Probability Density and Severity Probabilities

Why a Density Is Not a Probability

A severity model describes the size $X$ of a claim. For a continuous severity, it is often more useful to describe how probability is distributed across ranges of possible losses than to assign probabilities to individual loss amounts. The probability density function, or density, provides that description. Its value indicates probability per unit of loss near a given amount; probabilities for ranges are found from the area under the density across those ranges.

The distinction matters in calculations. A density value of $0.08$ does not mean that there is an 8% probability of exactly that loss. For a continuous model, the probability of any single exact value is zero. The density’s units also depend on how loss is measured: a density stated per thousand dollars is not numerically interchangeable with one stated per dollar.

Definition: For a continuous severity random variable $X$, a probability density function $f_X(x)$ is a nonnegative function such that, for any interval with endpoints $a$ and $b$, where $a<b$, $$ P(a<X\le b)=\int_a^b f_X(x)\,dx. $$ The density is zero outside the model’s support, and its integral over the full support is 1. For a continuous model, the probability at any single point is zero, so changing whether an interval includes either endpoint does not change its probability.

The integral has a direct area interpretation: probability is the area under the density curve across the loss range of interest. The density itself can be greater than 1 if loss is measured in sufficiently small units; this is not a problem, because it is the area—not the height—that must be a probability between 0 and 1.

If $x$ is measured in thousands of dollars, the density is measured per thousand dollars. The integration limits must use the same units as $x$. For example, a loss of $4,500 corresponds to $x=4.5$ when $x$ is recorded in thousands of dollars.

Why Integrating the Density Gives a Probability

A useful way to see the interval formula is to divide a loss interval into many short pieces. Over a sufficiently small piece of width $\Delta x$ near $x$, its probability is approximately $f_X(x)\Delta x$. Adding those approximations across the interval gives a sum of areas of thin rectangles. As the pieces become narrower, the sum approaches the integral.

P(a<X\le b) \approx \sum_j f_X(x_j)\,\Delta x \quad\longrightarrow\quad \int_a^b f_X(x)\,dx.

This also explains why a density value alone is not an interval probability. It is the height of the density at a point; an interval probability depends on both the density’s values and the width of the interval. Multiplying one density value by a width is exact only when the density is constant throughout that interval. Otherwise, integration accounts for the changing height.

Confirm the loss scale and support.
Translate the stated loss amounts into the units used by $x$, and identify where the density is nonzero.

Translate the event into intervals.
For example, losses between two thresholds form one interval; losses below one threshold or above another form two disjoint intervals.

Integrate over each relevant interval.
Split an integral wherever the density formula changes, then add probabilities for disjoint intervals.

Check the result.
Each probability must be between 0 and 1, and probabilities of disjoint events must add consistently with the model’s total probability of 1.

Worked Probability Calculations

The examples use hypothetical claim models. Each specifies the severity scale and density so that the integration limits, units, and probability calculation can be checked directly.

Worked Example: A Constant Density Across the Support

A hypothetical equipment policy has claim severity $X$, measured in thousands of dollars. Assume $X$ is continuous and has constant density $f_X(x)=1/12$ for $0\le x\le12$, and zero elsewhere. Calculate the probability that a claim is between $2,400 and $7,800, and the probability that it exceeds $10,000.

Method. Integrate the density over each event’s loss range. The stated density is valid because it is nonnegative and its total area is $12(1/12)=1$.

Between $2,400 and $7,800. In thousands of dollars, the limits are $2.4$ and $7.8$. Thus

P(2.4<X\le7.8) =\int_{2.4}^{7.8}\frac{1}{12}\,dx =\frac{7.8-2.4}{12} =\frac{5.4}{12} =0.45.

The probability is $0.4500$, or 45.00%.

Exceeds $10,000. Since the support ends at 12, integrate from $10$ to $12$:

P(X>10) =\int_{10}^{12}\frac{1}{12}\,dx =\frac{12-10}{12} =\frac{1}{6} =0.1667.

The probability is $0.1667$, or 16.67%. The two calculations use widths of $5.4$ and $2$ thousand dollars respectively, not widths in dollars. Common mistake avoided: using dollar limits such as 2,400 and 7,800 with a density defined per thousand dollars would mix scales and produce an invalid result.

As a check, the first interval and the interval above 10 are disjoint, and their probabilities sum to $0.45+1/6=0.6167$, which is less than 1.

Worked Example: An Event Crossing a Piecewise Density

For a hypothetical repair-claim model, let $X$ be severity in thousands of dollars, with density

f_X(x)= \begin{cases} cx, & 0\le x\le2,\\ c(4-x), & 2<x\le4,\\ 0, & \text{otherwise}. \end{cases}

First determine $c$, then calculate $P(X\le1\text{ or }X>2.5)$.

Step 1: Normalize the density. A density must have total area 1. Integrate both pieces:

\begin{aligned} 1 &=c\int_0^2 x\,dx+c\int_2^4(4-x)\,dx\\ &=c\left[\frac{x^2}{2}\right]_0^2 +c\left[4x-\frac{x^2}{2}\right]_2^4\\ &=2c+2c=4c. \end{aligned}

Therefore $c=1/4$.

