Loss Severity Random Variables And Their Supports

Loss Severity and Why Its Support Matters

A severity model describes the size of a loss from one occurrence. Before using a model to calculate probabilities, it is important to know which loss amounts the model can produce and where probability is concentrated. The support records this information, but it is not always identical to the list of values written in a model or to the values that have positive probability individually.

This distinction matters in actuarial work because a loss can be nonnegative without being strictly positive, bounded without attaining its upper bound, or strictly positive with a support that includes zero. These features affect the interpretation of probability statements and the assumptions used in later severity calculations.

Definition: A loss severity random variable $X$ represents the monetary amount of loss for one occurrence, under a specified loss model. A typical severity model assumes $X\geq 0$ with probability 1, although a zero loss may be included if the model permits it. The support of $X$ is the set of real numbers $x$ such that every open interval around $x$ has positive probability: $$ x\text{ is in the support of }X \quad\Longleftrightarrow\quad P(x-\varepsilon<X<x+\varepsilon)>0 \text{ for every }\varepsilon>0. $$ Equivalently, the support is the smallest closed set containing $X$ with probability 1.

The neighborhood definition is useful for both discrete and continuous models. A value need not itself have positive probability to be in the support: it is enough that every interval around it contains outcomes with positive probability. Conversely, a value outside the support has some neighborhood that the model assigns probability zero.

A model may describe a range of allowed outcomes, a list of candidate values, or the values attained by $X$ on a sample space. None of these descriptions automatically equals the support. In particular, points excluded from the support may still be listed in a model, and a support point may not be an outcome with positive probability at that exact point.

Finite Discrete Models and Continuous Ranges

For a finite discrete severity model, the support consists exactly of the listed loss amounts with positive probability. A listed value with probability zero is not in the support in this finite setting: a sufficiently small interval around it can exclude all the other listed values, leaving probability zero. This simple rule should not be extended without qualification to infinite discrete models, where positive-probability values can accumulate at a point.

In a continuous model, individual points usually have probability zero, including support endpoints. The support can nevertheless include those endpoints because every interval around an endpoint intersects a region of positive probability. Thus, “has probability zero” and “is outside the support” are not equivalent statements.

Worked Example: Support of a Finite Loss Model

A hypothetical equipment policy has a per-occurrence loss severity $X$, in dollars, with the following distribution: $P(X=0)=0.10$, $P(X=400)=0.55$, $P(X=1{,}200)=0.35$, and no other outcomes. Identify the support, determine whether losses are strictly positive, and state a finite upper bound.

Method. For a finite discrete model, retain exactly the listed values with positive probability. First check that the probabilities form a distribution: $$ 0.10+0.55+0.35=1.00. $$

Each of $0$, $400$, and $1{,}200$ has positive probability, so the support is $\{0,400,1{,}200\}$. The severity is nonnegative, but it is not strictly positive with probability 1, because $P(X=0)=0.10$. In fact, $$ P(X>0)=P(X=400)+P(X=1{,}200)=0.55+0.35=0.90. $$

The largest support value is $1{,}200$, so $X\leq 1{,}200$ with probability 1. As a check, the support includes all three outcomes whose probabilities sum to 1, and none exceeds the stated bound.

Common mistake avoided: Treating “severity” as necessarily greater than zero would incorrectly remove the zero-loss outcome. Nonnegative means $X\geq 0$; strictly positive means $X>0$. These are different assumptions.

Worked Example: Support Endpoints in a Continuous Model

A hypothetical repair-loss model assumes that $X$ is uniform over the interval from \$500 to \$3,500, inclusive. Identify its support and determine the probability that the loss is between \$1,000 and \$2,000, inclusive.

Method. Under a uniform model on an interval, probabilities within that interval are proportional to interval length. The total interval length is $$ 3{,}500-500=3{,}000. $$ Every open neighborhood of \$500 or \$3,500 contains a portion of the model’s interval with positive length. The same is true for each point between them. Therefore the support is the closed interval $[500,3{,}500]$.

The target interval has length $$ 2{,}000-1{,}000=1{,}000. $$ Its probability is therefore $$ P(1{,}000\leq X\leq 2{,}000) =\frac{1{,}000}{3{,}000} =\frac{1}{3} \approx 0.3333. $$

The answer is $0.3333$, rounded to four decimal places. The endpoints belong to the support even though each individual point has probability zero in this continuous model. Including or excluding either endpoint in this interval probability does not change the result. As a check, the target interval is one-third the length of the full interval, so its probability is one-third.

Common mistake avoided: Removing the endpoints from the support just because they have zero point probability confuses the probability of a single value with the probability of every neighborhood around that value.

Accumulation Points in Infinite Discrete Models

For an infinite discrete distribution, the support is the closure of the values with positive point probability. Closure adds any limit points of those values. A limit point can itself have probability zero and still belong to the support.

Result: If a discrete random variable places all its probability on a countable set $M=\{x:P(X=x)>0\}$, then its support is the closure of $M$. For finite discrete models, $M$ is already closed, so the support is just the positive-probability values.

Here is why the closure rule holds. If $x$ is in the closure of $M$, every open interval around $x$ contains some $y\in M$. Since $P(X=y)>0$, that interval has positive probability, so $x$ is in the support. If $x$ is outside the closure, some open interval around it contains no point of $M$. The distribution places all its probability on $M$, so that interval has probability zero, and $x$ is outside the support.

