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Zero-Claim Probabilities Across Frequency Models

Learn to calculate and compare zero-claim probabilities across three frequency models with matched means, while checking how dispersion changes the result.

Intermediate 9 min read

What You'll Learn

  • Calculate the probability of zero claims under binomial, Poisson, and negative binomial models.
  • Match the three models by their expected claim count.
  • Explain why the standard models have different zero-claim probabilities.
  • Connect each comparison to the model’s variance and dispersion.
  • Avoid parameterization and approximation errors in exam-style problems.

Why Zero Claims Need a Model

The probability of no claims is often an important quantity in a frequency model. It can be used to describe the chance that a policy, portfolio, or hypothetical group experiences no claims during a specified period. But the expected number of claims alone does not determine that probability. Different frequency models can have the same mean and still assign different probabilities to zero.

This tutorial compares the binomial, Poisson, and negative binomial models when their means are matched. Their variances indicate whether claim counts are less or more dispersed than the Poisson model, and their zero-claim probabilities follow a corresponding ordering. That ordering is a result for these particular models, not a universal rule for every distribution with the same mean and variance.

Definition: For a nonnegative integer-valued claim count \(N\), the zero-claim probability is \(P(N=0)\). A model’s mean is \(E[N]\), and its variance is \(\operatorname{Var}(N)\). In a matched-mean comparison, the models are chosen to have the same value of \(E[N]\), while their distributions and variances may differ.

Zero-Claim Formulas and Matched Means

For a binomial count, \(N\sim\operatorname{Bin}(n,p)\), there are \(n\) independent trials, each with claim probability \(p\). The mean is \(np\), the variance is \(np(1-p)\), and zero claims means that all \(n\) trials have no claim. For a Poisson count with parameter \(\lambda\), the mean and variance are both \(\lambda\). For a negative binomial count, use the convention that \(N\) counts failures before the \(r\)th success, with success probability \(p\). Under this convention, the mean is \(r(1-p)/p\), and \(P(N=0)=p^r\).

Write \(m\) for the common mean. Matching the models gives \(np=m\) for the binomial, \(\lambda=m\) for the Poisson, and \(r(1-p)/p=m\) for the negative binomial. Solving these equations gives the following formulas.

$$ \begin{aligned} \text{Binomial:}\quad &p=\frac{m}{n}, \qquad P(N=0)=\left(1-\frac{m}{n}\right)^n, &\operatorname{Var}(N)&=m\left(1-\frac{m}{n}\right), \quad n\ge m,\\ \text{Poisson:}\quad &\lambda=m, \qquad P(N=0)=e^{-m}, &\operatorname{Var}(N)&=m,\\ \text{Negative binomial:}\quad &p=\frac{r}{r+m}, \qquad P(N=0)=\left(\frac{r}{r+m}\right)^r, &\operatorname{Var}(N)&=m+\frac{m^2}{r}. \end{aligned} $$

For a nondegenerate binomial model with \(0<m<n\), its variance is less than \(m\). The Poisson variance equals \(m\), while a finite-\(r\) negative binomial variance exceeds \(m\). The zero-claim probabilities have the same ordering:

Theorem: For a common positive mean \(m\), a nondegenerate binomial model with \(n>m\), a Poisson model, and a negative binomial model with finite \(r>0\) satisfy \(P_{\text{Bin}}(N=0)<P_{\text{Pois}}(N=0)<P_{\text{NB}}(N=0)\). The binomial model has the smallest zero probability, and the negative binomial model has the largest. If \(m=0\), each model assigns probability 1 to zero.

Why the Ordering Holds

The comparison follows by rewriting each probability using an exponential and applying the logarithm inequalities \(\log(1-x)<-x\) for \(0<x<1\), and \(\log(1+x)<x\) for \(x>0\). These inequalities are strict under the stated conditions.

