim severities are represented on an integer-valued monetary scale, such as dollars or thousands of dollars. If a severity model originally has amounts between integer values, a discretization rule must first assign probabilities to integer amounts. The recursion then calculates probabilities for that discrete model; it does not remove any approximation introduced by discretization.

Binomial Parameters for Panjer Recursion

A count distribution is in the $(a,b,0)$ class when its probabilities $p_n=P(N=n)$ satisfy $p_n=(a+b/n)p_{n-1}$ for positive $n$ in the support. For a binomial count, the probability ratio is

\frac{p_n}{p_{n-1}} =\frac{\binom{m}{n}\pi^n(1-\pi)^{m-n}} {\binom{m}{n-1}\pi^{n-1}(1-\pi)^{m-n+1}} =\frac{m-n+1}{n}\frac{\pi}{1-\pi}.

Rearranging the last expression gives $-\pi/(1-\pi)+(m+1)\pi/((1-\pi)n)$. Therefore, the Panjer parameters are

a=-\frac{\pi}{1-\pi}, \qquad b=\frac{(m+1)\pi}{1-\pi}.

These parameters apply when $0<\pi<1$. If $\pi=0$, no trial produces a claim and $S=0$ with probability one. If $\pi=1$, there are exactly $m$ claims; the displayed parameters are undefined, so calculate the aggregate distribution directly as the distribution of a sum of $m$ severities.

Theorem: For $N\sim\operatorname{Bin}(m,\pi)$ with $0<\pi<1$, and independent, identically distributed nonnegative integer severities with probabilities $f_j$, the aggregate probabilities satisfy $$ g_0=(1-\pi+\pi f_0)^m, \qquad g_k=\frac{1}{1-a f_0}\sum_{j=1}^{k}\left(a+\frac{bj}{k}\right)f_jg_{k-j},\quad k\geq 1, $$ where $a=-\pi/(1-\pi)$ and $b=(m+1)\pi/(1-\pi)$. If severities are strictly positive, $f_0=0$, so $g_0=(1-\pi)^m$ and the denominator in the recursion is 1.

Why the Aggregate Recursion Holds

A short generating-function argument explains the recursion. Let $G_N(z)=E[z^N]$ be the probability generating function of the claim count, and let $F_X(z)=E[z^X]$ be that of a severity. Independence gives the aggregate generating function $H(z)=G_N(F_X(z))$. The count probability ratio implies

$$ (1-az)G_N'(z)=(a+b)G_N(z). $$

Substitute $F_X(z)$ for $z$ and use the chain rule on $H(z)=G_N(F_X(z))$. This gives

$$ (1-aF_X(z))H'(z)=(a+b)F_X'(z)H(z). $$

The coefficient of $z^{k-1}$ in $H'(z)$ is $kg_k$. In the product $F_X(z)H'(z)$, the term involving $f_0$ is $f_0kg_k$; the other terms at that power are $\sum_{j=1}^{k} f_j(k-j)g_{k-j}$. The coefficient of $z^{k-1}$ in $F_X'(z)H(z)$ is $\sum_{j=1}^{k}j f_jg_{k-j}$. Equating coefficients and collecting the $g_k$ terms yields

k(1-af_0)g_k =\sum_{j=1}^{k}\bigl(ak+bj\bigr)f_jg_{k-j}.

Dividing by $k(1-af_0)$ gives the stated recursion. Also, $g_0=G_N(f_0)$: conditional on $N=n$, all $n$ severities must equal zero for the aggregate to be zero, which has probability $f_0^n$. For a binomial count, $G_N(z)=(1-\pi+\pi z)^m$, giving the displayed initial probability.

Applying the Recursion

Start with $g_0$, then calculate $g_1,g_2,\ldots$ in order. At step $k$, only severity probabilities $f_1,\ldots,f_k$ and previously calculated aggregate probabilities are needed. For strictly positive severities, $f_0=0$, which simplifies the calculation. When the severity has a finite maximum $c$, no aggregate loss can exceed $mc$, because there are at most $m$ claims.

