Negative Binomial Probability Calculations

Why Negative Binomial Probability Calculations Need Care

A claim count can be modeled with a negative binomial distribution when variation in counts is greater than a simple Poisson model would represent. Once a negative binomial model and its parameters have been specified, exam calculations usually require an exact count probability, a cumulative probability, or a tail probability. The arithmetic is manageable; the main risks are using the wrong parameterization and including or excluding the boundary count incorrectly.

This tutorial focuses on calculating probabilities for a specified negative binomial model. It does not assume that the count must arise from literal repeated trials: the “failures before successes” definition gives a precise probability function, while a fitted claim-count model may use the same distribution as a statistical model for counts.

Definition: Let $N$ be the number of failures observed before the $r$th success in independent Bernoulli trials, each with success probability $p$, where $r$ is a positive integer and $0<p<1$. Then $N$ has a negative binomial distribution with parameters $r$ and $p$. Its support is $n=0,1,2,\ldots$, and $$ P(N=n)=\binom{n+r-1}{n}p^r(1-p)^n. $$ Here $p$ is the probability of a success, not the probability of a failure.

Write $q=1-p$ when it makes calculations shorter. The probability of $n$ failures is $q^n$, and the probability of the required $r$ successes is $p^r$. The binomial coefficient counts the possible arrangements of the first $n+r-1$ trials, which must contain $n$ failures and $r-1$ successes. The last trial must be a success, so that the $r$th success occurs at the end. This gives $\binom{n+r-1}{n}$ arrangements, each with probability $p^r q^n$.

For claim-frequency applications, $N$ can simply be interpreted as the modeled number of claims in the stated observation period. A count of zero is possible, and there is no upper bound on the support. Always use the parameterization actually given in the problem rather than relying on the word “negative binomial” alone.

Building Probabilities Efficiently

The probability function can be evaluated separately for each count. For a sequence of counts, a recurrence avoids recalculating the binomial coefficient from scratch. Let $P_n=P(N=n)$. Dividing the probability at $n+1$ by the probability at $n$ gives

\frac{P_{n+1}}{P_n} = \frac{\binom{n+r}{n+1}p^r q^{n+1}} {\binom{n+r-1}{n}p^r q^n} = \frac{n+r}{n+1}q.

The ratio of the two binomial coefficients is $(n+r)/(n+1)$, while the powers of $p$ cancel and the remaining power of $q$ contributes one factor of $q$. Since $P_0=p^r$, the recurrence is

P_0=p^r,\qquad P_{n+1}=P_n\frac{n+r}{n+1}q.

This recurrence is useful for exact point probabilities and for summing a run of probabilities. It does not change the meaning of $N$: the subscript $n$ is still the number of failures, or the modeled claim count, not the total number of trials.

Tail rule: For an integer $k\geq 1$, $P(N\geq k)=1-P(N\leq k-1)$. For an integer $k\geq 0$, $P(N>k)=1-P(N\leq k)$. These events differ by the boundary value $N=k$.

A cumulative probability is a sum over the requested support values: $P(N\leq k)=\sum_{n=0}^{k}P_n$. An upper tail can be found either by summing all probabilities from its lower bound onward or by subtracting the complementary cumulative probability from one. The complement is often shorter, but only if its endpoint is chosen correctly.

Worked Probability Calculations

Worked Example: A Point Probability and a Claim-Count Tail

A hypothetical insurer models the number $N$ of claims on a specified group of policies during one year using a negative binomial distribution. The parameters are $r=3$ and $p=0.60$, with $N$ defined as the number of failures before the third success. Calculate (a) the probability of exactly two claims and (b) the probability of at least three claims. Give probabilities to four decimal places.

Method. Use the failures-before-$r$-successes probability function, and calculate the lower cumulative probability for the tail complement. The common mistake avoided here is treating $p$ as the failure probability.

First, $q=1-p=1-0.60=0.40$. For exactly two claims,

P(N=2) =\binom{2+3-1}{2}(0.60)^3(0.40)^2 =\binom{4}{2}(0.216)(0.16) =6(0.03456) =0.20736.

Thus $P(N=2)=0.2074$ to four decimal places.

For the tail, $P(N\geq3)=1-P(N\leq2)$. Calculate the probabilities at 0, 1, and 2. The recurrence starts at

$$ P_0=(0.60)^3=0.216. $$

Then

P_1=0.216\frac{3}{1}(0.40)=0.2592, \qquad P_2=0.2592\frac{4}{2}(0.40)=0.20736.

Therefore $P(N\leq2)=0.216+0.2592+0.20736=0.68256$, and

P(N\geq3)=1-0.68256=0.31744\approx0.3174.

The results are consistent: the event $N=2$ is included in $N\leq2$, but not in $N\geq3$. The answer is a probability of a claim count, not a probability of a trial total.

Worked Example: An Interval and an Upper Tail

For a hypothetical book of policies, let $N$ have a negative binomial distribution with $r=2$ and $p=0.50$, where $N$ is the modeled claim count. Calculate $P(N\geq4)$ and $P(2\leq N\leq5)$. Give each result to four decimal places.

Method. Generate the probabilities from $P_0$ using the recurrence. For the upper tail, subtract the cumulative probability through 3, not through 4. This avoids excluding the boundary count 4 from the tail.

