Zero-Truncated Negative Binomial Counts

Why Zero Truncation Changes the Boundary

In some hypothetical portfolios, the count under study is recorded only when at least one event occurs. A negative binomial model can describe the count before this restriction, but conditioning on a positive count removes its zero probability and renormalizes every positive probability. The resulting distribution has a different boundary probability, even though its ratios of successive positive probabilities remain unchanged.

This distinction matters in the Panjer framework: zero-truncated negative binomial counts belong to the $(a,b,1)$ class. The class parameters describe how probabilities from count two onward relate to the preceding probability; the probability at count one must be calculated separately. The treatment is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Definition: Let $N$ have a negative binomial distribution with shape parameter $r>0$ and success probability $\theta$, where $0<\theta<1$. Write $q=1-\theta$. Under the convention that $N$ counts failures before $r$ successes, $$ f_n=P(N=n)=\frac{\Gamma(n+r)}{\Gamma(r)n!}\theta^r q^n,\qquad n=0,1,2,\ldots. $$ The zero-truncated count has probabilities $p_n=P(N=n\mid N>0)$ for $n\geq 1$. Thus $p_n=f_n/(1-f_0)$, and it assigns probability zero to count zero.

Here $\theta$ is the success probability, not the probability of a zero count: $f_0=\theta^r$. For integer $r$, the coefficient in the probability formula is the usual binomial coefficient $\binom{n+r-1}{n}$. The gamma-function form also covers noninteger $r>0$.

Deriving the Boundary and Recursion

The original negative binomial probabilities have a simple ratio. For $n\geq1$, divide the probability at $n$ by the probability at $n-1$:

\frac{f_n}{f_{n-1}} =\frac{\Gamma(n+r)}{\Gamma(n+r-1)} \frac{(n-1)!}{n!}q =q\frac{n+r-1}{n} =q+\frac{(r-1)q}{n}.

For the zero-truncated probabilities, the same conditioning factor $1-f_0$ divides both probabilities in a ratio. It therefore cancels whenever $n\geq2$. The result is the $(a,b,1)$ recursion, while the first positive probability is obtained directly by conditioning:

p_n=\left(a+\frac{b}{n}\right)p_{n-1},\qquad n\geq2, \qquad a=q,\quad b=(r-1)q,

p_1=\frac{f_1}{1-f_0} =\frac{r\theta^r q}{1-\theta^r}.

This is the key reason the recursion is preserved: conditioning rescales every positive mass by the same constant, so it leaves their successive ratios unchanged. It does not mean the original boundary formula can be carried over without adjustment. In the $(a,b,1)$ class, the recursion applies from $n=2$, and $p_1$ is specified separately.

Theorem: If $N$ is negative binomial with $r>0$ and $0<\theta<1$, then $N\mid N>0$ has $(a,b,1)$ parameters $$ a=1-\theta,\qquad b=(r-1)(1-\theta),\qquad p_1=\frac{r\theta^r(1-\theta)}{1-\theta^r}. $$ For $n\geq2$, its probabilities follow from $p_n=(a+b/n)p_{n-1}$. The factor $a+b/n=(1-\theta)(n+r-1)/n$ is positive for every $n\geq2$, including when $0<r<1$.

Repeated use of the recursion gives a useful expression for any positive count:

p_n=p_1\prod_{j=2}^{n}\left(a+\frac{b}{j}\right),\qquad n\geq2.

Equivalently, the original negative binomial mass can be conditioned directly:

p_n=\frac{\Gamma(n+r)}{\Gamma(r)n!} \frac{\theta^r q^n}{1-\theta^r},\qquad n\geq1.

The product form is convenient for generating probabilities one at a time. The direct form is useful for checking a value or calculating a probability without generating every intervening count. Both require the same parameterization and the same conditioning event $N>0$.

