Modeling Individual Claim Severity as a Random Variable

Introduction

A claim severity model describes the amount of an individual claim. That sounds simple, but actuarial calculations can go wrong when the amount for one claim is confused with the number of claims or the total losses for a group of policies. The first step is to identify exactly what one observation represents.

This tutorial defines ground-up claim severity and establishes how to interpret its probabilities. The central convention is that severity is measured per claim. A probability about severity therefore uses claims as its denominator, unless the question explicitly asks for a probability per policy or exposure.

Definition: Ground-up claim severity $X$ is a nonnegative random variable representing the loss amount for one individual claim, before any coverage modification such as a deductible or limit. Its value is expressed in a stated monetary unit. A probability such as $P(X>c)$ refers to the probability that one claim has severity greater than $c$.

Three Different Quantities

A loss model for a portfolio may involve several random quantities. Their units and meanings must not be interchanged.

Severity $X$: the amount of one claim, measured in dollars per claim.
Frequency $N$: the number of claims in a specified group or period, measured in claims.
Aggregate loss $S$: the total amount of all claims in that group or period, measured in dollars.

If there are $N$ claims, write their severities as $X_1,\ldots,X_N$. The aggregate loss is the sum of those individual amounts:

S=\sum_{j=1}^{N}X_j.

When there are no claims, the sum is empty and aggregate loss is $S=0$. The expression does not make severity and aggregate loss the same quantity: each $X_j$ is one claim amount, while $S$ combines all amounts in the group. In particular, $P(X>c)$ is not the probability that the portfolio's aggregate loss exceeds $c$.

A severity model is usually applied to claims that occur. A policy with no claim contributes to the frequency count but does not supply a realized claim severity for that period. This distinction matters whenever a problem gives both a probability of making a claim and information about the amounts of claims.

Worked Example: Interpreting Claim-Level Severity Probabilities

A hypothetical bicycle insurer models the ground-up severity of an individual reported claim. The model assigns probability 0.2200 to a severity at most $600, probability 0.3100 to a severity greater than $600 and at most $1,800, probability 0.2700 to a severity greater than $1,800 and at most $5,000, and probability 0.2000 to a severity greater than $5,000. For a set of 400 claims, find the modeled probability that a claim exceeds $1,800 and the expected number of such claims.

Step 1: Identify the unit. The random variable describes one claim, so the required probability is a claim-level probability. The event $X>1{,}800$ consists of the last two non-overlapping severity ranges.

Step 2: Add the probabilities for those ranges.

$$ P(X>1{,}800)=0.2700+0.2000=0.4700. $$

Step 3: Convert to an expected claim count. Among 400 claims with this same model, the expected number exceeding $1,800 is $400(0.4700)=188$ claims.

The answer is a probability of 0.4700 for one claim and an expected count of 188 claims in the stated set. The count is an expectation, not a guarantee that exactly 188 claims will exceed the threshold. This method avoids treating a probability as though it were itself a number of claims.

Why a Claim-Level Probability Gives an Expected Count

Suppose there are $m$ claims, and each has the same probability $p$ of satisfying a severity condition. Define $I_j=1$ if claim $j$ satisfies the condition and $I_j=0$ otherwise. Then the number $K$ of claims satisfying it is $K=\sum_{j=1}^{m}I_j$.

E[K] =E\!\left[\sum_{j=1}^{m}I_j\right] =\sum_{j=1}^{m}E[I_j] =\sum_{j=1}^{m}p =mp.

The justification uses linearity of expectation and $E[I_j]=p$. Independence among the claims is not required for this expected-count formula, though it may matter for other questions about the distribution of $K$. If claim probabilities differ, the corresponding result is $E[K]=\sum_{j=1}^{m}p_j$, where $p_j$ is the probability for claim $j$.

The denominator still needs attention. If a problem starts with policies rather than claims, first account for the chance that a policy produces a claim. The next example shows why multiplying a severity probability by the number of policies alone can be wrong.

Worked Example: Moving from Policies to Large Claims

A hypothetical insurer covers 1,500 commercial devices for one year. Each device has probability 0.0800 of producing a claim during the year. Conditional on a claim occurring, its ground-up severity has probability 0.3000 of exceeding $4,000. Assume the stated probabilities apply uniformly across devices. Find the probability that a randomly selected device produces a claim exceeding $4,000 and the expected number of such claims.

