Two Prediction Outputs, Two Different Questions
A prediction can describe what is expected for one entity or what is expected across a collection of entities. These outputs are related, but they answer different questions. A reviewer deciding which applications to examine needs information about particular applications. A manager setting a portfolio budget needs an estimate of the total cost across the portfolio. One output cannot always substitute for the other.
The unit of prediction is the entity for which a prediction is made, such as a policy, claim, customer, or branch. The aggregate unit is the collection over which predictions are combined, such as all policies renewing next month. The population and time period must be specified: an estimate of claims next quarter for 500 policies is not the same target as an estimate of claims next year for those policies.
The form of the individual prediction depends on the target. For a binary event, the prediction for entity \(i\) may be its probability \(p_i\) of experiencing the event during the horizon. For a numeric target, it may be an expected amount \(m_i\), such as expected claim cost. A portfolio forecast could then be the expected event count or expected total cost, respectively.
The appropriate output follows from the intended use. Individual predictions preserve differences between entities and can support a review of specific cases. An aggregate forecast summarizes the group and can support planning for the group as a whole. If both questions matter, retain the individual predictions and also calculate the relevant aggregate.
From Individual Predictions to an Expected Total
Suppose \(Y_i\) is a binary indicator equal to 1 if entity \(i\) experiences an event during the horizon and 0 otherwise. If the predicted probability for that entity is \(p_i=P(Y_i=1)\), then its expected event count is \(E[Y_i]=p_i\). For \(n\) entities, the portfolio event count is \(N=\sum_{i=1}^n Y_i\). By linearity of expectation:
This result does not require the entities’ outcomes to be independent. It follows because an expected value of a sum is the sum of the expected values. Independence matters for other quantities, such as the variance of the total or the probability of no events, but not for this expected count.
For a numeric outcome \(X_i\), such as cost for entity \(i\), the same principle gives \(E[\sum_i X_i]=\sum_i E[X_i]\), when the expectations exist. This produces an expected aggregate cost when each prediction is an expected cost for the same horizon and the entities are counted consistently. It does not say that the realized total will equal that expected value.
Worked Example: Case-Level Event Predictions and a Portfolio Count
A hypothetical insurer is reviewing five policies for a specified one-year period. Its estimates of the probability of at least one claim for policies A through E are 0.08, 0.12, 0.20, 0.35, and 0.05. Calculate the expected number of policies with a claim, the average predicted probability, and explain which result answers a case-review question.
Method. Keep the probabilities attached to their policies for case-level use, and sum the probabilities to obtain the expected event count.
The individual predictions are \(p_A=0.08\), \(p_B=0.12\), \(p_C=0.20\), \(p_D=0.35\), and \(p_E=0.05\). The expected claim count is
The average predicted probability is
Thus the portfolio has an expected 0.80 policies with a claim, while the mean probability across policies is 0.16, or 16%. The probabilities for A through E are the relevant output if a reviewer needs to examine particular policies. The value 0.80 summarizes the expected count for the group; it does not identify which policy will have a claim.
Common mistake and correction. A mistaken calculation treats the average probability, 0.16, as the expected number of claims. That confuses a per-policy average with a portfolio total. The expected count is \(5(0.16)=0.80\), equivalently the sum of the five probabilities. As a check, the expected count is between 0 and 5, as it should be.
Aggregating Expected Costs Across Groups
Budget planning often concerns a numeric total rather than an event count. If entities have different expected costs, multiplying one average by the number of entities can be misleading unless that average correctly represents the group being forecast. One safe approach is to calculate each group’s contribution separately, then add the contributions.
Worked Example: Forecasting a Portfolio’s Expected Cost
A hypothetical service portfolio contains 40 contracts in Group 1, 30 in Group 2, and 10 in Group 3. For the same upcoming year, their estimated expected cost per contract is $125, $210, and $480, respectively. Calculate the portfolio’s expected cost and the expected cost per contract.
Method. Multiply each group’s contract count by its expected cost per contract, then sum the group totals.
Group 1 contributes \(40\times \$125=\$5{,}000\). Group 2 contributes \(30\times \$210=\$6{,}300\). Group 3 contributes \(10\times \$480=\$4{,}800\). Therefore,
There are \(40+30+10=80\) contracts. The expected cost per contract across the full portfolio is
The aggregate output for budget planning is an expected total of $16,100 for the stated year. The expected per-contract amount of $201.25 is an average across the entire portfolio; it is not the expected cost for every contract. For example, it does not replace the Group 3 estimate of $480 when information about that group is needed.
Common mistake avoided. Applying the Group 1 estimate of $125 to all 80 contracts would give \(80\times \$125=\$10{,}000\), understating the total because it ignores the other groups’ expected costs. The group-by-group calculation uses every contract once and checks back to 80 contracts in total.
