Uniform Models For Bounded Measurements

When a Uniform Model Is Useful

A measurement is sometimes modeled as equally spread across a known finite range. For example, a device might report a reading anywhere between two limits, or an event might occur at a location chosen at random along a fixed-length segment. A continuous uniform model represents this assumption with a constant density over the permitted range and zero density elsewhere.

The endpoints matter: they determine both the height of the density and the probabilities of intervals. The model is appropriate only when treating all equal-length subintervals within the range as equally likely is a reasonable assumption for the application. It is a probability model, not a claim that every real measurement in a bounded range is automatically uniform.

Definition: Let $a$ and $b$ be finite constants with $a<b$. A continuous random variable $X$ has a uniform distribution on $[a,b]$, written $X\sim\operatorname{Uniform}(a,b)$, if its density is $$ f(x)= \begin{cases} \dfrac{1}{b-a}, & a\leq x\leq b,\\ 0, & \text{otherwise}. \end{cases} $$ The interval $[a,b]$ is the support, and $b-a$ is its width.

The density is constant across the support, and its height is the reciprocal of the support width. Thus a wider support gives a lower density. If $X$ is measured in minutes, for instance, the density has units of inverse minutes. The endpoints $a$ and $b$ must be expressed in the same units as $X$.

Checking the Density and Reading the Support

A density must be nonnegative and have total integral one. Here $b-a>0$, so the density is nonnegative. Its integral is

\int_{-\infty}^{\infty} f(x)\,dx =\int_a^b \frac{1}{b-a}\,dx =\frac{b-a}{b-a} =1.

The density is zero outside $[a,b]$, so the model assigns probability zero to measurements below $a$ or above $b$. In applications, the support describes the range allowed by the model; it need not describe every value that a real-world instrument could produce if the model assumptions fail.

Worked Example: A Waiting Time on a Fixed Range

Suppose a waiting time $T$, in minutes, is modeled as $T\sim\operatorname{Uniform}(0,12)$. Find its density and the probability that the wait is between 3 and 8 minutes.

The support endpoints are $a=0$ and $b=12$, so its width is $12-0=12$ minutes. The density is $$ f(t)= \begin{cases} \dfrac{1}{12}, & 0\leq t\leq 12,\\ 0, & \text{otherwise}. \end{cases} $$

The interval from 3 to 8 has length $8-3=5$ minutes. Since it lies within the support, $$ P(3\leq T\leq 8)=\int_3^8 \frac{1}{12}\,dt =\frac{8-3}{12} =\frac{5}{12}\approx 0.417. $$

As further checks, the probability of a wait no longer than 4 minutes is $4/12=1/3$, and the probability of a wait longer than 10 minutes is $(12-10)/12=1/6$. Each calculation uses an interval length divided by the full support width.

Event Probabilities From Interval Lengths

Within the support, the density is constant. Integrating it over an interval therefore multiplies that constant height by the interval’s length. This gives a direct rule for common events.

Theorem: If $X\sim\operatorname{Uniform}(a,b)$ and $a\leq r\leq s\leq b$, then $$ P(r\leq X\leq s)=\frac{s-r}{b-a}. $$ More generally, for any interval $I$, $P(X\in I)$ is the length of the intersection $I\cap[a,b]$, divided by $b-a$.

Proof. For an interval $[r,s]$ inside the support, the probability is the integral of the density over that interval:

P(r\leq X\leq s) =\int_r^s f(x)\,dx =\int_r^s \frac{1}{b-a}\,dx =\frac{s-r}{b-a}.

If an interval extends beyond the support, the density is zero outside $[a,b]$. Its probability is therefore the integral over only the part inside the support, whose length is the length of the intersection. This proves the general statement as well. The same probabilities result if either or both interval endpoints are excluded: a continuous uniform model assigns probability zero to a single endpoint. $\square$

Worked Example: An Interval That Crosses the Support Boundary

Let $X\sim\operatorname{Uniform}(2,10)$. Calculate $P(7\leq X\leq 12)$.

The support is $[2,10]$, so values above 10 are outside the model’s support. The intersection of the event interval $[7,12]$ with the support is $[7,10]$, which has length $10-7=3$. The full support has width $10-2=8$. Therefore, $$ P(7\leq X\leq 12) =\frac{3}{8} =0.375. $$

Using the entire event-interval length $12-7=5$ would be incorrect: the portion from 10 to 12 lies outside the support and contributes no probability.

The Cumulative Distribution Function

The cumulative distribution function (CDF) is $F(x)=P(X\leq x)$. For the uniform model, it increases linearly across the support and is constant outside it.

Result: If $X\sim\operatorname{Uniform}(a,b)$, its CDF is $$ F(x)= \begin{cases} 0, & x<a,\\ \dfrac{x-a}{b-a}, & a\leq x\leq b,\\ 1, & x>b. \end{cases} $$

Derivation. If $x<a$, the event $X\leq x$ is outside the support and has probability zero. If $a\leq x\leq b$, integrate the density from the lower support endpoint $a$ to $x$:

F(x)=P(X\leq x)=\int_a^x \frac{1}{b-a}\,dt =\frac{x-a}{b-a}.

