Endpoint Probabilities and Uniform-Model Pitfalls

Why Endpoints Cause Errors

Uniform probabilities are often calculated by comparing lengths: the probability of an interval is the portion of the model’s support covered by that interval. The calculation is simple, but several errors recur. A density value can be mistaken for a probability, an endpoint can be treated as if it had positive probability, or a requested interval can be counted beyond the distribution’s support.

These errors matter in actuarial work because a distribution model describes probabilities, not just a formula to substitute into. Before calculating, identify the support, determine which part of it belongs to the event, and distinguish an interval from a single value.

Definition: If $X\sim\operatorname{Uniform}(a,b)$, where $a<b$, then $X$ has support $[a,b]$ and density $f_X(x)=1/(b-a)$ for $a\le x\le b$, and $f_X(x)=0$ outside that interval. For any event interval, probability is its length inside $[a,b]$, divided by $b-a$. A single point has probability zero.

The support $[a,b]$ is the range of values the model allows, apart from events of probability zero at its boundaries. The density has units inverse to those of $X$. For example, if $X$ is measured in days, its density is expressed per day. A density is not itself the probability of observing that exact value.

For an interval with lower endpoint $l$ and upper endpoint $r$, where $l\le r$, the probability can be calculated by first restricting the interval to the support:

P(l\le X\le r) = \frac{\max\{0,\ \min(r,b)-\max(l,a)\}}{b-a}.

The numerator is the length of the overlap between the requested interval and the support. If there is no overlap, that length is zero. Whether either interval endpoint is included does not change the probability for a continuous uniform variable.

Density Is Not Point Probability

The density $f_X(x)$ describes probability per unit of measurement near $x$. Probabilities are obtained by integrating the density across a set of values. For a uniform density, this integration is the same as multiplying the constant density by the interval’s length.

P(c\le X\le d) = \int_c^d \frac{1}{b-a}\,dx = \frac{d-c}{b-a}, \qquad a\le c\le d\le b.

A point has length zero. Integrating the density over a single point therefore gives zero probability, even when the density at that point is positive:

P(X=c)=\int_c^c f_X(x)\,dx=0.

This also applies when $c=a$ or $c=b$, the endpoints of the support. A density may be defined using a different value at an endpoint without changing any probability, because changing a density at one point does not change its integral. The usual uniform-density convention assigns $1/(b-a)$ at both endpoints.

Worked Example: Density, a Short Interval, and a Support Endpoint

A hypothetical device’s calibration time $X$, measured in minutes, is modeled as uniform on $[12,72]$. Find the density at 12, the probability that the calibration time is between 12 and 42 minutes, and the probability that it is exactly 12 minutes. No interest rate or mortality assumption is involved; the model is $X\sim\operatorname{Uniform}(12,72)$.

Method. Find the support length first, use density times interval length for an interval probability, and treat a single value separately. This avoids treating the density at an endpoint as the probability of that endpoint.

Support length and density. The support length is $72-12=60$ minutes. Therefore the density is $f_X(x)=1/60=0.016666\ldots$ per minute for $12\le x\le72$. In particular, $f_X(12)=1/60$ per minute.

Interval probability. The interval $[12,42]$ is fully inside the support and has length $42-12=30$ minutes. Thus $P(12\le X\le42)=30/60=0.5$.

Exact endpoint probability. The event $X=12$ contains only one value, so its interval length is zero. Hence $P(X=12)=0$, not $1/60$. The density value $1/60$ is measured per minute; the point probability is dimensionless.

Check. The interval probability $0.5$ is between zero and one, and the requested interval covers half of the 60-minute support. The point probability is zero as required for a continuous model.

Open and Closed Endpoints

For a continuous variable, including or excluding a finite number of endpoints does not alter an interval probability. For example, under $X\sim\operatorname{Uniform}(a,b)$, the events $c<X<d$, $c\le X<d$, $c<X\le d$, and $c\le X\le d$ differ only at $c$ and $d$. Each of those points has probability zero, so the four events have equal probability, provided the same interval limits are used.

This is a property of the continuous model, not a general rule for every random variable. A discrete random variable may assign positive probability to an individual value, so changing whether an endpoint is included can change its interval probability. Here, the focus is the continuous uniform model.

The cumulative distribution function makes the endpoint result explicit. For $X\sim\operatorname{Uniform}(a,b)$, $F_X(x)=P(X\le x)$ is zero below $a$, increases linearly from zero to one on $[a,b]$, and is one above $b$:

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

This function is continuous, including at $a$ and $b$. For any $c$ in the support, $P(X=c)=F_X(c)-\lim_{x\uparrow c}F_X(x)=0$. The same reasoning holds at either support endpoint.

Worked Example: A Claim Interval That Crosses the Support

For a hypothetical portfolio exercise, a claim amount $X$, in dollars, is modeled as uniform on $[$250,$1,050]$. Find the probability that $200\le X\le650$ or $1000<X\le1200$. Also find the probability that $X=250$. No interest rate or mortality assumption is involved; the specified claim model is $X\sim\operatorname{Uniform}(250,1050)$.

Method. Clip each requested interval to the support, calculate the resulting lengths, and add the probabilities for the disjoint events. This avoids counting values below the minimum or above the maximum claim allowed by the model.