Worked Example: A Zero-Probability Limit Point

A hypothetical severity model has $X=1/n$ with probability $2^{-n}$, for each positive integer $n$, and no other outcomes. Find the support, decide whether zero is an outcome or a support point, and calculate $P(X<1/5)$.

Method. First verify that the assigned probabilities sum to 1: $$ \sum_{n=1}^{\infty}2^{-n} =\frac{1/2}{1-1/2} =1. $$ The positive-probability values are $1, 1/2, 1/3,\ldots$. They approach zero as $n$ increases. Their closure is $$ \left\{\frac{1}{n}:n=1,2,3,\ldots\right\}\cup\{0\}. $$

Zero is not an outcome of this model: $1/n$ is positive for every positive integer $n$, so $P(X=0)=0$. Yet zero is in the support, because every interval around zero contains $1/n$ for sufficiently large $n$, and each such value has positive probability.

To calculate $P(X<1/5)$, solve the inequality $1/n<1/5$. Since $n$ is positive, this is equivalent to $n>5$. Thus $$ P(X<1/5) =\sum_{n=6}^{\infty}2^{-n} =\frac{2^{-6}}{1-1/2} =2^{-5} =\frac{1}{32} =0.03125. $$

The probability is $0.03125$, or $0.0313$ to four decimal places. As a check, the excluded terms $n=1,\ldots,5$ have total probability $1-1/32=31/32$, so the two parts sum to 1.

Common mistake avoided: Calling the support only $\{1/n:n\geq1\}$ omits its limit point. The support uses neighborhoods, not just point probabilities.

Positivity, Boundedness, and Probability Statements

For a loss severity, nonnegativity is a common modeling assumption: $P(X\geq0)=1$. It does not imply that every loss is positive, because there may be a probability mass at zero. Nor does it by itself imply that the support has a strictly positive lower endpoint. For example, positive loss amounts can approach zero, making zero a support point even if $P(X=0)=0$.

A severity is bounded above by $M$ if $P(X\leq M)=1$ for some finite $M$. For a nonnegative severity, this means there is a finite ceiling on losses under the model. The support is then contained in $[0,M]$. The bound need not be attained: a model can have losses below $M$ that approach $M$ while assigning zero probability to $M$ itself. Conversely, an unbounded support means that no finite upper bound contains the entire support.

The support helps identify which values or regions need attention in a probability calculation, but it does not give their probabilities. Knowing that $X$ has support $[0,1{,}000]$, for example, does not by itself say whether losses near zero are more likely than losses near the upper end. A distributional model or other probability information is still needed.

State the model.
Identify the loss variable and whether the model is discrete, continuous, or a mixture. Record any given probability assumptions.

Find candidate support points or regions.
For a finite discrete model, keep values with positive probability. For a continuous model, check which neighborhoods intersect regions of positive probability.

Check limits and endpoints.
For an infinite discrete model, include accumulation points. For continuous models, test whether neighborhoods at each endpoint have positive probability.

State positivity and bounds separately.
Check whether zero is an outcome, whether $X>0$ with probability 1, and whether there is a finite upper bound. These are distinct properties.

Common Traps

Equating support with positive point probabilities. This works for finite discrete models, but not for every infinite discrete model. A zero-probability limit point can still be in the support.
Assuming a zero point probability rules out support membership. In a continuous distribution, support endpoints commonly have zero point probability but positive-probability neighborhoods.
Confusing nonnegative with strictly positive. Nonnegative severity allows a zero loss. Strict positivity requires $P(X>0)=1$.
Assuming a bound must be attained. A finite upper bound can hold even when the severity never equals that bound.
Using support alone to calculate a probability. Support identifies where probability can occur locally; it does not specify how probability is distributed across that support.

Key takeaway: Define support through neighborhoods of positive probability. For finite discrete models, it is the set of positive-probability values; for an infinite discrete model, include their accumulation points. Keep support, point probabilities, positivity, and boundedness conceptually separate.

Check Your Understanding

For each hypothetical model, identify the requested support or property. Answers are provided for checking.

A discrete severity $X$ takes values $0$, $250$, and $900$ with probabilities $0.20$, $0.50$, and $0.30$, respectively. State the support and determine whether $X$ is strictly positive with probability 1. (Answer: Support $\{0,250,900\}$; no, since $P(X=0)=0.20$.)
A continuous severity is uniform on $[200,1{,}400]$. State its support and calculate the probability that it lies in $[500,800]$. (Answer: Support $[200,1{,}400]$; probability $300/1{,}200=0.25$.)
A discrete severity takes value $1/n$ with probability $2^{-n}$ for every positive integer $n$. Is zero an outcome, and is it in the support? (Answer: Zero is not an outcome, but it is in the support as an accumulation point.)
A hypothetical model assigns probability $0.25$ to a loss of zero and probability $0.75$ to a severity uniform on $(0,1{,}000)$. State the support and the probability of a loss exactly equal to $1{,}000$. (Answer: Support $[0,1{,}000]$; $P(X=1{,}000)=0$.)
A discrete severity has $P(X=100n)=2^{-n}$ for positive integers $n$. State whether it is bounded above and whether its support is bounded. (Answer: It is not bounded above; its support $\{100,200,300,\ldots\}$ is unbounded.)

What You'll Learn