Two-Point And Finite-Support Severity Models

Why Finite-Support Severity Models Matter

A loss severity model describes the amount of an individual loss, denoted by $X$. In some hypothetical insurance settings, the loss can take only a small number of specified values. For example, a simplified travel policy might pay only one of two fixed benefits, or a small portfolio model might assign losses to a few possible amounts. These models are useful for direct calculations and for checking whether a more elaborate model has been implemented correctly.

The calculation method is simple, but accuracy depends on using every possible outcome and its probability. A loss amount of zero is still a possible outcome and must be included in sums for moments, even though it contributes zero to those sums. It can also affect event probabilities, such as $P(X>0)$.

Definition: A severity random variable $X$ has finite support if it can take one of a finite list of distinct values $x_1,\ldots,x_n$, with probabilities $p_i=P(X=x_i)$. A valid probability table has $p_i\geq 0$ for each $i$, and $\sum_{i=1}^n p_i=1$. A two-point severity model is the special case $n=2$.

Probabilities and Moments From a Table

For a finite-support model, the probability of an event is the sum of the probabilities of the support values that satisfy the event. The mean is a probability-weighted average: multiply each possible value by its probability and add. The same rule applies to raw moments, after raising each value to the required power.

P(X\in B)=\sum_{i:x_i\in B}p_i,\qquad E[X]=\sum_{i=1}^n p_i x_i,\qquad E[X^2]=\sum_{i=1}^n p_i x_i^2.

The variance follows from the standard identity $\operatorname{Var}(X)=E[X^2]-(E[X])^2$. This identity uses the raw second moment, not the square of the mean alone. If loss amounts are measured in dollars, $E[X]$ is in dollars and $\operatorname{Var}(X)$ is in squared dollars.

Derivation: For a finite-support random variable, the expected value of any function $g(X)$ is $\sum_i p_i g(x_i)$, because $g(X)$ takes the value $g(x_i)$ with probability $p_i$. Taking $g(X)=X$ gives $E[X]=\sum_i p_i x_i$; taking $g(X)=X^2$ gives $E[X^2]=\sum_i p_i x_i^2$. Finally, expanding the definition of variance gives $E[(X-E[X])^2]=E[X^2]-2E[X]E[X]+(E[X])^2=E[X^2]-(E[X])^2$.

For a two-point model with possible losses $a$ and $b$, where $P(X=a)=p$ and $P(X=b)=1-p$, this becomes

E[X]=pa+(1-p)b,\qquad \operatorname{Var}(X)=p(1-p)(a-b)^2.

The variance expression can be obtained by substituting the two-point mean into the raw-moment formula. It also shows that variance is zero if either outcome has probability zero, or if the two possible loss amounts are equal. In an ordinary two-point model with distinct values and probabilities strictly between zero and one, the variance is positive.

Worked Example: A Two-Point Benefit Model

A hypothetical benefit model assigns a loss of 2 thousand dollars with probability 0.25 and a loss of 6 thousand dollars with probability 0.75. Calculate the probability that the loss exceeds 3 thousand dollars, the mean loss, and the variance. Give the mean in dollars and the variance in squared thousands of dollars.

Method. First verify the table: $0.25+0.75=1.00$, and both probabilities are nonnegative. Since only the 6-thousand-dollar outcome exceeds 3 thousand dollars, $P(X>3)=0.75$.

Using thousands of dollars as the unit, the weighted mean is $E[X]=0.25(2)+0.75(6)=0.50+4.50=5$ thousand dollars, or $5,000.

The raw second moment is $E[X^2]=0.25(2^2)+0.75(6^2)=0.25(4)+0.75(36)=1+27=28$ squared thousands of dollars.

Therefore, $\operatorname{Var}(X)=28-5^2=28-25=3$ squared thousands of dollars. The two-point formula provides the same check: $0.25(0.75)(6-2)^2=0.1875(16)=3$.

Check. The mean of 5 thousand dollars lies between the two possible losses, 2 and 6 thousand dollars, and the variance is positive. The common mistake this method avoids is using the probability of the smaller outcome for the event $X>3$; that event includes the larger outcome, so its probability is 0.75.

Finite-Support Models With Several Outcomes

With more than two outcomes, the same procedure applies one row at a time. A table can make the arithmetic transparent: list each possible loss, its probability, the contribution $p_i x_i$ to the mean, and the contribution $p_i x_i^2$ to the raw second moment. The last two columns should be summed separately.

If amounts are large, measuring losses in thousands of dollars can simplify the arithmetic. This changes the units of the reported quantities, not the method. A mean of 3.6 in thousands of dollars is $3,600; a variance of 12.24 in squared thousands of dollars is $12.24\times 1{,}000{,}000$ dollars squared.

Worked Example: A Four-Outcome Severity Table

A hypothetical equipment-loss model has the following support, with loss amounts stated in thousands of dollars. Calculate $P(X\geq 4)$, $E[X]$, and $\operatorname{Var}(X)$.

Loss $x_i$ (thousands of dollars)	Probability $p_i$	$p_i x_i$	$p_i x_i^2$
0	0.10	0	0
1	0.40	0.40	0.40
4	0.30	1.20	4.80
10	0.20	2.00	20.00

Step 1: Check the probabilities. Their total is $0.10+0.40+0.30+0.20=1.00$, so the table is valid.

