Introduction
The geometric distribution is a useful frequency model when the count can be represented as a sequence of failures before a success. In the negative binomial family, it is the one-success case. That simple specialization has an equally simple Panjer recursion: the ratio between successive probabilities is constant.
This connection matters in frequency modeling because the Panjer parameters describe how a probability sequence can be generated recursively. Correctly identifying them requires care about the count convention, the starting probability, and the distinction between a success probability and a failure probability.
From the Negative Binomial to the Geometric Count
Use the negative binomial convention in which \(N\) counts failures before the \(r\)th success. For a positive integer \(r\), the probability of exactly \(n\) failures before that success is $$ P(N=n)=\binom{n+r-1}{n}\theta^r q^n,\qquad n=0,1,2,\ldots. $$ There are \(n+r-1\) trials before the final success: among those trials, \(n\) are failures and \(r-1\) are successes. The final trial is a success. This gives the binomial coefficient, the factor \(\theta^r\) for the \(r\) successes, and the factor \(q^n\) for the failures.
Setting \(r=1\) makes the binomial coefficient equal to one and gives $$ P(N=n)=\binom{n}{n}\theta q^n=\theta q^n. $$ Thus, the geometric count is precisely the negative binomial case with one required success. The model has support beginning at zero because the first trial can be a success: in that event, the number of failures is zero.
Worked Example: Probabilities and a Tail for a Geometric Count
Suppose a hypothetical annual claim count \(N\) is modeled as the number of failures before the first success, with success probability \(\theta=0.40\) on each independent trial. Find \(P(N=n)\) for \(n=0,1,2,3\), and find \(P(N\geq 3)\). Identify the Panjer parameters and check the tail calculation.
Method and assumptions. The trials are independent, each has success probability \(\theta=0.40\), and the modeled count is the number of failures before the first success. Therefore \(q=1-\theta=0.60\), and \(p_n=\theta q^n\).
The initial probability is $$ p_0=\theta=0.40. $$ For successive counts, multiply by \(q=0.60\): $$ p_1=0.40(0.60)=0.24, $$ $$ p_2=0.24(0.60)=0.144, $$ $$ p_3=0.144(0.60)=0.0864. $$ The Panjer recursion is \(p_n=(a+b/n)p_{n-1}\). Here the ratio is \(p_n/p_{n-1}=0.60\) for every \(n\geq1\), so \(a=0.60\) and \(b=0\).
For \(N\geq3\), the first three trials must all be failures. Independence gives $$ P(N\geq3)=q^3=(0.60)^3=0.216. $$ As a check, \(p_3=0.0864\), and the probability of four or more failures is \(q^4=0.1296\). Their sum is \(0.0864+0.1296=0.216\), as required. The common mistake is to use \(\theta\) as the multiplier between probabilities; it is the failure probability \(q\) that multiplies \(p_{n-1}\).
Deriving the Panjer Parameters
The Panjer \((a,b,0)\) recursion describes a probability sequence with \(p_0>0\) and $$ p_n=\left(a+\frac{b}{n}\right)p_{n-1},\qquad n\geq1. $$ To find the parameters for a negative binomial count, compare the ratio of successive probabilities with this form. For the negative binomial probability formula above, $$ \frac{p_n}{p_{n-1}} = \frac{\binom{n+r-1}{n}}{\binom{n+r-2}{n-1}}q = \frac{n+r-1}{n}q = q+\frac{q(r-1)}{n}. $$ The second equality follows by cancelling factorials in the two binomial coefficients. Therefore the negative binomial parameters in the Panjer recursion are \(a=q\) and \(b=q(r-1)\). In the one-success case \(r=1\), the term involving \(1/n\) vanishes. The geometric parameters are $$ a=q=1-\theta,\qquad b=0,\qquad p_0=\theta. $$
The initial probability and the recursion agree with the full probability formula. Starting at \(p_0=\theta\), repeated use of \(p_n=q p_{n-1}\) yields \(p_n=\theta q^n\). For \(0\leq q<1\), the resulting probabilities sum to one: $$ \sum_{n=0}^{\infty}p_n = \theta\sum_{n=0}^{\infty}q^n = \frac{\theta}{1-q} = 1, $$ because \(1-q=\theta\). This also shows why a proper geometric count requires \(\theta>0\). If \(\theta=0\), success never occurs and there is no finite count of failures before the first success.
The nondegenerate case has \(0<\theta<1\), so \(0<a<1\). If \(\theta=1\), then \(q=0\), \(a=b=0\), \(p_0=1\), and \(p_n=0\) for \(n\geq1\). The model is still a valid limiting, degenerate count distribution, although there are no positive counts to generate.
Confirm that the modeled variable counts failures before the first success and therefore has support \(0,1,2,\ldots\).
Write \(\theta\) for success and \(q=1-\theta\) for failure. The initial probability is \(p_0=\theta\), while the recursion multiplier is \(q\).
Since \(p_n/p_{n-1}=q\), compare with \(a+b/n\) to obtain \(a=q\) and \(b=0\).
Use \(p_0=\theta\), generate terms by the recursion, and verify the geometric sum is one when \(\theta>0\).
Worked Example: Recovering the Model From a Probability Ratio
A hypothetical count model has \(p_0=0.30\), and its stated recursion is \(p_n=0.70p_{n-1}\) for \(n\geq1\). Determine whether it is a geometric count distribution, find \(p_3\), and calculate \(P(N\geq4)\). Check that the first four probabilities and the tail account for all probability.
