Discrete Severity Models from Probability Tables

Representing Severity With a Probability Table

A severity model describes the size of an individual loss $X$. In a discrete model, $X$ can take values only from a finite or countable set. A probability table lists those possible values and the probability assigned to each one. It is a compact model of the entire distribution: once the support and masses are specified, probabilities of events can be found by identifying which support values satisfy the event and adding their masses.

The table must distinguish the loss amounts from the probabilities. A probability belongs to a particular loss amount; it is not itself a loss amount, and the rows do not describe a sequence of claims. The order of the rows does not affect the distribution, although listing the loss amounts in increasing order makes event calculations easier to check.

Definition: A discrete severity model has a support $x_1,x_2,\ldots$ of possible loss amounts and probability masses $p_j=P(X=x_j)$. Each mass must satisfy $p_j\ge 0$, and the masses over all possible values must sum to one: $$ \sum_j p_j=1. $$ For any set of loss amounts $A$, its probability is $$ P(X\in A)=\sum_{j:x_j\in A}p_j. $$

A table is a valid probability distribution only if the support values and probability masses are clearly paired, each mass is nonnegative, and the total mass is one. If the listed amounts are all the possible outcomes, a total below one means some probability has not been assigned; a total above one cannot describe probabilities as written. Do not silently change a probability table to make it sum to one. Normalizing is appropriate only when the entries are explicitly given as relative weights.

From a Table to Event Probabilities

To calculate an event probability, first translate its wording into a condition on $X$. Then select precisely the rows whose loss amounts satisfy that condition and add their probabilities. Because distinct rows represent mutually exclusive outcomes, their probabilities add. The inclusion symbols matter: “at least” includes the boundary, while “greater than” excludes it.

For a discrete severity distribution, the cumulative distribution function at $x$ is the total mass at or below $x$:

F_X(x)=P(X\le x)=\sum_{j:x_j\le x}p_j.

This gives a useful check on event calculations. For example, $P(X>b)=1-F_X(b)$, because the events $X\le b$ and $X>b$ are disjoint and cover every possible outcome. But endpoints still need care: $F_X(b)-F_X(a)$ gives $P(a<X\le b)$, not $P(a\le X\le b)$, because the mass at $a$ is removed.

Why the Addition Rule Works

The event probability formula follows directly from the fact that each outcome in the support is a separate, mutually exclusive possibility. To see this, suppose $A$ contains exactly support values $x_2$ and $x_4$. The event $X\in A$ is the union of $X=x_2$ and $X=x_4$. These events cannot occur together, so their probabilities add. The same argument applies to any finite collection of included rows.

\begin{aligned} \{X\in A\}&=\bigcup_{j:x_j\in A}\{X=x_j\},\\ P(X\in A)&=\sum_{j:x_j\in A}P(X=x_j) =\sum_{j:x_j\in A}p_j. \end{aligned}

This justification also explains the main practical rule: do not add probabilities for rows that fail the event condition, and do not omit a row that satisfies it. For a countably infinite support, the sum is the corresponding infinite sum of nonnegative masses.

Worked Example: Checking a Table and Applying Inclusive Boundaries

A hypothetical policy has a discrete loss $X$, measured in dollars, with the following proposed severity distribution. Check the table and calculate $P(400\le X\le 900)$, $P(X>900)$, and $P(X\le 900)$.

Loss amount $x_j$	Probability $p_j$
$0	0.10
$400	0.25
$900	0.40
$1,800	0.25

Method: validate the masses, then select rows using each event’s exact boundary symbols.

First check the total: $0.10+0.25+0.40+0.25=1.00$. All four masses are nonnegative, so this is a valid probability table.

For $400\le X\le 900$, both the $400 and $900 outcomes are included. Therefore, $P(400\le X\le 900)=0.25+0.40=0.65$.

For $X>900$, the $900 outcome is excluded and only the $1,800 outcome qualifies. Thus, $P(X>900)=0.25$.

For $X\le 900$, the $0, $400, and $900 outcomes are included: $P(X\le 900)=0.10+0.25+0.40=0.75$. As a check, the last two events, $X\le900$ and $X>900$, are complementary, and $0.75+0.25=1.00$.

Common mistake avoided: treating “greater than $900” as “at least $900.” Including the $900 row would give $0.40+0.25=0.65$, which is $P(X\ge900)$, not the requested probability.

Using Counts and Relative Weights

A problem may provide counts or relative weights rather than probabilities. These are not automatically probability masses. If a model explicitly assigns probabilities in proportion to nonnegative weights $w_j$, convert each weight to a probability by dividing it by the total weight. The resulting probabilities sum to one because the denominator is the sum of all the weights.

p_j=\frac{w_j}{\sum_k w_k}.

By contrast, if the table says that its entries are already probabilities, use them as given and check their sum. Normalizing an invalid probability table can conceal a missing or incorrectly transcribed value. The distinction between a probability and a relative weight is therefore a modeling detail, not just a calculation choice.

Worked Example: Converting Hypothetical Relative Weights

A hypothetical equipment-loss model assigns relative weights 1, 2, 3, and 4 to losses of $1,000, $2,000, $4,000, and $7,000, respectively. The weights are not probabilities. Convert them to a probability table and calculate $P(X\ge4{,}000)$ and $P(1{,}000\le X<7{,}000)$.