1
Compare binomial with Poisson.
Set \(x=m/n\), so \(0<x<1\). Then \(n\log(1-m/n)<-m\). Exponentiating both sides gives \((1-m/n)^n<e^{-m}\).
2
Compare Poisson with negative binomial.
For the negative binomial, \(\log P(N=0)=-r\log(1+m/r)\). Since \(\log(1+m/r)<m/r\), this gives \(\log P(N=0)>-m\), and therefore \(P(N=0)>e^{-m}\).
3
Interpret the result within these models.
The binomial model is underdispersed relative to Poisson, while the negative binomial model is overdispersed. With their means matched, the former assigns less probability to zero and the latter more. The inequalities do not establish a general ordering for arbitrary distributions.

Both comparisons also approach equality in appropriate limits: as \(n\) grows with \(m\) fixed, the binomial zero probability approaches \(e^{-m}\); as \(r\) grows with \(m\) fixed, the negative binomial zero probability approaches \(e^{-m}\). These limits help explain why models with less pronounced dispersion differences can have zero probabilities close to the Poisson value.

Worked Example: Three Models With a Mean of Two

A hypothetical portfolio is modelled in three alternative ways, each with expected claim count 2. The binomial model has \(n=10\) trials, the Poisson model has parameter \(\lambda=2\), and the negative binomial model has \(r=4\). Calculate each zero-claim probability and compare the variances.

Method. First match each model’s parameters to the mean, then use its probability mass at zero. This avoids treating equal means as though they implied equal distributions.

For the binomial model, \(p=2/10=0.2\). Thus \(P(N=0)=(1-0.2)^{10}=0.8^{10}=0.1073741824\), which is \(0.1074\) to four decimal places. Its variance is \(10(0.2)(0.8)=1.6\).

For the Poisson model, \(\lambda=2\), so \(P(N=0)=e^{-2}=0.1353352832\), or \(0.1353\) to four decimal places. Its variance is 2.

For the negative binomial model, solve \(4(1-p)/p=2\), giving \(4-4p=2p\), \(p=2/3\), and \(1-p=1/3\). Therefore \(P(N=0)=p^4=(2/3)^4=16/81=0.1975308642\), or \(0.1975\) to four decimal places. Its variance is \(2+2^2/4=3\).

The three variances are \(1.6\), \(2\), and \(3\), respectively, and the zero probabilities increase in the same model order: \(0.1074<0.1353<0.1975\). The result is consistent with the underdispersed binomial and overdispersed negative binomial models.

Worked Example: Comparing Models at a Mean of Four

A hypothetical claim count has mean 4. Compare a binomial model with \(n=20\), a Poisson model, and a negative binomial model with \(r=8\). Give each model’s zero-claim probability and variance. Round probabilities to four decimal places.

Method. Use the common mean to determine the binomial claim probability and the negative binomial success probability before evaluating \(P(N=0)\). This avoids mixing parameter conventions.

For the binomial model, \(p=4/20=0.2\). Its zero probability is \((1-0.2)^{20}=0.8^{20}=0.0115292150\), or \(0.0115\). Its variance is \(20(0.2)(0.8)=3.2\).

For the Poisson model, \(\lambda=4\), so \(P(N=0)=e^{-4}=0.0183156389\), or \(0.0183\), and the variance is 4.

For the negative binomial model, \(8(1-p)/p=4\), so \(8-8p=4p\), \(p=2/3\), and \(P(N=0)=(2/3)^8=256/6561=0.0390184423\), or \(0.0390\). Its variance is \(4+4^2/8=6\).

The zero-claim probabilities follow \(0.0115<0.0183<0.0390\), while the variances are \(3.2<4<6\). The calculation checks the expected ordering for these three matched-mean models.

Worked Example: Recovering a Negative Binomial Parameter

A hypothetical negative binomial claim count uses the convention of counting failures before the \(r\)th success. Its mean is 2 and its zero-claim probability is \(0.25\). Find \(r\), its variance, and compare its zero probability with a Poisson model of the same mean.

