Converting Negative Binomial Parameters

Why Parameter Conversions Matter

A negative binomial frequency model can be written using a size parameter and a success probability, or using its mean and a dispersion parameter. Both describe the same family of count distributions, but the parameters answer different questions. The first form is convenient for specifying the probability distribution; the second makes the expected count and extra variation more visible.

The most important convention to keep straight is the meaning of “success.” Here, $N$ counts failures before the $r$-th success. Thus $p$ is the probability of success on each trial, and $1-p$ is the probability of failure. If a different convention is used, the formulas for the mean and parameter conversions change.

Definition: Under the failures-before-$r$-successes convention, a negative binomial random variable $N$ has size $r>0$ and success probability $0<p<1$. Its support is $0,1,2,\ldots$, and its mean and variance are $$ E[N]=\frac{r(1-p)}{p}, \qquad \operatorname{Var}(N)=\frac{r(1-p)}{p^2}. $$ For integer $r$, this has the literal interpretation of counting failures before $r$ successes. The same probability model is defined for noninteger $r>0$, though that trial-count interpretation no longer applies.

The alternative mean-dispersion form uses $\mu=E[N]$ and $\alpha>0$, with variance $\mu+\alpha\mu^2$. In this form, $\alpha$ controls the amount of variance above the Poisson variance $\mu$. It is a parameter of the model, not the failure probability.

Definition: In the mean-dispersion parameterization, $\mu>0$ is the mean and $\alpha>0$ is the dispersion parameter. The negative binomial variance is $$ \operatorname{Var}(N)=\mu+\alpha\mu^2. $$ The equivalent size-success-probability parameters are $r=1/\alpha$ and $p=1/(1+\alpha\mu)$.

Deriving the Conversion Formulas

The conversion follows by equating the mean and variance in the two parameterizations. This also gives a useful check: a proposed set of parameters should reproduce both moments, not just the mean.

Start with the size-probability moments.
Write $\mu=r(1-p)/p$ and $V=\operatorname{Var}(N)=r(1-p)/p^2$, using the failures-before-successes convention.

Find the success probability.
Dividing the variance by the mean cancels $r(1-p)$, so $V/\mu=1/p$. Therefore $p=\mu/V$.

Find size and dispersion.
Since $1-p=(V-\mu)/V$, substitution into the mean gives $r=\mu^2/(V-\mu)$. Its reciprocal is $\alpha=(V-\mu)/\mu^2$.

Convert in the other direction.
From $r=1/\alpha$ and $\mu=r(1-p)/p$, solve for $p$: $\mu\alpha=(1-p)/p$, so $p=1/(1+\alpha\mu)$.

The complete conversions are

\begin{aligned} \mu&=\frac{r(1-p)}{p}, & \alpha&=\frac{1}{r}, & p&=\frac{1}{1+\alpha\mu}, \\[4pt] r&=\frac{1}{\alpha}, & \mu&=\frac{r(1-p)}{p}, & \operatorname{Var}(N)&=\mu+\alpha\mu^2. \end{aligned}

If the mean $\mu$ and variance $V$ are given instead, the moment-based conversion is

p=\frac{\mu}{V}, \qquad r=\frac{\mu^2}{V-\mu}, \qquad \alpha=\frac{V-\mu}{\mu^2}.

These expressions require $\mu>0$ and $V>\mu$ for a finite negative binomial with $r>0$ and $0<p<1$. If $V=\mu$, the moments are at the Poisson boundary: no finite positive $r$ and $\alpha$ represent that case. If $V<\mu$, the moments are underdispersed and cannot be matched by this negative binomial parameterization.

Worked Example: Converting a Size and Success Probability

A hypothetical portfolio’s claim count for a fixed observation period is modeled as negative binomial, counting failures before the sixth success, with success probability $p=0.75$. Convert to the mean-dispersion form and verify the variance.

Method. Use the failures-before-successes formulas, then set $\alpha=1/r$. The common mistake to avoid is treating $1-p$ as the success probability.

Here $r=6$, $p=0.75=3/4$, and the failure probability is $1-p=0.25=1/4$. The mean is

\mu=\frac{6(1/4)}{3/4} =6\left(\frac{1}{3}\right) =2.

The dispersion parameter is

\alpha=\frac{1}{r}=\frac{1}{6}.

The variance from the size-probability form is

V=\frac{6(1/4)}{(3/4)^2} =\frac{3/2}{9/16} =\frac{3}{2}\cdot\frac{16}{9} =\frac{8}{3}.

Check using the mean-dispersion form:

\mu+\alpha\mu^2 =2+\frac{1}{6}(2^2) =2+\frac{2}{3} =\frac{8}{3}.

The two forms agree: $\mu=2$, $\alpha=1/6$, and $V=8/3$. The check confirms that the converted dispersion parameter reproduces the variance, not just the mean.

Converting from Mean and Dispersion

When a model is specified by $\mu$ and $\alpha$, convert to $r$ and $p$ directly. First take the reciprocal of $\alpha$ to find the size. Then calculate $p=1/(1+\alpha\mu)$. Since $0<\alpha\mu$, this gives a success probability strictly between zero and one.

Worked Example: Converting Mean and Dispersion Parameters

For a hypothetical equipment-protection plan, the claim count in the stated period has mean $\mu=12$ and dispersion $\alpha=1/8$. Find its size $r$, success probability $p$, and variance, then check the moments using the size-probability form.