Method: normalize the weights before evaluating the events. The total weight is $1+2+3+4=10$. Dividing each weight by 10 gives the probability masses:

Loss amount	Relative weight	Probability
$1,000	1	0.10
$2,000	2	0.20
$4,000	3	0.30
$7,000	4	0.40

The probability check is $0.10+0.20+0.30+0.40=1.00$.

For $X\ge4{,}000$, include the $4,000 and $7,000 rows: $P(X\ge4{,}000)=0.30+0.40=0.70$.

For $1{,}000\le X<7{,}000$, include $1,000, $2,000, and $4,000, but exclude $7,000: $P(1{,}000\le X<7{,}000)=0.10+0.20+0.30=0.60$.

Check: the event and its complement partition the support. The complement of $1{,}000\le X<7{,}000$ consists of the $7,000 outcome, whose probability is 0.40; $0.60+0.40=1.00$.

Common mistake avoided: treating the relative weights as probabilities. Their total is 10, not one, so using the weights directly would produce an impossible event probability greater than one.

Worked Example: Building a Table From a Specified Count Model

For a hypothetical model, a claim severity takes one of four values: $250, $750, $1,500, or $3,000. The model specifies that the probability masses are proportional to counts 18, 30, 24, and 8 in that order. Construct and check the table, then calculate $P(X\ge750)$ and $P(250<X<3{,}000)$.

Method: use the stated proportional-count assumption to obtain probabilities. The total count is $18+30+24+8=80$. Divide every count by 80:

\begin{aligned} P(X=250)&=\frac{18}{80}=0.225,\\ P(X=750)&=\frac{30}{80}=0.375,\\ P(X=1{,}500)&=\frac{24}{80}=0.300,\\ P(X=3{,}000)&=\frac{8}{80}=0.100. \end{aligned}

The check is $0.225+0.375+0.300+0.100=1.000$. For $X\ge750$, include $750, $1,500, and $3,000: $P(X\ge750)=0.375+0.300+0.100=0.775$. For $250<X<3{,}000$, exclude both endpoint amounts, leaving $750 and $1,500: $P(250<X<3{,}000)=0.375+0.300=0.675$.

Check: the complement of $250<X<3{,}000$ consists of the $250 and $3,000 outcomes. Their total probability is $0.225+0.100=0.325$, and $0.675+0.325=1.000$.

Common mistake avoided: reading strict inequalities as inclusive. The $250 and $3,000 rows do not belong to the second event.

Common Table and Boundary Errors

A disciplined calculation separates model validation from event evaluation. First confirm that the table describes a probability distribution; only then calculate the requested probability. Next, rewrite the event in words or symbols and test each support value against it. This simple process prevents several common errors.

Forgetting the total-mass check. A table of plausible-looking values is not necessarily a valid distribution. Add every probability, not just the rows relevant to a later event.
Normalizing probabilities without permission. Normalize relative weights when the problem says they are weights. Do not alter entries explicitly labeled as probabilities to force their sum to one.
Misreading endpoints. $X\ge a$ includes a possible mass at $a$; $X>a$ does not. The same distinction applies to upper endpoints.
Adding values instead of probabilities. Event probabilities are formed from the masses of qualifying outcomes. The loss amounts identify which rows qualify; they are not added to obtain a probability.
Ignoring the rest of the support in a complement. The complement includes every possible outcome that fails the original condition, including outcomes on either side of an interval.

The same table can answer many event questions without changing the model. The only thing that changes is which support points satisfy the event. This is why carefully recording units, support values, and boundary conditions is as important as doing the arithmetic.

Key takeaway: A discrete severity table pairs each possible loss with a nonnegative probability mass, and all masses must sum to one. To find an event probability, include exactly the support values allowed by the event and add their masses.

Check Your Understanding

For each hypothetical severity model, check the table or weights as needed, then evaluate the requested event. Pay close attention to whether each boundary is included.

A severity table assigns probabilities 0.20, 0.50, and 0.30 to losses of $0, $600, and $1,400. Is the table valid, and what is $P(X\ge600)$? (Answer: Valid; 0.8000)
A discrete loss takes values $100, $500, $1,000, and $2,000 with probabilities 0.15, 0.35, 0.30, and 0.20. Find $P(100<X\le1{,}000)$ and $P(X>1{,}000)$. (Answer: 0.6500 and 0.2000)
A model gives relative weights 2, 1, and 5 to losses of $400, $1,200, and $2,400. Convert the weights to probabilities and find $P(X<2{,}400)$. (Answer: Probabilities 0.2500, 0.1250, and 0.6250; event probability 0.3750)
A proposed table assigns masses 0.10, 0.25, 0.30, and 0.20 to losses of $0, $300, $900, and $1,500. Is it a valid distribution as written? If not, by how much does its total differ from one? (Answer: Invalid; total 0.8500, short by 0.1500)
A severity model has losses of $250, $750, $1,250, and $2,500 with probabilities 0.10, 0.20, 0.45, and 0.25. Find $P(250\le X<2{,}500)$ and verify it using the complement. (Answer: 0.7500; complement 0.2500)

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