Discrete Uniform Severity

Introduction

A discrete uniform severity model represents a loss that can take one of a finite set of consecutive integer values, with every value equally likely. It is useful when a problem explicitly assigns equal probability to each permitted amount, or as a simple model for practicing probability, moment, and quantile calculations. The key is to count the possible values carefully: an interval with integer endpoints includes both endpoints.

The formulas are compact, but their application depends on the support. In particular, the number of outcomes from $a$ through $b$ is $b-a+1$, not $b-a$. Quantiles also require care because a discrete CDF jumps in increments rather than increasing continuously.

Definition: Let $X$ be a discrete uniform loss on the consecutive nonnegative integer values $a,a+1,\ldots,b$, where $a$ and $b$ are integers and $0\leq a\leq b$. Write $n=b-a+1$ for the number of supported values. Then each value has probability $1/n$: $$ P(X=x)= \begin{cases} \dfrac{1}{n}, & x\in\{a,a+1,\ldots,b\},\\ 0, & \text{otherwise}. \end{cases} $$

Probability Function and Cumulative Probabilities

The probability mass function assigns the same probability to every supported value and zero probability elsewhere. Thus, for any event that selects $r$ of the $n$ possible values, its probability is $r/n$. This counting rule works because the selected outcomes are mutually exclusive and equally likely.

For a real number $x$, the cumulative distribution function is $F_X(x)=P(X\leq x)$. Below the support it is zero, and at or above the largest supported value it is one. Between the endpoints, it counts how many integers in the support are at most $x$:

F_X(x)= \begin{cases} 0, & x<a,\\[2pt] \dfrac{\lfloor x\rfloor-a+1}{n}, & a\leq x<b,\\[6pt] 1, & x\geq b. \end{cases}

The floor function $\lfloor x\rfloor$ is the greatest integer less than or equal to $x$. For example, if $x=7.6$, then losses equal to $7$ are included in $P(X\leq x)$, but losses equal to $8$ are not. When calculating probabilities over a range, it is often clearer to count the included integers directly. If $c\leq d$ are supported integers, then $P(c\leq X\leq d)=(d-c+1)/n$.

Worked Example: Counting Outcomes in a Claim Range

A hypothetical warranty model assigns equal probability to each whole-dollar claim amount from $120$ through $125$, inclusive. Let $X$ be the claim amount in dollars. Calculate the probability that the claim is at least $122$ and at most $124$, and then calculate $F_X(123.7)$.

Method. List or count the integer outcomes in the support, retaining both endpoints. There are $n=125-120+1=6$ possible amounts, so each has probability $1/6$.

The event $122\leq X\leq124$ contains $122,123,124$, which is three supported values. Therefore, $P(122\leq X\leq124)=3/6=1/2$.

For the CDF, $X\leq123.7$ is equivalent to $X\leq123$, since $X$ only takes integer values. The included amounts are $120,121,122,123$, giving $F_X(123.7)=4/6=2/3$.

Check. The first probability is $0.5$, and the CDF value is larger because its event includes the additional outcomes $120$ and $121$. A common mistake is to count the range from $122$ to $124$ as $124-122=2$ outcomes; the inclusive count is $124-122+1=3$.

Mean and Variance

For a discrete uniform loss, the mean is the midpoint of the smallest and largest possible values. The variance depends only on the number $n$ of supported values, not on the starting value $a$:

Theorem: If $X$ is discrete uniform on $a,a+1,\ldots,b$, with $n=b-a+1$, then $$ E[X]=\frac{a+b}{2}=a+\frac{n-1}{2}, \qquad \operatorname{Var}(X)=\frac{n^2-1}{12}. $$

The midpoint formula follows by pairing outcomes from opposite ends of the support. The values $a$ and $b$ average to $(a+b)/2$; the next pair $a+1$ and $b-1$ has the same average. If there is an unpaired middle value, it is itself the midpoint. Since all outcomes are equally likely, their overall average is that midpoint.

