Checking a Severity Model Against Basic Probability Rules

Why a Severity Model Must Pass Basic Checks

Before using a severity model to calculate probabilities or moments, verify that it could describe a probability distribution at all. A formula may look plausible on part of its stated range yet assign negative probability, exceed probability one, decrease as losses grow, or place loss probability below zero. Calculations based on such a model are not meaningful until the problem is resolved.

Let \(X\) denote a claim severity: the amount of a loss, measured in a specified monetary unit. For an ordinary gross claim severity, \(X\) must be nonnegative with probability one. A distribution can be mathematically valid for a general random variable and still be unsuitable as a gross severity model if it assigns positive probability to negative losses.

Definition: A cumulative distribution function (CDF) of \(X\) is \(F_X(x)=P(X\le x)\). A valid CDF must satisfy \(0\le F_X(x)\le1\), be nondecreasing and right-continuous, and have limits \(0\) as \(x\) tends to negative infinity and \(1\) as \(x\) tends to positive infinity. For a nonnegative severity, it must also satisfy \(F_X(x)=0\) for \(x<0\).

These conditions allow discrete, continuous, and mixed distributions. In particular, a valid CDF need not be smooth or strictly increasing. A jump at \(x\) represents probability at the single value \(x\): the size of the jump is \(P(X=x)\). A flat interval means there is no probability in its interior.

Using the CDF to Check Probability Statements

The CDF is more than a function to inspect: its differences must produce legitimate probabilities. For any real numbers \(a<b\), the event \(\{a<X\le b\}\) has probability \(F_X(b)-F_X(a)\). This difference cannot be negative.

P(a<X\le b)=F_X(b)-F_X(a)\ge 0.

The identity follows because \(\{X\le b\}\) consists of the disjoint events \(\{X\le a\}\) and \(\{a<X\le b\}\). Therefore \(F_X(b)=F_X(a)+P(a<X\le b)\). Since probabilities are nonnegative, \(F_X(b)\ge F_X(a)\) whenever \(a<b\); this is why a CDF must be nondecreasing.

Endpoint conventions matter when a distribution has a point mass. The formula above excludes \(a\) and includes \(b\). In contrast, \(P(a\le X\le b)=F_X(b)-F_X(a-)\), where \(F_X(a-)\) is the limit of \(F_X(x)\) as \(x\) approaches \(a\) from below. For a continuous distribution the distinction has no effect, but for a discrete or mixed distribution it can change the answer.

Worked Example: Finding a Decrease in a Proposed CDF

A hypothetical claim model proposes

F_X(x)= \begin{cases} 0, & x<0,\\ x/100, & 0\le x<40,\\ 0.55, & 40\le x<80,\\ (x-40)/100, & 80\le x<120,\\ 1, & x\ge120. \end{cases}

Determine whether this is a valid CDF for a nonnegative severity. If it is not, identify a probability statement that demonstrates the problem and give one possible repair.

Step 1: Check the range and the support. The proposed values lie between 0 and 1, and \(F_X(x)=0\) for \(x<0\). Those checks pass. They do not, however, establish that the function is a valid CDF.

Step 2: Check the joins. At \(x=40\), the value is 0.55, while the limit from the left is \(40/100=0.40\). This is an upward jump, which is permitted. At \(x=80\), the value is \((80-40)/100=0.40\), but values immediately below 80 are 0.55. The function has dropped from 0.55 to 0.40, which is not permitted.

Step 3: Exhibit the impossible probability. Since \(70<90\), a valid CDF must have \(F_X(90)\ge F_X(70)\). Instead,

F_X(90)-F_X(70)=\frac{90-40}{100}-0.55 =0.50-0.55=-0.05.

This would imply \(P(70<X\le90)=-0.05\), an impossible negative probability. The proposed function is not a valid CDF.

One possible repair. Replace the third branch by \(0.55+(x-80)/200\) for \(80\le x<120\). It starts at 0.55, increases to a left limit of 0.75 at 120, and the value 1 at 120 is an allowed upward jump. This is one possible repair, not the only one; any repaired model must still meet all CDF conditions.