Method. Use the mean relation to express the success probability in terms of \(r\), substitute into the zero-probability formula, and check the resulting parameter.

From \(r(1-p)/p=2\), solve \(r-rp=2p\), so \(p=r/(r+2)\). The zero-probability condition becomes \(\left(r/(r+2)\right)^r=0.25\). The value \(r=2\) satisfies it, since \(p=2/(2+2)=1/2\) and \(p^r=(1/2)^2=1/4\). Thus \(r=2\) is a parameter value consistent with both given quantities.

The variance is \(2+2^2/2=4\). The same-mean Poisson model has zero probability \(e^{-2}=0.1353352832\), or \(0.1353\) to four decimal places. The negative binomial probability \(0.2500\) is larger, as expected for this overdispersed model. The substitution verifies the answer: the fitted model has mean \(2(1/2)/(1/2)=2\) and zero probability \(0.25\).

Common Exam Trap

A frequent error is to use \(e^{-m}\) for every model merely because the mean is \(m\). That is the zero probability for a Poisson model, not a general consequence of the mean. For example, with binomial \(n=10\) and \(p=0.2\), the mean is 2, but the correct calculation is \(0.8^{10}=0.1074\), not \(e^{-2}=0.1353\). The Poisson value is larger in this case.

Another trap is switching negative binomial conventions. Here \(N\) counts failures before the \(r\)th success, so the mean is \(r(1-p)/p\), and zero failures requires all \(r\) trials to be successes, giving \(P(N=0)=p^r\). If a problem defines a different count or parameter, derive the zero event from that definition rather than carrying over these formulas unchanged.

Finally, the variance comparison helps explain the result but should not be used to assert a universal zero-probability ordering. The ordering proved here depends on the binomial, Poisson, and negative binomial probability formulas. A different distribution with the same mean or variance need not follow it.

Key takeaway: When the common mean is \(m>0\), the standard matched-mean models give \(P_{\text{Bin}}(N=0)<e^{-m}<P_{\text{NB}}(N=0)\) under the stated parameterizations and nondegenerate conditions. Calculate from the model’s zero-count event, then use variance to interpret—not replace—the calculation.

Check Your Understanding

For each problem, calculate the requested probability and use the model definitions stated here.

  1. A binomial claim count has \(n=5\) and \(p=0.2\). Compare its mean, variance, and zero probability with a Poisson model of the same mean and a negative binomial model with \(r=2\) and that same mean. (Answer: Common mean \(1\); variances \(0.8,1,1.5\); zero probabilities \(0.3277,0.3679,0.4444\), respectively.)
  2. Compare matched-mean models with mean 4: binomial \(n=20\), Poisson, and negative binomial \(r=8\). Give the zero probabilities to four decimal places and state which is largest. (Answer: \(0.0115,0.0183,0.0390\); the negative binomial probability is largest.)
  3. A negative binomial count has \(r=3\) and mean 3. Find its success probability, zero probability, and variance, then compare its zero probability with the matched-mean Poisson model. (Answer: \(p=0.5000\); negative binomial zero probability \(0.1250\), variance \(6\); Poisson zero probability \(e^{-3}=0.0498\).)
  4. A binomial count has \(n=6\) and mean 3. Find its zero probability and variance, then compare its zero probability with the matched-mean Poisson and negative binomial models with \(r=3\). (Answer: Binomial \(P(N=0)=1/64=0.0156\), variance \(1.5\); Poisson zero probability \(0.0498\); negative binomial zero probability \(0.1250\).)
  5. A hypothetical negative binomial count has mean 5 and \(r=5\). Calculate the success probability, zero probability, and variance, and compare its zero probability with the same-mean Poisson value. (Answer: \(p=5/(5+5)=0.5\); negative binomial \(P(N=0)=0.5^5=0.0313\), variance \(10\); Poisson \(P(N=0)=e^{-5}=0.0067\).)

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