Diagnosing Severity Distribution Calculation Errors

Why Severity Calculations Need a Diagnostic Check

A severity calculation can be numerically tidy and still be wrong. Common causes include using a formula outside its support, confusing a rate with a scale, treating a density value as a probability, reversing a CDF complement, overlooking a point mass, or reporting a moment that does not exist. These errors matter because a calculation is meaningful only when it matches the stated model.

For a loss $X$, keep three questions in view: what values can $X$ take, what does each parameter mean, and is the requested quantity a probability, a density, or a moment? The examples below use hypothetical loss models and make these checks explicit.

Definition: The support of $X$ is the set of values that $X$ can take with positive probability, or on which a continuous model has positive density. The cumulative distribution function is $F_X(x)=P(X\le x)$, and the survival function is $S_X(x)=P(X>x)=1-F_X(x)$. For a continuous model with density $f_X$, probabilities over intervals are integrals of the density; a density value by itself is not a probability.

Start With Support, Units, and the Requested Quantity

A severity model for an ordinary nonnegative loss generally has support $x\ge 0$. Some models have a stricter lower bound. For example, a Pareto Type I model with minimum parameter $\theta$ has support $x\ge\theta$. A formula should not be extended below that boundary just because its algebraic expression can be evaluated there.

Parameter names alone can be ambiguous, particularly for gamma distributions. A rate has units of inverse loss, while a scale has units of loss. Always check the density or the stated convention, then verify the mean against the parameterization. A scale $\theta$ and rate $\lambda$ describe the same gamma distribution when $\theta=1/\lambda$.

For a continuous random variable, $f_X(x)$ is probability per unit of loss, not the probability of exactly $x$. In particular, $P(X=x)=0$ at every single point in a purely continuous model. For an interval, integrate the density or use a difference of CDF values.

Worked Example: Rate, Density, and a CDF Difference

A hypothetical loss $X$, measured in dollars, has an exponential distribution with rate $\lambda=0.002$ per dollar and support $x\ge0$. Find $F_X(500)$, $P(500<X\le1000)$, and the density at $x=500$. Explain why the density is not the probability of a loss of exactly $500.

Method. For an exponential model with rate $\lambda$, $S_X(x)=e^{-\lambda x}$, $F_X(x)=1-e^{-\lambda x}$ for $x\ge0$, and $f_X(x)=\lambda e^{-\lambda x}$. The rate’s units are inverse dollars, so $\lambda x$ is dimensionless.

At $x=500$, $\lambda x=0.002(500)=1$. Therefore,

F_X(500)=1-e^{-1}=0.6321206\ldots

At $x=1000$, $\lambda x=0.002(1000)=2$, so $S_X(1000)=e^{-2}=0.1353353\ldots$. The requested interval probability is

P(500<X\le1000)=F_X(1000)-F_X(500) =(1-e^{-2})-(1-e^{-1}) =e^{-1}-e^{-2} =0.2325442\ldots

The density at $500 is

f_X(500)=0.002e^{-1}=0.0007358\ldots\text{ per dollar}.

Rounded to four decimals, the answers are $F_X(500)=0.6321$, $P(500<X\le1000)=0.2325$, and $f_X(500)=0.0007$ per dollar. The last number is a density, not a point probability: for this continuous model $P(X=500)=0$. A common mistake is to report $f_X(500)$ as the probability of a $500 loss. The correction is to distinguish probability at a point from density at a point, and integrate the density when an interval probability is needed.

Check. The interval probability is positive and less than $S_X(500)=e^{-1}$, as it must be because $500<X\le1000$ is a subset of $X>500$.

Rate and Scale Are Not Interchangeable

For a gamma distribution with shape $\alpha$ and rate $\lambda$, the density is

f_X(x)=\frac{\lambda^\alpha x^{\alpha-1}e^{-\lambda x}}{\Gamma(\alpha)},\qquad x\ge0,

and its mean is $\alpha/\lambda$. In the scale parameterization, the density uses $\theta$ in the form $x^{\alpha-1}e^{-x/\theta}/(\Gamma(\alpha)\theta^\alpha)$, and the mean is $\alpha\theta$. Thus, $\theta=1/\lambda$. The exponent must be dimensionless: $\lambda x$ for a rate and $x/\theta$ for a scale.

Worked Example: Diagnosing a Gamma Parameterization Error

A hypothetical claim amount $X$, in dollars, has a gamma distribution with shape $\alpha=2$ and rate $\lambda=1/1000$ per dollar. Calculate its mean and $P(X\le1000)$. A calculation instead treats $1/1000$ as a scale in dollars. Identify and correct that error.

Method. Use the stated rate parameterization. The mean is shape divided by rate:

E[X]=\frac{\alpha}{\lambda} =\frac{2}{1/1000} =2000\text{ dollars}.

For shape 2, the rate-form density integrates to the CDF $F_X(x)=1-e^{-\lambda x}(1+\lambda x)$ for $x\ge0$. At $x=1000$, $\lambda x=(1/1000)(1000)=1$, giving

P(X\le1000)=1-e^{-1}(1+1) =1-\frac{2}{e} =0.2642411\ldots

The equivalent scale is $\theta=1/\lambda=1000$ dollars. The scale-form mean is therefore $\alpha\theta=2(1000)=2000$ dollars, agreeing with the rate-form result. If $1/1000$ is wrongly inserted as a scale measured in dollars, the calculated mean would be $2(1/1000)=0.002$ dollars. That result contradicts the model’s 1000-dollar scale and signals the parameterization error. Rounded to four decimals, the probability is 0.2642; the mean is $2,000.00.

