Why Compare Means and Variances?
A count model describes both the expected number of events and how much that number varies. The mean gives the center of the count distribution; the variance describes its spread. For binomial, Poisson, and negative binomial counts, the relationship between these two quantities is characteristic of the model. It is useful for interpreting variation and checking whether parameter calculations are internally consistent.
The comparison is meaningful only after the count and its model have been specified. A binomial count concerns a fixed number of independent trials with a common event probability. A Poisson count has a rate parameter. Here, a negative binomial count means the number of failures before a specified number of successes, with independent trials and a common success probability. Its parameterization matters: the formulas below use \(r\) for the target number of successes and \(p\) for the probability of success on each trial.
The Three Mean-Variance Signatures
The formulas in the table apply under the stated models. For a binomial count, \(n\) is a positive integer and each of the \(n\) independent trials has event probability \(p\). For a Poisson count, \(\lambda\) is its rate parameter. For a negative binomial failure count, \(r\) is a positive integer, \(0<p<1\), and \(X\) counts failures before the \(r\)th success.
| Model for \(X\) | Mean | Variance | Relationship when the mean is positive |
|---|---|---|---|
| Binomial, \(X\sim\operatorname{Bin}(n,p)\) | \(np\) | \(np(1-p)\) | \(\operatorname{Var}(X)<E[X]\) |
| Poisson, \(X\sim\operatorname{Poisson}(\lambda)\) | \(\lambda\) | \(\lambda\) | \(\operatorname{Var}(X)=E[X]\) |
| Negative binomial failure count, parameters \(r,p\) | \(\dfrac{r(1-p)}{p}\) | \(\dfrac{r(1-p)}{p^2}\) | \(\operatorname{Var}(X)>E[X]\) |
For a binomial count with positive mean, \(p>0\), so \(np(1-p)<np\). The Poisson variance equals its mean. For the stated negative binomial model, the variance is the mean divided by \(p\), and \(0<p<1\), so dividing a positive mean by \(p\) makes it larger. The binomial comparison needs its positive-mean qualification: if \(p=0\), both mean and variance are zero, so neither is strictly below the other.
Why the Negative Binomial Has These Moments
The failure-count definition gives a direct derivation. Divide the trials into \(r\) successive waiting periods: \(G_j\) is the number of failures after the \((j-1)\)th success and before the \(j\)th success. Each \(G_j\) has a geometric failure-count distribution with success probability \(p\). The periods are independent, and the total number of failures is their sum.
Linearity of expectation gives the mean of the sum. Independence makes its variance the sum of the component variances. Thus,
This derivation also shows why using a different negative binomial convention can produce a parameter error. A model that counts trials through the \(r\)th success includes the \(r\) successes as well as the failures; it is not the same random variable as the failure count defined here. Always identify what is being counted before applying the moment formulas.
Reading the Signatures
When the mean is positive, the variance-to-mean ratio summarizes the distinction particularly simply. It is \(1-p\) for a binomial count, \(1\) for a Poisson count, and \(1/p\) for the negative binomial failure count. These ratios describe variation relative to the mean, not the absolute size of the variance. Two models can have the same mean and very different variances.
A moment comparison is a description, not proof that observed data follow one of these distributions. In particular, the mean and variance alone do not determine a unique count distribution. The model assumptions and the full context still matter. Use the signatures to check consistency with a proposed model or to describe its characteristic variation, rather than treating a variance comparison as a complete model-selection procedure.
Worked Example: Equal Means, Different Variability
Three hypothetical portfolios each have an expected count of 4 claims during a specified period. Portfolio A is modeled by \(X_A\sim\operatorname{Bin}(10,0.4)\); Portfolio B by \(X_B\sim\operatorname{Poisson}(4)\); and Portfolio C by a negative binomial failure count with \(r=2\) and success probability \(p=1/3\). Calculate and compare the variances.
Method. Apply each model's mean and variance formulas, then compare the variances. The common trap is to assume that equal means imply equal variability.
For Portfolio A, the binomial mean is \(np=10(0.4)=4\). Its variance is \[ np(1-p)=10(0.4)(0.6)=4(0.6)=2.4. \]
For Portfolio B, the Poisson parameter is \(\lambda=4\), so \(E[X_B]=4\) and \(\operatorname{Var}(X_B)=4\).
