Why Negative Binomial Parameterizations Need Care
A negative binomial frequency distribution can be described using the probability of a success in repeated trials, or using the probability of the event that contributes to the count. These probabilities are complements, but confusing them changes the entire probability sequence. The distinction matters when identifying Panjer parameters: the parameter multiplying the probability ratio is the probability of the counted event, not necessarily the success probability named in a trial description.
This tutorial uses \(N\) for the nonnegative integer count. In the trial interpretation, \(N\) is the number of failures observed before the \(r\)th success. The counted events are failures. The same distribution can instead be written directly in terms of the probability of a counted event. Once the convention is stated, the conversion and the Panjer parameters follow from the probability ratio.
The \(r\) successes and \(k\) failures can occur in any order among the first \(r+k-1\) trials, while the final trial must be the \(r\)th success. There are \(\binom{k+r-1}{k}\) such arrangements. Each has probability \(p^r(1-p)^k\), which gives the stated probability mass function. The distribution is supported on every nonnegative integer when \(0<p<1\).
The trial story requires integer \(r\). The probability formula and the Panjer ratio also extend to real \(r>0\) by replacing the binomial coefficient with \(\binom{k+r-1}{k}=\Gamma(k+r)/(\Gamma(r)\,k!)\). Unless stated otherwise, the examples below use integer \(r\), so the trial interpretation applies directly.
From Either Convention to Panjer Parameters
A distribution belongs to the Panjer \((a,b,0)\) class when its probabilities satisfy \(P(N=0)>0\) and, for \(k\geq 1\), \(P(N=k)=(a+b/k)P(N=k-1)\). For the negative binomial distribution, divide its probability at \(k\) by its probability at \(k-1\). In the counted-event convention, the ratio is
Comparing this expression with \(a+b/k\) identifies the Panjer parameters. The initial probability is the probability of zero counted events: all of the first \(r\) trials must be successes.
The formula for \(b\) also covers \(r=1\): then \(b=0\), and the negative binomial becomes the geometric distribution for the number of failures before one success. A quick reverse conversion is possible whenever \(a>0\): \(q=a\), \(r=1+b/a\), and \(p=1-a\). A proposed pair must give an allowed shape and counted-event probability; matching a formula algebraically does not by itself make arbitrary parameters a valid negative binomial model.
Decide whether the random variable counts failures before successes, successes before failures, or another specified event. Do not infer the meaning of a symbol called \(p\) without reading the definition.
For failures before the \(r\)th success, the counted-event probability is \(q=1-p\).
Write the adjacent-probability ratio as \(q+q(r-1)/k\), then read \(a=q\) and \(b=q(r-1)\).
Verify \(P(N=0)=(1-q)^r\). When useful, compare the mean \(rq/(1-q)\) with the problem’s frequency interpretation.
Worked Conversions
Worked Example: Converting a Success Probability to Panjer Form
A hypothetical service plan models the number \(N\) of minor repairs before a customer completes four successful inspection cycles. Each inspection cycle succeeds independently with probability \(p=0.8\). Find the counted-event probability, the \((a,b,0)\) parameters, \(P(N=0)\), and \(P(N=1)\). Check the result using the mean and variance.
Method. \(N\) counts failures, so the counted-event probability is the failure probability, not the inspection success probability: \(q=1-p=1-0.8=0.2\). The shape is \(r=4\).
The Panjer parameters are \(a=q=0.2\) and \(b=q(r-1)=0.2(4-1)=0.6\). The zero-count probability is \(P(N=0)=p^r=0.8^4=0.4096\).
For \(k=1\), the probability ratio is \(a+b/1=0.2+0.6=0.8\). Therefore, \(P(N=1)=0.8P(N=0)=0.8(0.4096)=0.32768\), or \(0.3277\) to four decimal places.
As a check, the negative binomial mean is \(rq/(1-q)=4(0.2)/0.8=1\), and its variance is \(rq/(1-q)^2=4(0.2)/(0.8)^2=1.25\). These moments are consistent with a count of failures when success is relatively likely. The calculation avoids the common mistake of setting \(a=p=0.8\); the ratio uses the probability of a counted failure.
Worked Example: Starting With the Counted-Event Probability
A hypothetical equipment-monitoring model counts the number of warning events before a system records five successful diagnostic cycles. A warning event has probability \(q=0.3\) on each independent cycle. Find the success probability, the Panjer parameters, the first two probabilities \(P(N=0)\) and \(P(N=1)\), and the mean.
Method. The problem gives the probability of the counted event directly: \(q=0.3\). Thus the success probability is its complement, \(p=1-q=0.7\), and \(r=5\).
The Panjer parameters are \(a=q=0.3\) and \(b=q(r-1)=0.3(4)=1.2\). The zero-count probability is \(P(N=0)=(1-q)^r=0.7^5=0.16807\), or \(0.1681\) to four decimal places.