Worked Probabilities

Worked Example: Generate the First Four Positive Probabilities

A hypothetical count $N$ follows a negative binomial distribution with $r=2$ and success probability $\theta=0.5$. The recorded count is conditional on $N>0$. Find $p_1,p_2,p_3,p_4$, and then find $P(N\geq4\mid N>0)$.

Method. Calculate the truncated boundary probability, then use the $(a,b,1)$ recursion. Here $q=1-\theta=0.5$, $f_0=\theta^r=0.5^2=0.25$, and $1-f_0=0.75$. The original probability at one is $$ f_1=\binom{2}{1}(0.5)^2(0.5)^1=2(0.25)(0.5)=0.25. $$ Therefore $$ p_1=\frac{0.25}{0.75}=\frac{1}{3}. $$

The recursion parameters are $a=q=0.5$ and $b=(r-1)q=(2-1)(0.5)=0.5$. Applying the recursion: $$ p_2=\left(0.5+\frac{0.5}{2}\right)\frac{1}{3} =0.75\left(\frac{1}{3}\right)=\frac{1}{4}, $$ $$ p_3=\left(0.5+\frac{0.5}{3}\right)\frac{1}{4} =\frac{2}{3}\left(\frac{1}{4}\right)=\frac{1}{6}, $$ $$ p_4=\left(0.5+\frac{0.5}{4}\right)\frac{1}{6} =\frac{5}{8}\left(\frac{1}{6}\right)=\frac{5}{48}. $$

For the tail probability, the original probabilities at counts zero through three are $1/4,1/4,3/16,1/8$. Their sum is $13/16$, so $P(N\geq4)=1-13/16=3/16$. Conditioning gives $$ P(N\geq4\mid N>0)=\frac{3/16}{3/4}=\frac{1}{4}. $$ As a check, the four generated probabilities sum to $1/3+1/4+1/6+5/48=41/48$; the remaining tail is $7/48$ for counts greater than four, and $5/48+7/48=1/4$. The method avoids treating the original $f_1$ as the conditional $p_1$.

Worked Example: Compute a Cumulative Probability and Check Directly

A hypothetical positive count is obtained by conditioning a negative binomial count with $r=3$ and $\theta=0.5$ on being nonzero. Calculate $P(N\leq3\mid N>0)$ and the next probability $p_4$.

Method. First, $q=0.5$, $f_0=(0.5)^3=1/8$, and $1-f_0=7/8$. The original probability at one is $$ f_1=\binom{3}{1}(0.5)^3(0.5)=\frac{3}{16}, \qquad p_1=\frac{3/16}{7/8}=\frac{3}{14}. $$ The parameters are $a=0.5$ and $b=(3-1)(0.5)=1$. Thus $$ p_2=\left(\frac12+\frac12\right)\frac{3}{14}=\frac{3}{14}, \qquad p_3=\left(\frac12+\frac13\right)\frac{3}{14}=\frac{5}{28}. $$ Consequently, $$ P(N\leq3\mid N>0)=p_1+p_2+p_3 =\frac{3}{14}+\frac{3}{14}+\frac{5}{28} =\frac{17}{28}. $$

For the next probability, $p_4=(1/2+1/4)(5/28)=15/112$. Check directly: the original mass at four is $$ f_4=\binom{6}{4}(0.5)^3(0.5)^4=\frac{15}{128}, \qquad \frac{f_4}{1-f_0}=\frac{15/128}{7/8}=\frac{15}{112}. $$ The direct calculation agrees with the recursion. This check also confirms that the conditioning denominator is $7/8$, not the original probability of a positive count left unaccounted for.

Worked Example: Use the Geometric Special Case for a Tail

A hypothetical count has a negative binomial distribution with $r=1$ and success probability $\theta=0.25$, and only positive counts are retained. Find $p_1,p_2,p_3,p_4$ and $P(N\geq3\mid N>0)$.

