Probability Densities And Interval Probabilities

Introduction

For a continuous loss model, a density describes how probability is distributed across possible loss amounts. It is not the probability of any one exact loss. To find the probability that a loss falls in a range, integrate the density over that range. This distinction is fundamental in severity calculations: a density value has units of probability per unit of loss, while a probability has no units.

Throughout this tutorial, $X$ denotes a hypothetical loss amount, and we assume $X$ has a purely continuous distribution. Thus, its probability is described by a density $f_X$, with no probability mass concentrated at an individual loss amount. The support is the set of loss amounts that can occur; outside the support, the density is zero.

Definition: A probability density function $f_X(x)$ for a continuous loss $X$ is a nonnegative function whose integral over all possible loss amounts is one. For any interval from $a$ to $b$, where $a<b$, the probability that the loss falls in that interval is the integral of the density over the interval: $$ P(a<X\le b)=\int_a^b f_X(x)\,dx. $$ For a nonnegative loss, $f_X(x)=0$ for $x<0$. A density is measured in inverse units of loss, such as probability per dollar.

From Density to Interval Probability

The area under the density curve over a range gives the probability of a loss in that range. If the density is constant over an interval, the area is a rectangle: density multiplied by interval width. If the density varies, integration adds the contributions across the interval.

The formula can also be justified using the cumulative distribution function $F_X$. For a purely continuous loss with density $f_X$, the probability accumulated up to $x$ is the integral of the density up to $x$. Subtracting the accumulated probability at $a$ from that at $b$ leaves precisely the probability between them:

F_X(b)-F_X(a) =\int_{-\infty}^{b} f_X(x)\,dx-\int_{-\infty}^{a} f_X(x)\,dx =\int_a^b f_X(x)\,dx.

This subtraction works because probabilities on disjoint intervals add. In familiar cases with a continuous or piecewise-continuous density, the integral can be understood as the limit of sums of approximate areas over increasingly fine subintervals. That Riemann-sum explanation assumes sufficient regularity and suitable partitions and sample points. More generally, the integral defining a density is a Lebesgue integral; one should not assume that arbitrary sample-point sums converge for every integrable density.

For a purely continuous model, the probability at any one exact value is zero. In particular, $P(X=a)=0$. Consequently, for finite endpoints, $P(a<X<b)$, $P(a\le X\le b)$, and $P(a<X\le b)$ are equal. This does not mean the loss cannot take a value in the interval; it means that a single point has zero probability under this model.

Key distinction: $f_X(x)$ is a density value, not $P(X=x)$. Probabilities come from integrating the density over a range. A density can even be greater than one if it is concentrated over a sufficiently narrow range; the total integral, not each individual density value, must equal one.

Checking a Density Before Using It

Before calculating an interval probability, verify that the proposed function is a valid density for the stated model. It must be nonnegative on its support, zero outside the support if none is otherwise specified, and have total integral one. If the total integral is not one, the function as given is not a probability density.

The limits of integration must match the event in the question, not automatically the entire support. For example, the probability of a loss between 2 and 5 is found by integrating from 2 to 5, even if the model permits losses from 0 to 10. Units also matter: if $x$ is measured in dollars, the density is per dollar and the integration variable contributes dollars, leaving a dimensionless probability.

Worked Example: A Density That Increases Across Its Support

A hypothetical loss $X$, measured in dollars, has density $f_X(x)=x/8$ for $0\le x\le4$, and zero otherwise. Find the probability that the loss is greater than 1 dollar and at most 3 dollars. Also interpret the density at 2 dollars.

Method and validity check. First check the total integral: $$ \int_0^4 \frac{x}{8}\,dx =\left[\frac{x^2}{16}\right]_0^4 =\frac{16}{16}-0 =1. $$ The density is nonnegative on the support, so it is a valid density.

Interval probability. The requested event uses the interval from 1 to 3, so integrate over those limits: $$ P(1<X\le3) =\int_1^3 \frac{x}{8}\,dx =\left[\frac{x^2}{16}\right]_1^3 =\frac{9}{16}-\frac{1}{16} =\frac{8}{16} =0.5000. $$ The probability is 0.5000, or 50.00%.

Interpret the density value. At $x=2$, $f_X(2)=2/8=0.25$ per dollar. This is not a 25% probability of a loss exactly equal to 2 dollars. The probability at that exact point is zero; probability is obtained by integrating over an interval.

Check and common mistake avoided. The integral over the full support is one, and the interval probability is between zero and one. Treating $f_X(2)=0.25$ as a point probability would confuse a density value with a probability.

Worked Example: Integrating a Piecewise-Constant Density

A hypothetical loss $X$, measured in thousands of dollars, has density $f_X(x)=1/8$ for $0\le x<4$, $f_X(x)=1/16$ for $4\le x\le12$, and zero otherwise. Find $P(2<X\le6)$, and compare the density values at 3 and 5.

Method and validity check. The support has two pieces, so check the total area by integrating each piece: $$ \int_0^4 \frac{1}{8}\,dx+\int_4^{12}\frac{1}{16}\,dx =\frac{4}{8}+\frac{8}{16} =\frac{1}{2}+\frac{1}{2} =1. $$ Both pieces are nonnegative, so the function is a valid density.

