Deriving The Negative Binomial Mean

Why the Negative Binomial Mean Has a Simple Form

A negative binomial failure count asks how many unsuccessful trials occur before a fixed number of successes has been reached. In a claim-frequency setting, a hypothetical process might count unsuccessful reviews before a specified number of successful reviews. The main calculation is simple once the count is viewed in stages: wait for the first success, then the next, and continue until the $r$th success.

This tutorial derives the expected count from those successive waits. The distinction between failures and total trials is essential: the total includes the $r$ successes as well as the failures.

Definition: Suppose trials are independent, each trial has the same probability $p$ of success, and $0<p\leq 1$. Write $q=1-p$. The negative binomial failure count $N$ is the number of failures observed before the $r$th success, where $r$ is a fixed positive integer. Its possible values are $0,1,2,\ldots$. The model assumes that the success probability does not change from trial to trial and that the outcomes of distinct trials are independent.

The number of successes is fixed at $r$, but the number of failures is not fixed. If the process stops at the $r$th success, its total number of trials is $T=N+r$. Consequently, once the mean of $N$ is known, the mean total number of trials follows by adding the $r$ successes.

Split the Count Into Successive Waiting Counts

For each $j=1,\ldots,r$, let $G_j$ be the number of failures after the $(j-1)$st success and before the $j$th success. For $j=1$, this means the failures before the first success. Each $G_j$ counts only failures in its own stage; it does not include the success that ends that stage.

The total failure count is exactly the sum of the stage counts:

N=G_1+G_2+\cdots+G_r.

Under independent trials with the same success probability $p$, each stage has the same waiting-count distribution. For any one stage, $G_j=k$ means that the next $k$ trials fail and the following trial succeeds. Thus, for $k=0,1,2,\ldots$,

$$ P(G_j=k)=q^k p. $$

The stages are independent: after any success, the outcomes of subsequent trials are independent of the outcomes already observed, and each subsequent trial still has success probability $p$. The mean of the total can therefore be obtained by adding the $r$ stage means. In fact, linearity of expectation does not require independence, but the stated model does make the stage counts independent and identically distributed.

Deriving the Mean of One Waiting Count

There are two useful ways to see the mean for one stage. Using its probability function, the mean is the sum of each possible failure count multiplied by its probability:

E[G_j]=\sum_{k=0}^{\infty} kq^k p.

For $0\leq q<1$, the geometric series is $\sum_{k=0}^{\infty}q^k=1/(1-q)$. Differentiating with respect to $q$ gives $\sum_{k=1}^{\infty}kq^{k-1}=1/(1-q)^2$. Multiplying by $q$ and then by $p$ yields:

E[G_j] =p\sum_{k=0}^{\infty}kq^k =p\frac{q}{(1-q)^2} =p\frac{q}{p^2} =\frac{q}{p}.

Here $1-q=p$. The result also has a useful tail-sum interpretation: $G_j\geq m$ exactly when the first $m$ trials in that stage all fail, which has probability $q^m$. For a nonnegative integer-valued count, its mean is the sum of its positive-integer tail probabilities, so $E[G_j]=\sum_{m=1}^{\infty}q^m=q/(1-q)=q/p$. Both arguments give the expected number of failures before one success.

Theorem: Under independent, identically distributed Bernoulli trials with success probability $p>0$, the negative binomial failure count before $r$ successes has mean $$ E[N]=r\frac{q}{p}=r\frac{1-p}{p}. $$ The expected total number of trials through the $r$th success is $$ E[T]=E[N+r]=r\frac{1-p}{p}+r=\frac{r}{p}. $$

The first formula counts failures only. The second counts all trials, including the $r$ successes. The total-trial result can also be seen by adding the $r$ expected stage lengths: each stage has mean $E[G_j+1]=q/p+1=1/p$.

Worked Applications

Worked Example: Expected Failures and Trials for a Target Count

A hypothetical quality-control process records a success with probability $p=0.75$ on each independent inspection. The process stops at the fourth success. Find the expected number of failures and the expected total number of inspections. Also find the expected failures before the first two successes.

Method and assumptions. This is a negative binomial failure-count model with fixed target $r=4$, constant success probability $p=0.75$, and independent inspections. The failure probability is $q=1-p=0.25$.

Expected failures through the fourth success. Apply $E[N]=rq/p$:

E[N]=4\left(\frac{0.25}{0.75}\right) =4\left(\frac{1}{3}\right) =\frac{4}{3}.

Expected total inspections. Add the four successes to the failure count, or use $E[T]=r/p$:

E[T]=\frac{4}{0.75}=\frac{16}{3} =4+\frac{4}{3}.

Expected failures before the first two successes. There are two waiting stages, so the expected failures in those stages are:

2\left(\frac{0.25}{0.75}\right)=\frac{2}{3}.

