Deriving The Binomial Mean

Why the Binomial Mean Is a Product

A binomial claim count describes the number of successes in a fixed set of trials. In an insurance setting, a success might mean that a policy produces at least one claim during a specified period. The expected claim count is useful as a summary of the model: it gives the long-run average number of claims across repetitions of the same portfolio and period.

The mean has a simple form, but the reasoning behind it is important. Instead of treating the total count as one complicated quantity, represent it as a sum of individual indicators. Each indicator records whether one particular trial is a success. Adding the expected values of those indicators gives the expected total.

Definition: Suppose there are \(n\) trials, each with two outcomes designated success and failure. If the trials are independent and each has the same success probability \(p\), where \(n\) is a nonnegative integer and \(0\leq p\leq 1\), the number of successes \(N\) has a binomial distribution with parameters \(n\) and \(p\), written \(N\sim\operatorname{Binomial}(n,p)\). Its possible values are the integers from \(0\) through \(n\).

For claim-count applications, define success carefully. If success means that a policy has at least one claim during the period, then \(p\) is the probability of at least one claim for an individual policy. The resulting binomial count is a count of policies with claims, not necessarily a count of every claim if one policy can have multiple claims.

Indicators: A Trial-by-Trial Representation

For each trial \(j\), define an indicator variable \(I_j\). It equals 1 when trial \(j\) is a success and 0 when it is a failure. Thus, \(I_j\) is a Bernoulli random variable with success probability \(p\). If \(N\) counts all successes, then each success contributes one to the total and each failure contributes zero.

I_j= \begin{cases} 1, & \text{if trial } j \text{ is a success},\\ 0, & \text{if trial } j \text{ is a failure}, \end{cases} \qquad N=\sum_{j=1}^{n} I_j.

The expected value of one indicator follows directly from its two possible values. It is 1 with probability \(p\) and 0 with probability \(1-p\). Therefore, its expected value is \(1\cdot p+0\cdot(1-p)=p\).

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

Expected value is linear: the expected value of a finite sum is the sum of the expected values. Applying this property to the indicator representation gives the binomial mean.

Theorem: If \(N\sim\operatorname{Binomial}(n,p)\), then \(E[N]=np\). The formula also holds at the boundary values \(p=0\) and \(p=1\).

Derivation of the Binomial Mean

The indicator representation explains why the formula holds. Each of the \(n\) trials contributes an expected amount \(p\) to the count. Adding those \(n\) contributions gives \(np\).

Write the total as a sum.
Let \(I_j\) indicate success on trial \(j\). Then \(N=I_1+\cdots+I_n\).

Take expected values.
By linearity of expectation, \(E[N]=E[I_1]+\cdots+E[I_n]\).

Use the common success probability.
Every indicator has expected value \(p\), so the sum has \(n\) terms each equal to \(p\).

Simplify.
Therefore, \(E[N]=p+\cdots+p=np\).

E[N] =E\left[\sum_{j=1}^{n} I_j\right] =\sum_{j=1}^{n}E[I_j] =\sum_{j=1}^{n}p =np.

Independence is part of the assumptions that make \(N\) binomial. However, the addition of expected values in this derivation does not itself require independence. It requires that the expected value of each indicator is \(p\). This distinction matters: one cannot automatically call a count binomial just because it is a sum of indicators, but the same linearity argument can still be useful for other counts.

Applying the Mean in Claim-Count Settings

Before using \(E[N]=np\), identify what one trial represents, what counts as success, and the period over which success is measured. Then confirm that the model uses a fixed number of trials, independent outcomes, and a common success probability. The result is an expected count; it need not be an integer and it does not predict the exact count in a particular period.

Worked Example: Expected Claims Among Service Plans

A hypothetical provider issues 18 one-year service plans. For each plan, the probability of at least one covered claim during the year is 0.10. Assume the plans’ claim indicators are independent and have the same probability. Find the expected number of plans with a claim and the expected number without a claim.

Model and method. Let \(N\) be the number of plans with at least one claim. There are \(n=18\) independent trials, and each has success probability \(p=0.10\), so \(N\sim\operatorname{Binomial}(18,0.10)\). Use the indicator derivation of the binomial mean.

Expected number with a claim. Each of the 18 indicators has expected value 0.10. Adding them gives

$$ E[N]=18(0.10)=1.80. $$

Expected number without a claim. The probability of no claim for one plan is \(1-0.10=0.90\). Equivalently, the number without a claim is \(18-N\). Thus its expected value is

$$ E[18-N]=18-E[N]=18-1.80=16.20. $$

Check. The two expected counts sum to the number of plans: \(1.80+16.20=18\). The decimal expected count is valid; an actual observed count must be a whole number.

Method and mistake avoided. The method is to sum the expected indicator for each plan. Do not round 1.80 to 2 before computing the expected number without a claim; rounding changes the model’s expected value.

Worked Example: Recovering the Claim Probability

A hypothetical portfolio contains 250 independent equipment-rental contracts for a fixed period. Each contract either has at least one claim or has no claim, with the same probability across contracts. The probability of no claim for one contract is 0.96. Find the expected number of contracts with a claim and the expected number without a claim.

Model and method. Success means at least one claim. Since the no-claim probability is 0.96, the success probability is its complement:

$$ p=1-0.96=0.04. $$

Let \(N\) be the number of contracts with a claim. Under the stated independence and common-probability assumptions, \(N\sim\operatorname{Binomial}(250,0.04)\). Therefore,

$$ E[N]=np=250(0.04)=10. $$

There are two equivalent ways to obtain the expected number without a claim. The complement probability is \(0.96\), giving \(250(0.96)=240\). Or use the total number minus the expected number with a claim:

$$ E[250-N]=250-E[N]=250-10=240. $$

Check. The expected counts add to \(10+240=250\), and the probabilities add to \(0.04+0.96=1\).