Step 1: Keep the events distinct. A device must first produce a claim, and that claim must then exceed $4,000. Thus the relevant probability per device is a product of a claim-occurrence probability and a conditional severity probability:

P(\text{claim and severity}>4{,}000) =0.0800(0.3000) =0.0240.

Step 2: Calculate the expected number across devices. There are 1,500 devices, each with probability 0.0240 of producing a claim above the threshold. By adding the indicator expectations, the expected count is

1{,}500(0.0240)=36 \text{ claims}.

As a check, the expected number of all claims is $1{,}500(0.0800)=120$. Applying the severity probability to that expected claim count gives $120(0.3000)=36$, consistent with the direct calculation. The answer is 0.0240 per device and 36 large claims in expectation, not 0.3000 per device.

Observed Claims and Model Probabilities

An observed proportion and a model probability are related but not identical. If a data set contains $m$ claim amounts and $k$ exceed a threshold, then $k/m$ is the observed proportion among those claims. It describes that particular set. A model probability $P(X>c)$ describes the chance for an individual claim under the assumed severity model. The observed proportion can be used as an estimate in an appropriate analysis, but it does not automatically equal the model probability.

The denominator is especially important when a data set includes policies with no claims. A fraction with all policies in the denominator is a policy-level rate; a fraction with claims in the denominator is a claim-level severity proportion. Those quantities answer different questions.

Worked Example: Separating Severity from Frequency and Aggregate Loss

A hypothetical claims file contains four ground-up claim amounts: $700, $1,200, $2,600, and $5,500. Find the number of claims, the aggregate loss, and the observed proportion of claims exceeding $2,000. State what each result measures.

Step 1: Count the individual claims. There are four recorded amounts, so the observed frequency in this file is $N=4$ claims.

Step 2: Add the amounts to obtain aggregate loss.

S=700+1{,}200+2{,}600+5{,}500=10{,}000 \text{ dollars}.

Step 3: Count claims above the severity threshold. The amounts greater than $2,000 are $2,600 and $5,500. There are two such claims out of four, so the observed claim-level proportion is $2/4=0.5000$.

The results have different units: frequency is 4 claims, aggregate loss is $10,000, and the observed severity proportion is 0.5000 of the claims in this file. The proportion is not, by itself, a model probability for a future claim; interpreting it that way would require an appropriate model or sampling assumption.

Common Exam Trap: Using the Wrong Denominator

Suppose a problem gives the number of policies, a probability of claim occurrence, and a severity probability conditional on a claim. Multiplying the severity probability directly by the number of policies skips the claim-occurrence step.

Common mistake: For 1,500 policies and conditional severity probability 0.3000, calculate $1{,}500(0.3000)=450$ large claims.

Correction: This treats every policy as if it produced a claim. If the claim probability is 0.0800, the per-policy probability of a large claim is $0.0800(0.3000)=0.0240$, giving $1{,}500(0.0240)=36$ expected large claims.

A reliable check is to name the population in the denominator before calculating: one claim, one policy, or one portfolio period. Then name the random quantity being requested: severity, frequency, or aggregate loss.

Key takeaway: Ground-up severity $X$ is the nonnegative amount for one claim. A probability $P(X>c)$ concerns an individual claim; frequency counts claims, and aggregate loss sums their amounts. When starting from policies, include the probability of claim occurrence before applying a claim-conditional severity probability.

Check Your Understanding

For each problem, identify the population and quantity before calculating.

A hypothetical insurer has 900 claims subject to the same severity model. A claim has probability 0.4500 of exceeding $1,000. Find the expected number of claims above $1,000. (Answer: 405 claims.)
A portfolio has 2,400 policies. Each policy has probability 0.0500 of producing a claim, and a claim has conditional probability 0.2000 of exceeding $7,500. Find the probability per policy of producing such a claim and the expected count. (Answer: 0.0100 per policy; 24 claims.)
A hypothetical claims file contains amounts of $500, $900, $1,300, $2,100, and $3,200. Find the frequency, aggregate loss, and observed proportion exceeding $1,000. (Answer: 5 claims; $8,000; 0.6000 of claims.)
A portfolio's expected claim count is 2.5. Each claim has probability 0.1600 of exceeding a stated severity threshold. Assuming the same claim-level probability applies to each claim, find the expected number exceeding the threshold. (Answer: 0.4 claims.)
For one year, a policy has a 0.0400 probability of producing a claim. Conditional on a claim, the probability that severity is at most $3,000 is 0.7000. Find the probability per policy of producing a claim with severity greater than $3,000. (Answer: 0.0120 per policy.)

What You'll Learn