Continuous Uniform Severity

Introduction

A continuous uniform severity model is useful when a loss is assumed to be equally likely anywhere in a specified finite interval. The assumption is deliberately simple: it gives a constant density over the interval, which makes probabilities, moments, and quantiles straightforward to calculate. Its value in an actuarial model depends on whether that equal-likelihood assumption is appropriate for the losses being represented.

This tutorial develops the model on an interval with lower endpoint $a$ and upper endpoint $b$, where $a<b$. The variable $X$ denotes the loss severity. The model is continuous, so the probability of an exact single value—including either endpoint—is zero. Probabilities are determined by interval lengths, not by counting possible values as in a discrete uniform model.

Definition and Distribution Functions

Definition: A loss severity $X$ has a continuous uniform distribution on $[a,b]$, written $X\sim\operatorname{Uniform}(a,b)$, if $a<b$ and its density is constant on that interval and zero outside it:

f_X(x)= \begin{cases} \dfrac{1}{b-a}, & a\le x\le b,\\ 0, & \text{otherwise}. \end{cases}

The interval $[a,b]$ is the support. Its length $b-a$ determines the density: a wider interval has a lower constant density.

The cumulative distribution function is the probability that the loss is no greater than $x$. Below the support that probability is zero; above the support it is one. Within the support, the probability accumulates in proportion to the length from $a$ to $x$.

F_X(x)= \begin{cases} 0, & x<a,\\[2pt] \dfrac{x-a}{b-a}, & a\le x\le b,\\[6pt] 1, & x>b. \end{cases}

For $a\le c\le d\le b$, the probability that the loss lies between $c$ and $d$ is the length of that subinterval divided by the total support length. Because the distribution is continuous, including or excluding either endpoint does not change the probability.

P(c\le X\le d)=\frac{d-c}{b-a}.

For a threshold $t$ inside the support, the upper-tail probability is the length from $t$ to $b$, divided by the total length. Thresholds outside the support require checking the support first: $P(X>t)=1$ when $t<a$, and $P(X>t)=0$ when $t\ge b$.

P(X>t)=\frac{b-t}{b-a},\qquad a\le t\le b.

Mean, Variance, and Quantiles

The mean is the midpoint of the support, while the variance depends only on its width. These results follow directly from integrating against the constant density.

E[X]=\frac{a+b}{2}, \qquad \operatorname{Var}(X)=\frac{(b-a)^2}{12}.

To see why the mean is the midpoint, integrate $x$ times the density over the support. For the variance, first calculate the second raw moment, then subtract the square of the mean.

E[X]=\int_a^b x\frac{1}{b-a}\,dx =\frac{1}{b-a}\left[\frac{x^2}{2}\right]_a^b =\frac{b^2-a^2}{2(b-a)} =\frac{a+b}{2}.

E[X^2]=\int_a^b x^2\frac{1}{b-a}\,dx =\frac{b^3-a^3}{3(b-a)} =\frac{a^2+ab+b^2}{3}.

Writing the variance as $E[X^2]-(E[X])^2$ gives the stated result:

\operatorname{Var}(X) =\frac{a^2+ab+b^2}{3}-\left(\frac{a+b}{2}\right)^2 =\frac{(b-a)^2}{12}.

The $p$-quantile, for $0\le p\le1$, is the value $x_p$ satisfying $F_X(x_p)=p$. For an interior percentile, solve the linear part of the CDF for $x_p$. This also shows why the median is the midpoint.

x_p=a+p(b-a),\qquad 0\le p\le1.

Key result: For $X\sim\operatorname{Uniform}(a,b)$, interval probability is proportional to interval length, $E[X]$ is the support midpoint, $\operatorname{Var}(X)$ is the squared support width divided by $12$, and the $p$-quantile is $a+p(b-a)$.

Worked Applications

Worked Example: Probabilities, Moments, and a Percentile

A hypothetical severity model assumes that a covered loss $X$, in dollars, is uniformly distributed from $\$200$ to $\$1{,}000$. Calculate $P(350<X\le750)$, the mean and variance, and the 90th percentile. Treat this as the stated continuous model, not as a claim about actual losses.

Method. Identify $a=200$, $b=1{,}000$, and the support width $b-a=800$. Apply the uniform formulas to the probability, moments, and quantile. Since $X$ is continuous, the strict or inclusive endpoint at 350 or 750 makes no difference.

Probability. The interval length is $750-350=400$, so

P(350<X\le750)=\frac{400}{800}=0.5000.

Mean. The midpoint is $(200+1{,}000)/2=1{,}200/2=\$600$.

Variance. The squared width is $800^2=640{,}000$, hence

\operatorname{Var}(X)=\frac{640{,}000}{12} =53{,}333.333\ldots\text{ dollars}^2.

90th percentile. With $p=0.90$,

x_{0.90}=200+0.90(800)=200+720=\$920.

Check. The probability is one-half because the interval from 350 to 750 is half the full support length. The mean is centered in the support, and the 90th percentile, $\$920$, lies near the upper endpoint. The variance is reported in dollars squared; rounding it to two decimal places gives $53{,}333.33\text{ dollars}^2$. A common mistake is to divide by the endpoint $b$ rather than the width $b-a$; the denominator must be the full support length.

