Uniform Severity: Probabilities and Moments

Uniform Severity as a Bounded Loss Model

A uniform severity model is useful when a hypothetical loss is treated as equally likely to fall anywhere in a specified range. Its two endpoints are not just parameters in a formula: they describe the smallest and largest possible losses in the model. Once those bounds are clear, probabilities are found by comparing interval lengths, while the mean and variance summarize the location and spread of the loss.

Let $X$ denote a nonnegative loss. Suppose its lower and upper bounds are $a$ and $b$, where $0\leq a<b$. The notation $X\sim\operatorname{Uniform}(a,b)$ means that $X$ has a continuous uniform distribution on that interval. There is no point mass at either endpoint, so including or excluding a single endpoint does not change a probability.

Definition: If $X\sim\operatorname{Uniform}(a,b)$, its density is $1/(b-a)$ for $a\leq x\leq b$, and zero elsewhere. The model assigns equal probability to intervals of equal length within $[a,b]$. The values $a$ and $b$ are loss bounds, not the mean and variance.

The density is constant over the support, and its total area must be one. The support has width $b-a$, so a rectangle of height $1/(b-a)$ over that width has area $(b-a)/(b-a)=1$. The density has units inverse to the units of $X$: if losses are measured in dollars, the density is per dollar.

f_X(x)= \begin{cases} \dfrac{1}{b-a}, & a\leq x\leq b,\\[4pt] 0, & \text{otherwise}. \end{cases}

Distribution Function and Interval Probabilities

The cumulative distribution function $F_X(x)=P(X\leq x)$ is the accumulated area under the density up to $x$. It is zero below the lower bound, increases linearly across the support, and equals one at and above the upper bound. The survival function is $S_X(x)=P(X>x)=1-F_X(x)$.

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

For an interval $(r,s]$ entirely within the support, its probability is its length divided by the full support length. If an interval extends beyond the support, only its overlap with $[a,b]$ contributes. This clipping matters: a loss outside the stated bounds is impossible under the model.

P(r<X\leq s) = \frac{\max\!\left(0,\min(s,b)-\max(r,a)\right)}{b-a}, \qquad r<s.

The same probability can be calculated as $F_X(s)-F_X(r)$. Because the model is continuous, probabilities at individual points are zero: for example, $P(X=a)=P(X=b)=0$. Thus, changing a strict inequality to a non-strict inequality at a single endpoint does not affect the result.

Worked Example: Probabilities and Moments in Dollars

A hypothetical repair-cost model assumes that the loss $X$ is uniform from $400 to $1,600. Find the density, the probability that the loss is between $700 and $1,300, the mean, and the variance. This is a severity calculation only; no interest or mortality assumption applies.

Model and support. Here $X\sim\operatorname{Uniform}(400,1600)$, with dollar-valued bounds $a=400$ and $b=1600$. The support width is $1600-400=1200$ dollars.

Density. The constant density is the reciprocal of the width:

f_X(x)=\frac{1}{1200}=0.0008333\text{ per dollar},\qquad 400\leq x\leq1600.

Interval probability. The interval length is $1300-700=600$ dollars. Therefore,

P(700<X\leq1300)=\frac{600}{1200}=0.5000.

Mean and variance. The mean is the midpoint of the bounds. The variance is the squared support width divided by 12:

E[X]=\frac{400+1600}{2}=1000\text{ dollars}, \qquad \operatorname{Var}(X)=\frac{(1600-400)^2}{12} =\frac{1{,}440{,}000}{12}=120{,}000\text{ dollars}^2.

The calculations check against the model: the probability is between zero and one, the mean lies between $400 and $1,600, and the variance has squared-dollar units. The reported probability is to four decimals; money is to the cent where applicable.

Mean and Variance: Why the Formulas Hold

The uniform density allows moments to be calculated directly by integration. For any positive integer $k$, the $k$th raw moment is the integral of $x^k f_X(x)$ over the support. The resulting expression gives the mean when $k=1$, and the second raw moment when $k=2$.

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

In particular, simplifying the $k=1$ result gives the midpoint. For the variance, use $\operatorname{Var}(X)=E[X^2]-(E[X])^2$. Substitution and simplification yield the squared width divided by 12.

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

A second way to see the variance formula is to center the loss at the midpoint $m=(a+b)/2$. The centered variable ranges symmetrically from $-(b-a)/2$ to $(b-a)/2$. Averaging its square over this symmetric interval gives $(b-a)^2/12$. This explains why the variance depends only on the width, not on the location of the interval.

Result: For $X\sim\operatorname{Uniform}(a,b)$, the mean is $(a+b)/2$, the variance is $(b-a)^2/12$, and the standard deviation is $(b-a)/\sqrt{12}$. These formulas require finite bounds with $a<b$; they do not describe a uniform distribution over an unbounded interval.

Worked Example: Losses Measured in Thousands

For a hypothetical equipment-loss model, let $X$, measured in thousands of dollars, be uniform from 2 to 10. Find the density, the probability of a loss above $8,000, the probability of a loss from $4,500 to $8,000, and the mean and variance. This model has no interest or mortality component.

Model and units. $X\sim\operatorname{Uniform}(2,10)$, with both bounds in thousands of dollars. Its width is $10-2=8$ thousand dollars, so the density is $1/8=0.125$ per thousand dollars.