Method. For $r=1$, the negative binomial distribution is geometric. Here $q=0.75$, $f_0=0.25$, and $1-f_0=0.75$. Since $f_1=\theta q=(0.25)(0.75)=0.1875$, $$ p_1=\frac{0.1875}{0.75}=0.25. $$ The Panjer parameters are $a=q=0.75$ and $b=(r-1)q=0$, so for every $n\geq2$, $p_n=0.75p_{n-1}$. Therefore $$ p_2=0.75(0.25)=0.1875,\quad p_3=0.75(0.1875)=0.140625,\quad p_4=0.75(0.140625)=0.10546875. $$

In the untruncated geometric model, $P(N\geq3)=q^3=(0.75)^3=0.421875$. Dividing by $P(N>0)=0.75$ gives $$ P(N\geq3\mid N>0)=\frac{0.421875}{0.75}=0.5625. $$ As a second check, the conditional tail beginning at three is a geometric series: $$ p_3+p_4+\cdots=\frac{p_3}{1-0.75} =\frac{0.140625}{0.25}=0.5625. $$ The two methods agree. The geometric case is especially simple because $b=0$, so every positive-count probability after the boundary is obtained by multiplying by the same factor $q$.

Common Boundary Mistakes

A common error is to use the original probability $f_1$ as the zero-truncated boundary $p_1$. In the first example, that incorrect calculation would set $p_1=0.25$. But the positive probabilities must sum to one after zero is removed, so the correct value is $$ p_1=\frac{f_1}{1-f_0}=\frac{0.25}{0.75}=\frac{1}{3}, $$ not $0.25$.

Another error is to apply the recursion at $n=1$ using a truncated $p_0=0$. That would incorrectly give $p_1=(a+b)p_0=0$. Zero truncation changes the support and uses a separately calculated boundary $p_1$; the recursion begins at $n=2$. The boundary and the recursion together define the probability sequence.

Specify the parameterization.
Write $r$, $\theta$, and $q=1-\theta$, with $f_0=\theta^r$.

Calculate the boundary.
Use $p_1=f_1/(1-f_0)=r\theta^r q/(1-\theta^r)$.

Generate further probabilities.
Set $a=q$, $b=(r-1)q$, then calculate $p_n=(a+b/n)p_{n-1}$ for $n\geq2$.

Check the result.
When useful, compare a generated probability or tail with the original negative binomial probability divided by $1-\theta^r$.

The central practical distinction is between the boundary and the ratios. Truncation changes the boundary because the original positive probabilities must be renormalized. It preserves the ratios among positive counts because that common renormalizing factor cancels. Keeping these two facts separate makes the $(a,b,1)$ representation straightforward to use and helps prevent off-by-one or missing-normalization errors.

Check Your Understanding

For each problem, condition the stated negative binomial count on being positive and use the $(a,b,1)$ recursion where appropriate.

A count has $r=2$ and success probability $\theta=0.4$. Find the truncated probabilities $p_1$ and $p_2$. (Answer: $p_1=16/70\approx0.2286,\ p_2=0.2057$)
A count has $r=4$ and $\theta=0.5$. Find $p_1,p_2,p_3$ and $P(N\leq3\mid N>0)$. (Answer: $p_1=2/15,\ p_2=1/6,\ p_3=1/6,\ P(N\leq3\mid N>0)=7/15$)
A geometric count has $\theta=0.2$. Find the zero-truncated probability $P(N\geq4\mid N>0)$. (Answer: $0.5120$)
A count has $r=2$ and $\theta=0.25$. Find $p_1,p_2,p_3$ using the boundary and recursion. (Answer: $p_1=0.1000,\ p_2=0.1125,\ p_3=0.1125$)
A count has $r=3$ and $\theta=0.4$. Find the truncated boundary $p_1$, the recursion parameters $a,b$, and $p_2$. (Answer: $p_1=8/65,\ a=0.6,\ b=1.2,\ p_2=48/325\approx0.1477$)

What You'll Learn