Here $q=0.50$, and $P_0=p^r=(0.50)^2=0.25$. Apply the recurrence in order:

\begin{aligned} P_1&=0.25\frac{2}{1}(0.50)=0.25,\\ P_2&=0.25\frac{3}{2}(0.50)=0.1875,\\ P_3&=0.1875\frac{4}{3}(0.50)=0.125,\\ P_4&=0.125\frac{5}{4}(0.50)=0.078125,\\ P_5&=0.078125\frac{6}{5}(0.50)=0.046875. \end{aligned}

For the first requested probability,

P(N\geq4) =1-(P_0+P_1+P_2+P_3) =1-(0.25+0.25+0.1875+0.125) =0.1875.

For the interval, add both endpoints and every count between them:

P(2\leq N\leq5) =P_2+P_3+P_4+P_5 =0.1875+0.125+0.078125+0.046875 =0.4375.

Both results are already exact to four decimal places. As a check on the first result, the probability excluded from the tail is precisely the sum for counts 0 through 3; the count 4 remains included.

Converting a Mean-and-Shape Specification

A problem may give the negative binomial model through $r$ and its mean $\mu=E[N]$, rather than through $r$ and $p$. For the failures-before-$r$-successes convention, the mean is $rq/p$. Therefore $p=r/(r+\mu)$ and $q=\mu/(r+\mu)$. Convert first, then use the probability function. This conversion is a parameterization step; it does not alter the meaning or support of the count.

Worked Example: Converting the Mean Before Finding a Tail

A hypothetical portfolio’s claim count $N$ is modeled as negative binomial with $r=2$ and mean $\mu=4$. Under the convention that $N$ counts failures before the second success, calculate $P(N=3)$ and $P(N\geq3)$. Give probabilities to four decimal places.

Method. Convert the mean-and-shape specification to $p$ and $q$, then calculate the point probability and cumulative complement. The common mistake avoided is substituting the mean directly for $p$.

From $\mu=rq/p$ and $p+q=1$,

\mu p=r(1-p) \quad\Longrightarrow\quad p(\mu+r)=r \quad\Longrightarrow\quad p=\frac{r}{r+\mu}=\frac{2}{2+4}=\frac13.

Thus $q=1-p=2/3$. The exact probability of three claims is

P(N=3) =\binom{3+2-1}{3}\left(\frac13\right)^2\left(\frac23\right)^3 =4\left(\frac19\right)\left(\frac{8}{27}\right) =\frac{32}{243} \approx0.1317.

For the tail, first calculate the probabilities at 0, 1, and 2:

P_0=\frac19,\qquad P_1=\frac19\frac{2}{1}\frac23=\frac{4}{27},\qquad P_2=\frac{4}{27}\frac{3}{2}\frac23=\frac{4}{27}.

Hence $P(N\leq2)=1/9+4/27+4/27=11/27$, so

P(N\geq3)=1-\frac{11}{27}=\frac{16}{27}\approx0.5926.

The point probability $0.1317$ is less than the tail probability $0.5926$, as it must be because $N=3$ is only one of the outcomes included in $N\geq3$.

Exam Traps and Practical Checks

A reliable calculation begins by writing down what the random variable counts and which parameter convention is in use. For this tutorial’s convention, $N$ counts failures before the $r$th success, $P(N=0)=p^r$, and the failure probability is $q=1-p$. These facts provide immediate checks before any longer calculation.

Swapping success and failure probabilities. The term $p^r$ represents the $r$ successes; $q^n$ represents the $n$ failures. Replacing $p$ by $q$ in both places changes the model.
Confusing failures with total trials. If a variable instead counts the total number of trials through the $r$th success, it equals $N+r$. The support and point-probability index shift accordingly.
Misreading a strict or inclusive tail. $P(N\geq k)$ excludes counts below $k$, so its complement ends at $k-1$. By contrast, $P(N>k)$ has complement ending at $k$.
Summing the wrong interval. For $P(a\leq N\leq b)$, include both $P_a$ and $P_b$. There are $b-a+1$ integer outcomes in the interval.
Rounding too early. Keep intermediate probabilities at full precision, sum them, and round the requested final probability to four decimal places.

For a final reasonableness check, each point probability must be between zero and one, cumulative probabilities cannot decrease as their upper endpoint increases, and $P(N\geq k)$ cannot exceed $P(N\geq k-1)$. These checks can catch an arithmetic or boundary error without replacing the calculation.

Key takeaway: Identify the count convention, establish $p$ and $q$, and use either the probability function or the recurrence. For a tail, write the event and its complement explicitly before summing.

Exact negative binomial calculations are useful whenever a model specifies the parameters but asks about a particular claim count or range of counts. Accurate work depends as much on translating the wording into the correct event as on evaluating the formula. The methods here are aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, use the failures-before-$r$-successes convention unless the problem asks you to convert a mean-and-shape specification. Round probabilities to four decimal places.

A hypothetical policy group has $N\sim\text{negative binomial}(r=2,p=0.40)$. Find $P(N=1)$ and $P(N\leq1)$. (Answer: $0.1920$ and $0.3520$)
A hypothetical claim count has $r=3$ and $p=0.50$. Find $P(N=2)$ and $P(N\geq2)$. (Answer: $0.1875$ and $0.6875$)
A negative binomial claim count has $r=2$ and mean $\mu=3$. Convert to $p$, then find $P(N=4)$. (Answer: $p=0.4000$; $P(N=4)=0.1037$)
For a hypothetical portfolio, $r=1$ and $p=0.25$. Find the probability of between three and five claims, inclusive. (Answer: $0.2439$)
A model has $r=2$ and $p=0.50$. Find the probability of more than three claims, taking care with the strict inequality. (Answer: $0.1875$)

What You'll Learn