Method. Use $r=1/\alpha$ and $p=1/(1+\alpha\mu)$. Do not substitute $\mu$ for $r$: the size is determined by dispersion.

First,

r=\frac{1}{\alpha}=\frac{1}{1/8}=8.

Next, $\alpha\mu=(1/8)(12)=12/8=3/2$, so

p=\frac{1}{1+3/2} =\frac{1}{5/2} =\frac{2}{5} =0.40.

The variance in mean-dispersion form is

V=12+\frac{1}{8}(12^2) =12+\frac{144}{8} =12+18 =30.

Verify the mean from $r=8$ and $p=2/5$. The failure probability is $1-p=3/5$, hence

E[N]=\frac{8(3/5)}{2/5} =8\cdot\frac{3}{2} =12.

The variance from the size-probability form is

V=\frac{8(3/5)}{(2/5)^2} =\frac{24}{5}\cdot\frac{25}{4} =30.

Both moments check: the converted parameters give mean $12$ and variance $30$, as required.

Recovering Parameters from Moments

Sometimes the mean and variance are supplied, rather than either parameterization. The moment formulas determine the corresponding negative binomial parameters when the variance exceeds the mean. In an applied setting, this conversion describes the model whose moments match the supplied values; it does not by itself establish that the negative binomial is an appropriate model.

Worked Example: Matching Given Mean and Variance

A hypothetical claim-count model is to have mean $8$ and variance $20$. Find the corresponding $\alpha$, $r$, and success probability $p$, and verify both moments. State whether the parameters define a valid finite negative binomial model.

Method. Set $\mu=8$ and $V=20$ in the moment conversion formulas. The common mistake to avoid is reporting the failure probability $1-p$ as the success probability.

Since $20>8$, the variance exceeds the mean, as required. First,

\alpha=\frac{V-\mu}{\mu^2} =\frac{20-8}{8^2} =\frac{12}{64} =\frac{3}{16}.

Thus

r=\frac{1}{\alpha}=\frac{16}{3}.

The success probability is

p=\frac{\mu}{V} =\frac{8}{20} =\frac{2}{5} =0.40.

The failure probability is $1-p=3/5=0.60$, not $0.40$. Check the mean:

E[N]=\frac{(16/3)(3/5)}{2/5} =\frac{16}{3}\cdot\frac{3}{2} =8.

Check the variance:

V=\frac{(16/3)(3/5)}{(2/5)^2} =\frac{16}{5}\cdot\frac{25}{4} =20.

Therefore $\alpha=3/16$, $r=16/3$, and $p=0.40$ reproduce both stated moments. The size is positive and $p$ lies strictly between zero and one, so these are valid finite negative binomial parameters.

Common Parameterization Traps

The same symbol $p$ is not used consistently across every presentation of the negative binomial distribution. Before converting, identify whether the model counts failures before a fixed number of successes, or uses a different convention. In this tutorial, $p$ always means the success probability in the failures-before-$r$-successes convention.

For the moments $\mu=8$ and $V=20$, a tempting but incorrect answer is $p=0.50$. This does not follow from the conversion: the correct calculation is $p=\mu/V=8/20=0.40$. If the associated size $r=16/3$ is used with $p=0.50$, the implied mean is

\frac{r(1-p)}{p} =\frac{16}{3}\cdot\frac{0.50}{0.50} =\frac{16}{3},

which is not $8$. The direct check catches the inconsistency. Also remember that $1-p$ is the failure probability; substituting it where the success probability belongs reverses the conversion.

Key takeaway: Under the failures-before-successes convention, $r=1/\alpha$ and $p=1/(1+\alpha\mu)$. From moments, $p=\mu/V$, $r=\mu^2/(V-\mu)$, and $\alpha=(V-\mu)/\mu^2$. Always substitute the converted values back into both the mean and variance formulas.

Parameter conversion is useful because different model descriptions emphasize different features. Size and success probability specify the negative binomial distribution directly, while mean and dispersion make the expected count and excess variability explicit. For work aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus, state the convention, show the conversion, and verify the result rather than relying on parameter names alone.

Check Your Understanding

For each problem, use the failures-before-$r$-successes convention and check that success and failure probabilities are not interchanged.

A count has size $r=5$ and success probability $p=0.80$. Find its mean, variance, and dispersion parameter. (Answer: $\mu=1.25$, $V=1.5625$, $\alpha=0.20$.)
A negative binomial count has $\mu=15$ and $\alpha=0.20$. Find $r$, $p$, and the variance. (Answer: $r=5$, $p=0.25$, $V=60$.)
A proposed negative binomial model has mean $10$ and variance $16$. Find $\alpha$, $r$, and the success probability, and verify that the moments permit a finite model. (Answer: $\alpha=0.06$, $r=50/3$, $p=0.625$; a finite model is permitted because $16>10$.)
A count is specified to have mean $4$ and variance $4$. Can finite positive $r$ and $\alpha$ match these moments? Describe the boundary case. (Answer: No finite positive $r$ and $\alpha$; this is the Poisson boundary, with $\alpha\to0$ and $r\to\infty$.)
A proposed count model has mean $8$ and variance $6$. Can it be represented by the mean-dispersion negative binomial form with $\alpha>0$? Explain using the variance condition. (Answer: No; $V<\mu$, which would give $\alpha=(6-8)/8^2<0$, outside the model.)

What You'll Learn