Choosing Severity Support and Measurement Units

Why Support and Units Matter

A severity model is not fully specified by naming a distribution. Its possible values must also make sense for the loss being modeled, and its measurement unit must be clear. A model for a positive claim amount, for example, should not assign positive probability to a negative amount. A model expressed in thousands of dollars can be convenient, but its moments must be interpreted in those same units.

These choices affect more than notation. Support determines which loss outcomes the model permits; units determine the numerical scale of values and moments. Both should be checked before calculating or interpreting results.

Definition: The support of a severity model is the set or range of values that the model permits for the random variable $X$. For a discrete model, this consists of the values with positive probability. For a continuous model described by a density, it is the range on which the density can be positive, with boundary conventions depending on the model. Values outside the support have probability zero. The support must be stated together with the context and units of $X$.

For a continuous model, an individual point in the support need not have positive probability. For instance, a continuous loss model can permit every value from zero to $6,000 even though the probability of the loss being exactly $3,000 is zero. If the endpoints have no point masses, including or excluding an endpoint does not change probabilities or moments. In contrast, endpoints matter in a discrete model if they are possible outcomes with positive probability.

A model’s support is not the same as the smallest and largest values in a finite sample. If observed claims range from $400 to $8,000, that alone does not show that the underlying severity model excludes losses below $400 or above $8,000. The range of observations describes what happened in the sample; support describes what the model permits.

Choosing a Meaningful Support

For an individual claim severity, $X$ commonly represents the amount of a loss conditional on a claim occurring. That modeling choice usually gives $X$ a positive support. If instead the random variable represents an amount over an exposure that may have no claim, zero can be a possible value. These are different definitions of the random variable, not just different ways to write the same support.

Physical or contractual constraints can also determine support. A loss amount cannot be negative under the usual definition of severity. A particular hypothetical product might have a maximum covered payment, but that does not automatically impose the same maximum on the underlying loss $X$. The variable must be identified first: a limit on a payment does not, by itself, establish a limit on the loss.

Support restrictions also matter when interpreting a probability statement. If a model’s support is nonnegative, then it assigns probability zero to $X<0$. If the support is bounded above by $b$, then values greater than $b$ are impossible under that model. These statements follow from the chosen model, not from the absence of such outcomes in a small sample.

Worked Example: A Continuous Severity in Dollars or Thousands

A hypothetical repair-cost severity $X$, conditional on a repair claim, is modeled as uniform over the interval from $0 to $6,000. Identify its support and calculate its mean and variance first in dollars and then in thousands of dollars. No interest or mortality assumption is relevant. The distribution is uniform with lower endpoint 0 and upper endpoint 6,000 dollars.

Support in dollars. The model permits losses from $0 to $6,000, so the support is $[0,6000]$ dollars. It rules out negative losses and losses above $6,000. Because this is a continuous model, the probability of any one exact value, including either endpoint, is zero.

For a uniform distribution on $[a,b]$, the mean is $(a+b)/2$ and the variance is $(b-a)^2/12$. Here $a=0$ and $b=6000$, so

E[X]=\frac{0+6000}{2}=3000\text{ dollars}

\operatorname{Var}(X)=\frac{(6000-0)^2}{12} =\frac{36{,}000{,}000}{12} =3{,}000{,}000\text{ dollars}^2.

Convert the model to thousands of dollars. Define $Z=X/1000$, where $Z$ is measured in thousands of dollars. Dividing both support endpoints by 1,000 gives support $[0,6]$ for $Z$. Its mean is $3000/1000=3$ thousand dollars. Its variance is $3{,}000{,}000/1{,}000{,}000=3$ squared thousands of dollars. As a check, $Z$ is uniform on $[0,6]$, whose variance is $6^2/12=3$.

The support and mean describe compatible scales in each version: $[0,6000]$ dollars with mean $3,000, or $[0,6]$ thousand dollars with mean 3 thousand dollars. The model and its loss probabilities have not changed; only the numerical unit has changed.

How Moments Change When Units Change

A change of units rescales the numerical values of a random variable. If $Z=X/c$ for a positive conversion constant $c$, then the support endpoints are divided by $c$, and the mean is divided by $c$. The variance is divided by $c^2$, because variance is measured in squared units.

Result: If $Z=X/c$, where $c>0$, then $$ E[Z]=\frac{E[X]}{c}, \qquad \operatorname{Var}(Z)=\frac{\operatorname{Var}(X)}{c^2}, \qquad \operatorname{SD}(Z)=\frac{\operatorname{SD}(X)}{c}. $$ The support of $Z$ is the support of $X$ with each value divided by $c$. Probabilities of corresponding events are unchanged; for example, $P(X\leq x)=P(Z\leq x/c)$.

The variance rule follows directly from its definition. For $Z=X/c$, the mean is $E[X]/c$. Subtracting this mean from $Z$ gives $Z-E[Z]=(X-E[X])/c$. Squaring the centered quantity introduces a factor of $1/c^2$; taking its expectation therefore gives $\operatorname{Var}(Z)=\operatorname{Var}(X)/c^2$. The standard deviation is the nonnegative square root of the variance, so it is divided by $c$, not $c^2$.

A shift changes support and mean but does not change variance: if $W=X+k$, its support is shifted by $k$, its mean is $E[X]+k$, and its variance remains $\operatorname{Var}(X)$. For severity unit conversions, the main practical issue is scaling, but this distinction helps prevent treating every transformation as a variance scaling.

Worked Example: Checking Support and Moments in a Discrete Model

A hypothetical claim severity $X$, measured in dollars, has the following discrete distribution conditional on a claim. No interest or mortality assumption is relevant. Its support consists of $400, $1,000, and $2,400, with respective probabilities 0.2, 0.5, and 0.3. Find the mean and variance, then express them in thousands of dollars.

