Why Limiting Relationships Matter
Several standard frequency models describe counts using different assumptions. A binomial model counts successes in a fixed number of independent trials. A Poisson model describes counts with a single rate parameter. A negative binomial model allows the count to vary according to a success-and-failure process. Relationships among these models help explain when one model can approximate another—and when two models are exactly the same.
The key distinction is between an exact special case and a limit. A geometric distribution is exactly a negative binomial distribution with shape parameter one under the convention used here. A binomial distribution, by contrast, is generally not exactly Poisson: it approaches a Poisson distribution as the number of trials grows, the success probability shrinks, and the expected count stays fixed.
Deriving the Poisson Limit of the Binomial
Write \(p_n=\lambda/n\) first, so the binomial mean is exactly \(np_n=\lambda\) for each \(n\). For a fixed count \(k\), the binomial probability is
To see the limit, separate the factors that have different roles. The binomial coefficient can be written as a product, giving
For fixed \(k\), the finite product tends to 1 as \(n\to\infty\), and \((1-\lambda/n)^{-k}\) also tends to 1. The standard exponential limit gives \((1-\lambda/n)^n\to e^{-\lambda}\). Therefore,
This is the probability mass function of a Poisson random variable with mean \(\lambda\). The same result holds when \(np_n\) merely approaches \(\lambda\), rather than equaling it for every \(n\). The assumptions express the practical idea behind the approximation: there are many opportunities for an event, each with a small probability, but the expected number of events remains stable.
The binomial mean is \(np_n\), which approaches \(\lambda\), and its variance is \(np_n(1-p_n)\), which also approaches \(\lambda\). These matching limiting moments support the probability argument, but matching means alone is not enough to establish a Poisson approximation. The small-probability, many-trial conditions matter.
Worked Example: Comparing a Binomial Count With Its Poisson Limit
A hypothetical inspection process checks 20 independent components. Each component has probability \(0.10\) of requiring adjustment. Let \(X\) be the number requiring adjustment. Use a Poisson model with the same mean to approximate \(P(X=0)\) and \(P(X=2)\), then compare with the exact binomial probabilities. State the method and identify the trap avoided.
Model and method. The exact model is \(X\sim\operatorname{Bin}(20,0.10)\). Its mean is \(np=20(0.10)=2\), so the Poisson approximation is \(Z\sim\operatorname{Poisson}(2)\). This is a comparison at finite \(n\), not an assertion that the two distributions are identical.
Exactly zero adjustments. Under the binomial model, \(P(X=0)=\binom{20}{0}(0.10)^0(0.90)^{20}=0.1215767\). Under the Poisson model, \(P(Z=0)=e^{-2}2^0/0!=e^{-2}=0.1353353\).
Exactly two adjustments. The exact binomial probability is \(P(X=2)=\binom{20}{2}(0.10)^2(0.90)^{18}\). The intermediate values are \(\binom{20}{2}=190\), \((0.10)^2=0.01\), and \((0.90)^{18}=0.1500946\). Thus \(P(X=2)=190(0.01)(0.1500946)=0.2851798\). The Poisson approximation is \(P(Z=2)=e^{-2}2^2/2!=2e^{-2}=0.2706706\).
Check and trap. Rounded to four decimal places, the exact probabilities are \(0.1216\) and \(0.2852\), while the approximations are \(0.1353\) and \(0.2707\). The approximation is plausible but not exact at only 20 trials with success probability \(0.10\). The trap is to report the Poisson values as exact binomial probabilities merely because the means match.
The Negative Binomial Family and the Geometric Case
Use the following convention for a negative binomial count \(N\): \(N\) is the number of failures observed before the \(r\)th success in independent Bernoulli trials, each with success probability \(p\). Here \(r\) is a positive integer and \(0<p<1\). The count can be \(0,1,2,\ldots\); it does not count the success that completes the process.