Step 2: Calculate the event probability. Losses at least 4 thousand dollars are 4 and 10, so $P(X\geq 4)=0.30+0.20=0.50$.

Step 3: Calculate the mean. Summing the weighted losses gives $E[X]=0+0.40+1.20+2.00=3.60$ thousand dollars, or $3,600.

Step 4: Calculate the variance. The raw second moment is $E[X^2]=0+0.40+4.80+20.00=25.20$ squared thousands of dollars. Thus $\operatorname{Var}(X)=25.20-(3.60)^2=25.20-12.96=12.24$ squared thousands of dollars.

Check. The zero-loss outcome contributes 0 to both moments but remains part of the probability total. The variance is positive, and the calculated event probability is 0.50. The common mistake to avoid is omitting the zero-loss probability when checking that the probabilities sum to one.

When a Probability Is Missing

A finite-support table may give all but one probability. Since the probabilities must sum to one, the missing probability is one minus the sum of the probabilities already given. After finding it, check that it is between zero and one, then use the completed table for all further calculations. The missing probability must not be inferred from the mean unless the problem explicitly supplies a mean condition; probability total and expected value are separate pieces of information.

Worked Example: Completing a Table Before Finding Moments

A hypothetical repair-cost model has losses of 0, 2, and 8 thousand dollars, with probabilities 0.20, $p$, and 0.30, respectively. Find $p$, the mean, the variance, and the probability that the loss is at least 2 thousand dollars.

Method. Use the probability total first: $0.20+p+0.30=1$, so $p=1-0.50=0.50$. The completed probabilities, 0.20, 0.50, and 0.30, are nonnegative and sum to 1.

The mean is $E[X]=0.20(0)+0.50(2)+0.30(8)=0+1.00+2.40=3.40$ thousand dollars.

The raw second moment is $E[X^2]=0.20(0^2)+0.50(2^2)+0.30(8^2)=0+2.00+19.20=21.20$ squared thousands of dollars.

Therefore, $\operatorname{Var}(X)=21.20-(3.40)^2=21.20-11.56=9.64$ squared thousands of dollars. The event $X\geq 2$ includes the 2- and 8-thousand-dollar outcomes, so its probability is $0.50+0.30=0.80$.

Check. The mean is between the minimum and maximum losses, 0 and 8 thousand dollars, and the variance is positive. The common mistake this method avoids is treating the unspecified probability as zero rather than completing the table.

Common Traps and a Reliable Workflow

Finite-support calculations are often short enough that candidates skip the setup. That is risky: an omitted outcome, a reversed inequality, or a unit conversion can change the answer even when the arithmetic is correct. Use a fixed sequence so each part can be checked independently.

List the complete support.
Include every possible loss, including zero, and combine duplicate outcomes if they are listed separately.

Validate the probabilities.
Check that each probability is nonnegative and that their sum is exactly 1, allowing only for stated rounding.

Translate the event.
Identify the support values that satisfy the inequality, paying attention to whether it is strict, such as $X>c$, or inclusive, such as $X\geq c$.

Calculate moments and units.
Use $p_i x_i$ for the mean and $p_i x_i^2$ for the raw second moment, then subtract the squared mean. State the resulting units.

A useful consistency check is that the mean of a finite-support variable must lie between its smallest and largest possible values. The variance cannot be negative. If a computed variance is negative, the usual cause is an arithmetic error, often confusing $E[X^2]$ with $(E[X])^2$. Also, squaring a loss amount measured in thousands produces squared thousands, not thousands.

Key takeaway: For a finite-support severity model, probabilities, the mean, and raw moments are all obtained by summing the relevant probability weights. Validate the table first, include every outcome, and apply $\operatorname{Var}(X)=E[X^2]-(E[X])^2$ with the units carried through the calculation. These discrete calculations are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, check the probability total before calculating. Loss amounts are stated in the units specified.

A hypothetical appliance policy has loss $X=500$ dollars with probability 0.60 and $X=2{,}500$ dollars with probability 0.40. Find $P(X>1{,}000)$ and $E[X]$. (Answer: $0.4000$ and $\$1{,}300$.)
A model has losses 0, 3, and 7 thousand dollars with probabilities 0.25, 0.50, and 0.25. Find $P(X\geq 3)$, $E[X]$, and $\operatorname{Var}(X)$. (Answer: $0.7500$, 3.25 thousand dollars, and 6.1875 squared thousands of dollars.)
A hypothetical cargo model has losses 1, 4, and 9 thousand dollars with probabilities 0.10, 0.60, and 0.30. Find the mean and variance. (Answer: 5.20 thousand dollars and 6.96 squared thousands of dollars.)
A model has losses 0, 2, and 5 thousand dollars with probabilities 0.20, $p$, and 0.30. Find $p$, $P(X\geq 2)$, and the variance. (Answer: $p=0.50$, $0.8000$, and 3.25 squared thousands of dollars.)
A hypothetical claim model has losses 0, 1, 3, and 6 thousand dollars with probabilities 0.10, 0.20, 0.40, and 0.30. Find $P(X>1)$, the mean, and the variance. (Answer: $0.7000$, 3.20 thousand dollars, and 4.36 squared thousands of dollars.)

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