Deriving the Panjer Probability Recursion

Why Derive a Probability Recursion?

A count distribution is often given by a probability function, but calculating every probability independently can be repetitive. A recursion replaces each new probability with a calculation based on the preceding one. The key is to express the ratio of consecutive probabilities in a form that makes this step systematic.

Here $N$ is a nonnegative integer-valued count, and $p_n=P(N=n)$. The focus is the probability recursion for counts, not the recursion for aggregate claim amounts. The derivation starts from the affine probability-ratio form introduced in the earlier material on recognizing an affine probability ratio.

Definition: Suppose that, for the relevant integers $n\geq 1$, consecutive probabilities satisfy $$ \frac{p_n}{p_{n-1}}=a+\frac{b}{n}. $$ Then multiplying by $p_{n-1}$ gives the Panjer probability recursion $$ p_n=\left(a+\frac{b}{n}\right)p_{n-1}. $$ The recursion is used together with the boundary probability $p_0$. The ratio form requires $p_{n-1}>0$; apply it only at indices where the ratio is defined.

From Ratio to Recursion

The derivation is short, but the index matters. At a particular $n$, the ratio compares the probability at $n$ with the probability at $n-1$. Substitute the affine expression for that same ratio and multiply both sides by $p_{n-1}$:

\frac{p_n}{p_{n-1}}=a+\frac{b}{n} \quad\Longrightarrow\quad p_n=\left(a+\frac{b}{n}\right)p_{n-1}.

Thus $a$ and $b$ do not by themselves determine the probabilities: the starting value $p_0$ is also needed. Once $p_0$ and the applicable recursion are known, probabilities can be calculated in sequence. For a distribution with a finite maximum count, the recurrence is used only through that count; it is not a license to calculate probabilities beyond the distribution's support.

The form is useful because important count models have probability ratios that are affine in $1/n$. The two verifications below begin with the probability functions themselves, so the coefficients are derived rather than assumed.

Write the ratio.
Start from $p_n/p_{n-1}$, using the probability function for the specified count model.

Separate constant and $1/n$ terms.
Match the ratio to $a+b/n$ and identify $a$ and $b$.

Multiply by the preceding probability.
Obtain $p_n=(a+b/n)p_{n-1}$, then use $p_0$ as the starting value.

Check against the model.
Compare at least one recursively calculated probability with the original probability function.

Verification for Poisson Counts

Assume $N\sim\operatorname{Poisson}(\lambda)$, where $\lambda>0$. Its probability function is $p_n=e^{-\lambda}\lambda^n/n!$ for $n=0,1,2,\ldots$. For $n\geq1$, divide the probability at $n$ by the one at $n-1$:

\frac{p_n}{p_{n-1}} = \frac{e^{-\lambda}\lambda^n/n!}{e^{-\lambda}\lambda^{n-1}/(n-1)!} = \frac{\lambda}{n}.

This is $0+\lambda/n$, so $a=0$ and $b=\lambda$. Consequently, $p_n=(\lambda/n)p_{n-1}$, starting from $p_0=e^{-\lambda}$. The ratio is valid for every integer $n\geq1$ in this model.

Worked Example: Building Poisson Probabilities Recursively

A hypothetical portfolio has a Poisson-distributed annual claim count with parameter $\lambda=2$. Find $p_0,p_1,p_2,p_3$ using the recursion, and check $p_3$ against the Poisson probability function. Give probabilities to four decimal places.

Method. For a Poisson count, the ratio gives $a=0$, $b=\lambda=2$, and the starting probability is $p_0=e^{-2}$. Retain exact expressions during the recursion, then round the reported probabilities.

