Representing a Discrete Loss
A loss does not have to take every value in an interval. It may instead have a set of distinct possible outcomes, such as a fixed deductible amount, a particular repair cost, or no loss at all. A table of those outcomes and their probabilities gives a direct way to describe uncertainty and to calculate quantities such as the probability of a specified event and the average loss.
Here, \(X\) denotes the loss over a stated period, measured in a consistent unit. Unless stated otherwise, the examples use one-period losses, no discounting, and nonnegative dollar amounts. Each table describes the complete distribution of \(X\), not a sample of observed losses.
A loss table is a compact representation of the support and pmf. Each row describes one possible outcome and its probability. The outcomes are mutually exclusive: in a single period, \(X\) cannot equal two different values at once. The rows must also be exhaustive: one of the listed outcomes must occur. An omitted outcome or an incorrect probability can make every later calculation wrong.
| Possible loss | Probability |
|---|---|
| \(x_1\) | \(P(X=x_1)\) |
| \(x_2\) | \(P(X=x_2)\) |
| \(\vdots\) | \(\vdots\) |
| \(x_n\) | \(P(X=x_n)\) |
A value with probability zero can be included for clarity, but it is not part of the positive-probability support. In a loss model, a row with loss zero is often important: it represents the possibility that no covered loss occurs. Do not omit it simply because no payment or loss is incurred in that outcome.
Checking and Using a Probability Mass Function
Before using a table, check that every listed probability is between zero and one and that all the probabilities add to one. If one probability is unknown, use the total-probability condition to solve for it. If the probabilities sum to a value other than one, the table is not yet a valid pmf; do not silently proceed as if it were.
An event is a specified collection of outcomes. Its probability is the sum of the probabilities for the outcomes that satisfy the event. This follows because those outcomes are mutually exclusive, so their probabilities add. Translate inequality signs carefully: for example, \(X<40\) excludes an outcome equal to 40, whereas \(X\leq40\) includes it.
Worked Example: Completing a Loss Table and Finding Its Mean
A hypothetical one-year portfolio loss \(X\), measured in thousands of dollars, has possible values \(0,12,48,\) and \(96\). The assigned probabilities are \(0.55,0.25,0.15,\) and \(q\), respectively. Find \(q\), calculate \(P(12\leq X<96)\), and calculate the expected loss.
Method. First enforce the pmf total, then add the masses for the requested event, and finally calculate the probability-weighted average. The distribution is finite and fully described by these four outcomes and their probabilities; no frequency or severity assumption is needed.
Step 1: Solve for the missing probability. Since the probabilities must total one,
The completed probabilities sum to \(0.55+0.25+0.15+0.05=1.00\), and each is nonnegative, so the table is a valid pmf.
Step 2: Calculate the event probability. The condition \(12\leq X<96\) includes losses of 12 and 48, but excludes both 0 and 96. Thus
Step 3: Calculate expected loss. Weight each loss, expressed in thousands of dollars, by its probability:
The expected loss is 15 thousand dollars, or $15,000.00. As a check, all contributions are nonnegative, and the result lies between the smallest and largest possible losses, 0 and 96 thousand dollars. The common mistake to avoid is misadding the weighted terms: here they total 15, not 22.2 thousand dollars.
Expected Loss as a Weighted Average
For a finite discrete loss, expected loss is the average of the possible outcomes weighted by their probabilities. It is not generally the ordinary average of the listed loss amounts: outcomes with larger probabilities contribute more to the result. The same calculation applies when outcomes and probabilities are stated in a table rather than in a named parametric distribution.
The formula is justified by viewing each outcome as contributing its value in proportion to how often it occurs. For a finite model, imagine \(m\) independent periods generated under the same distribution. If outcome \(x_j\) occurs about \(mP(X=x_j)\) times, the total loss over those periods is approximately \(\sum_j x_jmP(X=x_j)\). Dividing by \(m\), the average loss is approximately \(\sum_jx_jP(X=x_j)\). The exact expectation follows from the linearity of expectation, without requiring independence among different periods for this single-period formula.
Identify every possible loss amount and make sure the unit is consistent across the table.
Confirm that each probability is nonnegative and that the probabilities sum to one.
For an event, include exactly the outcomes satisfying its stated condition, including or excluding boundary values as indicated.
Add event masses or multiply each loss by its mass and sum. Check the unit and whether the result is plausible given the possible losses.
Worked Example: Reading Strict and Inclusive Boundaries
A hypothetical risk register assigns a one-period loss \(X\), in thousands of dollars, the following distribution: \(X=0\) with probability \(0.18\), \(X=10\) with probability \(0.32\), \(X=25\) with probability \(0.27\), \(X=60\) with probability \(0.15\), and \(X=100\) with probability \(0.08\). Calculate \(P(10<X\leq60)\) and \(E[X]\).
