Binomial Moments From Indicator Variables

Why Indicator Variables Clarify Binomial Moments

A claim count can be viewed as a total across a fixed set of exposures. For each exposure, define whether the specified claim event occurred; the total count is then the sum of those yes-or-no outcomes. This representation makes the mean and variance transparent: the mean adds across exposures, while independence determines whether the variances can also be added without extra terms.

The key is to define an exposure and a claim event precisely. An exposure might be one policy during one year, or one policy during one quarter. The event might be “at least one claim during that period.” The probability \(p\) must refer to that event and that exposure period. A binomial model is appropriate only under the assumptions below.

Definition: A count \(N\) has a binomial distribution with parameters \(n\) and \(p\) if it is the sum of \(n\) independent indicator variables \(I_1,\ldots,I_n\), where each \(I_j\) equals 1 when the specified event occurs on trial \(j\), equals 0 otherwise, and satisfies \(P(I_j=1)=p\). Thus \(N=\sum_{j=1}^n I_j\), with \(N\sim\operatorname{Bin}(n,p)\). The number \(n\) is fixed, the probability \(p\) is the same on every trial, and the trials are independent.

In an insurance application, a trial is often an exposure unit rather than a claim. If \(I_j=1\) when policy \(j\) has at least one claim during the stated period, then \(N\) counts policies with at least one claim, not necessarily the total number of claims. Those counts coincide only when the problem’s event and assumptions make them coincide.

Deriving the Mean and Variance

An indicator variable has only two possible values. It equals 1 with probability \(p\) and 0 with probability \(1-p\). Its expectation and variance follow directly from these two outcomes.

E[I_j]=1\cdot p+0\cdot(1-p)=p

E[I_j^2]=1^2\cdot p+0^2\cdot(1-p)=p, \qquad \operatorname{Var}(I_j)=E[I_j^2]-(E[I_j])^2=p-p^2=p(1-p)

Since \(N=\sum_{j=1}^n I_j\), linearity of expectation gives the mean, whether or not the indicators are independent. For the binomial model, the indicators are independent, so every covariance between distinct indicators is zero. Expanding the variance of a sum then gives the binomial variance.

Write the count as a sum.
\(N=I_1+\cdots+I_n\), where each indicator records the specified event on one trial.

Add the expectations.
\(E[N]=\sum_{j=1}^n E[I_j]=np\), because each trial has event probability \(p\).

Expand the variance.
\(\operatorname{Var}(N)=\sum_{j=1}^n\operatorname{Var}(I_j)+2\sum_{j<k}\operatorname{Cov}(I_j,I_k)\). Independence makes every covariance zero.

Add the indicator variances.
\(\operatorname{Var}(N)=n p(1-p)\), since each of the \(n\) indicators has variance \(p(1-p)\).

The derivation shows why both conditions matter. Equal event probabilities give the common value \(p\) in each indicator’s expectation and variance. Independence removes the covariance terms. If the trials have different probabilities, the count is generally not binomial; if the trials are dependent, the covariance terms may not be zero.

Theorem: If \(N\sim\operatorname{Bin}(n,p)\), then \(E[N]=np\) and \(\operatorname{Var}(N)=np(1-p)\). The standard deviation is \(\sqrt{np(1-p)}\). The variance is measured in squared count units; the standard deviation is in count units.

Worked Applications to Exposure Counts

Before substituting into the formulas, translate the scenario into the trial definition, the fixed number of trials \(n\), and the probability \(p\) of the event being counted. That step prevents a correct formula from being applied to the wrong exposure or to the wrong event.

Worked Example: Quarterly Claimant Count

A hypothetical insurer covers 80 independent policies for one quarter. For each policy, the probability of at least one claim during that quarter is 0.05. Let \(N\) be the number of policies with at least one claim. Find the mean, variance, and standard deviation of \(N\).

Method and model. One trial is one policy during the quarter, and the event is at least one claim on that policy during the quarter. The problem states a common probability and independence, so \(N\sim\operatorname{Bin}(80,0.05)\).

The mean is \[ E[N]=np=80(0.05)=4. \] Thus the expected number of policies with at least one claim is 4.

The variance is \[ \operatorname{Var}(N)=np(1-p)=80(0.05)(1-0.05)=80(0.05)(0.95)=4(0.95)=3.8. \] The standard deviation is \[ \sqrt{3.8}\approx1.9494. \]

Check. The mean lies between 0 and 80, as a count should. The variance is below the mean because \(p(1-p)<p\) for \(p>0\). The common mistake avoided here is interpreting the count as the number of claims; the defined \(N\) counts policies with at least one claim.