Method and common mistake. Check ordering and evaluate CDF differences over intervals. A common mistake is to check only that all displayed values are between 0 and 1. That misses a decrease, which directly creates a negative interval probability.

Checking a Density and Its Total Probability

For a purely continuous distribution, a density \(f_X\) must be nonnegative except possibly on a set of length zero, and its integral over the entire real line must equal one. The CDF is then obtained by integrating the density. A negative density on an interval would produce negative probability there; an integral other than one would give the model the wrong total probability.

Density check: If the entire distribution of \(X\) is absolutely continuous with density \(f_X\), then \(f_X(x)\ge0\) almost everywhere and \(\int_{-\infty}^{\infty} f_X(x)\,dx=1\). For a mixed distribution, point masses must also be included: the total of the density’s integral and all point masses must equal one.

The qualification about a purely continuous distribution is important. A mixed model may have a density whose integral is less than one because some probability is concentrated at one or more points. It is the combined probability, not necessarily the continuous part alone, that must total one.

Worked Example: Normalizing a Proposed Continuous Density

A hypothetical severity model proposes \(f_X(x)=\frac{1}{3}(1+x)\) for \(0\le x\le2\), and \(f_X(x)=0\) otherwise, with \(x\) measured in dollars. Check whether it is a valid density, then normalize the same functional form and use it to find \(P(0.5<X\le1.5)\).

Step 1: Check nonnegativity and support. On \(0\le x\le2\), \(1+x\ge1\), so the proposed density is nonnegative. It is zero outside the stated nonnegative support. These checks pass.

Step 2: Integrate the proposed density.

\int_0^2 \frac{1}{3}(1+x)\,dx =\frac{1}{3}\left[x+\frac{x^2}{2}\right]_0^2 =\frac{1}{3}(2+2)=\frac{4}{3}.

The integral is \(4/3\), not 1, so the proposed function assigns too much total probability and is not a valid density.

Step 3: Find the required constant. Write the density as \(k(1+x)\) on the same support. Normalization requires

1=\int_0^2 k(1+x)\,dx =k\left(2+\frac{2^2}{2}\right)=4k, \qquad k=\frac14.

The normalized density is \(f_X(x)=\frac14(1+x)\) for \(0\le x\le2\), and zero otherwise. Its CDF on the support is

F_X(x)=\int_0^x \frac14(1+t)\,dt =\frac{x+x^2/2}{4},\qquad 0\le x\le2.

Step 4: Calculate the requested probability.

F_X(1.5)=\frac{1.5+(1.5)^2/2}{4} =\frac{2.625}{4}=0.65625, $$ $$ F_X(0.5)=\frac{0.5+(0.5)^2/2}{4} =\frac{0.625}{4}=0.15625, $$ $$ P(0.5<X\le1.5)=0.65625-0.15625=0.5000.

Check. The normalized density is nonnegative, integrates to one, and gives a probability between zero and one. The answer is reported to four decimal places. The original density’s failure was its total integral, not its shape or support.

Method and common mistake. Integrate over the full support before using a proposed density to calculate probabilities. A common mistake is to assume that a nonnegative formula is a density without checking its total integral.

Support Is Part of the Severity Model

The support describes the values the model can take with positive probability. A gross claim severity cannot be negative. This does not mean every mathematically valid distribution with negative values is defective as a random-variable model; it means that the model does not represent a nonnegative gross claim severity as stated. The context determines whether the support is appropriate.

Worked Example: A Valid Probability Model With Invalid Severity Support

A hypothetical model assigns probability 0.10 to \(X=-10\) dollars, 0.50 to \(X=100\) dollars, and 0.40 to \(X=200\) dollars. Check the probability assignments, decide whether the model is a suitable gross severity model, and find the conditional probabilities among positive outcomes if the negative outcome is excluded by conditioning.

