Why Use an Indicator for Each Risk?
A portfolio claim count can be built from simple yes-or-no outcomes. For each insured risk, ask whether it produces at least one claim during a specified period. The answer is either 1 or 0, so it can be represented by an indicator random variable. This representation makes each risk’s contribution to the total count explicit, even when different risks have different claim probabilities.
The event must be defined carefully. If a risk can produce several claims during the period, an indicator for “at least one claim” records only whether the risk had any claim; it is not the number of claims. The portfolio count formed from such indicators counts risks with claims, not individual claims.
The claim event could mean, for example, “this hypothetical vehicle has at least one covered claim during the calendar year.” The period and event definition should be the same for all risks being counted. The probabilities \(p_j\) need not be equal. Also, defining the indicators does not itself imply that the risks are independent.
Mean and Variance of One Indicator
An indicator has just two possible values. It equals 1 with probability \(p_j\) and 0 with probability \(1-p_j\). Its expected value is therefore its probability of being 1:
Since \(I_j^2=I_j\) for both possible values, \(E[I_j^2]=p_j\). Applying \(\operatorname{Var}(I_j)=E[I_j^2]-E[I_j]^2\) gives:
This also has a useful interpretation. The variance is 0 if \(p_j=0\) or \(p_j=1\), because the outcome is then certain. It is largest when \(p_j=0.5\): a yes-or-no outcome is most variable when either result is equally likely.
Combining Indicators Into a Portfolio Count
For a portfolio of \(m\) risks, the count \(N=\sum_{j=1}^{m}I_j\) adds one for each risk that has the specified claim event. By the linearity of expectation, the count’s mean is the sum of the individual claim probabilities. This result does not require independence.
Variance requires more care. For any collection of indicators,
The covariance term captures whether two risks’ claim events tend to occur together. For a pair \(j,k\),
If the indicators are independent, every covariance is zero, so the count variance is the sum of the individual variances. If the risks are dependent, simply adding those variances is generally wrong. Positive covariance increases the count variance; negative covariance decreases it. The mean formula remains valid in either case.
The variance formula follows by expanding the square in the definition of variance. Write \(N-E[N]=\sum_j(I_j-p_j)\), then square this sum. Each squared term has expectation \(p_j(1-p_j)\). Each cross-product has expectation \(\operatorname{Cov}(I_j,I_k)\), and each pair appears twice in the square. Adding these terms gives the displayed expression.
Worked Example: Independent Risks With Different Claim Probabilities
A hypothetical insurer tracks four equipment risks over one year. Their probabilities of at least one covered claim are \(0.04\), \(0.08\), \(0.12\), and \(0.20\). Assume the four indicators are independent. Find the expected number of risks with a claim, the variance of that count, and the probability that no risk has a claim.
Method. Represent each risk by an indicator \(I_j\). Use linearity for the mean and add individual variances because independence is given.
Expected count. Add the four claim probabilities:
Count variance. For each risk, calculate \(p_j(1-p_j)\):
Independence makes every covariance zero, so
Probability of no risk having a claim. For each risk the no-claim probability is \(1-p_j\). Independence lets us multiply these probabilities:
To four decimal places, \(P(N=0)=0.6218\). As a check, the expected count \(0.44\) is less than one, which is consistent with the relatively high probability of no claims. The method avoids treating these unequal probabilities as though they were all the same.
Worked Example: A Shared Cause Creates Dependence
Two hypothetical commercial properties have claim indicators \(I_1\) and \(I_2\). Their marginal claim probabilities are \(p_1=0.20\) and \(p_2=0.30\), and \(P(I_1=1,I_2=1)=0.10\). A third risk has claim probability \(0.10\), and its indicator is independent of the pair. Find the mean and variance of the total count \(N=I_1+I_2+I_3\).
Method. Use the marginal probabilities for the mean. For the variance, calculate the covariance of the first two indicators and include it; the third indicator contributes no covariance because it is independent of the pair.
Mean.
Individual variances.
Covariance. The joint claim probability is not the product \(p_1p_2\):
Thus,
The check is that omitting dependence would give \(0.46\), which is lower by \(0.08\), exactly twice the covariance. The common mistake is to add the individual variances without checking whether the indicators are independent.
Worked Example: Measuring the Effect of Risk Mitigation
A hypothetical three-risk portfolio has independent indicators with claim probabilities \(0.10\), \(0.03\), and \(0.07\). A prevention measure changes the first risk’s probability from \(0.10\) to \(0.04\), leaving the others unchanged. Find the change in the expected count and the variance.
Method. Calculate the portfolio mean and variance before and after the change. Because the risks are independent, each variance is \(p(1-p)\), and the portfolio variance is their sum.
Before mitigation:
After mitigation:
The expected count decreases by \(0.20-0.14=0.06\), and the variance decreases by \(0.1842-0.1326=0.0516\). The variance change is not simply the decrease in claim probability: the first risk’s variance changes from \(0.0900\) to \(0.0384\). The result avoids confusing a change in the mean with a change in variability.
Common Mistakes and Interpretation
- Counting the wrong event. An indicator for “at least one claim” is 1 even if the risk has several claims. Summing those indicators counts affected risks, not total claims.
- Assuming independence without evidence. Different claim probabilities do not establish independence. If shared conditions or other dependence are present, include the covariance terms in the variance.
- Using \(p_j\) as the variance. The indicator variance is \(p_j(1-p_j)\), not \(p_j\). The value \(p_j\) is its mean.
- Adding probabilities to find a probability of any claim. The sum \(\sum_j p_j\) is the expected count, not generally the probability that at least one risk has a claim. Under independence, the latter is \(1-\prod_j(1-p_j)\).
Indicators are useful because they make heterogeneous risks straightforward to combine: each contributes its own probability to the expected count. For a variance calculation, however, the dependence structure matters. This distinction is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.
Check Your Understanding
For each problem, define the relevant indicator or indicators before calculating. Probabilities are for the stated period.
- A single hypothetical risk has probability \(0.07\) of a claim. Find the indicator’s mean and variance. (Answer: mean \(0.07\); variance \(0.0651\))
- Three independent risks have claim probabilities \(0.10\), \(0.20\), and \(0.40\). Find the expected count and variance. (Answer: mean \(0.70\); variance \(0.49\))
- Two indicators have marginal claim probabilities \(0.25\) and \(0.40\), with joint claim probability \(0.15\). Find the mean and variance of their count. (Answer: mean \(0.65\); variance \(0.5275\))
- Three independent risks have probabilities \(0.05\), \(0.10\), and \(0.15\). A measure reduces the second probability to \(0.04\). Find the change in the expected count and variance. (Answer: expected count decreases by \(0.06\); variance decreases by \(0.0516\))
- Two indicators have probabilities \(0.20\) and \(0.30\), with joint claim probability \(0.08\). A third indicator, independent of both, has probability \(0.10\). Find the mean and variance of the three-risk count. (Answer: mean \(0.60\); variance \(0.50\))