Negative Binomial Counts as an (a,b,0) Distribution

Why the Negative Binomial Fits the (a,b,0) Class

A frequency model is useful in a recursive calculation when each probability can be obtained from the preceding one using a simple rule. The negative binomial distribution has this property: under a specified count convention, the ratio of successive probabilities is an affine function of the reciprocal count. That puts the distribution in the Panjer $(a,b,0)$ class.

The count convention matters. Here, $N$ is the number of failures observed before the $r$th success in independent Bernoulli trials. Each trial has success probability $p$, so its failure probability is $q=1-p$. This convention assigns positive probability to zero failures and has an unbounded count support.

Definition: For a positive integer $r$ and $0<p<1$, let $N$ count failures before the $r$th success in independent trials, each with success probability $p$. Write $\pi_n=P(N=n)$. Then, for $n=0,1,2,\ldots$, $$ \pi_n=\binom{n+r-1}{n}p^r q^n,\qquad q=1-p. $$ The probabilities sum to one. An $(a,b,0)$ distribution has $\pi_0>0$ and satisfies $\pi_n=(a+b/n)\pi_{n-1}$ for every integer $n\geq 1$.

The negative binomial probability formula counts the arrangements of the first $n+r-1$ trials that contain $n$ failures and $r-1$ successes, followed by a success on the final trial. There are $\binom{n+r-1}{n}$ such arrangements. Each has probability $q^np^r$, giving the stated formula. The initial probability is the probability that the first $r$ trials are all successes:

\pi_0=P(N=0)=p^r.

Deriving the Probability Recursion

To identify the Panjer parameters, compare the probability of $n$ failures with the probability of $n-1$ failures. For $n\geq 1$, divide the negative binomial formula for $\pi_n$ by the formula for $\pi_{n-1}$. The factors $p^r$ cancel, and the ratio of the combinatorial coefficients is $(n+r-1)/n$.

\begin{aligned} \frac{\pi_n}{\pi_{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}. \end{aligned}

Multiplying by $\pi_{n-1}$ gives the recursion. Matching its two terms to the $(a,b,0)$ form identifies the class parameters directly.

Theorem: Under the failure-count convention above, the negative binomial distribution is in the $(a,b,0)$ class with $$ a=q=1-p,\qquad b=q(r-1)=(1-p)(r-1),\qquad \pi_0=p^r. $$ Thus, for $n\geq 1$, $$ \pi_n=\left(q+\frac{q(r-1)}{n}\right)\pi_{n-1}. $$

This derivation explains why the reciprocal-count term appears: the combinatorial ratio contributes $(n+r-1)/n$, which separates into a constant and a term proportional to $1/n$. It is the probability ratio—not a moment approximation—that establishes class membership.

For the stated model, $r$ is a positive integer and $0<p<1$. These conditions make $\pi_0>0$, $a>0$, and $b\geq 0$. The count support includes every nonnegative integer. Normalization can also be checked using the negative binomial series:

\sum_{n=0}^{\infty}\pi_n =p^r\sum_{n=0}^{\infty}\binom{n+r-1}{n}q^n =p^r(1-q)^{-r} =p^r p^{-r} =1.

Using the Recursion

Once $a$, $b$, and $\pi_0$ are known, calculate probabilities in order. At each step, use the current index $n$ in the factor $a+b/n$, multiply by $\pi_{n-1}$, and retain adequate precision until the requested result is obtained. The recurrence does not replace the initial probability: it needs $\pi_0$ to start.

Fix the count convention.
Confirm that the variable counts failures before $r$ successes, rather than counting trials or successes.

Find the model inputs.
Set $q=1-p$, then calculate $a=q$, $b=q(r-1)$, and $\pi_0=p^r$.

Apply the recurrence at the correct index.
For each $n\geq1$, calculate $\pi_n=(a+b/n)\pi_{n-1}$.

Check the result.
Confirm probabilities are nonnegative, the requested event uses the correct count range, and any direct probability calculation agrees.

Worked Example: Finding the Parameters and Early Probabilities

A hypothetical equipment-monitoring process is modeled by the number $N$ of unsuccessful checks before the fourth successful check. Assume independent checks, each successful with probability $p=0.6$. Find the $(a,b,0)$ parameters and calculate $\pi_0,\pi_1,\pi_2,$ and $\pi_3$. Give probabilities to four decimal places.

Method. Use the failures-before-successes parameterization, set $r=4$, and apply the derived recursion. The failure probability is $q=1-p=1-0.6=0.4$.

The class parameters and initial probability are $$ a=q=0.4,\qquad b=q(r-1)=0.4(4-1)=1.2,\qquad \pi_0=p^r=0.6^4=0.1296. $$

For $n=1$, $$ \pi_1=\left(0.4+\frac{1.2}{1}\right)(0.1296) =1.6(0.1296)=0.20736. $$

For $n=2$, $$ \pi_2=\left(0.4+\frac{1.2}{2}\right)(0.20736) =(0.4+0.6)(0.20736)=1(0.20736)=0.20736. $$

For $n=3$, $$ \pi_3=\left(0.4+\frac{1.2}{3}\right)(0.20736) =(0.4+0.4)(0.20736)=0.8(0.20736)=0.165888. $$

Rounded to four decimal places, the results are $\pi_0=0.1296$, $\pi_1=0.2074$, $\pi_2=0.2074$, and $\pi_3=0.1659$. As a check, the direct formula gives $\pi_2=\binom{5}{2}(0.6)^4(0.4)^2=10(0.1296)(0.16)=0.20736$, matching the recursion. A common mistake is to use $p$, rather than $q$, as the multiplier for each failure; the count concerns failures, so the factor for each counted trial is $q$.