First interval. The event $200\le X\le650$ extends below the support, which begins at 250. Its overlap with $[250,1050]$ is $[250,650]$, of length $650-250=400$ dollars. The total support length is $1050-250=800$ dollars, so this event has probability $400/800=0.5$.

Second interval. The event $1000<X\le1200$ extends above the support, which ends at 1050. Its overlap has length $1050-1000=50$ dollars. Whether 1000 is included does not affect the probability. Thus this event has probability $50/800=0.0625$.

The two intervals are disjoint, so the probability of their union is $0.5+0.0625=0.5625$. Values above $1,050 contribute no probability.

Exact endpoint. The event $X=250$ is a single point, so $P(X=250)=0$, even though $250$ is the lower endpoint of the support and the density there is $1/800$ per dollar.

Check. The first interval covers 400 of the 800 dollars in the support, and the second covers 50 of those dollars. Their combined probability is $450/800=0.5625$, agreeing with the sum above.

A Reliable Support-and-Endpoint Procedure

When a question contains several inequalities or unions of intervals, work from the event rather than guessing which formula to use. The following procedure makes the support restriction and endpoint treatment explicit.

Write the model and support.
State $X\sim\operatorname{Uniform}(a,b)$, with $a<b$, so the allowed range is $[a,b]$ and its length is $b-a$.

Translate the event into intervals and points.
Keep separate any interval pieces and exact-value events. Check whether pieces overlap before adding their probabilities.

Restrict intervals to the support.
Discard portions below $a$ or above $b$. If an interval does not overlap the support, its probability is zero.

Calculate lengths and handle points.
Divide the total length of the relevant, nonoverlapping parts by $b-a$. A finite collection of single points contributes zero probability.

Check the answer.
Confirm the result is between zero and one and that no part outside the support was counted.

Worked Example: A Composite Event with Outside Values and a Point

A hypothetical repair duration $X$, measured in days, is modeled as uniform on $[30,90]$. Find the probability of $X<25$, or $35\le X\le55$, or $X=70$, or $X>90$. State the density at 70 as well. No interest rate or mortality assumption is involved; $X\sim\operatorname{Uniform}(30,90)$.

Method. Separate the event into outside-support pieces, an interval inside the support, and a single point. This avoids giving positive probability either to an impossible value or to an exact point.

The support length is $90-30=60$ days, so the density inside the support is $1/60$ per day. Since 70 lies inside the support, $f_X(70)=1/60=0.016666\ldots$ per day.

The event $X<25$ is entirely below the support $[30,90]$, so its probability is zero. The event $X>90$ is above the support, so its probability is also zero.

The interval $35\le X\le55$ lies inside the support and has length $55-35=20$ days. Its probability is $20/60=1/3$.

The event $X=70$ is a single value, so $P(X=70)=0$, although $f_X(70)=1/60$ per day. The union probability is therefore $0+1/3+0+0=1/3$.

Check. Only the 20-day interval contributes probability. It covers $20/60=1/3$ of the model’s 60-day support, and no portion outside that support has been counted.

Common Mistakes and Why They Matter

A useful final check is to ask whether the calculation describes an interval’s share of the support or mistakenly treats a formula value as an event probability. These errors can produce answers that look plausible if the units and support are ignored.

Using the density as a point probability. The incorrect calculation is $P(X=c)=f_X(c)=1/(b-a)$. The correction is $P(X=c)=0$; a density must be integrated over a set of positive length to produce a positive probability.
Counting beyond the support. For a requested interval extending below $a$ or above $b$, the incorrect calculation divides its full length by $b-a$. The correction is to calculate only the overlap with $[a,b]$.
Changing an endpoint inequality without considering the model. In a continuous uniform model, adding or removing a finite endpoint does not change probability. In a discrete model, a point may have positive probability, so this shortcut would not be safe.
Adding overlapping pieces twice. Before adding probabilities for a union, check that the pieces are disjoint. If they overlap, calculate the length of the union rather than the sum of the individual lengths.
Ignoring units. A density has inverse measurement units, whereas a probability has no units. If a purported point probability is reported “per day” or “per dollar,” a density has likely been mistaken for a probability.

Key takeaway: For a continuous uniform model, an interval’s probability is the length of its overlap with the support divided by the support length. A single value—including a support endpoint—has probability zero, even when the density there is positive.

Check Your Understanding

For each problem, identify the support before calculating. The answers are given at the end of each question for checking.

Let $X\sim\operatorname{Uniform}(10,30)$. Find $f_X(10)$ and $P(X=10)$. (Answer: $f_X(10)=0.05$ per unit; $P(X=10)=0$.)
A hypothetical measurement $X$ is uniform on $[50,150]$. Find $P(70\le X\le150)$ and $P(150<X<190)$. (Answer: $0.8$ and $0$, respectively.)
Let $X\sim\operatorname{Uniform}(100,500)$. Find $P(80\le X<260)$ and $P(X=260)$. (Answer: $0.4$ and $0$, respectively.)
Suppose $X\sim\operatorname{Uniform}(2,8)$. Find $P(X<1\text{ or }X=5\text{ or }7<X\le10)$. (Answer: $1/6$.)
A hypothetical duration $X$ is uniform on $[0,20]$. Find $P(4\le X\le12\text{ and }X\ne8)$. (Answer: $0.4$.)

What You'll Learn