If $x>b$, the event includes the entire support, whose probability is one. These cases give the stated CDF. In particular, the formula reaches $0$ at $a$ and $1$ at $b$, as required. $\square$

For $a\leq r\leq s\leq b$, the CDF gives the same interval rule by subtraction: $P(r<X\leq s)=F(s)-F(r)=(s-r)/(b-a)$. The integral argument above also shows why endpoint inclusion does not change the answer.

Worked Example: A Bounded Laboratory Measurement

A concentration $C$, measured in milligrams per litre, is modeled as $C\sim\operatorname{Uniform}(4,10)$. Find its density and calculate $P(5\leq C\leq 8.5)$.

The support width is $10-4=6$ milligrams per litre. Thus $$ f(c)= \begin{cases} \dfrac{1}{6}, & 4\leq c\leq 10,\\ 0, & \text{otherwise}. \end{cases} $$ The density’s units are inverse milligrams per litre. The requested interval has length $8.5-5=3.5$ milligrams per litre, so $$ P(5\leq C\leq 8.5) =\frac{3.5}{6} =\frac{7}{12} \approx 0.583. $$

As a CDF check, $F(8.5)=(8.5-4)/6=4.5/6$, while $F(5)=(5-4)/6=1/6$. Their difference is $4.5/6-1/6=3.5/6=7/12$, agreeing with the interval-length calculation.

Changing the Measurement Scale

Uniformity is preserved when a measurement is shifted and multiplied by a positive constant. This is useful when converting units or describing a measurement on a new scale.

Result: If $X\sim\operatorname{Uniform}(a,b)$, $d>0$, and $Y=c+dX$, then $Y\sim\operatorname{Uniform}(c+da,c+db)$.

Proof. Since $d>0$, the event $Y\leq y$ is equivalent to $X\leq (y-c)/d$. For $c+da\leq y\leq c+db$, the transformed value $(y-c)/d$ lies in $[a,b]$. Using the CDF just derived,

P(Y\leq y) =P\left(X\leq\frac{y-c}{d}\right) =\frac{(y-c)/d-a}{b-a} =\frac{y-(c+da)}{d(b-a)}.

The transformed support has width $(c+db)-(c+da)=d(b-a)$, so this is exactly the linear CDF of a uniform distribution on $[c+da,c+db]$. Below this support the CDF is zero, and above it the CDF is one. Hence $Y$ has the stated uniform distribution. $\square$

Worked Example: Rescaling a Uniform Measurement

Suppose $X\sim\operatorname{Uniform}(0,1)$ records a location as a fraction of a segment’s length. The segment is 5 metres long, and the location in metres is $Y=5X$. Find the model for $Y$ and the probability that the location is between 1 and 3 metres.

Here $c=0$, $d=5$, $a=0$, and $b=1$. The transformed support is $[0+5(0),\,0+5(1)]=[0,5]$, so $Y\sim\operatorname{Uniform}(0,5)$. Its density on that support is $1/(5-0)=1/5$ per metre.

The requested interval has length $3-1=2$ metres, while the full support has length 5 metres. Therefore, $$ P(1\leq Y\leq 3)=\frac{3-1}{5-0}=\frac{2}{5}=0.4. $$ Equivalently, $1\leq 5X\leq 3$ means $0.2\leq X\leq 0.6$, an interval of length $0.4$ within $[0,1]$. This gives the same probability.

Common Mistake and Practical Checks

A frequent error is to divide by an endpoint instead of by the support width. For $X\sim\operatorname{Uniform}(4,10)$, the density is not $1/10$: the interval begins at 4, and its width is $10-4=6$, so the density is $1/6$. Likewise, an event probability is based on the length of the event’s overlap with the support, not simply on its upper endpoint.

Before calculating, check three things: the lower endpoint is less than the upper endpoint; the density is the reciprocal of their difference; and the event interval has been restricted to the support. The resulting probability must lie between zero and one. If the event interval covers the entire support, its probability is one; if it does not intersect the support, its probability is zero.

Key takeaway: For $X\sim\operatorname{Uniform}(a,b)$, the density is $1/(b-a)$ on the support, and an interval’s probability is its overlap length with $[a,b]$, divided by $b-a$.

Check Your Understanding

Use the support endpoints to answer each question.

If $X\sim\operatorname{Uniform}(3,15)$, what is its density on the support?
For that same model, calculate $P(6\leq X\leq 9)$.
If $W\sim\operatorname{Uniform}(2,8)$, calculate $P(7\leq W\leq 10)$. Which part of the event interval contributes probability?
Write the CDF of $Z\sim\operatorname{Uniform}(-1,3)$ as a piecewise function.
If $X\sim\operatorname{Uniform}(0,1)$ and $Y=2+3X$, give the support and density of $Y$.