Worked Example: A Two-Trial Model With Two Possible Claim Sizes

A hypothetical portfolio has two independent claim opportunities. Each produces one claim with probability $\pi=0.5$. A claim is 1 unit with probability $f_1=0.5$, or 2 units with probability $f_2=0.5$. Calculate the full aggregate distribution using Panjer recursion, and verify that its probabilities sum to 1.

Method and parameters. Here $m=2$, $a=-0.5/0.5=-1$, and $b=3(0.5)/0.5=3$. The severities are strictly positive, so $g_0=(1-\pi)^m=0.5^2=0.25$.

For $k=1$, only $j=1$ contributes: $g_1=(a+b)f_1g_0=2(0.5)(0.25)=0.25$.

For $k=2$, the $j=1$ coefficient is $a+b/2=-1+1.5=0.5$, and the $j=2$ coefficient is $a+b=2$. Thus, $g_2=(0.5)(0.5)(0.25)+2(0.5)(0.25)=0.0625+0.25=0.3125$.

For $k=3$, the coefficient for $j=1$ is $-1+3/3=0$, and for $j=2$ it is $-1+6/3=1$. Since $f_3=0$, $g_3=0(0.5)g_2+1(0.5)g_1=0.125$.

For $k=4$, the coefficient for $j=1$ is $-1+3/4=-0.25$, and for $j=2$ it is $-1+6/4=0.5$. Therefore, $g_4=(-0.25)(0.5)(0.125)+(0.5)(0.5)(0.3125)=-0.015625+0.078125=0.0625$.

The loss cannot exceed $2\times2=4$ units. The check is $g_0+g_1+g_2+g_3+g_4=0.25+0.25+0.3125+0.125+0.0625=1$. The results also agree with conditioning on whether there are zero, one, or two claims. Common mistake avoided: the factor $a+bj/k$ already includes the division by $k$; dividing the resulting sum by $k$ again would be incorrect.

Worked Example: Probability of a Small Aggregate Loss

A hypothetical group has three independent claim opportunities, each with claim probability $\pi=0.25$. Each claim is 1 unit with probability $0.75$, or 2 units with probability $0.25$. Use the recursion to calculate $P(S\leq 2)$, and check the result by conditioning on the claim count.

Method and parameters. Here $m=3$, $a=-0.25/0.75=-1/3$, and $b=4(0.25)/0.75=4/3$. Since severity is positive, $g_0=(0.75)^3=0.421875$.

For $k=1$, the coefficient is $a+b=1$. Hence $g_1=(1)(0.75)(0.421875)=0.31640625$.

For $k=2$, the coefficient for $j=1$ is $-1/3+(4/3)(1/2)=1/3$, and for $j=2$ it is $a+b=1$. Thus, $g_2=(1/3)(0.75)(0.31640625)+(1)(0.25)(0.421875)=0.0791015625+0.10546875=0.1845703125$.

Therefore, $P(S\leq2)=g_0+g_1+g_2=0.421875+0.31640625+0.1845703125=0.9228515625$, or 0.9229 to four decimal places.

For a direct check, $P(N=0)=0.421875$. Exactly one claim has probability $3(0.25)(0.75)^2=0.421875$; its severity is at most 2, so all of this probability is included. Two claims have probability $3(0.25)^2(0.75)=0.140625$. Their total severity is at most 2 only if both equal 1, with probability $0.75^2=0.5625$, contributing $0.140625(0.5625)=0.0791015625$. Adding these contributions gives $0.421875+0.421875+0.0791015625=0.9228515625$, as required. Common mistake avoided: counting every two-claim outcome as having aggregate loss at most 2 ignores outcomes where one or both claims are 2 units.

Worked Example: A Deterministic Severity and the Claim-Count Cap

A hypothetical system has four independent opportunities, each producing a claim with probability $\pi=0.2$. Every claim costs exactly 2 units. Calculate $P(S\leq4)$ using the recursion, then interpret the result in terms of the fixed number of trials.

