Negative Binomial Probability Recursion

Why Use a Probability Recursion?

A negative binomial probability can be calculated directly from its probability mass function, but repeated factorials and binomial coefficients are easy to mishandle. A recursion replaces each calculation with a multiplication of the previous probability by a simple ratio. This is especially useful when a problem asks for several adjacent point probabilities or a cumulative probability over a range of counts.

The parameterization must be stated before using the recursion. Here, the count is the number of failures observed before the $r$th success. Other conventions count total trials or use a different parameter as the probability parameter; their formulas must not be substituted without translating the definition.

Definition: Suppose independent Bernoulli trials are repeated, each with success probability $p$, where $0<p<1$. Let $N$ be the number of failures before the $r$th success, where $r$ is a positive integer. Write $q=1-p$. Then $N$ has a negative binomial distribution with probability mass function $$ P(N=n)=\binom{n+r-1}{n}p^r q^n,\qquad n=0,1,2,\ldots. $$ The support starts at zero because the $r$ successes can occur with no failures.

Deriving the Consecutive-Probability Ratio

The recursion follows by dividing the probability at count $n+1$ by the probability at count $n$. This comparison is valid for every $n\geq 0$, because each point probability is positive when $0<p<1$.

Write adjacent probabilities.
Use the stated probability mass function at $n+1$ and at $n$, with the same $r$, $p$, and $q$.

Form the ratio.
The factors $p^r$ cancel, and the powers of $q$ leave one factor of $q$.

Simplify the combinatorial terms.
The ratio of binomial coefficients is $\binom{n+r}{n+1}/\binom{n+r-1}{n}=(n+r)/(n+1)$.

\frac{P(N=n+1)}{P(N=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.

Thus, beginning with $P(N=0)=p^r$, the probabilities can be generated successively:

Theorem: For the negative binomial definition above, $$ P(N=0)=p^r,\qquad P(N=n+1)=P(N=n)\frac{n+r}{n+1}(1-p),\quad n\geq 0. $$ The factor for moving from $n$ to $n+1$ depends on the current index $n$; it is not a constant multiplier.

The starting probability has a direct interpretation: $N=0$ means that the first $r$ trials are all successes, which has probability $p^r$. The ratio then accounts for how many sequences of successes and failures produce each next count. In particular, the first step is $P(N=1)/P(N=0)=rq$, not simply $q$.

Worked Examples

Worked Example: Building Point Masses and a Cumulative Probability

A hypothetical claim-monitoring process is modeled by a negative binomial count $N$: the number of nonclaim periods before the third claim period, with independent periods and claim-period probability $p=0.6$. Find $P(N=2)$ and $P(N\leq 3)$.

Method and parameters. This is the failures-before-$r$-successes parameterization, with $r=3$, $p=0.6$, and $q=1-p=0.4$. Start at zero and apply the recursion.

Start at zero. $P(N=0)=p^r=0.6^3=0.216.$

Calculate $P(N=1)$. At $n=0$, the ratio is $(0+3)/(0+1)\times 0.4=1.2$. Therefore, $P(N=1)=0.216\times 1.2=0.2592.$

Calculate $P(N=2)$. At $n=1$, the ratio is $(1+3)/(1+1)\times 0.4=0.8$. Therefore, $P(N=2)=0.2592\times 0.8=0.20736.$

Continue through $N=3$. At $n=2$, the ratio is $(2+3)/(2+1)\times 0.4=2/3$. Thus, $P(N=3)=0.20736\times 2/3=0.13824.$

Combine the requested probabilities. $P(N\leq 3)=0.216+0.2592+0.20736+0.13824=0.8208.$ Hence $P(N=2)=0.20736$, and $P(N\leq3)=0.8208$. Both results are already exact to the displayed decimal places. As a check, the cumulative probability is between zero and one. The method avoids recalculating each mass with a separate binomial coefficient.

Worked Example: A Tail Probability and a Specified Point Mass

For a hypothetical portfolio count model, let $N$ be the number of nonclaim periods before the fourth claim period. Assume independent periods with claim-period probability $p=0.5$. Find $P(N\geq3)$ and $P(N=4)$.

Method and parameters. Use $r=4$, $p=0.5$, and $q=0.5$. Generate the masses starting at $P(N=0)=0.5^4=0.0625$. For the tail, it is efficient to sum the complementary probabilities through 2.

At $n=0$, the ratio is $4/1\times0.5=2$, so $P(N=1)=0.0625\times2=0.125.$

At $n=1$, the ratio is $5/2\times0.5=1.25$, so $P(N=2)=0.125\times1.25=0.15625.$

The complementary cumulative probability is $P(N\leq2)=0.0625+0.125+0.15625=0.34375.$ Therefore, $P(N\geq3)=1-0.34375=0.65625.$

For the point mass at 4, continue the recursion. At $n=2$, the ratio is $6/3\times0.5=1$, giving $P(N=3)=0.15625$. At $n=3$, the ratio is $7/4\times0.5=0.875$, giving $P(N=4)=0.15625\times0.875=0.13671875.$

Thus $P(N\geq3)=0.65625$ and $P(N=4)=0.13671875$. The check is that the tail includes $N=3$ as well as all larger counts; subtracting only through 3 would instead give $P(N\geq4)$.

