The Negative Binomial Limit To The Poisson

Why a Negative Binomial Can Approach a Poisson Model

A Poisson frequency model has equal mean and variance. A negative binomial model allows variance to exceed the mean, which can represent additional variability in claim counts. The negative binomial includes the Poisson as a limiting case: as its dispersion decreases while its mean stays fixed, each fixed-count probability approaches the corresponding Poisson probability. The key is to change the negative binomial parameters together, not to hold its success probability fixed.

Definition: Let \(N_r\) have a negative binomial distribution with shape parameter \(r>0\) and fixed mean \(\lambda>0\), using the parameterization \[ P(N_r=k)=\frac{\Gamma(r+k)}{\Gamma(r)k!} \left(\frac{r}{r+\lambda}\right)^r \left(\frac{\lambda}{r+\lambda}\right)^k,\qquad k=0,1,2,\ldots. \] Its mean is \(E[N_r]=\lambda\), and its variance is \(\operatorname{Var}(N_r)=\lambda+\lambda^2/r\). The excess of the variance over the mean is \(\lambda^2/r\), which tends to zero as \(r\) increases.

For integer \(r\), this parameterization can also describe the number of failures before \(r\) successes in independent Bernoulli trials, with success probability \(p_r=r/(r+\lambda)\). The expected number of failures is then \(r(1-p_r)/p_r=\lambda\). The gamma-function form extends the count model to noninteger \(r>0\). In either form, the mean remains fixed only when the success probability changes along with \(r\).

Deriving the Poisson Limit

For fixed \(k\), rewrite the negative binomial probability using a product:

P(N_r=k)=\frac{1}{k!} \left(\prod_{j=0}^{k-1}(r+j)\right) \left(1+\frac{\lambda}{r}\right)^{-r} \left(\frac{\lambda}{r+\lambda}\right)^k.

The first and last factors involving \(r\) can be grouped. For each fixed \(k\),

\left(\prod_{j=0}^{k-1}(r+j)\right) \left(\frac{\lambda}{r+\lambda}\right)^k = \lambda^k \frac{\prod_{j=0}^{k-1}(1+j/r)} {(1+\lambda/r)^k} \longrightarrow \lambda^k.

Each factor in the finite product tends to 1, as does the denominator. Also, the standard exponential limit gives \((1+\lambda/r)^{-r}\to e^{-\lambda}\). Therefore:

\lim_{r\to\infty}P(N_r=k)=e^{-\lambda}\frac{\lambda^k}{k!}.

The expression on the right is the probability that a Poisson random variable with mean \(\lambda\) equals \(k\). Thus the negative binomial count converges to \(\operatorname{Poisson}(\lambda)\) when \(r\) tends to infinity with \(\lambda\) fixed. For a fixed integer cutoff \(m\), the probability of \(N_r\leq m\) is a finite sum of the probabilities for \(k=0,\ldots,m\), so it converges to the corresponding Poisson cumulative probability as well.

Theorem: Under the fixed-mean parameterization above, as \(r\to\infty\), the negative binomial probability at every fixed count \(k\) tends to the Poisson probability \(e^{-\lambda}\lambda^k/k!\). The variance excess \(\lambda^2/r\) tends to zero at the same time. This is a limiting statement; it does not say that each finite-\(r\) probability is always above or always below its Poisson counterpart.

Comparing Probabilities in Practice

The negative binomial probabilities can be computed directly from the probability formula. A useful ratio for successive counts follows by dividing the probability at \(k\) by the probability at \(k-1\):

\frac{P(N_r=k)}{P(N_r=k-1)} =\frac{r+k-1}{k}\frac{\lambda}{r+\lambda},\qquad k\geq1.

This ratio is a convenient way to calculate low-count probabilities without repeatedly evaluating gamma functions. For the Poisson model, the corresponding ratio is \(\lambda/k\). In either model, calculate with unrounded values and round only the final result.