Method and parameters. Here $m=4$, $a=-0.2/0.8=-0.25$, $b=5(0.2)/0.8=1.25$, and $f_2=1$, with all other severity probabilities zero. The initial probability is $g_0=0.8^4=0.4096$.

At $k=2$, only $j=2$ contributes, with coefficient $a+b=-0.25+1.25=1$. Thus, $g_2=(1)(1)(0.4096)=0.4096$.

At $k=4$, only $j=2$ contributes, now with coefficient $a+2b/4=-0.25+0.625=0.375$. Therefore, $g_4=(0.375)(1)(0.4096)=0.1536$. No odd aggregate values are possible, because each claim costs exactly 2 units.

It follows that $P(S\leq4)=g_0+g_2+g_4=0.4096+0.4096+0.1536=0.9728$. As a check, the event is equivalent to $N\leq2$, and the binomial probabilities are $P(N=0)=0.4096$, $P(N=1)=4(0.2)(0.8)^3=0.4096$, and $P(N=2)=6(0.2)^2(0.8)^2=0.1536$. Their sum is 0.9728. The largest possible aggregate loss is $4(2)=8$ units, not an unbounded amount. Common mistake avoided: overlooking that four trials cap the number of claims at four, even if the recursion is calculated only for small loss values.

Interpreting and Checking Results

The binomial recursion is particularly useful when claim opportunities are naturally limited and repeated conditioning on every possible count would be cumbersome. Its negative $a$ is not an error: it reflects the decline in the binomial probability ratio as the count approaches its fixed maximum. For strictly positive severities, the probability of no aggregate loss is exactly the probability of no successful trial, $(1-\pi)^m$.

A finite count does not necessarily mean a finite aggregate-loss support. If each claim severity can be arbitrarily large, the sum can also be arbitrarily large, even though $N\leq m$. A finite maximum aggregate loss follows only when claim severity itself has a finite maximum. Conversely, if severity is bounded above by $c$, the condition $S\leq mc$ is a useful support check.

Before relying on calculated probabilities, check the model assumptions and the indexing. The count must be binomial with fixed $m$ and common claim probability $\pi$; severities must be identically distributed and independent of the count. The severity probabilities must be indexed on the same integer scale as the aggregate probabilities. Useful checks include the initial probability, zero probabilities at impossible lattice values, and the total probability over the full support when that support is finite.

Key takeaway: For a binomial claim count, use $a=-\pi/(1-\pi)$ and $b=(m+1)\pi/(1-\pi)$. Initialize with $g_0=(1-\pi+\pi f_0)^m$, then apply the Panjer recursion in increasing order of aggregate amount. The fixed value of $m$ caps claim count, but caps aggregate severity only when individual claim sizes are also bounded.

Check Your Understanding

For each problem, use the stated discrete severity model and the binomial claim-count assumptions. Probabilities in the answers are rounded to four decimal places where needed.

A model has $m=2$, $\pi=0.2$, and deterministic claim severity of 1 unit. Find $P(S=0)$ and $P(S=1)$. (Answer: $P(S=0)=0.6400,\;P(S=1)=0.3200$)
A group has $m=3$, $\pi=0.5$, and deterministic claim severity of 1 unit. Find $P(S=2)$, and state why aggregate losses above 3 are impossible. (Answer: $P(S=2)=0.3750$; at most three trials can produce claims.)
Let $m=2$ and $\pi=0.5$. Each claim is 1 unit with probability 0.25 and 2 units with probability 0.75. Calculate $P(S=2)$ by considering the possible claim counts and severities. (Answer: $P(S=2)=0.3906$)
A model has $m=4$, $\pi=0.25$, and deterministic claim severity of 3 units. Calculate $P(S\leq3)$. (Answer: $P(S\leq3)=0.7383$)
Let $m=3$, $\pi=0.4$, and let severity equal 1 unit with probability 0.6 or 2 units with probability 0.4. Calculate $P(S=3)$, accounting for both two-claim and three-claim outcomes. (Answer: $P(S=3)=0.1521$)

What You'll Learn