Method and assumptions. A constant ratio identifies the geometric form if the starting probability normalizes the sequence. Here \(a=0.70\), \(b=0\), and the implied success probability is \(\theta=1-a=0.30\). This agrees with the given \(p_0\).
Apply the recursion one count at a time: $$ p_1=0.70(0.30)=0.21,\qquad p_2=0.70(0.21)=0.147,\qquad p_3=0.70(0.147)=0.1029. $$ For at least four failures, the first four trials must all be failures, so $$ P(N\geq4)=(0.70)^4=0.2401. $$ The first four probabilities sum to \(0.30+0.21+0.147+0.1029=0.7599\). Adding the tail gives \(0.7599+0.2401=1.0000\), checking the result. A common error is to set \(p_0=a\); the correct relation is \(p_0=1-a\), because \(a\) is the failure probability.
Support Conventions and the Memoryless Property
A second convention counts the trial on which the first success occurs. Let \(K\) be that trial number. Then \(K=N+1\), so \(K\) has support \(1,2,3,\ldots\), with $$ P(K=k)=\theta q^{k-1},\qquad k\geq1. $$ This is also commonly called a geometric distribution, but it is not the same count variable as the frequency \(N\) used in the Panjer recursion. To use the \((a,b,0)\) form, first express the quantity as \(N=K-1\); its probability sequence then starts at \(n=0\).
The geometric count also has a memoryless tail property. Since \(P(N\geq m)=q^m\) for \(m\geq0\), for integers \(m,n\geq0\) with \(P(N\geq m)>0\), $$ P(N\geq m+n\mid N\geq m) = \frac{q^{m+n}}{q^m} = q^n = P(N\geq n). $$ In words, conditional on having observed at least \(m\) failures, the probability of at least \(n\) additional failures is unchanged. This follows from the independent-trial model; it is not an assumption that can be applied to every count distribution.
Worked Example: Translating a Trials-Until-Success Convention
In a hypothetical inspection process, each inspection is independently acceptable with probability \(\theta=0.25\). Let \(K\) be the number of inspections through the first acceptable one. Find \(P(K=4)\) and \(P(K\leq3)\), then express the corresponding failure count in the Panjer form.
Method and assumptions. The event \(K=4\) requires three failures followed by one success. Here \(q=1-0.25=0.75\), so $$ P(K=4)=q^3\theta=(0.75)^3(0.25)=0.421875(0.25)=0.10546875. $$ For \(K\leq3\), at least one success occurs in the first three inspections. The complementary event is three failures, hence $$ P(K\leq3)=1-q^3=1-0.421875=0.578125. $$
Define the failure count \(N=K-1\). Then \(P(N=0)=0.25\), and the Panjer parameters for \(N\) are \(a=q=0.75\) and \(b=0\). For instance, \(p_1=0.75(0.25)=0.1875\), \(p_2=0.75(0.1875)=0.140625\), and \(p_3=0.75(0.140625)=0.10546875\), which agrees with \(P(K=4)\). The common mistake is to assign \(K\) support starting at zero; it counts the successful inspection too, whereas \(N\) does not.
What the Special Case Tells You
The one-success specialization is easy to recognize, but the conditions behind it matter. A geometric count has a constant successive-probability ratio because each added failure contributes the same factor \(q\). By contrast, the negative binomial ratio for \(r>1\) is \(q+q(r-1)/n\), which depends on \(n\); it is not the same constant-ratio case.
When a problem gives a probability sequence, check both the ratio and the starting probability. A constant ratio alone gives a proportional sequence, but normalization requires the matching value \(p_0=1-a\) when \(0\leq a<1\). Also verify that the variable being modeled starts at zero. These checks distinguish a geometric frequency model from a shifted waiting-time variable and help prevent a failure probability from being mistaken for a success probability. This treatment is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.
Check Your Understanding
For each problem, keep the support convention explicit and check the relationship between the initial probability and the recursion.
- A geometric count has success probability \(\theta=0.35\). Give its Panjer parameters, \(p_0\), and \(p_4\). (Answer: \(a=0.65\), \(b=0\), \(p_0=0.35\), \(p_4=0.35(0.65)^4=0.0624771875\).)
- A count sequence has \(p_0=0.20\) and \(p_n=0.80p_{n-1}\) for \(n\geq1\). Identify the geometric success probability and calculate \(P(N\geq3)\). (Answer: \(\theta=0.20\), \(a=0.80\), \(b=0\), and \(P(N\geq3)=0.80^3=0.512\).)
- A hypothetical frequency model has \(p_0=0.40\), \(p_1=0.24\), and \(p_2=0.144\). If it is geometric, find \(p_4\) and its Panjer parameters. (Answer: the ratio is \(0.60\), so \(\theta=0.40\), \(a=0.60\), \(b=0\), and \(p_4=0.40(0.60)^4=0.05184\).)
- Each independent inspection is acceptable with probability \(0.30\). If \(K\) is the inspection number of the first acceptable inspection, find \(P(K=5)\) and state the Panjer parameters for \(N=K-1\). (Answer: \(P(K=5)=0.30(0.70)^4=0.07203\); for \(N\), \(a=0.70\) and \(b=0\).)
- A negative binomial count has \(r=3\) required successes and success probability \(\theta=0.40\). Find \(p_n/p_{n-1}\) at \(n=1\) and \(n=4\), and decide whether it is the geometric one-success case. (Answer: \(q=0.60\), the ratios are \(0.60(1+2/1)=1.8\) at \(n=1\) and \(0.60(1+2/4)=0.9\) at \(n=4\); it is not the geometric case because \(r=3\), and the ratio is not constant.)