Check the model. The probabilities sum to $0.2+0.5+0.3=1$, and each is positive. Thus the support is exactly $\{400,1000,2400\}$ dollars. Other values, including zero and $2,000, have probability zero in this discrete model.

Calculate the mean.

E[X]=400(0.2)+1000(0.5)+2400(0.3) =80+500+720 =1300\text{ dollars}.

Calculate the variance. First calculate the second moment:

E[X^2]=400^2(0.2)+1000^2(0.5)+2400^2(0.3) =32{,}000+500{,}000+1{,}728{,}000 =2{,}260{,}000\text{ dollars}^2.

Using $\operatorname{Var}(X)=E[X^2]-(E[X])^2$,

\operatorname{Var}(X)=2{,}260{,}000-1300^2 =2{,}260{,}000-1{,}690{,}000 =570{,}000\text{ dollars}^2.

Convert to thousands of dollars. The support becomes $\{0.4,1.0,2.4\}$. The mean becomes 1.3 thousand dollars, and the variance becomes $570{,}000/1{,}000{,}000=0.57$ squared thousands of dollars. As a check, the second moment in thousands is $0.4^2(0.2)+1^2(0.5)+2.4^2(0.3)=2.26$; subtracting $1.3^2=1.69$ gives $0.57$.

The method avoids treating the support as a continuous interval: this model allows only three possible severities, even though values between the smallest and largest losses might seem plausible.

A Restricted Range Can Mean a Different Model

A severity model’s support is not automatically changed just because an analyst focuses on a subset of losses. If attention is restricted to claims satisfying a condition, the relevant distribution is conditional on that condition. Its probabilities are renormalized over the retained outcomes, and its moments can differ from those of the original severity.

This distinction is important when a data set or question concerns only claims above a threshold. A mean calculated among those claims is not generally the mean for all claims. The restriction may describe a selected population rather than a revised support for the original model.

Worked Example: Mean Severity Above a Threshold

For a hypothetical claim portfolio, severity $X$ in dollars has support $\{500,1500,3500\}$ with probabilities $0.25, 0.50, 0.25$, respectively. A report includes only claims of at least $1,500. Find the mean severity among the included claims and compare it with the original mean. No interest or mortality assumption is relevant. The distribution is discrete with the stated support and probabilities.

Original mean.

E[X]=500(0.25)+1500(0.50)+3500(0.25) =125+750+875 =1750\text{ dollars}.

Probability of inclusion. The included outcomes are $1,500 and $3,500, so the probability of meeting the threshold is $0.50+0.25=0.75$. Conditional on inclusion, the probabilities are $0.50/0.75=2/3$ for $1,500 and $0.25/0.75=1/3$ for $3,500. They sum to one.

Conditional mean.

E[X\mid X\geq1500] =1500\left(\frac{2}{3}\right)+3500\left(\frac{1}{3}\right) =1000+1166.666\ldots =2166.666\ldots\text{ dollars}.

Rounded to the nearest cent, the mean among included claims is $2,166.67, compared with the original mean of $1,750.00. The conditional support is $\{1500,3500\}$, but the original model still permits $500 claims. The calculation avoids the mistake of simply discarding the $500 outcome while leaving the remaining probabilities at 0.50 and 0.25; those probabilities must be rescaled to sum to one.

Common Unit and Support Traps

A common unit error is to divide a variance by the unit conversion factor rather than its square. In the uniform example, converting dollars to thousands uses $c=1000$. The wrong calculation $3{,}000{,}000/1000=3000$ does not give the variance in squared thousands. The correct calculation is $3{,}000{,}000/1000^2=3$. A quick dimensional check catches the error: variance has squared units.

Other frequent errors include assuming that the observed sample minimum and maximum are the model’s support, confusing the support of a loss with the possible values of a payment, and treating a conditional mean as the original mean. Before using a severity model, check the following:

What exactly does $X$ represent: a claim severity, an amount that may be zero, or another loss quantity?
What values can it take under the model, and which values are impossible?
Are the values stated in dollars, thousands of dollars, or another unit?
Are any reported moments in squared units, and has the conversion factor been squared where required?
Is the calculation for the original severity distribution or for a population selected by a condition?

Key takeaway: State the severity variable, its support, and its measurement unit together. A support restriction changes which values the model permits; a unit conversion changes the numerical scale. Conditioning on a restricted range changes the distribution used for the calculation and may change its moments.

Check Your Understanding

For each problem, identify the relevant support or conversion before calculating.

A hypothetical continuous severity is uniform from $2,000 to $10,000. State its support in dollars and its mean in dollars. (Answer: Support $[2000,10000]$ dollars; mean $6,000.)
A severity model has discrete support $\{0.5,1.5,2.5\}$ in thousands of dollars, with probabilities $0.2,0.5,0.3$. Find its mean in dollars. (Answer: $1,600.)
A model has variance $800{,}000$ dollars squared. Convert the variance to squared thousands of dollars. (Answer: $0.8$ squared thousands of dollars.)
A hypothetical severity takes values $1,000 and $4,000 with probabilities 0.6 and 0.4. Claims are included only when $X\geq4000$. Find the conditional mean and state the conditional support. (Answer: Mean $4,000; support $\{4000\}$ dollars.)
A discrete severity has support $\{300,900,2100\}$ dollars and probabilities $0.2,0.5,0.3$. An analyst observes only claims of at least $900. Find the inclusion probability and the mean among included claims. (Answer: Inclusion probability $0.8$; conditional mean $1,350.)

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