Why Compare Variance With the Mean?
A frequency model describes a random count, such as the number of hypothetical claims in a period. Its mean gives the expected count, while its variance describes how much the count fluctuates around that expectation. Comparing these quantities is a useful way to understand the model’s spread. In particular, a binomial count has variance below its mean except in a degenerate zero-mean case. This is called underdispersion relative to the Poisson benchmark, because a Poisson count has variance equal to its mean.
The comparison is meaningful only after specifying the model. The binomial result depends on a fixed number of trials, independent outcomes, and the same success probability on every trial. If these conditions do not hold, the binomial variance formula need not apply.
For a hypothetical portfolio, a trial might represent one policyholder’s claim status during a specified period: success means at least one claim, and failure means none. A binomial model is appropriate only if each policyholder is represented by one trial, the trial outcomes are independent, and all have the same claim probability. This tutorial concerns the spread of the resulting count, not the probability of a particular count.
The Binomial Mean and Variance
The binomial moments follow directly by writing the count as a sum of indicator variables. For trial \(j\), let \(I_j=1\) if it is a success and \(I_j=0\) otherwise. Then \(P(I_j=1)=p\), \(E[I_j]=p\), and \(\operatorname{Var}(I_j)=p(1-p)\). Since \(N=I_1+\cdots+I_n\), linearity of expectation gives the mean. Independence makes the covariance terms zero when finding the variance.
The variance comparison can now be made without approximation. Subtracting the binomial variance from its mean gives \(np-np(1-p)=np^2\), which is nonnegative. If \(n>0\) and \(p>0\), it is positive, so the binomial variance is strictly less than the mean. If \(p=0\) (or \(n=0\)), the count is always zero and both moments are zero.
A Poisson random variable with parameter \(\lambda\) has mean and variance both equal to \(\lambda\). Therefore, if a Poisson model is used as a reference with the same mean as a binomial count, set \(\lambda=np\). The Poisson variance is then \(np\), whereas the binomial variance is \(np(1-p)\); the binomial variance is smaller by \(np^2\). This is a comparison of two models with a common mean, not a claim that a binomial count is Poisson.
How the Fixed Trial Structure Affects Dispersion
For a binomial count with positive mean \(\mu=np\), the variance-to-mean ratio is \(1-p\). Thus a larger success probability means a smaller ratio, when the mean is held fixed by changing the number of trials as needed. The model’s fixed number of opportunities also limits the count to at most \(n\), unlike a Poisson count, which has no finite upper bound. The variance result itself, however, follows from the Bernoulli trial variances and independence; a finite upper bound alone is not enough to establish the binomial formula.
The formula is especially useful in two directions. Given \(n\) and \(p\), it gives the mean and variance directly. Conversely, if a count is assumed to be binomial and both moments are supplied, their relationship can help recover the parameters. It can also rule out a proposed binomial model: a variance greater than the mean is impossible under this model. A variance below the mean is consistent with binomial underdispersion, but does not by itself prove that the count is binomial.
Worked Example: Measuring the Binomial–Poisson Variance Gap
A hypothetical group contains 20 independent insured units. Each has probability \(0.10\) of at least one claim during the year. Let \(N\) be the number of units with a claim. Find the binomial mean and variance, then compare the variance with that of a Poisson model having the same mean.
Method and assumptions. There are exactly \(n=20\) trials, each with the same \(p=0.10\), and the outcomes are independent. Therefore \(N\sim\operatorname{Bin}(20,0.10)\).
The mean is \(E[N]=np=20(0.10)=2\). The variance is \(\operatorname{Var}(N)=np(1-p)=20(0.10)(0.90)=1.8\).
The variance-to-mean ratio is \(1.8/2=0.9\). The gap is \(2-1.8=0.2\), which also equals \(np^2=20(0.10)^2=0.2\). A Poisson model with the same mean has parameter \(\lambda=2\) and variance \(2\). Thus the binomial variance is \(0.2\) lower.
Check. The calculated variance is below the mean, as required for a nondegenerate binomial count. A common mistake would be to assign variance \(2\) merely because the mean is \(2\); that would use the Poisson variance rule for a binomial model.
Worked Example: Comparing Two Binomial Designs With the Same Mean
Two hypothetical designs count successes among independent trials. Design A has \(n=10\), \(p=0.40\); Design B has \(n=40\), \(p=0.10\). Calculate and compare their means, variances, and variance-to-mean ratios. Which is more underdispersed relative to a Poisson count with the same mean?
Method and assumptions. For each design, outcomes are independent and have a common success probability within that design, so the binomial formulas apply.