First, $$ p_0=e^{-2}=0.135335\ldots. $$ For $n=1$, the multiplier is $2/1$, so $$ p_1=\frac{2}{1}p_0=2e^{-2}=0.270671\ldots. $$ For $n=2$, the multiplier is $2/2=1$, giving $$ p_2=\frac{2}{2}p_1=2e^{-2}=0.270671\ldots. $$ For $n=3$, the multiplier is $2/3$, so $$ p_3=\frac{2}{3}p_2=\frac{4}{3}e^{-2}=0.180447\ldots. $$

Check using the probability function. Directly, $$ P(N=3)=e^{-2}\frac{2^3}{3!} =e^{-2}\frac{8}{6} =\frac{4}{3}e^{-2} =0.180447\ldots, $$ which agrees with the recursive result. Rounded to four decimal places, the probabilities are $p_0=0.1353$, $p_1=0.2707$, $p_2=0.2707$, and $p_3=0.1804$. The common mistake avoided is to use the recursion multiplier without its $n$ denominator.

Verification for Binomial Counts

Assume $N\sim\operatorname{Binomial}(m,r)$, where $m$ is a positive integer and $0<r<1$ is the probability of success on each of $m$ independent trials. The probability function is $$ p_n=\binom{m}{n}r^n(1-r)^{m-n},\qquad n=0,1,\ldots,m. $$ For $1\leq n\leq m$, form the successive ratio. Using $\binom{m}{n}/\binom{m}{n-1}=(m-n+1)/n$ gives $$ \frac{p_n}{p_{n-1}} =\frac{m-n+1}{n}\frac{r}{1-r} =-\frac{r}{1-r}+\frac{(m+1)r}{(1-r)n}. $$

Therefore the coefficients are $a=-r/(1-r)$ and $b=(m+1)r/(1-r)$. The resulting recursion is applied for $n=1,\ldots,m$, beginning with $p_0=(1-r)^m$. Unlike the Poisson case, this binomial model has a finite maximum count $m$; the ratio derivation does not define probabilities for larger counts.

Worked Example: Applying the Binomial Recursion Through Its Support

A hypothetical inspection program checks $m=6$ components independently, with probability $r=1/4$ that each component requires repair. Let $N$ be the number requiring repair. Calculate $p_0$ through $p_6$ recursively, verify the probabilities sum to one, and check $p_3$ directly. Give probabilities to four decimal places.

Method and coefficients. Here $N\sim\operatorname{Binomial}(6,1/4)$, so $1-r=3/4$. Thus $$ a=-\frac{1/4}{3/4}=-\frac13,\qquad b=\frac{(6+1)(1/4)}{3/4}=\frac73,\qquad p_0=\left(\frac34\right)^6=\frac{729}{4096}. $$ At each step the multiplier is $-1/3+7/(3n)$.

For $n=1$, the multiplier is $2$, giving $p_1=1458/4096$. For $n=2$, it is $5/6$, giving $p_2=(5/6)(1458/4096)=1215/4096$. For $n=3$, it is $4/9$, giving $p_3=(4/9)(1215/4096)=540/4096$.

For $n=4$, the multiplier is $1/4$, so $p_4=135/4096$. For $n=5$, it is $2/15$, so $p_5=18/4096$. For $n=6$, it is $1/18$, so $p_6=1/4096$.

Checks. The numerators sum to $729+1458+1215+540+135+18+1=4096$, so the probabilities sum to one. Directly, $$ P(N=3)=\binom63\left(\frac14\right)^3\left(\frac34\right)^3 =20\cdot\frac{1}{64}\cdot\frac{27}{64} =\frac{540}{4096}, $$ matching the recursion. To four decimal places, $p_0,\ldots,p_6$ are $0.1780, 0.3560, 0.2966, 0.1318, 0.0330, 0.0044, 0.0002$. The common mistake avoided is extending the calculation past $n=6$, where the binomial count has no possible outcomes.

Using Ratios to Recover Binomial Parameters

The ratio form can also be used in reverse. If a binomial model is known and two successive probability ratios are available, those ratios can identify $a$ and $b$. The binomial coefficient formulas then connect those values to $r$ and $m$. This is useful when probabilities are supplied as ratios rather than as a full probability table.