Method. Check the table, interpret the strict and inclusive boundaries separately, and then use the weighted-average formula. The probabilities total \(0.18+0.32+0.27+0.15+0.08=1.00\), so the table is a valid finite pmf.
For \(10<X\leq60\), the value 10 is excluded by the strict lower boundary. The values 25 and 60 are included, while 0 and 100 are excluded. Therefore,
Now multiply each outcome, in thousands of dollars, by its probability:
Thus the expected loss is $26,950.00. The event probability is between zero and one, and the expected loss is between the minimum and maximum possible losses of 0 and $100,000. A common boundary error is to include the 10-thousand-dollar outcome: that would give \(0.32+0.27+0.15=0.74\), but it violates \(10<X\). The correct probability is 0.4200.
Probabilities Given as Relative Weights
A table may give relative weights rather than probabilities. Weights are not automatically probabilities: they must first be divided by their total. If the outcomes are mutually exclusive and exhaustive, and the weights are nonnegative, this normalization produces a valid pmf. It can be useful to retain exact fractions during the calculation and round only the final reported result.
Worked Example: Normalizing Weights Before Calculating
A hypothetical planning model assigns losses \(0,25,75,\) and \(150\) thousand dollars relative weights \(6,4,3,\) and \(1\), respectively. Treat the weights as proportional to the probabilities. Find the normalized pmf, calculate \(P(X\geq75)\), and find the expected loss.
Method. Normalize each weight by the total weight, verify the resulting masses, and then calculate the event probability and weighted average. The outcomes and relative weights fully specify this finite distribution once normalized.
The total weight is \(6+4+3+1=14\). Dividing each weight by 14 gives
These probabilities are nonnegative and sum to \((6+4+3+1)/14=1\). The event \(X\geq75\) includes the 75- and 150-thousand-dollar outcomes, so
The expected loss, in thousands of dollars, is
Rounding the final monetary amount to the nearest cent gives $33,928.57. A check using the unrounded value confirms it is between the smallest and largest possible losses. The common mistake is to use the raw weight 4 as an event probability; weights must be normalized, so the event probability is \(4/14\), not 4.
Common Checks Before Reporting a Result
A calculation can be arithmetically correct and still answer the wrong question. Before reporting a result, check the distribution, event definition, and units together.
- Check completeness. The listed outcomes must cover every possibility, including no loss if it can occur.
- Check normalization. Probabilities sum to one; relative weights must be converted into probabilities first.
- Check the event wording. Translate \(<,\leq,>,\geq\) exactly. A boundary outcome may change the answer substantially.
- Check units. If losses are stated in thousands of dollars, the weighted average is also in thousands until converted to dollars.
- Check plausibility. For a finite distribution, the expected loss must lie between the minimum and maximum possible losses.
These checks matter because the pmf is the foundation for further loss analysis. Event probabilities depend on selecting the right rows, while an expected value depends on both the outcomes and their assigned masses. A missing zero-loss outcome, an unnormalized weight, or a boundary misread changes the model being calculated.
Check Your Understanding
For each problem, treat the listed outcomes as the complete one-period loss distribution. Amounts are in thousands of dollars unless stated otherwise.
- A loss \(X\) takes values \(0,8,30,80\) with probabilities \(0.50,0.25,0.15,q\). Find \(q\), \(P(8\leq X<80)\), and \(E[X]\). (Answer: \(q=0.10\); event probability \(0.4000\); expected loss \($14,500.00\))
- A loss \(X\) takes values \(0,15,40,90\) with probabilities \(0.20,0.35,0.30,0.15\). Find \(P(15<X\leq90)\) and \(E[X]\). (Answer: \(0.4500\); \($30,750.00\))
- A model assigns relative weights \(2,5,3\) to losses \(5,20,55\), respectively. Find the normalized probability of a loss at least 20 and the expected loss. (Answer: probability \(0.8000\); \($27,500.00\))
- A loss has outcomes \(0,12,36,120\) with probabilities \(0.45,0.30,0.20,0.05\). Find \(P(X<36)\) and \(E[X]\). (Answer: \(0.7500\); \($16,800.00\))
- A loss \(X\) takes values \(0,25,100\), with probabilities \(0.30,0.50,q\). Find \(q\), \(P(X\geq25)\), and \(E[X]\). (Answer: \(q=0.20\); event probability \(0.7000\); expected loss \($32,500.00\))