For A, \(E[N_A]=10(0.40)=4\), and \(\operatorname{Var}(N_A)=10(0.40)(0.60)=2.4\). Its variance-to-mean ratio is \(2.4/4=0.60\), and its gap from the mean is \(4-2.4=1.6\).
For B, \(E[N_B]=40(0.10)=4\), and \(\operatorname{Var}(N_B)=40(0.10)(0.90)=3.6\). Its variance-to-mean ratio is \(3.6/4=0.90\), and its gap from the mean is \(4-3.6=0.4\).
Both designs have mean 4, so the equal-mean Poisson benchmark has variance 4 for each. Design A has the smaller variance and the larger gap from 4; it is more underdispersed relative to that benchmark. This agrees with the ratios \(1-p\): \(0.60\) for A versus \(0.90\) for B.
Check. Both variances are less than their common mean. A common mistake is to assume that equal means imply equal variances; the trial probabilities differ, so the binomial variances differ.
Using Moments to Test a Binomial Fit
When the binomial model is assumed and the mean \(\mu\) and variance \(v\) are known, with \(\mu>0\), the variance-to-mean relationship gives \(v/\mu=1-p\). Hence \(p=1-v/\mu\), and then \(n=\mu/p\) if \(p>0\). These relationships are useful for checking a proposed parameter pair. They do not establish that a real count follows a binomial distribution: the equal-probability and independence assumptions must still be justified.
Worked Example: Recovering Binomial Parameters From Moments
A hypothetical count is modelled as binomial. Its stated mean is 6 and its stated variance is 4.8. Find the parameters \(n\) and \(p\), and check that the result is a valid binomial model.
Method and assumptions. Assume a binomial model with independent, equal-probability trials. With \(\mu=6>0\) and \(v=4.8\), first use \(v/\mu=1-p\).
The ratio is \(v/\mu=4.8/6=0.8\), so \(1-p=0.8\) and \(p=0.2\). Now use \(\mu=np\): \(n=\mu/p=6/0.2=30\).
The implied parameters are \(n=30\) and \(p=0.2\). Checking them, \(np=30(0.2)=6\) and \(np(1-p)=30(0.2)(0.8)=4.8\), exactly the stated moments. The trial count is a positive integer and \(p\) lies between 0 and 1, so the parameters are valid.
Check. A common mistake is to calculate \(p=v/\mu=0.8\), omitting the complement in \(v/\mu=1-p\). The correct success probability is \(0.2\); using it also produces the integer trial count \(30\).
Common Mistakes and Practical Interpretation
- Applying the Poisson variance rule to a binomial count. A Poisson count has variance equal to its mean. A binomial count has variance \(np(1-p)\), which is smaller when \(p>0\).
- Forgetting the complement. The binomial variance contains \(1-p\). Omitting it incorrectly makes the variance equal to the mean.
- Using the formula without checking assumptions. A fixed number of independent trials with a common probability is required. If trial probabilities differ or outcomes are dependent, the binomial result cannot be applied as stated.
- Treating underdispersion as proof of a binomial model. A variance below the mean is consistent with binomial underdispersion, but other models or data-generating processes may also have variance below the mean.
- Forgetting feasibility constraints. Under a binomial model, \(0\le\operatorname{Var}(N)\le E[N]\). A proposed variance greater than the mean cannot be matched by any binomial parameters.
Check Your Understanding
For each question, use the stated model and assumptions, and compare variance with mean where requested.
- A hypothetical portfolio has 25 independent units, each with probability \(0.08\) of a claim in the period. Find the mean, variance, variance-to-mean ratio, and gap between mean and variance. (Answer: Mean \(2\); variance \(1.84\); ratio \(0.92\); gap \(0.16\).)
- A binomial count has mean 5 and variance 4. Find its success probability and number of trials, assuming the moments are exact. (Answer: \(p=0.20\), \(n=25\).)
- Compare two binomial counts: one has \(n=8,p=0.50\), and the other has \(n=20,p=0.20\). State each mean and variance, and identify which is more underdispersed relative to a same-mean Poisson model. (Answer: Both means are \(4\); variances are \(2\) and \(3.2\), respectively; the first is more underdispersed.)
- A hypothetical model uses 15 independent trials with common success probability \(0.40\). Calculate the binomial variance and its difference from the mean; compare with the variance of a Poisson count having the same mean. (Answer: Mean \(6\); binomial variance \(3.6\); gap \(2.4\); same-mean Poisson variance \(6\).)
- A proposed binomial model is said to have mean 7 and variance 7.7. Can any binomial parameters produce these moments? Explain using the variance bound. (Answer: No. A binomial variance cannot exceed its mean, but \(7.7>7\).)