Worked Example: Comparing Probabilities at a Fixed Mean

A hypothetical portfolio has a negative binomial annual claim count with mean \(\lambda=2\) and shape \(r=4\). Calculate the probabilities of zero claims and one claim, then compare the probability of at most one claim with the Poisson model having the same mean.

Method. Use the negative binomial probability at zero and the successive-probability ratio. Then calculate the Poisson probabilities with mean 2. This keeps the mean identical in both models.

For the negative binomial model,

P(N_4=0)=\left(\frac{4}{4+2}\right)^4 =\left(\frac{2}{3}\right)^4 =\frac{16}{81} =0.197530864\ldots.

The ratio for the probability of one claim to the probability of zero claims is \(4(2)/(4+2)=4/3\). Hence:

P(N_4=1)=\frac{16}{81}\cdot\frac{4}{3} =\frac{64}{243} =0.263374485\ldots.

Adding the unrounded probabilities gives \(P(N_4\leq1)=16/81+64/243=112/243=0.460905349\ldots\), or 0.4609 to four decimal places.

For a Poisson count \(N\) with mean 2, \(P(N=0)=e^{-2}=0.135335283\ldots\) and \(P(N=1)=2e^{-2}=0.270670566\ldots\). Therefore \(P(N\leq1)=3e^{-2}=0.406005849\ldots\), or 0.4060 to four decimal places. At this finite shape, the negative binomial probability of at most one claim is higher by approximately \(0.460905349-0.406005850=0.054899499\), or 0.0549.

Check. The negative binomial variance is \(2+2^2/4=3\), greater than its mean of 2. This is consistent with positive excess dispersion; it does not by itself determine the direction of every individual probability difference.

Worked Example: An Upper-Tail Comparison

A hypothetical service program models the number of annual incidents by a negative binomial distribution with mean \(\lambda=3\) and shape \(r=3\). Find the probability of at least two incidents and compare it with the Poisson probability of at least two incidents at the same mean.

Method. Compute the complement of zero or one incident in each model. This avoids summing an infinite upper tail directly.

For the negative binomial model,

P(N_3=0)=\left(\frac{3}{3+3}\right)^3=\frac18.

The successive-probability ratio at one is \(3(3)/(3+3)=3/2\), so \(P(N_3=1)=(1/8)(3/2)=3/16\). Thus:

P(N_3\geq2) =1-\frac18-\frac{3}{16} =\frac{11}{16} =0.6875.

The negative binomial variance is \(3+3^2/3=6\). For the Poisson model, the probabilities at zero and one are \(e^{-3}\) and \(3e^{-3}\), respectively. Therefore:

P(N\geq2)=1-e^{-3}-3e^{-3} =1-4e^{-3} =0.800851726\ldots

The Poisson upper-tail probability is 0.8009 to four decimal places, compared with 0.6875 for this negative binomial model. The models have equal means but different distributions and variances, so their probabilities for a particular event need not match.

Check. The negative binomial probabilities at zero and one sum to \(1/8+3/16=5/16\); their complement is \(11/16\), as required.

Worked Example: A Finite Shape Can Be Close Without Matching

A hypothetical count model has mean \(\lambda=1\) and negative binomial shape \(r=4\). Calculate the probability of at most one count and compare it with the limiting Poisson probability.

Method. Use the zero-count probability and the ratio for one count, then calculate the same cumulative probability under the Poisson model.

The negative binomial zero-count probability is \[ P(N_4=0)=\left(\frac{4}{5}\right)^4=\frac{256}{625}=0.4096. \] The ratio \(P(N_4=1)/P(N_4=0)\) is \(4/(4+1)=4/5\). Therefore \(P(N_4=1)=0.4096(0.8)=0.32768\), and:

P(N_4\leq1)=0.4096+0.32768=0.73728.

The Poisson model with mean 1 has \(P(N=0)=e^{-1}\) and \(P(N=1)=e^{-1}\), giving \(P(N\leq1)=2e^{-1}=0.735758882\ldots\). The negative binomial result is higher by about \(0.001521118\), or 0.0015 to four decimal places. The two results are close, but not equal at \(r=4\).