The probability formula follows by considering the sequence of trials. To have exactly \(n\) failures before the \(r\)th success, the first \(n+r-1\) trials must contain \(n\) failures and \(r-1\) successes, in any order. There are \(\binom{n+r-1}{n}\) such arrangements. Each arrangement has probability \(p^{r-1}(1-p)^n\), and the final trial must be a success, contributing another factor of \(p\). Multiplying gives the stated probability.
At \(r=1\), there are no earlier successes to arrange: the count is simply the number of failures before the first success. The model is therefore geometric exactly, not approximately. Parameter conventions differ across sources; some geometric models count the trial on which success occurs. That version takes values \(1,2,\ldots\) and is \(N+1\), not \(N\).
Worked Example: Using Shape One to Obtain a Geometric Count
A hypothetical device is tested repeatedly until it passes a test. Each test passes independently with probability \(p=0.40\). Let \(N\) be the number of failures before the first pass. Find \(P(N=0)\), \(P(N=2)\), \(P(N\leq 2)\), the mean, and the variance. Name the method and the convention trap avoided.
Model and method. This is a negative binomial model with shape \(r=1\) and success probability \(p=0.40\), hence also a geometric model on \(0,1,2,\ldots\). The failure probability is \(1-p=0.60\).
Zero failures. \(P(N=0)=\binom{0}{0}(0.40)^1(0.60)^0=0.40\). A pass on the first test gives zero failures.
Two failures. \(P(N=2)=\binom{2}{2}(0.40)(0.60)^2=1(0.40)(0.36)=0.144\).
At most two failures. The probabilities for zero and one failure are \(P(N=0)=0.40\) and \(P(N=1)=\binom{1}{1}(0.40)(0.60)=0.24\). Therefore \(P(N\leq2)=0.40+0.24+0.144=0.784\). As a check, \(P(N\geq3)\) requires the first three tests all to fail, so it is \((0.60)^3=0.216\), and \(0.784+0.216=1\).
Moments. The mean is \(r(1-p)/p=1(0.60)/0.40=1.5\). The variance is \(r(1-p)/p^2=1(0.60)/(0.40)^2=0.60/0.16=3.75\).
Check and trap. The probabilities through two failures sum to \(0.784\), consistent with the remaining tail \(0.216\). The common mistake is to assign \(P(N=0)=1-p\), as if zero meant an immediate failure. Under this convention, zero failures means immediate success, so \(P(N=0)=p=0.40\).
A Poisson Limit for Negative Binomial Counts
The negative binomial family also approaches a Poisson distribution when its shape grows while its mean stays fixed. Fix \(\lambda>0\), let \(r\) grow, and choose
This choice gives \(r(1-p_r)/p_r=\lambda\). For a fixed nonnegative integer \(n\), substitute this probability into the negative binomial mass function:
Using \(\binom{n+r-1}{n}=\prod_{j=0}^{n-1}(r+j)/n!\), rearrange the expression as
For fixed \(n\), the finite product tends to 1, and the final factor tends to 1. Also, \((1+\lambda/r)^{-r}\to e^{-\lambda}\). Thus \(P(N=n)\to e^{-\lambda}\lambda^n/n!\), the Poisson probability. The negative binomial variance under this fixed-mean choice is \(\lambda+\lambda^2/r\), which approaches the Poisson variance \(\lambda\). At finite \(r\), however, the variance exceeds the mean by \(\lambda^2/r\).
Worked Example: Tracking a Fixed-Mean Negative Binomial Limit
For hypothetical claim counts, let \(N_r\) be negative binomial with mean \(\lambda=2\) and integer shape \(r\). Calculate \(P(N_r=0)\) for \(r=1,10,100\), and compare with the Poisson probability at mean 2. Use the fixed-mean method and explain what the comparison does—and does not—show.
Set the probability parameter. From \(r(1-p_r)/p_r=2\), solve \(r= p_r(r+2)\), so \(p_r=r/(r+2)\). At zero failures, the negative binomial mass is \(P(N_r=0)=p_r^r\).