Worked Example: Combining a Tail Probability and a Quantile

For a hypothetical severity model, suppose $X\sim\operatorname{Uniform}(50,350)$, with losses measured in dollars. Find the probability that a loss is at most $\$125$, the probability that it is between $\$175$ and $\$275$, and the 80th percentile.

Method. The support width is $350-50=300$. Each probability is a subinterval length divided by 300; the percentile is found using the inverse CDF.

For $X\le125$, the relevant length is $125-50=75$:

P(X\le125)=\frac{75}{300}=0.2500.

For the interval from 175 to 275, the length is $275-175=100$:

P(175<X<275)=\frac{100}{300}=\frac{1}{3}=0.3333

to four decimal places. For the 80th percentile,

x_{0.80}=50+0.80(300)=50+240=\$290.

Check. The first probability is one-quarter because 75 is one-quarter of the support width. The second is one-third because its interval is 100 of the 300 dollars in the support. The percentile is 80% of the way from 50 to 350, giving $\$290$. A common mistake is to calculate a percentile as $0.80b$; the percentile is measured from the lower endpoint, so the correct calculation is $a+0.80(b-a)$.

Worked Example: Recovering the Support from Moments

A hypothetical continuous uniform severity model has mean $\$900$ and variance $120{,}000\text{ dollars}^2$. Find its support, the probability that $X>1{,}200$, and its 35th percentile. Assume the stated mean and variance are exact.

Method. Let the support be $[a,b]$. The mean determines its midpoint, and the variance determines its width. For a uniform model, $(b-a)^2=12\operatorname{Var}(X)$.

First calculate the width:

b-a=\sqrt{12(120{,}000)} =\sqrt{1{,}440{,}000} =1{,}200.

The midpoint is 900, so the endpoints are 600 dollars below and above 900:

a=900-\frac{1{,}200}{2}=300, \qquad b=900+\frac{1{,}200}{2}=1{,}500.

Thus $X\sim\operatorname{Uniform}(300,1{,}500)$. The upper-tail interval above 1,200 has length $1{,}500-1{,}200=300$, and the full support has length 1,200:

P(X>1{,}200)=\frac{300}{1{,}200}=0.2500.

The 35th percentile is

x_{0.35}=300+0.35(1{,}200)=300+420=\$720.

Check. The recovered endpoints have midpoint $(300+1{,}500)/2=900$ and width 1,200, giving variance $1{,}200^2/12=120{,}000\text{ dollars}^2$. Both match the supplied moments. A common mistake is to take the square root of the variance alone; the uniform variance formula requires multiplying by 12 before taking the square root.

Assumptions and Common Exam Traps

The continuous uniform model makes a strong assumption: every equal-length subinterval within the support has the same probability. For example, an interval of length 20 has twice the probability of an interval of length 10, provided both lie within the support. This model does not put probability outside $[a,b]$, and it assigns zero probability to any exact point.

It is important to keep the continuous model distinct from a discrete uniform model. A discrete uniform variable assigns equal probability to each member of a finite set; a continuous uniform variable assigns constant density per unit of length. The probability of an interval is therefore calculated using lengths, not by counting integer values.

Using the wrong denominator. Divide a subinterval length by $b-a$, not by $b$, unless $a=0$ and those happen to be equal.
Ignoring the support. A proposed threshold below $a$ or above $b$ cannot be handled by the interior formula without checking the boundary case.
Confusing a percentile with a tail probability. The $p$-quantile has cumulative probability $p$ to its left; its upper-tail probability is $1-p$ for this continuous distribution.
Misreporting the variance units. If losses are in dollars, the variance is in dollars squared. The standard deviation, if needed, is the square root of that variance and is in dollars.

The model is especially convenient when a problem explicitly specifies uniform severity or asks for calculations under that assumption. In applications, equal likelihood across a finite interval may be a rough approximation rather than a realistic description. Model choice should not be inferred merely because the loss is bounded.

Key takeaway: First identify the support endpoints. Then use its width $b-a$ consistently: interval probability is interval length divided by support length, the mean is the midpoint, the variance is width squared divided by 12, and a percentile is measured from the lower endpoint.

Check Your Understanding

For each problem, assume the stated severity model is continuous uniform and losses are measured in dollars unless noted otherwise.

A hypothetical severity $X\sim\operatorname{Uniform}(40,240)$. Find $P(100\le X<190)$, $E[X]$, and $\operatorname{Var}(X)$. (Answer: $0.4500$, $\$140$, and $3{,}333.33\text{ dollars}^2$.)
A loss severity is uniform on $[0,800]$. Find the 92.5th percentile and $P(X>600)$. (Answer: $\$740$ and $0.2500$.)
A uniform severity has mean $\$500$ and variance $30{,}000\text{ dollars}^2$. Recover its support and calculate $P(350<X\le650)$. (Answer: $[200,800]$ and $0.5000$.)
For $X\sim\operatorname{Uniform}(150,450)$, calculate $P(X<225\text{ or }X>375)$ and the 60th percentile. (Answer: $0.5000$ and $\$330$.)
The 25th and 75th percentiles of a continuous uniform severity are $\$300$ and $\$700$, respectively. Find the support, mean, variance, and $P(X>700)$. (Answer: support $[100,900]$, mean $\$500$, variance $53{,}333.33\text{ dollars}^2$, and probability $0.2500$.)

What You'll Learn