Continuous Loss Distributions and Density Functions

Introduction

For a discrete loss, probability can be assigned directly to particular possible values. A continuous loss is described differently: its density indicates how probability is distributed across the loss scale. A density value is not the probability of one loss amount. To obtain a probability, integrate the density over a range of losses.

This distinction matters in actuarial modeling because loss amounts are often represented on a continuous scale, such as dollars or thousands of dollars. Correct interpretation requires keeping track of both the integration limits and the units of the density.

Definition: A continuous random loss $X$ has probability density function $f_X(x)$ on a support $A$ if $f_X(x)\geq 0$, the integral of $f_X$ over the support is 1, and the probability that $X$ falls in a region $B$ is the integral of $f_X$ over that region. In particular, for an interval with endpoints $a$ and $b$, $$ P(a<X<b)=\int_a^b f_X(x)\,dx. $$ For a continuous loss, including or excluding either endpoint does not change the probability.

What a Density Tells You

The density describes probability per unit of loss, locally along the loss scale. If $X$ is measured in thousands of dollars, for example, $f_X(x)$ has units of probability per thousand dollars. Integrating a density over a loss interval multiplies those inverse-loss units by loss units, producing a dimensionless probability.

A density can be greater than 1. That does not violate the requirement that probabilities lie between 0 and 1: a density is not a probability, and its numerical value depends on the units used to measure loss. The probability of an interval is the area under the density over that interval. The total area over the support must equal 1.

f_X(x)\geq 0,\qquad \int_A f_X(x)\,dx=1,\qquad P(a<X<b)=\int_a^b f_X(x)\,dx.

The integral formula also explains why a density value is not a point probability. For a short interval of width $h$ around a point $x$, the probability is the area under the density across that interval. If the density is approximately constant there, that area is approximately $f_X(x)h$. As $h$ shrinks toward zero, the area shrinks toward zero as well. Thus, for any single value $c$, a continuous loss satisfies $P(X=c)=0$, even when $f_X(c)$ is positive.

Key distinction: $f_X(c)$ is a density at a point, with units of probability per unit of loss. $P(X=c)$ is the probability of one exact value, which is zero for a continuous loss. An interval probability is an area, calculated by integrating the density.

Checking and Using a Density

A proposed function is a valid density only if it is nonnegative on its support and integrates to 1 over that support. If it contains an unknown constant, use the normalization condition to determine the constant before calculating probabilities. Then integrate over the loss interval requested, using the same loss units as the density.

Identify the model and units.
Record the support, the expression for $f_X(x)$, and the units in which $x$ is measured.

Check or determine normalization.
Confirm nonnegativity and set the integral over the full support equal to 1. Solve for any unknown parameter.

Set the interval limits.
Translate the requested loss range into the model’s units. Split the integral at any point where the density’s formula changes.

Integrate and check.
Evaluate the area over the requested interval. Confirm that the result is between 0 and 1 and that it is consistent with the total area of 1.

Worked Example: A Uniform Loss on a Finite Range

A hypothetical insurer models one policy’s annual covered loss $X$, measured in thousands of dollars, as uniform on the interval from 0 to 120. Find the probability that the loss is between 30 and 75 thousand dollars. The model is continuous and has no probability mass at the endpoints. No interest or mortality assumptions are involved; this is a one-period loss model.

Method and density. For a uniform distribution over an interval of width $120-0=120$, the density is constant on the support and zero outside it. Write $f_X(x)=k$ for $0\leq x\leq120$. Normalization gives

1=\int_0^{120} k\,dx=120k, \qquad k=\frac{1}{120}.

The density is therefore $1/120=0.008333\ldots$ per thousand dollars. Check: its total area is $120(1/120)=1$.

Interval probability. The requested event has limits 30 and 75 in the stated units:

P(30<X<75)=\int_{30}^{75}\frac{1}{120}\,dx =\frac{75-30}{120} =\frac{45}{120} =\frac{3}{8} =0.3750.

The answer is a probability of 0.3750. The density value $1/120$ is not the probability of a loss of, for example, 50 thousand dollars; that exact loss has probability zero. The calculation avoids the common mistake of reporting a density as though it were an interval probability.

Integrating Piecewise Densities

A density need not have one formula throughout its support. When the formula changes within the requested interval, divide the interval at each change point and add the resulting areas. The pieces must still be nonnegative, and their combined integral over the entire support must equal 1.

Worked Example: A Piecewise Density for Claim Severity

For a hypothetical portfolio, let the annual claim severity $X$ be measured in thousands of dollars and have density

f_X(x)= \begin{cases} kx, & 0\leq x<2,\\ k(4-x), & 2\leq x\leq4,\\ 0, & \text{otherwise}. \end{cases}

Find $k$, then calculate the probability that the claim is between 1 and 3 thousand dollars. Assume this is a continuous severity model on the stated support, with no point masses. No interest or mortality assumption is required.

Step 1: Normalize. Integrate each piece over its own support. The first area is

\int_0^2 kx\,dx =k\left[\frac{x^2}{2}\right]_0^2 =2k.

The second area is

\int_2^4 k(4-x)\,dx =k\left[4x-\frac{x^2}{2}\right]_2^4 =k(8-6) =2k.

Thus the total area is $2k+2k=4k$. Setting it equal to 1 gives $k=1/4$. The check is $4(1/4)=1$, and both pieces are nonnegative on their supports.