Worked Example: A Short Upper Tail from Recursive Masses

A hypothetical frequency model uses $N$, the number of failure periods before the second success period. Period outcomes are independent, with success probability $p=0.7$. Find $P(N\leq4)$ and $P(N\geq5)$, reporting probabilities to four decimal places.

Method and parameters. Here $r=2$, $p=0.7$, and $q=0.3$. Begin with $P(N=0)=0.7^2=0.49$.

For $n=0$, the ratio is $2/1\times0.3=0.6$, so $P(N=1)=0.49\times0.6=0.294.$

For $n=1$, the ratio is $3/2\times0.3=0.45$, so $P(N=2)=0.294\times0.45=0.1323.$

For $n=2$, the ratio is $4/3\times0.3=0.4$, so $P(N=3)=0.1323\times0.4=0.05292.$

For $n=3$, the ratio is $5/4\times0.3=0.375$, so $P(N=4)=0.05292\times0.375=0.019845.$

Adding the masses gives $P(N\leq4)=0.49+0.294+0.1323+0.05292+0.019845=0.989065.$ The complementary event is $N\geq5$, so $P(N\geq5)=1-0.989065=0.010935.$

Rounded to four decimal places, $P(N\leq4)=0.9891$ and $P(N\geq5)=0.0109$. Their sum is 1 before rounding, which checks the complementary-event calculation. Rounding each mass early could produce a slightly different final digit, so retain the unrounded values until the end.

Using Recursion for Count Events

A point probability is one mass, while a cumulative probability requires adding the relevant masses. For integer-valued $N$, the endpoint matters: $P(N\leq k)$ includes $P(N=k)$, whereas $P(N<k)$ stops at $k-1$. Likewise, $P(N\geq k)=1-P(N\leq k-1)$. The recursion computes the masses; it does not change the event being summed.

A practical calculation table can help prevent index errors. Each row should record the count, the ratio used to move from the preceding row, and the resulting probability.

Count being computed	Ratio applied	Probability update
$n+1$	$\frac{n+r}{n+1}q$	$P(N=n+1)=P(N=n)\frac{n+r}{n+1}q$
Zero	Not applicable	$P(N=0)=p^r$
Upper tail from $k$	Compute through $k-1$	$P(N\geq k)=1-\sum_{n=0}^{k-1}P(N=n)$

The recursion is useful when a requested probability involves a run of adjacent counts, because every new mass uses the previous one. It is not a replacement for checking the model definition: if a problem defines the count as the total number of trials until the $r$th success, that variable equals $N+r$, not $N$. Its support and point-probability indices are shifted.

Common Exam Traps

Switching the parameterization. In this tutorial $p$ is the success probability and $N$ counts failures. Confirm both facts before calculating.
Starting at the wrong mass. The zero-count probability is $p^r$, since the first $r$ trials must all be successes.
Using the wrong index in the ratio. To find $P(N=n+1)$, multiply $P(N=n)$ by $(n+r)q/(n+1)$. The index $n$ refers to the mass already known.
Confusing strict and inclusive tails. $P(N\geq k)$ is one minus the sum through $k-1$, not through $k$.
Rounding at every step. Keep full precision in intermediate probabilities and round only the final result.

Check Your Understanding

For each problem, state the failures-before-successes convention and apply the recursion from $P(N=0)=p^r$. Probabilities in final answers are rounded to four decimal places when needed.

A hypothetical process has $N$ failures before the second success, with independent trials and success probability $p=0.8$. Find $P(N=2)$. (Answer: 0.0768)
Let $N$ count failures before the third success, with independent trial success probability $p=0.4$. Find $P(N\leq2)$. (Answer: 0.3174)
A model counts failures before the second success, with $p=0.5$. Find $P(N\geq4)$ by a complement. (Answer: 0.1875)
For failures before the fourth success with $p=0.7$, find $P(N\leq3)$. (Answer: 0.8740)
For failures before the third success with $p=0.25$, find $P(2\leq N\leq4)$. (Answer: 0.1928)

Count being computed	Ratio applied	Probability update
\(n+1\)	\(\frac{n+r}{n+1}q\)	\(P(N=n+1)=P(N=n)\frac{n+r}{n+1}q\)
Zero	Not applicable	\(P(N=0)=p^r\)
Upper tail from \(k\)	Compute through \(k-1\)	\(P(N\geq k)=1-\sum_{n=0}^{k-1}P(N=n)\)

What You'll Learn