Lesson 34 Uniform Distribution

Motivation

How do we model a random variable that is equally likely to take on any real number between \(a\) and \(b\)?

Theory

Definition 34.1 (Uniform Distribution) A random variable \(X\) is said to follow a \(\text{Uniform}(a, b)\) distribution if its p.d.f. is \[ f(x) = \begin{cases} \frac{1}{b-a} & a \leq x \leq b \\ 0 & \text{otherwise} \end{cases}, \] or equivalently, if its c.d.f. is \[ F(x) = \begin{cases} 0 & x < a \\ \frac{x-a}{b-a} & a \leq x \leq b \\ 1 & x > b \end{cases}. \]

The p.d.f. and c.d.f. are graphed below.

PDF and CDF of the Uniform Distribution

Figure 34.1: PDF and CDF of the Uniform Distribution

Why is the p.d.f. of a \(\text{Uniform}(a, b)\) random variable what it is?

  • Since we want all values between \(a\) and \(b\) to be equally likely, the p.d.f. must be constant between \(a\) and \(b\).
  • This constant is chosen so that the total area under the p.d.f. (i.e., the total probability) is 1. Since the p.d.f. is a rectangle of width \(b-a\), the height must be \(\frac{1}{b-a}\) to make the total area 1.
Example 34.1 You buy fencing for a square enclosure. The length of fencing that you buy is uniformly distributed between 0 and 4 meters. What is the probability that the enclosed area will be larger than \(0.5\) square meters?
Solution. Let \(L\) be a \(\text{Uniform}(a=0, b=4)\) random variable. Note that \(L\) represents the perimeter of the square enclosure, so \(L/4\) is the length of a side and the area is \[ A = \left( \frac{L}{4} \right)^2 = \frac{L^2}{16}. \] We want to calculate \(P(\frac{L^2}{16} > 0.5)\). To do this, we rearrange the expression inside the probability to isolate \(L\), at which point we can calculate the probability by integrating the p.d.f. of \(L\) over the appropriate range. \[\begin{align*} P(\frac{L^2}{16} > 0.5) &= P(L > \sqrt{8}) \\ &= \int_{\sqrt{8}}^{4} \frac{1}{4 - 0}\,dx \\ &= \frac{4 - \sqrt{8}}{4} \\ &\approx .293. \end{align*}\]

Essential Practice

  1. A point is chosen uniformly along the length of a stick, and the stick is broken at that point. What is the probability the left segment is more than twice as long as the right segment?

    (Hint: Assume the length of the stick is 1. Let \(X\) be the point at which the stick is broken, and observe that \(X\) is the length of the left segment.)

  2. You inflate a spherical balloon in a single breath. If the volume of air you exhale in a single breath (in cubic inches) is \(\text{Uniform}(a=36\pi, b=288\pi)\) random variable, what is the probability that the radius of the balloon is less than 5 inches?