Step 1: Check the discrete probabilities. Each assigned probability is nonnegative, and

$$ 0.10+0.50+0.40=1.00. $$

Thus the assignments define a valid probability distribution for a general random variable. Its CDF is 0 below \(-10\), 0.10 from \(-10\) up to 100, 0.60 from 100 up to 200, and 1 at and above 200. These steps are nondecreasing and the endpoint limits are correct.

Step 2: Check suitability for gross severity. The model assigns \(P(X=-10)=0.10>0\). Because a gross claim severity must be nonnegative, this is not a suitable gross severity model as stated. The issue is not a failure of the probability axioms; it is the model’s support in the claim context.

Step 3: Condition on a positive outcome. The probability of a positive outcome is \(0.50+0.40=0.90\). Given \(X>0\),

P(X=100\mid X>0)=\frac{0.50}{0.90}=\frac59, \qquad P(X=200\mid X>0)=\frac{0.40}{0.90}=\frac49.

These conditional probabilities sum to \(5/9+4/9=1\) and give a nonnegative distribution. Conditioning is a deliberate change of model; simply deleting the negative outcome without renormalizing would leave total probability 0.90.

Method and common mistake. Separate mathematical validity from suitability for the stated severity. A common mistake is to call the original distribution invalid because it includes a negative number, without noticing that its probabilities and CDF are valid for a general random variable.

A Practical Validation Sequence

Use the following order when a problem supplies a proposed severity model. It reduces the chance of overlooking a basic contradiction before undertaking more involved calculations.

Read the context and support.
For gross severity \(X\), verify that the model assigns probability one to nonnegative amounts, unless the problem defines a different quantity.

Check probability bounds and ordering.
A CDF must stay in \([0,1]\), be nondecreasing, and have limits zero and one at the two ends of the real line.

Inspect boundaries and jumps.
Check the value at each piecewise boundary against its left- and right-hand behavior. Upward jumps are allowed; downward jumps are not.

Check the probability total.
For a pure density, verify nonnegativity and integral one. For a discrete model, add the masses. For a mixed model, include both continuous probability and point masses.

Test any requested probability.
Use the correct CDF difference and endpoint convention. A negative result or a result above one signals an inconsistency that must be resolved.

These checks are basic but consequential: a model that fails them cannot support reliable calculations of claim probabilities or moments. When a proposed model fails, state precisely which condition fails and whether the problem is mathematical validity, severity support, or both. Do not silently adjust a model unless the question asks for a correction and the correction is justified.

Key takeaway: A valid severity model must obey probability rules and fit the meaning of severity. For a nonnegative loss, check its support; for a CDF, check bounds, monotonicity, right-continuity, and endpoint limits; for a pure density, check nonnegativity and total integral one.

Check Your Understanding

For each problem, first identify the relevant probability rule, then check the model or calculate the requested quantity.

A proposed CDF has \(F_X(15)=0.32\) and \(F_X(25)=0.29\). What probability would the difference imply for \(15<X\le25\), and what does this say about the model? (Answer: It implies \(-0.03\), so the proposed function cannot be a CDF.)
A proposed pure density is \(f_X(x)=c\) for \(0\le x\le5\), and zero otherwise. Find \(c\), then find \(P(1<X\le4)\). (Answer: \(c=0.2\); \(P(1<X\le4)=0.6000\).)
A discrete severity model assigns masses 0.25 at \(0\), 0.45 at \(80\), and 0.30 at \(160\) dollars. Does it define a probability distribution, and is its support compatible with nonnegative severity? (Answer: Yes; the masses total 1 and all support values are nonnegative.)
A proposed CDF has \(F_X(x)=0\) for \(x<0\), \(F_X(x)=0.4\) for \(0\le x<10\), and \(F_X(x)=1\) for \(x\ge10\). Find \(P(0<X\le10)\) and \(P(X=0)\). (Answer: \(P(0<X\le10)=0.6000\) and \(P(X=0)=0.4000\).)
A pure density is proposed as \(f_X(x)=k(2-x)\) for \(0\le x\le2\), and zero otherwise. Find \(k\) and \(P(0.5<X\le1.5)\). (Answer: \(k=0.5\); \(P(0.5<X\le1.5)=0.5000\).)

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