For Portfolio C, \(1-p=2/3\). Its mean is \[ \frac{r(1-p)}{p} =\frac{2(2/3)}{1/3}=4, \] and its variance is \[ \frac{r(1-p)}{p^2} =\frac{2(2/3)}{(1/3)^2} =\frac{4/3}{1/9}=12. \]
The variances are therefore \(2.4\), \(4\), and \(12\), respectively. All three models have mean 4, but the binomial model has the least variance, the Poisson model has variance equal to its mean, and the negative binomial model has the greatest variance. Each calculated mean agrees with the stated value, which checks the parameter substitutions.
Worked Example: Recovering Binomial Parameters From Moments
A hypothetical fixed-trial count is known to follow a binomial model. Its mean is 6 and its variance is 4.2. Recover \(n\) and \(p\), and verify the resulting moments.
Method. Use the ratio of variance to mean to recover \(1-p\), then use the mean to recover \(n\). This avoids confusing the binomial variance with \(np\).
From \(\operatorname{Var}(X)=np(1-p)\) and \(E[X]=np=6\), \[ \frac{\operatorname{Var}(X)}{E[X]}=\frac{4.2}{6}=0.7=1-p. \] Thus \(p=1-0.7=0.3\). Since \(np=6\), \[ n=\frac{6}{0.3}=20. \]
Check the mean: \(np=20(0.3)=6\). Check the variance: \(np(1-p)=20(0.3)(0.7)=6(0.7)=4.2\). The recovered parameters reproduce both given moments, and \(n=20\) is a positive integer as required by the binomial model.
Worked Example: Recovering Negative Binomial Parameters
A hypothetical negative binomial failure count has mean 24 and variance 72. Find its success probability \(p\) and target number of successes \(r\). State the dispersion relationship and verify both moments.
Method. For this failure-count parameterization, dividing variance by mean gives \(1/p\). Then solve the mean formula for \(r\). The common mistake is to switch the roles of \(p\) and \(1-p\), or to use a different negative binomial convention.
The variance-to-mean ratio is \[ \frac{72}{24}=3=\frac{1}{p}, \] so \(p=1/3\), and \(1-p=2/3\). Substitute these into the mean formula: \[ 24=\frac{r(2/3)}{1/3}=2r, \qquad r=12. \]
The variance exceeds the mean, since \(72>24\), so the moments show overdispersion relative to the mean. Check the mean using \(r=12\): \[ \frac{12(2/3)}{1/3}=\frac{8}{1/3}=24. \] Check the variance: \[ \frac{12(2/3)}{(1/3)^2} =\frac{8}{1/9}=72. \] Both moments are reproduced. The recovered \(r=12\) is a valid positive integer target number of successes.
Boundary Cases and Common Traps
The strict comparisons apply under their stated conditions. A Poisson count with \(\lambda=0\) is degenerate at zero, so its mean and variance are both zero. A binomial count with \(p=0\) is also degenerate at zero; with \(p=1\), it is degenerate at \(n\), with variance zero and positive mean. For the negative binomial failure count, the formulas above assume \(r>0\) and \(0<p<1\); in that range its mean is positive and its variance is strictly greater than its mean.
When using moments to recover parameters, state the model first. For example, a positive mean equal to the variance is consistent with a Poisson model, but it does not establish that the count is Poisson. Likewise, a variance below the mean is consistent with a nondegenerate binomial model, but the binomial assumptions—fixed trial count, independence, and common event probability—must also be appropriate.
Check Your Understanding
For each problem, calculate or interpret the moments using the specified model and parameterization.
- A hypothetical count follows \(\operatorname{Bin}(15,0.2)\). Find its mean and variance, and state whether its variance is below, equal to, or above its mean. (Answer: mean \(3\), variance \(2.4\); below)
- A hypothetical Poisson count has parameter \(\lambda=5.5\). Give its mean and variance and state their relationship. (Answer: mean \(5.5\), variance \(5.5\); equal)
- A negative binomial failure count has \(r=3\) and \(p=0.5\). Find its mean and variance. (Answer: mean \(3\), variance \(6\))
- A binomial count has mean 10 and variance 7.5. Recover \(p\) and \(n\), and verify the mean. (Answer: \(p=0.25\), \(n=40\); \(40(0.25)=10\))
- A negative binomial failure count has mean 12 and variance 36. Recover \(p\) and \(r\), and classify its variance relative to its mean. (Answer: \(p=1/3\), \(r=6\); overdispersed, since \(36>12\))