An Expected Total Is Not a Certain Total
An expected count or cost is a summary of uncertainty, not a promise about the realized outcome. A portfolio may have an expected count of 1.5 events, although the actual count must be a whole number. A budget forecast of $16,100 likewise does not imply that actual spending will equal $16,100. The expected value describes a long-run average over comparable repetitions, or a probability-weighted average of possible outcomes.
The distinction can be made concrete for a small group. Under independence, a binary event count has a variance that depends on the probabilities as well as its expected value. For entity \(i\), a binary indicator with event probability \(p_i\) has variance \(p_i(1-p_i)\). If the indicators are independent, the variance of their sum is the sum of their variances. If they are dependent, covariance terms must also be included.
This formula explains why an expected total alone does not describe the full range of possible totals. Dependence can change how likely unusually high or low totals are, even though it does not change the sum of the individual expected counts. Do not report an uncertainty calculation that assumes independence unless that assumption is justified for the application.
Worked Example: Expected Count and Probability of No Events
A hypothetical monitoring team tracks four independent sites over the same month. Their estimated probabilities of an incident are 0.10, 0.25, 0.40, and 0.75. Calculate the expected incident count, its variance, and the probability of no incidents. State what independence is used for.
Method. Sum probabilities for the expected count. Under the stated independence assumption, sum the Bernoulli variances for the count variance and multiply the no-incident probabilities for the probability of zero incidents.
The expected count is
The variance contributions are \(0.10(0.90)=0.09\), \(0.25(0.75)=0.1875\), \(0.40(0.60)=0.24\), and \(0.75(0.25)=0.1875\). Therefore,
For no incidents, each site must have no incident. Independence allows the probabilities to be multiplied:
The probability of no incidents is 0.1013 to four decimals. The expected count is 1.50, but the possible realized counts are 0, 1, 2, 3, or 4. Independence was needed for the variance sum and the product for \(P(N=0)\), not for the expected count. The calculation is internally consistent: a no-incident probability near 0.10 is possible even with an expected count of 1.50 because the expected value averages over all possible counts.
Choosing the Output That Supports the Decision
Before producing a prediction, identify whether the decision concerns separate entities, a group total, or both. The following distinction is useful:
| Decision question | Useful prediction output | What the output does not provide by itself |
|---|---|---|
| Which specific cases may warrant review? | Individual predictions that remain linked to the cases | A guarantee that a selected case will experience the outcome |
| What total event count should be planned for? | Expected count formed from compatible individual event probabilities | The identities of the entities that will experience events |
| What overall cost should be budgeted? | Expected aggregate cost for the defined group and horizon | A certainty that realized spending will equal the estimate |
| How does the portfolio total vary? | An uncertainty summary using suitable assumptions about joint outcomes | A valid uncertainty estimate if dependence is ignored without justification |
An aggregate forecast can hide substantial variation among entities. Conversely, a list of individual predictions may be cumbersome when the immediate question is the total expected amount. A sound analysis preserves the level of detail needed for the decision rather than replacing every individual prediction with a single average.
Check the following before interpreting an aggregate: the individual predictions refer to the same outcome definition; they use the same forecast horizon; the entities form the intended group without accidental duplication; and the total is expressed in a suitable unit. If one group’s prediction is a monthly cost and another’s is an annual cost, they must be made comparable before being added.
Check Your Understanding
For each problem, distinguish the individual-level quantities from the group-level quantity and check that the time period and units are consistent.
- A hypothetical claims team has six policies with one-year claim probabilities of 0.05, 0.10, 0.15, 0.20, 0.25, and 0.30. Find the expected number of policies with claims and the average predicted probability. (Answer: expected count \(=1.05\) policies; average probability \(=0.175\), or 17.5%.)
- A hypothetical portfolio has 25 units with an expected annual cost of $84 per unit and 15 units with an expected annual cost of $140 per unit. Calculate the expected annual total. (Answer: \(25(\$84)+15(\$140)=\$4{,}200\).)
- Three independent hypothetical devices have probabilities 0.20, 0.50, and 0.80 of failing during a specified period. Find the expected failure count, the variance of that count, and the probability of no failures. (Answer: expected count \(=1.50\); variance \(=0.16+0.25+0.16=0.57\); probability of no failures \(=0.80(0.50)(0.20)=0.0800\).)
- A hypothetical network has 12 branches with an expected annual cost of $300 each, 8 branches with an expected annual cost of $450 each, and 5 branches with an expected annual cost of $700 each. Find the total expected annual cost and the mean expected cost per branch. (Answer: total \(=\$10{,}700\); mean \(=\$10{,}700/25=\$428.00\) per branch.)
- Ten hypothetical cases each have an event probability of 0.10, but their outcomes may be dependent. What is the expected event count, and can the variance be calculated by adding ten Bernoulli variances without further information? (Answer: expected count \(=10(0.10)=1.00\), regardless of dependence; no, the variance also depends on the covariances between cases.)