Above $8,000. In the model's units, $8,000 is $8$. The interval from 8 to 10 has length 2, giving

P(X>8)=\frac{10-8}{10-2}=\frac{2}{8}=0.2500.

Between $4,500 and $8,000. These amounts correspond to 4.5 and 8 thousand dollars. The interval length is $8-4.5=3.5$, so

P(4.5<X\leq8)=\frac{8-4.5}{10-2}=\frac{3.5}{8}=0.4375.

Moments. The mean is $(2+10)/2=6$ thousand dollars, or $6,000. The variance is $8^2/12=64/12=5.3333$ thousand-dollars-squared. If expressed in dollars squared, it is $5.3333\times1{,}000{,}000$, or $5,333,333.33 dollars squared.

The probability checks are consistent with the interval lengths: the whole support has length 8, while the two requested intervals have lengths 2 and 3.5. The unit conversion for the variance is squared because variance squares the measurement unit.

Using Bounds Carefully

The endpoints specify what losses the model allows, so first check whether a requested interval is wholly inside the support. For an interval that only partly overlaps, use its overlap length. For example, if the interval begins below $a$, the part below $a$ contributes no probability. If it lies entirely above $b$, the probability is zero.

A useful exam workflow is to write the model and bounds, calculate the support width, then identify the requested interval's overlap with the support. Only after that should you divide lengths or apply the moment formulas. Keeping the variable's units visible helps catch errors, especially when the loss is given in thousands but thresholds are stated in dollars.

State the support.
Write $X\sim\operatorname{Uniform}(a,b)$, confirm $a<b$, and note the units of both endpoints.

Find the width.
Calculate $b-a$. The density is its reciprocal, with units inverse to those of the loss.

Handle probabilities by overlap.
Intersect the requested interval with $[a,b]$, then divide the overlap length by $b-a$.

Calculate and check moments.
Use the midpoint for the mean and the squared width divided by 12 for the variance; check the bounds and units.

Worked Example: Recovering an Unknown Upper Bound

A hypothetical loss severity is uniform from $200 to an unknown upper bound $b$. Its mean is $650. Find $b$, the probability of a loss exceeding $920, and the variance. This is a loss model without discounting or mortality.

Method and model. Let $X\sim\operatorname{Uniform}(200,b)$. Apply the midpoint formula to the stated mean:

650=\frac{200+b}{2} \quad\Longrightarrow\quad 1300=200+b \quad\Longrightarrow\quad b=1100.

The support is therefore $[200,1100]$, with width $1100-200=900$ dollars. The requested tail interval runs from $920 to $1,100, with length $1100-920=180$ dollars. Thus,

P(X>920)=\frac{180}{900}=0.2000.

The variance is

\operatorname{Var}(X)=\frac{900^2}{12} =\frac{810{,}000}{12} =67{,}500\text{ dollars}^2.

Check. The recovered upper bound exceeds the lower bound, and the midpoint $(200+1100)/2=650$ matches the given mean. The tail probability is the fraction of the support above $920, not a fraction of the dollar value itself.

Common Exam Trap

A frequent error is to divide an interval length by one of the endpoints rather than by the total support width. For the model $X\sim\operatorname{Uniform}(400,1600)$, the probability from $700 to $1,300 is not $600/1600=0.3750$. The $1,600 endpoint is a bound, not the width of the support. The correct denominator is $1600-400=1200$, so the probability is $600/1200=0.5000$.

Another related error is to calculate an interval's full length even when part of it is outside the support. For example, under that same model, $P(300<X\leq700)$ uses only the overlap from $400 to $700. Its probability is $(700-400)/1200=0.2500$, not $(700-300)/1200$. Always interpret the endpoints as the full range of possible losses in the model.

Key takeaway: For a uniform severity on $[a,b]$, the density is $1/(b-a)$, an interval probability is its overlap length divided by $b-a$, the mean is the midpoint, and the variance is the squared width divided by 12. Check support and units before substituting.

Check Your Understanding

For each hypothetical severity model, identify the support and keep the stated units consistent. Probabilities are rounded to four decimals where needed.

A repair-loss severity is uniform from $100 to $500. Find its density, $P(200<X\leq350)$, the mean, and variance. (Answer: density $0.0025$ per dollar; probability $0.3750$; mean $300; variance $13{,}333.33$ dollars squared.)
A loss $X$, measured in thousands of dollars, is uniform from 1 to 7. Find $P(X>5)$, the mean, and variance. (Answer: $0.3333$; 4 thousand dollars; 3 thousand-dollars-squared.)
A hypothetical claim severity is uniform from $250 to $1,450. Find the probability of a loss between $500 and $1,100, and find the variance. (Answer: probability $0.5000$; variance $120{,}000$ dollars squared.)
A severity is uniform from $200 to an unknown upper bound and has mean $650. Find the upper bound, $P(X>920)$, and the variance. (Answer: upper bound $1,100; probability $0.2000$; variance $67{,}500$ dollars squared.)
A loss is uniform from zero to an unknown upper bound and has mean $750. Find the upper bound, $P(400<X\leq1000)$, and the variance. (Answer: upper bound $1,500; probability $0.4000$; variance $187{,}500$ dollars squared.)

What You'll Learn