Calculate each value. For \(r=1\), \(p_1=1/3\), so \(P(N_1=0)=(1/3)^1=0.3333\). For \(r=10\), \(p_{10}=10/12=5/6\), so \(P(N_{10}=0)=(5/6)^{10}=0.1615\). For \(r=100\), \(p_{100}=100/102=50/51\), so \(P(N_{100}=0)=(50/51)^{100}=0.1380\).
Compare and check. The limiting Poisson value is \(P(Z=0)=e^{-2}=0.1353\). The negative binomial zero probability moves closer to this value as \(r\) increases: \(0.3333,\ 0.1615,\ 0.1380\), compared with \(0.1353\). Values are rounded to four decimals. This illustrates convergence, not equality at any of the displayed finite shapes; in particular, the \(r=1\) case is geometric and is far from the Poisson limit.
Trap avoided. Increasing \(r\) while holding \(p\) fixed would also change the mean. To study this limit, \(p\) must change with \(r\) according to \(p_r=r/(r+2)\).
Choosing and Interpreting the Relationship
These relationships answer different modeling questions. For a binomial count, a Poisson approximation is motivated by many independent opportunities, a small event probability, and a stable expected count. For a negative binomial count, the geometric model is obtained by setting the shape to one, while the Poisson limit requires increasing the shape and adjusting the success probability to maintain a fixed mean.
A useful summary is:
| Relationship | Condition | Conclusion |
|---|---|---|
| Binomial to Poisson | \(n\to\infty,\ p_n\to0,\ np_n\to\lambda\) | Binomial probabilities approach Poisson probabilities with mean \(\lambda\) |
| Negative binomial to geometric | \(r=1\), with failures counted before success | Exact geometric distribution with \(P(N=n)=p(1-p)^n\) |
| Negative binomial to Poisson | \(r\to\infty,\ p_r=r/(r+\lambda)\) | Negative binomial probabilities approach Poisson probabilities with mean \(\lambda\) |
A frequent exam trap is to identify a model from one feature alone. A mean of \(\lambda\) does not make a count Poisson: a binomial, negative binomial, and Poisson count can share the same mean while having different variances and probabilities. State the model, its parameter convention, and the limiting conditions before using an approximation. These distinctions are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.
Check Your Understanding
For each problem, identify the model relationship before calculating. Probabilities in the answers are rounded to four decimals.
- A count is \(X\sim\operatorname{Bin}(80,0.025)\). Give its mean and the Poisson approximation to \(P(X=0)\) and \(P(X=1)\). (Answer: mean \(2\); \(P(X=0)\approx0.1353\), \(P(X=1)\approx0.2707\))
- A process has \(X\sim\operatorname{Bin}(30,0.10)\). Calculate the exact \(P(X=0)\) and its Poisson approximation using the same mean. (Answer: exact \(0.9^{30}\approx0.0424\); Poisson \(e^{-3}\approx0.0498\))
- Let \(N\) count failures before the first success, with success probability \(0.25\). Find \(P(N=2)\), \(E[N]\), and \(\operatorname{Var}(N)\). (Answer: \(P(N=2)=0.1406\), \(E[N]=3\), \(\operatorname{Var}(N)=12\))
- Let \(N\) be negative binomial with \(r=4\), \(p=0.50\), counting failures before the fourth success. Find \(P(N=0\text{ or }N=2)\). (Answer: \(0.50^4+\binom{5}{2}(0.50)^4(0.50)^2=0.2188\))
- A negative binomial count has fixed mean 2 and shape \(r=40\). Find \(p\) and \(P(N=0)\), then compare that probability with the Poisson limit. (Answer: \(p=40/42=20/21\); \(P(N=0)=(20/21)^{40}\approx0.1419\), compared with \(e^{-2}\approx0.1353\))