Method and mistake avoided. First convert the no-claim probability to the claim probability. Using 0.96 directly in \(np\) would calculate the expected number without a claim, not the expected number with a claim.

Worked Example: Combining Identical Blocks of Accounts

A hypothetical insurer reviews four separate blocks of 35 accounts each. Each account independently has probability 0.06 of producing at least one claim during the review period, and the probability is the same for all accounts. Find the expected total number of accounts with a claim. Then find how much the expected count would increase if a fifth block of 35 accounts with the same assumptions were added.

Model and method. The four blocks contain \(4(35)=140\) accounts. Since every account has the same success probability and the outcomes are independent, the total count \(N\) has distribution \(\operatorname{Binomial}(140,0.06)\). Its expected value is

$$ E[N]=140(0.06)=8.40. $$

The additional block adds 35 accounts, each with expected indicator value 0.06. Its expected contribution is

$$ 35(0.06)=2.10. $$

The five-block portfolio has \(5(35)=175\) accounts, so its expected count is \(175(0.06)=10.50\). The increase is \(10.50-8.40=2.10\), agreeing with the expected contribution from the added block.

Check. The original portfolio’s mean is also \(4[35(0.06)]=4(2.10)=8.40\). The five-block mean is \(5(2.10)=10.50\).

Method and mistake avoided. Count the accounts across all blocks before multiplying by the common probability, or calculate each block’s mean and add. Do not multiply by four twice: 140 already includes all four blocks.

Worked Example: Finding a Common Probability From a Target Mean

A hypothetical monitoring program covers 80 identical devices for one quarter. Assume each device independently has probability \(p\) of generating at least one alert in the quarter. The expected number of devices with an alert is 6. Find \(p\) and the expected number of devices with no alert.

Model and method. Let \(N\) be the alert count. Under the given fixed-size, independent, common-probability assumptions, \(N\sim\operatorname{Binomial}(80,p)\). Set the mean formula equal to the stated mean and solve:

E[N]=80p=6, \qquad p=\frac{6}{80}=0.075.

The probability of no alert for one device is \(1-0.075=0.925\). The expected number without an alert is therefore

$$ 80(0.925)=74. $$

Check. The mean number with alerts is \(80(0.075)=6\), as specified, and \(6+74=80\).

Method and mistake avoided. Solve \(np=E[N]\) for the per-device probability, then use the complement for the no-alert count. Do not interpret the mean 6 as a probability; it is an expected count.

Interpreting and Checking the Result

The binomial mean is governed by two quantities: the number of trials and the probability of success on each trial. Holding \(p\) fixed, increasing the number of trials increases the mean in direct proportion. Holding \(n\) fixed, increasing \(p\) also increases the mean in direct proportion. The formula is a statement about the average across repetitions of the model, not a guarantee about any single observed count.

A useful check is to calculate the expected number of failures as \(n(1-p)\). The expected successes and failures must add to \(n\):

$$ np+n(1-p)=n. $$

This check can reveal a complement error, such as using a no-claim probability when the question asks for claim counts. It can also reveal premature rounding or an incorrect number of trials.

Common Exam Traps

The most frequent errors concern the meaning of \(p\), the number of trials, and the interpretation of an expectation.

Using the complement probability for the requested count. If \(p\) is the probability of a claim, then \(np\) is the expected number with a claim. The expected number without a claim is \(n(1-p)\).
Changing what a trial counts. If each trial is one policy and success means at least one claim, \(N\) counts policies with claims. It does not count multiple claims from the same policy as separate successes.
Assuming that a mean is an observed outcome. A mean such as 8.4 does not mean 8.4 accounts can have claims in one observation. It summarizes the model’s expected count over repeated applications.
Using the formula without checking the model. The binomial model requires a fixed number of trials, two outcomes for each trial, independence, and a common success probability. If these assumptions do not hold, \(np\) is not automatically the mean of a binomial count.
Rounding too early. Keep the probability and the product \(np\) at their stated precision until the calculation is complete. An expected count need not be rounded to an integer.

Key takeaway: Write the count as a sum of success indicators. Each indicator has expected value \(p\), so \(n\) trials contribute \(n\) terms of size \(p\), giving \(E[N]=np\). Confirm what one trial and one success represent before substituting values.

Check Your Understanding

For each problem, identify the binomial parameters where requested and use the indicator-based mean. Treat all scenarios as hypothetical.

A group of 90 independent warranty contracts each has probability 0.03 of at least one claim during the coverage period. What is the expected number of contracts with a claim? (Answer: 2.70)
For each of 160 independent rental units, the probability of no claim in a month is 0.94. Find the expected number of units with a claim. (Answer: 9.60)
Three independent batches each contain 25 accounts. All accounts have the same independent probability, 0.08, of at least one claim. Find the expected total number with claims and the expected number without claims. (Answer: 6 with claims; 69 without claims)
A portfolio of 120 independent policies has a common probability \(p\) of at least one claim. If the expected number with claims is 9, find \(p\) and the expected number without claims. (Answer: \(p=0.075\); 111 without claims)
Compare two hypothetical independent portfolios with a common claim probability within each portfolio: Portfolio A has 80 policies with probability 0.05, while Portfolio B has 50 policies with probability 0.08. Find each expected claim count and state whether the two binomial models have the same parameters. (Answer: both expected counts are 4; the models do not have the same parameters)

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