Check. The negative binomial variance is \(1+1/4=1.25\), compared with the Poisson variance of 1. The positive variance excess has not vanished, although it is smaller than it would be at a lower shape parameter.

Common Exam Trap: Changing Shape but Not Holding the Mean Fixed

A common error is to use the number-of-failures parameterization and increase the number of successes \(r\) while leaving the success probability \(p\) unchanged. In that parameterization, the mean is \(r(1-p)/p\), so this changes the mean. It does not demonstrate the fixed-mean Poisson limit.

Common mistake and correction: Suppose \(r=4\) and \(\lambda=2\). The required success probability is \(p_4=4/(4+2)=2/3\). If \(r\) is increased to 8 while \(p=2/3\) is mistakenly kept fixed, the mean becomes \(8(1/3)/(2/3)=4\), not 2. To preserve \(\lambda=2\), instead use \(p_8=8/(8+2)=4/5\). Then the variance is \(2+2^2/8=2.5\), closer to the Poisson variance of 2. The fixed-mean limit requires adjusting the probability parameter as the shape changes.

A second trap is to treat “approaches the Poisson distribution” as a guarantee about the direction of a finite-\(r\) probability difference. The zero-count probability may be above the Poisson value while another count probability is below it, as the first worked example shows. Compare the event actually asked about rather than inferring it from the variance alone.

Using the Limit Correctly

The limit is useful for understanding the relationship between two frequency models and for checking calculations. It also gives a sensible benchmark: if \(r\) is large at a fixed mean, the negative binomial probabilities should be close to their Poisson counterparts. For a specific finite shape, however, compute the required negative binomial probability if an exact result is requested; do not silently replace the model by a Poisson distribution.

A reliable comparison follows these steps:

Identify the parameterization.
Confirm that the negative binomial is expressed using shape \(r\) and mean \(\lambda\), or convert the supplied parameters to this form.

Check what is held fixed.
For the Poisson limit, keep \(\lambda\) fixed while \(r\) increases. If using a success probability, set \(p_r=r/(r+\lambda)\).

Calculate the requested event.
Use the appropriate count probabilities, ratios, or complements. Do not infer a tail probability solely from the mean and variance.

Check and round.
Verify that complements and sums are consistent, retain unrounded intermediate values, and round only the final probability as requested.

The limiting result is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus. The central conditions to remember are a negative binomial model, a fixed mean, and a shape parameter increasing without bound.

Check Your Understanding

For each problem, keep the stated mean fixed when comparing the negative binomial and Poisson models. Give probabilities to four decimal places where numerical rounding is needed.

A negative binomial count has mean \(\lambda=4\) and shape \(r=4\). Find \(P(N_r\leq1)\), then find the corresponding Poisson probability. (Answer: Negative binomial \(3/16=0.1875\); Poisson \(5e^{-4}=0.0916\).)
A negative binomial count has mean \(\lambda=2\) and shape \(r=2\). Find \(P(N_r\geq2)\) and the Poisson probability of at least two. (Answer: Negative binomial \(1/2=0.5000\); Poisson \(1-3e^{-2}=0.5940\).)
A negative binomial count has mean \(\lambda=3\) and shape \(r=12\). Find its variance and its zero-count probability, then compare the latter with the Poisson zero-count probability. (Answer: Variance \(3.75\); negative binomial \(P(N_r=0)=(4/5)^{12}=0.0687\); Poisson \(P(N=0)=e^{-3}=0.0498\).)
A negative binomial count has mean \(\lambda=1\) and shape \(r=1\). Find the probability of exactly one count and compare it with the Poisson limit probability. (Answer: Negative binomial \(P(N_1=1)=1/4=0.2500\); Poisson \(P(N=1)=e^{-1}=0.3679\).)
A negative binomial count has mean \(\lambda=4\) and shape \(r=2\). Find \(P(N_r\leq1)\) and compare it with the Poisson probability of at most one. (Answer: Negative binomial \(7/27=0.2593\); Poisson \(5e^{-4}=0.0916\).)

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