The variance formula can be justified by shifting the support. Write $X=a+J$, where $J$ is uniform on $0,1,\ldots,n-1$. Adding the constant $a$ does not change variance, so it is enough to calculate $\operatorname{Var}(J)$. The standard finite sums are $\sum_{j=0}^{n-1}j=n(n-1)/2$ and $\sum_{j=0}^{n-1}j^2=n(n-1)(2n-1)/6$. Using $E[J^2]-(E[J])^2$ gives the result:

\begin{aligned} \operatorname{Var}(X) &=\operatorname{Var}(J)\\ &=\frac{1}{n}\sum_{j=0}^{n-1}j^2- \left(\frac{1}{n}\sum_{j=0}^{n-1}j\right)^2\\ &=\frac{(n-1)(2n-1)}{6}-\left(\frac{n-1}{2}\right)^2\\ &=\frac{(n-1)(n+1)}{12} =\frac{n^2-1}{12}. \end{aligned}

This derivation also explains why shifting every possible loss by the same amount changes the mean but leaves the variance unchanged. The formula requires consecutive integer outcomes with equal probabilities; it does not apply just because a severity model has a finite support.

Worked Example: Mean, Variance, and an Event Probability

A hypothetical equipment policy uses a claim severity $X$ that is discrete uniform on the whole-dollar amounts $8,9,\ldots,14$. Calculate $E[X]$, $\operatorname{Var}(X)$, $P(X>11)$, and the probability that the claim is between $9$ and $12$, inclusive.

Method. Treat the model as uniform over all integers from $a=8$ to $b=14$, inclusive. The support size is $n=14-8+1=7$, and each amount has probability $1/7$.

The mean is the midpoint: $E[X]=(8+14)/2=22/2=11$.

The variance is $\operatorname{Var}(X)=(7^2-1)/12=(49-1)/12=48/12=4$.

For $X>11$, the supported outcomes are $12,13,14$, so $P(X>11)=3/7$. For $9\leq X\leq12$, the outcomes are $9,10,11,12$, so $P(9\leq X\leq12)=4/7$.

Check. The event probabilities have numerators equal to their respective counts of supported outcomes, and their denominators are both seven. In particular, $X>11$ excludes $11$; treating it as $X\geq11$ would incorrectly count four outcomes.

Quantiles of a Discrete Uniform Loss

For $0<p\leq1$, define the $p$-quantile as the smallest value $x$ for which $F_X(x)\geq p$. This inverse-CDF definition is important for a discrete distribution: a quantile is a supported value, and the CDF may jump past $p$ rather than equal it.

Definition: For a discrete uniform loss on $a,a+1,\ldots,b$, the $p$-quantile, for $0<p\leq1$, is $$ Q_X(p)=a+\lceil np\rceil-1, $$ where $\lceil z\rceil$ is the least integer greater than or equal to $z$. Equivalently, it is the outcome at rank $\lceil np\rceil$ when the $n$ supported values are listed in increasing order.

To see why, the first $k$ outcomes have total probability $k/n$. The smallest rank $k$ that reaches probability $p$ must satisfy $k/n\geq p$, or $k\geq np$. The smallest integer meeting that condition is $k=\lceil np\rceil$, so the outcome at that rank is $a+k-1$. When $np$ is already an integer, the quantile is at that rank exactly; it is not the next outcome.

Worked Example: Applying the Quantile Rule at a Boundary

Let $X$ be a hypothetical whole-dollar loss, discrete uniform on $40,41,\ldots,50$. Find the $0.6$-quantile and the $0.75$-quantile, and verify each using the CDF.

Method. The support size is $n=50-40+1=11$. The probability at each value is $1/11$. For the $0.6$-quantile, $np=11(0.6)=6.6$, so the required rank is $\lceil6.6\rceil=7$. The seventh supported value is $40+7-1=46$. Thus $Q_X(0.6)=46$.

For the $0.75$-quantile, $np=11(0.75)=8.25$, so the required rank is $\lceil8.25\rceil=9$. The ninth supported value is $40+9-1=48$, hence $Q_X(0.75)=48$.

Verification. There are seven outcomes from $40$ through $46$, so $F_X(46)=7/11\geq0.6$, while $F_X(45)=6/11<0.6$. Similarly, $F_X(48)=9/11\geq0.75$, while $F_X(47)=8/11<0.75$. Each result is therefore the smallest value whose CDF reaches the target probability.

Common quantile trap. Rounding $np$ to the nearest integer is not the rule. For $p=0.6$, rounding $6.6$ to $7$ happens to give the correct rank, but a value such as $np=6.2$ would round to $6$ even though rank $6$ does not reach the target. Always use the ceiling.

Inferring a Support and Checking the Model

Sometimes the support is not stated directly. A probability below a threshold can identify the number of outcomes below that threshold, while the mean relates the endpoints. These facts can be combined to solve for the support. After finding candidate endpoints, check that the threshold lies in the support and that the resulting count gives the stated probability.