Worked Example: Recovering a Binomial Model from Ratios

A hypothetical count is known to have a binomial distribution with unknown trial count $m$ and success probability $r$, where $0<r<1$. The ratios are $p_1/p_0=5/4$ and $p_2/p_1=1/2$. Determine $m$ and $r$, then calculate $p_3$ by recursion and verify it from the binomial probability function.

Method. Since $p_n/p_{n-1}=a+b/n$, the first two ratios give $a+b=5/4$ and $a+b/2=1/2$. Subtracting the second equation from the first gives $b/2=3/4$, hence $b=3/2$. Then $a=5/4-3/2=-1/4$.

For a binomial model, $a=-r/(1-r)$. Therefore $1/4=r/(1-r)$, so $1-r=4r$, $5r=1$, and $r=1/5$. Also $b=(m+1)r/(1-r)=(m+1)/4$. Setting this equal to $3/2$ gives $m+1=6$, hence $m=5$.

The starting probability is $p_0=(4/5)^5=1024/3125$. At $n=1$, the given ratio gives $p_1=(5/4)p_0=1280/3125$. At $n=2$, $p_2=(1/2)p_1=640/3125$. At $n=3$, the multiplier is $$ a+\frac{b}{3}=-\frac14+\frac12=\frac14, $$ so $p_3=(1/4)(640/3125)=160/3125=0.0512$.

Check using the probability function. With $m=5$ and $r=1/5$, $$ P(N=3)=\binom53\left(\frac15\right)^3\left(\frac45\right)^2 =10\cdot\frac{1}{125}\cdot\frac{16}{25} =\frac{160}{3125}=0.0512. $$ The recursive and direct calculations agree. This check also confirms that the inferred integer trial count and probability parameter are consistent with the supplied ratios.

Common Index Trap

The $n$ in the denominator belongs to the probability at the new count $n$, not the preceding count $n-1$. For example, with a Poisson count having $\lambda=2$, the correct multiplier for moving from $p_1$ to $p_2$ is $2/2=1$. Using $2/(n-1)$ instead would give $2/1=2$, and the incorrect calculation would be $p_2=2p_1=4e^{-2}$. The correct value is $p_2=(2/2)p_1=2e^{-2}$. The ratio from the probability function, $p_2/p_1=2/2=1$, immediately exposes the error.

A second practical safeguard is to keep the boundary probability separate from the ratio. The recursion advances from $p_0$; it does not supply $p_0$ on its own. For the models above, $p_0$ comes from the corresponding probability function.

Key takeaway: Derive the ratio from the model, write it as $a+b/n$, and only then multiply by $p_{n-1}$. For Poisson counts, $a=0$ and $b=\lambda$; for binomial counts, $a=-r/(1-r)$ and $b=(m+1)r/(1-r)$. Check the index, the boundary probability, and the model's range of counts. This derivation is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, use the probability-ratio method where applicable and keep the count index explicit.

A Poisson count has parameter $\lambda=3$. State $a$ and $b$, and give the recursion multiplier for calculating $p_2$ from $p_1$. (Answer: $a=0$, $b=3$; multiplier $3/2$.)
A binomial count has $m=4$ and $r=1/3$. Find $a$, $b$, and $p_0$. (Answer: $a=-1/2$, $b=5/2$, $p_0=16/81$.)
For $N\sim\operatorname{Binomial}(3,1/2)$, use the recursion to find $p_2$ from $p_1$. (Answer: $p_1=3/8$, the $n=2$ multiplier is $1/2$, and $p_2=3/16$.)
A binomial count with unknown $m,r$ has $p_1/p_0=2$ and $p_2/p_1=3/4$. Recover $m,r$. (Answer: $a=1/2$, $b=3/2$, so $r=-1$, which is incompatible with a binomial success probability; the supplied ratios cannot describe the stated binomial model.)
For a Poisson count with $\lambda=4$, calculate $p_0$ and then $p_1$ recursively; give each to four decimal places. (Answer: $p_0=e^{-4}=0.0183$, $p_1=4p_0=0.0733$.)

What You'll Learn