Check. The expected failures in the first two stages, $2/3$, plus the expected failures in the remaining two stages, also $2/3$, equal $4/3$. The expected total $16/3$ is exactly four successes plus $4/3$ failures. A common mistake would be to report $4/3$ as the expected total inspections; that omits the four successful inspections.

Worked Example: Recovering the Success Probability From the Mean

A hypothetical screening process continues until the sixth successful result. Each independent screening has the same success probability $p$, and the expected number of failures before the sixth success is 18. Find $p$, the failure probability $q$, and the expected total number of screenings.

Method and assumptions. The fixed target is $r=6$, and the process follows the independent, constant-probability model. Use the mean equation and solve for the unknown success probability rather than treating the expected failure count as a total-trial count.

From $E[N]=r(1-p)/p$, the given mean implies:

18=6\frac{1-p}{p}.

Divide both sides by 6:

3=\frac{1-p}{p}.

Multiply by $p$, then solve:

3p=1-p,\qquad 4p=1,\qquad p=0.25.

Therefore $q=1-p=0.75$. The expected total number of screenings is:

E[T]=\frac{r}{p}=\frac{6}{0.25}=24.

Check. The expected total is the six successes plus the 18 failures, giving $6+18=24$, consistent with $r/p=24$. A common mistake is to set $18=6/p$; that uses the total-trial mean formula for a failure-only count and gives the wrong parameter.

Worked Example: Updating the Target After Successes Have Occurred

A hypothetical audit has independent trials with success probability $p=0.40$. The audit ends after five successes. Find the expected total failures from the start, and then find the expected additional failures if two successes have already occurred and the audit continues until the fifth success.

Method and assumptions. The success probability remains $0.40$ on every trial, including after the first two successes. Thus $q=0.60$. From the start, the target is five successes; after two successes have been recorded, only three more successes are needed.

From the start. With $r=5$, the expected failure count is:

E[N]=5\frac{0.60}{0.40} =5(1.5)=7.5.

After two successes. The completed successes do not need to be earned again. The remaining target is $5-2=3$, so the expected additional failures are:

E[N_{\text{additional}}]=3\frac{0.60}{0.40} =3(1.5)=4.5.

Check. The first two waiting stages have expected failures $2(1.5)=3$, while the final three stages have expected failures $3(1.5)=4.5$. Their sum is $3+4.5=7.5$, the total expected failures from the start. A common mistake is to use $r=5$ again after two successes have occurred, which counts the already-completed stages twice.

Conditions, Interpretation, and Exam Traps

The mean depends on the model, not just on the phrase “until a target is reached.” The calculation $r(1-p)/p$ requires a fixed number $r$ of successes, independent trials, and the same positive success probability $p$ on every trial. If the probability changes over time, or trial outcomes are dependent, the stage means need not all equal $q/p$, and this formula cannot be applied without further analysis.

Also distinguish this model from one that fixes the number of trials in advance. Here the number of successes is fixed and the number of failures varies; in a fixed-trial binomial setting, the number of trials is fixed instead. A related unit check helps prevent errors: $\frac{q}{p}$ is an expected number of failures per success, so multiplying by $r$ gives failures before $r$ successes. The expression $r/p$ gives total trials, not failures.

Confirm the stopping rule.
Check that the process stops at a fixed positive number $r$ of successes, rather than after a fixed number of trials.

Confirm the trial assumptions.
Verify independence and a constant success probability $p>0$, then calculate $q=1-p$.

Match the requested count.
For failures use $rq/p$. For total trials use $r/p$, or add $r$ to the expected failures.

Adjust for completed successes.
If some successes have already occurred, use only the remaining number of successes as the target for additional failures.

The decomposition into successive waits is what makes the result transparent: each required success contributes an expected $q/p$ failures, so $r$ such stages contribute $r q/p$. This reasoning is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus and is especially useful when a problem changes the requested quantity from failures to total trials or asks about a process partway through.

Check Your Understanding

For each situation, identify the relevant number of remaining successes and distinguish failures from total trials.

A hypothetical process has independent trials with success probability $0.60$ and stops at the third success. Find the expected number of failures. (Answer: 2 failures)
Independent trials have success probability $0.80$, and the process stops at the eighth success. Find the expected failures and expected total trials. (Answer: 2 expected failures; 10 expected total trials)
A process stops at the fifth success and has expected failure count 15. Under the constant-probability independent-trial model, find $p$ and the expected total trials. (Answer: $p=0.25$; 20 expected total trials)
Each independent review succeeds with probability $0.40$. A process ends at the sixth success. Find the expected failures from the start and, after two successes have occurred, the expected additional failures. (Answer: 9 from the start; 6 additional failures)
A hypothetical sequence has independent trials with success probability $0.20$ and stops at the fourth success. Find the expected failures before the first success, before the fourth success, and the expected total trials. (Answer: 4 before the first success; 16 before the fourth success; 20 total trials)

What You'll Learn