Step 2: Split the requested interval at the formula change. The interval from 1 to 3 includes the change at 2, so

\begin{aligned} P(1<X<3) &=\int_1^2 \frac{x}{4}\,dx+\int_2^3 \frac{4-x}{4}\,dx\\ &=\left[\frac{x^2}{8}\right]_1^2 +\left[x-\frac{x^2}{8}\right]_2^3\\ &=\frac{4-1}{8}+\left(\frac{15}{8}-\frac{12}{8}\right)\\ &=\frac{3}{8}+\frac{3}{8} =\frac{3}{4} =0.7500. \end{aligned}

The interval probability is 0.7500, which is within the required range from 0 to 1. The method avoids applying one piece of the density across an interval where the formula changes.

Why the Density Is Not a Probability

The distinction is easiest to see by comparing a density at one point with the area over an interval. In the piecewise example, $f_X(1.5)=1.5/4=0.375$ per thousand dollars. That number has inverse-loss units and is not a probability. The probability from 1.5 to 2 thousand dollars is instead

\int_{1.5}^{2}\frac{x}{4}\,dx =\left[\frac{x^2}{8}\right]_{1.5}^{2} =\frac{4-2.25}{8} =0.21875 \approx 0.2188.

The area uses the density over a positive-width interval. Multiplying the density at one point by an interval width can be an approximation when the density changes little over that interval, but the integral is the exact calculation.

Worked Example: A Quadratic Density and an Interval Probability

A hypothetical analyst models a one-year loss $X$, in thousands of dollars, with density $f_X(x)=k(3-x)^2$ for $0\leq x\leq3$, and zero otherwise. Find $k$ and calculate the probability that $X$ lies between 0.5 and 2. The model is continuous on its support; no interest or mortality assumptions are part of this loss model.

Step 1: Find the normalizing constant. The total probability must be 1:

\begin{aligned} 1 &=k\int_0^3(3-x)^2\,dx\\ &=k\left[-\frac{(3-x)^3}{3}\right]_0^3\\ &=k\left(0-\left(-9\right)\right) =9k. \end{aligned}

Therefore $k=1/9$. Substitution verifies normalization: the integral over the support is $9(1/9)=1$. The density is nonnegative because it is a positive constant multiplied by a square.

Step 2: Integrate over the requested range.

\begin{aligned} P(0.5<X<2) &=\frac{1}{9}\int_{0.5}^{2}(3-x)^2\,dx\\ &=\frac{1}{9}\left[-\frac{(3-x)^3}{3}\right]_{0.5}^{2}\\ &=\frac{1}{9}\left(\frac{(2.5)^3-(1)^3}{3}\right)\\ &=\frac{1}{9}\left(\frac{15.625-1}{3}\right) =\frac{4.875}{9} =\frac{13}{24} \approx 0.5417. \end{aligned}

The result is a valid probability, and the exact fraction confirms the decimal: $13/24=0.541666\ldots$, which rounds to 0.5417. The calculation uses the density over the full requested interval rather than evaluating it at a representative point.

Common Mistakes and Checks

Treating $f_X(x)$ as $P(X=x)$. A density is probability per unit of loss. For a continuous model, the probability at one exact loss amount is zero.
Skipping normalization. A nonnegative function is not automatically a density. Its integral over the support must equal 1.
Using inconsistent units. If the density uses thousands of dollars, use thousands of dollars for the integration limits. Converting the limits to dollars without adjusting the density’s units changes the calculation.
Ignoring a formula change. For a piecewise density, split the integral wherever the formula changes and use the appropriate expression on each part.
Using a probability density’s value as an area. A density alone has units; the interval integral produces a probability. Check that a reported probability is between 0 and 1.

A quick reasonableness check is to compare the requested interval with the full support. Its probability cannot exceed the total probability of 1. Also, for disjoint intervals, their probabilities add because their areas under the density add.

Key takeaway: For a continuous loss, check that the density is nonnegative and integrates to 1. Then calculate the probability of a loss interval by integrating the density over that interval. A density value is a local rate, not a probability.

Check Your Understanding

For each problem, identify the density’s units where relevant, normalize if needed, and integrate over the requested interval. The loss models are hypothetical and continuous on their stated supports.

A loss $X$, in thousands of dollars, is uniform from 0 to 80. Calculate $P(10<X<34)$. (Answer: 0.3000)
On $0\leq x\leq1$, a proposed density is $f_X(x)=k(1+x)$, and it is zero elsewhere. Find $k$, then find $P(0.25<X<0.75)$ and $f_X(0.5)$. (Answer: $k=2/3$, interval probability $=0.5000$, and $f_X(0.5)=1$ per unit of loss)
On $0\leq x\leq3$, let $f_X(x)=kx$, and let the density be zero elsewhere. Determine $k$ and calculate $P(1<X<2)$. (Answer: $k=2/9$ and probability $=0.3333$)
A loss in thousands of dollars has density $k$ for $0\leq x<2$, density $k(4-x)$ for $2\leq x\leq4$, and zero elsewhere. Find $k$ and $P(1<X<3)$, splitting the interval at the change point. (Answer: $k=0.25$ and probability $=0.6250$)
A loss $X$, in thousands of dollars, is uniform on $[0,50]$. Find its density, $P(X=20)$, and $P(20<X<25)$. (Answer: density $=0.0200$ per thousand dollars, $P(X=20)=0$, and interval probability $=0.1000$)

What You'll Learn