Worked Example: Converting a Claim-Free Probability

A hypothetical portfolio has 600 independent one-year policy exposures. For each exposure, the probability of no claim during the year is 0.97. Let \(N\) count the exposures with at least one claim. Calculate \(E[N]\), \(\operatorname{Var}(N)\), and the standard deviation.

Method and model. The stated probability is for no claim, but \(N\) counts exposures with at least one claim. These are complementary events, so the claim probability for one exposure is \[ p=1-0.97=0.03. \] With a fixed 600 exposures, a common event probability, and independence, \(N\sim\operatorname{Bin}(600,0.03)\).

The mean is \[ E[N]=600(0.03)=18. \] The variance is \[ \operatorname{Var}(N)=600(0.03)(1-0.03)=600(0.03)(0.97)=18(0.97)=17.46. \] The standard deviation is \[ \sqrt{17.46}\approx4.1785. \]

Check. The expected number of claim-free exposures is \(600(0.97)=582\), and \(600-582=18\), matching the calculated mean. The common mistake avoided is using 0.97 as the claim probability, which would count the opposite event.

Worked Example: Counting Policy-Period Exposures

A hypothetical study follows 40 policies over two six-month periods. Assume that, for every policy-period, the probability of at least one claim is 0.04, and that all 80 policy-period outcomes are independent, including the two periods for the same policy. Let \(N\) count policy-periods with at least one claim. Find the mean and variance.

Method and model. The trial is a policy-period, not a policy over the full year. There are \(n=40(2)=80\) trials. The stated independence assumption applies across all these trials, and every trial has \(p=0.04\). Therefore \(N\sim\operatorname{Bin}(80,0.04)\).

The mean is \[ E[N]=80(0.04)=3.2. \] The variance is \[ \operatorname{Var}(N)=80(0.04)(1-0.04)=80(0.04)(0.96)=3.2(0.96)=3.072. \] The standard deviation is \[ \sqrt{3.072}\approx1.7527. \]

Check. The modeled count ranges from 0 to 80 because it counts 80 policy-period trials. It does not count the number of distinct policies with a claim over the full year. The common mistake avoided is using 40 as \(n\) when the defined trial is a six-month policy-period.

Common Errors and Model Boundaries

A frequent error is to use \(np^2\) for the variance. That expression is not the variance of a binomial count. For example, with \(n=80\) and \(p=0.05\), the incorrect calculation gives \(80(0.05)^2=0.2\). The correct calculation is \[ 80(0.05)(1-0.05)=3.8. \] The factor \(1-p\) comes from the variance of each Bernoulli indicator, \(p(1-p)\); it does not come from squaring the probability.

Another error is to substitute a probability for the wrong outcome. If \(p\) is given as the probability of no claim but the count records exposures with claims, use \(1-p\) as the event probability. Also check the time unit: a quarterly claim probability applies to quarterly trials, not automatically to annual trials. The number of trials must match the exposure unit used to define the probability.

Finally, a fixed count of exposures does not by itself make the count binomial. The event probability must be the same on every trial, and the outcomes must be independent. For example, policyholders with different claim probabilities cannot in general be represented by one binomial model using a simple average probability. If outcomes for different exposures are dependent, the covariance terms in the variance expansion need not vanish. Use \(np\) and \(np(1-p)\) only when the binomial assumptions are appropriate.

Key takeaway: Define one trial and the event being counted before identifying \(n\) and \(p\). For a fixed number of independent trials with a common event probability, represent the count as a sum of indicators to obtain \(E[N]=np\) and \(\operatorname{Var}(N)=np(1-p)\).

Check Your Understanding

For each question, identify the relevant exposure count and event probability before calculating.

A hypothetical group has 150 independent one-year policy exposures, each with probability 0.02 of at least one claim. Find the mean and variance of the number of exposures with a claim. (Answer: mean \(3\); variance \(2.94\).)
There are 240 independent monthly exposures. Each has probability 0.985 of being claim-free during the month. Find the mean and variance of the number with at least one claim. (Answer: mean \(3.6\); variance \(3.546\).)
A hypothetical portfolio has 75 independent quarterly exposures, each with probability 0.08 of at least one claim in the quarter. Find the mean, variance, and standard deviation of the claimant count. (Answer: mean \(6\); variance \(5.52\); standard deviation \(\sqrt{5.52}\approx2.3495\).)
A study counts claim-bearing policy-periods for 90 policies over two periods. Assume all 180 policy-period outcomes are independent and each has probability 0.025 of at least one claim. Find the mean and variance of the count. (Answer: mean \(4.5\); variance \(4.3875\).)
A hypothetical set of 300 independent annual exposures has probability 0.01 of at least one claim per exposure. Find the mean, variance, and standard deviation. (Answer: mean \(3\); variance \(2.97\); standard deviation \(\sqrt{2.97}\approx1.7234\).)

What You'll Learn