Step 2: Integrate over each part of the event. The events $X\le1$ and $X>2.5$ are disjoint. Their ranges lie in different pieces of the density:

\begin{aligned} P(X\le1) &=\frac14\int_0^1 x\,dx =\frac14\left[\frac{x^2}{2}\right]_0^1 =\frac18=0.125,\\ P(X>2.5) &=\frac14\int_{2.5}^4(4-x)\,dx\\ &=\frac14\left[4x-\frac{x^2}{2}\right]_{2.5}^4\\ &=\frac14\bigl(8-6.875\bigr) =0.28125. \end{aligned}

Add the disjoint probabilities:

P(X\le1\text{ or }X>2.5)=0.125+0.28125=0.40625.

To four decimal places, the answer is $0.4063$. As a check, the omitted interval from 1 to 2.5 has probability $1-0.40625=0.59375$. Direct integration gives $0.25\int_1^2x\,dx+0.25\int_2^{2.5}(4-x)\,dx=0.375+0.21875=0.59375$. Common mistake avoided: integrating one formula across the entire event would be wrong because the density changes at $x=2$.

Worked Example: Comparing a Density Height With an Interval Probability

A hypothetical severity model has $X$, in thousands of dollars, with density $f_X(x)=3x^2/8$ for $0\le x\le2$, and zero elsewhere. Calculate the probability that severity is between $500 and $1,500, the probability it exceeds $1,500, and the density value at $x=1$.

Step 1: Check normalization.

\int_0^2\frac{3x^2}{8}\,dx =\left[\frac{x^3}{8}\right]_0^2 =\frac{8}{8}=1.

Step 2: Calculate the interval probability. The dollar amounts correspond to $x=0.5$ and $x=1.5$:

P(0.5<X\le1.5) =\left[\frac{x^3}{8}\right]_{0.5}^{1.5} =\frac{1.5^3-0.5^3}{8} =\frac{3.375-0.125}{8} =0.40625.

Step 3: Calculate the upper-tail interval and density value.

P(X>1.5) =\int_{1.5}^2\frac{3x^2}{8}\,dx =\frac{2^3-1.5^3}{8} =\frac{8-3.375}{8} =0.578125, \qquad f_X(1)=\frac{3}{8}=0.375.

The interval probability is $0.4063$ and the probability above $1,500 is $0.5781$, both rounded to four decimal places. The density at $x=1$ is $0.375$ per thousand dollars; it is not the probability of a claim being exactly $1,000. As a check, the remaining probability below $0.5$ is $\int_0^{0.5}3x^2/8\,dx=0.015625$, and $0.40625+0.578125+0.015625=1$. Common mistake avoided: treating the density height at $x=1$ as a point probability or as the probability of the interval from $0.5$ to $1.5$.

Common Interpretation Traps

Using a height instead of an area. In general, $f_X(x)$ is not a probability. Integrate across the stated range.
Forgetting the loss units. Convert thresholds before integrating. A density per thousand dollars requires limits in thousands of dollars.
Ignoring changes in the formula. Split an interval at every point where the density changes, then add the resulting integrals.
Applying a constant-density shortcut to a changing density. The product of a density value and interval width is exact only when the density is constant across that interval.
Worrying about endpoints in a continuous model. Including or excluding one endpoint does not alter the probability, because a single point has probability zero.

Key takeaway: For a continuous severity model, probability is area under the density over the relevant loss range. Confirm the support and units, integrate every piece of the range, and interpret the result—not the density height—as the probability.

Check Your Understanding

For each hypothetical continuous severity model, use the density’s stated support and loss units. Give probabilities to four decimal places where needed.

A claim amount $X$, in thousands of dollars, has constant density $1/8$ on $0\le x\le8$ and zero elsewhere. Find the probability that $X$ is greater than $1,500 but no more than $5,500. (Answer: 0.5000)
A severity $X$, in thousands of dollars, has constant density $0.04$ on $0\le x\le25$ and zero elsewhere. Find the probability of a claim from $10,000 through $18,000. (Answer: 0.3200)
A continuous severity has density $f_X(x)=x/8$ on $0\le x\le4$ and zero elsewhere, with $x$ measured in thousands of dollars. Find $P(1<X<3)$, and first verify that the density integrates to 1. (Answer: normalization integral $=1$; probability $=0.5000$)
A model has density $c$ on $0\le x\le1$, density $c(3-x)$ on $1<x\le3$, and zero elsewhere. Find $c$, then find the probability that $X<0.5$ or $2<X<3$. (Answer: $c=1/3$; probability $=0.3333$)
A severity in thousands of dollars has density $3x^2/8$ on $0\le x\le2$ and zero elsewhere. Find the probability that the loss is between $500 and $1,500, and state $f_X(1)$ with its units. (Answer: probability $=0.4063$; $f_X(1)=0.375$ per thousand dollars)

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