Check. The rate-times-loss product at the evaluation point is dimensionless, and the two correct parameterizations give the same mean and distribution. The common mistake is to recognize a gamma formula but not verify whether its parameter is a rate or scale.

Complements, Endpoints, and Atoms

The survival function is the complement of the CDF: $S_X(x)=1-F_X(x)$, because the events $X\le x$ and $X>x$ are disjoint and cover all outcomes. For any $a<b$, subtracting the nested events gives $P(a<X\le b)=F_X(b)-F_X(a)$. In a continuous model, changing a strict endpoint to an inclusive endpoint does not change the probability. For a model with an atom at an endpoint, it can change the answer.

A distribution with an atom has positive probability concentrated at a point. Its CDF jumps there by exactly that point probability. A density for the continuous part does not describe the size of this jump. A mixed model must therefore be handled using its probability masses and its continuous component, not by treating the whole CDF as the integral of an ordinary density.

Worked Example: Detecting a Point Mass in a Mixed Model

A hypothetical loss equals $1,000 with probability 0.25. With probability 0.75, it instead follows an exponential distribution with rate $1/1000$ per dollar. Find $F_X(1000-)$, $F_X(1000)$, and $P(X=1000)$. Here $F_X(1000-)$ means the probability of a loss strictly less than $1,000.

Method. Separate the fixed-loss component from the continuous component. The exponential component has probability $1-e^{-1}$ of being strictly below $1,000, since its rate times $1,000 is 1. Thus,

F_X(1000-)=0.75(1-e^{-1}) =0.75(0.6321206\ldots) =0.4740904\ldots

The fixed $1,000 outcome is included in $X\le1000$, so

F_X(1000)=0.75(1-e^{-1})+0.25 =0.7240904\ldots

The jump is the point mass:

$$ P(X=1000)=F_X(1000)-F_X(1000-)=0.25. $$

Rounded to four decimals, $F_X(1000-)=0.4741$, $F_X(1000)=0.7241$, and $P(X=1000)=0.2500$. The common mistake is to use only the continuous exponential component when calculating $F_X(1000)$, thereby omitting the atom. The difference between the left limit and the CDF at the point exposes that omission.

Check. The CDF jumps upward by 0.25, exactly the probability assigned to $1,000.

Check Whether Moments Exist Before Calculating Them

A formula for a moment may be valid only for certain parameter values. The Pareto Type I model illustrates why checking conditions matters. With minimum parameter $\theta>0$ and shape $\alpha>0$, its survival function is $S_X(x)=(\theta/x)^\alpha$ for $x\ge\theta$. Its $r$th raw moment is

E[X^r]=\alpha\theta^\alpha\int_\theta^\infty x^{r-\alpha-1}\,dx,

for $r>0$. This improper integral is finite exactly when $\alpha>r$. When it is finite, evaluating the integral gives $E[X^r]=\alpha\theta^r/(\alpha-r)$. In particular, the mean requires $\alpha>1$, while the second raw moment requires $\alpha>2$. A finite mean does not guarantee a finite variance.

For example, if $\alpha=1.5$, the mean is finite but the second moment is infinite. The variance is therefore infinite, not a finite number obtained by applying a variance formula outside its conditions. The key diagnostic is to test the tail or the stated moment-existence condition before substituting parameter values.

Read the support.
Check that the requested value lies in the model’s domain and that any lower or upper bound is respected.

Decode parameter meanings.
Confirm rate versus scale from the density, units, or stated convention; check that exponents are dimensionless.

Classify the quantity.
Use a CDF or survival function for a tail probability, a density for probability per unit, and a point mass for an atom.

Check endpoints and existence conditions.
Use the appropriate strict or inclusive event and verify that the requested moment is finite for the parameter values.

Common Exam Traps

Evaluating outside the support. A Pareto Type I model has no probability below its minimum parameter; do not apply its tail formula there as though it were an unrestricted loss model.
Reporting a density as a probability. A density has units of inverse loss. A probability comes from integrating it over an interval.
Using the wrong complement. $F_X(x)=P(X\le x)$, while $S_X(x)=P(X>x)$. State the event before selecting the function.
Ignoring a jump. At an atom, $P(X=x)=F_X(x)-F_X(x-)$, which can be positive.
Reporting an infinite moment as finite. Check the parameter restriction before using a moment formula.

Key takeaway: Before accepting a severity result, check the support, parameter units, probability event, endpoint convention, presence of atoms, and moment-existence condition. These checks are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus and prevent errors that can survive otherwise careful arithmetic.

Check Your Understanding

For each hypothetical model, identify the relevant diagnostic and calculate the requested quantity.

A loss has an exponential distribution with rate $0.0004$ per dollar. Find $P(1000<X\le3000)$. (Answer: 0.3693)
A gamma loss has shape 3 and rate $0.002$ per dollar. Find its mean and $P(X\le500)$. (Answer: Mean = $1,500.00; probability = 0.0803)
A loss equals $2,000 with probability 0.10; otherwise it is exponential with rate $1/2000$ per dollar. Find the jump in the CDF at $2,000. (Answer: 0.1000)
A Pareto Type I loss has $\theta=4000$ dollars and $\alpha=2.5$. Is its variance finite? If so, calculate it. (Answer: Yes; $\operatorname{Var}(X)=35,555,555.56$ dollars squared)
A gamma model is specified with shape 2 and scale $600. Find its mean and the equivalent rate. (Answer: Mean = $1,200.00; rate = $1/600$ per dollar, approximately 0.001667 per dollar)

What You'll Learn