Worked Example: Recovering Unknown Integer Endpoints

A hypothetical nonnegative integer loss $X$ is discrete uniform on an unknown consecutive support $a,a+1,\ldots,b$. It is known that $P(X\leq12)=3/5$ and $E[X]=11$. Find the support, variance, and $0.8$-quantile.

Method. Let $n=b-a+1$. Because exactly $3/5$ of the $n$ equally likely outcomes are at most $12$, write $n=5k$, where the count at or below $12$ is $3k$. That count is also $12-a+1=13-a$. Therefore $13-a=3k$, or $a=13-3k$.

The mean condition gives $(a+b)/2=11$, so $a+b=22$. Since $b=a+n-1=a+5k-1$, it follows that $2a+5k-1=22$, or $2a+5k=23$. Substitute $a=13-3k$: $2(13-3k)+5k=23$, so $26-6k+5k=23$, giving $k=3$.

Thus $n=5k=15$, $a=13-3(3)=4$, and $b=a+n-1=4+15-1=18$. The support is $4,5,\ldots,18$. Checking the given probability, the outcomes at most $12$ are $4$ through $12$, or nine outcomes; $9/15=3/5$. The mean check is $(4+18)/2=11$.

The variance is $(15^2-1)/12=(225-1)/12=224/12=56/3$, or approximately $18.6667$. For the $0.8$-quantile, $np=15(0.8)=12$, so the rank is exactly $12$. The twelfth value is $4+12-1=15$, giving $Q_X(0.8)=15$. Indeed, $F_X(15)=12/15=0.8$, while $F_X(14)=11/15<0.8$.

Check. The endpoints are nonnegative integers, the support has the required probability and mean, and the quantile is the first value whose cumulative probability reaches $0.8$. The exact integer value of $np$ is a useful reminder that no extra rank is added at a percentile boundary.

Common Errors and Practical Checks

The calculations become routine once the support is explicit, but several errors recur in problems involving integer outcomes:

Dropping an endpoint. The number of integers from $a$ through $b$, inclusive, is $b-a+1$. Use this count in the probability mass function, mean parameterization, and quantile rank.
Counting the wrong event. Translate inequalities into included outcomes before counting. For integer $X$, $X>11$ starts at $12$, while $X\geq11$ includes $11$.
Rounding a quantile rank incorrectly. The rank is $\lceil np\rceil$, not the nearest integer and not always $np+1$. If $np$ is an integer, use that rank.
Using the variance formula outside its model. The expression $(n^2-1)/12$ requires equally likely consecutive integer outcomes. A finite support with unequal probabilities needs its own weighted calculation.
Confusing location with spread. Moving every support value up by the same amount changes $E[X]$, but leaves $n$ and therefore the variance unchanged.

A reliable check is to verify the support size first, then confirm that probabilities sum to one. For a proposed quantile $q$, check both $F_X(q)\geq p$ and $F_X(q-1)<p$. These checks are especially helpful when the target percentile lies exactly on a CDF jump.

Key takeaway: For a discrete uniform loss on consecutive integers, count the outcomes first. With $n=b-a+1$, each has probability $1/n$, the mean is $(a+b)/2$, the variance is $(n^2-1)/12$, and the $p$-quantile is the outcome at rank $\lceil np\rceil$.

Check Your Understanding

For each problem, identify the support size before calculating. Quantiles use the smallest supported value whose cumulative probability is at least the stated level.

A hypothetical loss is discrete uniform on the whole-dollar amounts $3$ through $8$, inclusive. Find $P(X\geq6)$, $E[X]$, and $\operatorname{Var}(X)$. (Answer: $1/2,\ 5.5,\ 35/12$)
A discrete uniform loss has support $14,15,\ldots,22$. Find $P(X\leq17)$, its variance, and its $0.7$-quantile. (Answer: $4/9,\ 20/3,\ 20$)
A loss is discrete uniform on the integers $0$ through $15$. Find $P(4<X\leq10)$, the $0.625$-quantile, and the mean. (Answer: $3/8,\ 9,\ 7.5$)
A discrete uniform loss is supported on the integers $5$ through $16$. Find its $0.5$-quantile and its $0.9$-quantile. (Answer: $10,\ 15$)
A nonnegative integer loss is discrete uniform on unknown consecutive values. Its mean is $8$ and $P(X\leq10)=2/3$. Find the support, variance, and $0.8$-quantile. (Answer: $1,2,\ldots,15;\ 56/3;\ 12$)

What You'll Learn