Split the requested interval where the formula changes. The interval from 2 to 6 contains 2 units in the first piece and 2 units in the second: $$ P(2<X\le6) =\int_2^4\frac{1}{8}\,dx+\int_4^6\frac{1}{16}\,dx =\frac{2}{8}+\frac{2}{16} =\frac{1}{4}+\frac{1}{8} =\frac{3}{8} =0.3750. $$

At $x=3$, the density is $1/8=0.125$ per thousand dollars. At $x=5$, it is $1/16=0.0625$ per thousand dollars. These are local density values, not probabilities. The value assigned at the single boundary $x=4$ does not alter an integral or any interval probability in this continuous model.

Check and common mistake avoided. The interval probability is the sum of the two areas, not the density at either endpoint and not the area across the entire support. The result, 0.3750, is a valid probability.

Reading the Units and the Shape

A density has units that are the reciprocal of the loss unit. If $X$ is measured in dollars, then $f_X(x)$ is probability per dollar. If the same loss is expressed in thousands of dollars, the numerical density values change to match the new scale, while probabilities for the same actual loss range do not change. The integration variable and the density units combine to give a dimensionless result.

The density's height helps indicate where probability is concentrated, but its height alone does not give the probability of a range. Both height and width matter. Over a narrow interval, probability is approximately density times width when the density is nearly constant there; the exact probability is the integral. If the density varies substantially, split the interval where its formula changes or integrate its full expression.

Likewise, a density value may exceed one without violating the rules for a density. For example, a constant density of 2 over an interval of width 0.5 has total area $2(0.5)=1$. The requirement is that the total area be one, not that density values be at most one.

Worked Example: A Declining Density and a Probability Range

A hypothetical loss $X$, measured in dollars, has density $f_X(x)=(6-x)/18$ for $0\le x\le6$, and zero otherwise. Find the probability that the loss is between 2 and 5 dollars. Also state the density at 2 dollars and explain why it is not the requested probability.

Method and validity check. On the support, $6-x$ is nonnegative. The total integral is $$ \int_0^6 \frac{6-x}{18}\,dx =\left[\frac{6x-x^2/2}{18}\right]_0^6 =\frac{36-18}{18} =1. $$ Thus the function is a valid density.

Integrate over the requested range. An endpoint has probability zero in this continuous model, so the choice of strict or inclusive endpoints does not change the answer: $$ P(2<X<5) =\int_2^5\frac{6-x}{18}\,dx =\left[\frac{6x-x^2/2}{18}\right]_2^5. $$ At $x=5$, the antiderivative is $(30-25/2)/18=17.5/18$. At $x=2$, it is $(12-2)/18=10/18$. Therefore, $$ P(2<X<5) =\frac{17.5-10}{18} =\frac{7.5}{18} =\frac{5}{12} =0.4167. $$

At 2 dollars, $f_X(2)=(6-2)/18=2/9$ per dollar. That density value describes probability concentration locally at 2 dollars; it is not the probability of the entire interval from 2 to 5. The requested probability is the area across that interval.

Check and common mistake avoided. The integrated probability is between zero and one, and the full-support integral is one. Substituting a point into the density cannot replace integrating over the interval.

Common Exam Traps

Reporting a density as a probability. A value such as $f_X(3)=0.12$ is not $P(X=3)$, and it is not automatically a 12% chance of a nearby event. Integrate over the event's interval.
Using the wrong limits. The support describes all possible losses; the event determines the integration limits. For $P(2<X\le5)$, integrate from 2 to 5.
Forgetting a change in the formula. When a density is piecewise-defined, split the integral at every point where the formula changes.
Ignoring units. A density per thousand dollars is not numerically interchangeable with a density per dollar. Keep the loss scale consistent throughout the integration.
Assuming every function is a density. Check nonnegativity and total integral before using the function to calculate probabilities.
Overinterpreting endpoint notation. For a purely continuous model, including or excluding a finite endpoint leaves the probability unchanged. That conclusion depends on the no-point-mass assumption.

Key takeaway: A continuous severity density allocates probability across loss amounts. To find the probability of an interval, integrate the density over that interval, check the model's support and normalization, and keep density values distinct from probabilities.

Check Your Understanding

For each hypothetical continuous loss model, check what the density represents and integrate over the event stated. Answers are included for checking.

A loss $X$, in dollars, has density $1/20$ on $0\le x\le20$ and zero otherwise. Find $P(3<X\le11)$. (Answer: $0.4000$)
A loss $X$, in thousands of dollars, has density $0.06$ on $0\le x<10$, density $0.02$ on $10\le x\le30$, and zero otherwise. Verify the total integral and find $P(5<X<20)$. (Answer: Total integral $=1$; probability $=0.5000$)
A loss $X$, in dollars, has density $x/18$ on $0\le x\le6$ and zero otherwise. Find $P(2<X\le5)$. (Answer: $7/12=0.5833$)
A loss $X$, in dollars, has density $k$ on $0\le x<2$, density $2k$ on $2\le x\le6$, and zero otherwise. Find $k$, then find $P(1<X\le4)$ and $f_X(3)$. (Answer: $k=0.1$; probability $=0.5000$; $f_X(3)=0.2$ per dollar)
A loss $X$, in dollars, has density $2x/25$ on $0\le x\le5$ and zero otherwise. Find $P(1<X<3)$, $P(X=2)$, and $f_X(2)$. (Answer: $0.3200$; $0$; $0.16$ per dollar)

What You'll Learn