At \(k=1\), the ratio is \(a+b=0.3+1.2=1.5\). Hence \(P(N=1)=1.5(0.16807)=0.252105\), or \(0.2521\) to four decimal places. The mean is \(rq/(1-q)=5(0.3)/0.7=15/7\approx2.1429\).
The probability at one warning event can exceed the probability at zero warning events because the ratio at \(k=1\) is \(1.5\). This is not a problem: individual probabilities need not decrease as \(k\) increases. The calculation avoids treating the stated warning probability as a success probability and then complementing it a second time.
Worked Example: Recovering the Trial Parameters From \(a\) and \(b\)
A frequency model is reported to have Panjer parameters \(a=0.25\) and \(b=0.50\), with a negative binomial form. Recover \(r\), the counted-event probability \(q\), and the success probability \(p\). Then calculate \(P(N=0)\) and \(P(N=1)\), and verify the probability ratio at \(k=1\).
Method. For a negative binomial count, \(a=q\), so \(q=0.25\). Since \(b=q(r-1)\), \(r=1+b/a=1+0.50/0.25=3\). This is a positive integer, as required for the stated trial interpretation. The success probability is \(p=1-q=0.75\).
The boundary probability is \(P(N=0)=p^r=0.75^3=0.421875\), or \(0.4219\) to four decimal places. The Panjer recursion at \(k=1\) gives \(P(N=1)=(a+b)P(N=0)=(0.25+0.50)(0.421875)=0.31640625\), or \(0.3164\) to four decimal places.
Directly from the negative binomial formula, \(P(N=1)=\binom{3}{1}(0.75)^3(0.25)=3(0.421875)(0.25)=0.31640625\). The direct probability and the recursion agree. This check catches the common error \(r=b/a\), which would omit the \(1\) in \(r=1+b/a\).
Checks, Conditions, and an Exam Trap
The probability generating function, mean, and variance offer useful checks on a conversion. For \(N\) counting failures before \(r\) successes, the probability generating function is \(G_N(z)=[p/(1-qz)]^r\), for \(q=1-p\) and values of \(z\) for which the expression is finite. Differentiating at \(z=1\) gives \(E[N]=rq/p=rq/(1-q)\) and \(\operatorname{Var}(N)=rq/p^2=rq/(1-q)^2\). A candidate does not need to re-derive these quantities for every parameter conversion, but a mean inconsistent with the scenario is a warning that the probabilities may have been reversed.
A frequent trap is to confuse the count's event with the trial's success. Suppose \(r=3\) and the success probability is \(p=0.6\). The failures being counted have probability \(q=0.4\), so the correct Panjer parameters are \(a=0.4\) and \(b=0.4(3-1)=0.8\). A wrong calculation sets \(a=0.6\), \(b=1.2\), and \(P(N=0)=0.6^3\). Those values mix the counted-event probability in the ratio with the success probability in the boundary probability. The correction is to use \(q=0.4\) throughout the counted-event parameterization: \(P(N=0)=(1-q)^3=0.6^3\), while \(a=q\) and \(b=q(r-1)\).
In the success-probability convention, \(P(N=0)=p^r\); in the counted-event convention, it is \((1-q)^r\). These are the same quantity because \(p=1-q\). Do not use \(q^r\) as the zero-count probability: zero failures require \(r\) successes, not \(r\) failures. Also distinguish this model from one that counts successes before a fixed number of failures. That alternative can have a similar-looking formula, but its trial roles and boundary probability must be identified from its definition.
Check Your Understanding
For each problem, state the relevant convention before using the Panjer parameters. Answers are rounded to four decimal places where needed.
- A count \(N\) is the number of failures before the fourth success, with success probability \(p=0.7\). Find \(q\), \(a\), \(b\), and \(P(N=0)\). (Answer: \(q=0.3,\ a=0.3,\ b=0.9,\ P(N=0)=0.2401\))
- A count \(N\) records warning events before six successful tests; each warning has probability \(q=0.2\). Find \(p\), \(a\), \(b\), and \(P(N=0)\). (Answer: \(p=0.8,\ a=0.2,\ b=1.0,\ P(N=0)=0.2621\))
- For failures before the second success, \(p=0.5\). Use the Panjer recursion to find \(P(N=0), P(N=1), P(N=2)\), and \(P(N=3)\). (Answer: \(0.2500,\ 0.2500,\ 0.1875,\ 0.1250\))
- A negative binomial frequency has \(a=0.4\) and \(b=1.6\). Recover \(q,r,p\), then find \(P(N=0)\) and the mean. (Answer: \(q=0.4,\ r=5,\ p=0.6,\ P(N=0)=0.0778,\ E[N]=3.3333\))
- A hypothetical count records successes before the fourth failure, with success probability \(0.3\) on each independent trial. Express this using the counted-event probability for the count, then give the Panjer parameters and \(P(N=0)\). (Answer: counted-event probability \(q=0.3,\ a=0.3,\ b=0.9,\ P(N=0)=0.2401\))