Worked Example: Calculating a Count Range

A hypothetical quality-control process is modeled by the number $N$ of failed tests before the third passing test. Assume independent tests with pass probability $p=0.7$. Calculate $P(N\leq 2)$ using the recursion, and verify the probabilities through $\pi_2$ with the direct formula. Give the final probability to four decimal places.

Method. The event $N\leq2$ includes exactly $N=0,1,2$, so calculate and add those three probabilities. Here $r=3$ and $q=1-0.7=0.3$. Therefore $$ a=0.3,\qquad b=0.3(3-1)=0.6,\qquad \pi_0=(0.7)^3=0.343. $$

For $n=1$, $$ \pi_1=\left(0.3+\frac{0.6}{1}\right)(0.343) =0.9(0.343)=0.3087. $$

For $n=2$, $$ \pi_2=\left(0.3+\frac{0.6}{2}\right)(0.3087) =0.6(0.3087)=0.18522. $$

The direct formula checks these values: $\pi_1=\binom{3}{1}(0.7)^3(0.3)=3(0.343)(0.3)=0.3087$, and $\pi_2=\binom{4}{2}(0.7)^3(0.3)^2=6(0.343)(0.09)=0.18522$. Thus $$ P(N\leq2)=\pi_0+\pi_1+\pi_2 =0.343+0.3087+0.18522=0.83692. $$ The answer is $0.8369$ to four decimal places. The main trap is to interpret “at most two failures” as including $N=3$; the count range ends at two.

Worked Example: Recovering the Model Inputs From Class Parameters

A hypothetical failure-count model is known to be negative binomial under the failures-before-successes convention. Its recursion is $\pi_n=(0.4+1.2/n)\pi_{n-1}$, and the number of required successes is a positive integer. Recover $p$ and $r$, then find $\pi_0$ and $\pi_2$. Give probabilities to four decimal places.

Method. Match the given coefficients to $a=q$ and $b=q(r-1)$. Since $a=0.4$, the failure probability is $q=0.4$, so the success probability is $$ p=1-q=1-0.4=0.6. $$

Use $b=q(r-1)$ to recover $r$: $$ 1.2=0.4(r-1),\qquad r-1=\frac{1.2}{0.4}=3,\qquad r=4. $$

The initial probability is $$ \pi_0=p^r=(0.6)^4=0.1296. $$ For the first count probability, $$ \pi_1=\left(0.4+\frac{1.2}{1}\right)(0.1296) =1.6(0.1296)=0.20736. $$ Then $$ \pi_2=\left(0.4+\frac{1.2}{2}\right)(0.20736) =1(0.20736)=0.20736. $$ To four decimal places, $\pi_0=0.1296$ and $\pi_2=0.2074$. The recovered values satisfy $0<p<1$ and positive integer $r$, and substituting them back gives $a=0.4$ and $b=0.4(4-1)=1.2$, as required. A common error is to set $r=b/a$; the defining relation is $b=a(r-1)$, so the correct recovery is $r=1+b/a$.

Common Errors and Why the Parameterization Matters

The negative binomial name is used with different count conventions in probability work. A problem may count failures before a fixed number of successes, or it may define the count in a different way. The formula for the zero-count probability and the probability ratio must match the stated variable. Here they are fixed by the convention in the definition: $N$ counts failures, $p$ is the success probability, and each counted failure contributes a factor $q=1-p$.

Using $p$ where $q$ belongs. In $\pi_n=\binom{n+r-1}{n}p^rq^n$, the exponent $n$ belongs to the failure probability.
Dropping the $1/n$ in the recursion. The multiplier is $a+b/n$, not $a+b$. Its value changes with the current count $n$.
Starting at the wrong probability. The recursion begins with $\pi_0=p^r$, not with $\pi_1$ or $q^r$.
Confusing model parameters with count probabilities. Here $p$ is the success probability, while $\pi_n$ is the probability of a count of $n$.
Summing the wrong event. For example, $P(N\leq k)=\sum_{n=0}^{k}\pi_n$; the zero count is included.

The practical value of the class representation is that it replaces repeated evaluation of a combinatorial expression with a simple step-by-step probability calculation. It is particularly important to preserve the indexing and the initial probability when using that recurrence in a larger frequency calculation.

Key takeaway: For the number of failures before $r$ successes, $q=1-p$, $a=q$, $b=q(r-1)$, and $\pi_0=p^r$. The complete recursion is $\pi_n=(a+b/n)\pi_{n-1}$ for $n\geq1$.

Check Your Understanding

For each problem, use the failures-before-successes convention and the stated probability model.

A count $N$ follows a negative binomial model with $r=2$ and success probability $p=0.6$. Find $a$, $b$, $\pi_0$, and $\pi_1$. (Answer: $a=0.4,\ b=0.4,\ \pi_0=0.36,\ \pi_1=0.288$.)
For $r=4$ and $p=0.5$, use the recursion to find $\pi_2$. (Answer: $\pi_2=0.15625$.)
For $r=3$ and $p=0.4$, calculate $P(N\leq2)$ using the recursion. (Answer: $0.31744$.)
A negative binomial failure-count model has $a=0.4$ and $b=1.2$. Recover $p$ and $r$, then find $\pi_0$ and $\pi_2$. (Answer: $p=0.6,\ r=4,\ \pi_0=0.1296,\ \pi_2=0.20736$.)
For $r=2$ and $p=0.5$, calculate $P(N\geq2)$ from the initial probability and recursion. (Answer: $0.5$.)

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