S2 Chapter 5: The Uniform Distribution

The Purest Form of Randomness: A World of Equal Possibilities

We have seen that the exponential distribution describes the patient mathematics of “waiting,” what if we need to model something fundamentally different - complete ignorance?

A Philosophical Question

If you know a number lies between 0 and 10, but have absolutely no other information, what’s the most fair assumption you can make about its value?

The principle of maximum ignorance suggests: treat every equal-length interval as equally likely. This leads us naturally to the uniform distribution - the mathematics of “pure randomness.”

1 The Language of Uniform Distributions

1.1 Definition and Probability Density Function - The Rectangle of Certainty

Definition (Continuous Uniform Distribution): A continuous random variable $X$ follows a uniform distribution on the interval $[a, b]$ if it has constant probability density over that interval. We write:

$X \sim \text{Uniform}(a, b) \quad \text{or} \quad X \sim U(a, b)$

The key insight is that “uniform” means constant density - imagine a perfectly flat rectangle where height never changes.

Deriving the PDF through logical necessity:

Since the distribution is “uniform,” the PDF must be some constant $k$ over $[a, b]$:

$f(x) = \begin{cases} k & \text{if } a \leq x \leq b \\ 0 & \text{otherwise} \end{cases}$

But what is this constant $k$? We use the fundamental requirement: total probability equals 1.

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

Therefore: $k = \frac{1}{b - a}$

The Uniform PDF

For $X \sim \text{Uniform}(a, b)$:

$f(x) = \begin{cases} \frac{1}{b - a} & \text{if } a \leq x \leq b \\ 0 & \text{otherwise} \end{cases}$

Geometric Interpretation: This creates a rectangle with width $(b-a)$ and height $\frac{1}{b-a}$, giving area = 1.

Example 1:

Consider $X \sim \text{Uniform}(1, 5)$.

(a) Write down the PDF and sketch it

$f(x) = \begin{cases} \frac{1}{5-1} = \frac{1}{4} & \text{if } 1 \leq x \leq 5 \\ 0 & \text{otherwise} \end{cases}$

This creates a rectangle from $x = 1$ to $x = 5$ with height $\frac{1}{4}$.

(b) Find $P(2 < X < 3.5)$

Using the “rectangular property”:

$P(2 < X < 3.5) = \frac{\text{interval length}}{\text{total length}} = \frac{3.5 - 2}{5 - 1} = \frac{1.5}{4} = 0.375$

(c) Find $P(X = 3)$

For any continuous distribution: $P(X = 3) = 0$. Individual points have zero probability.

(d) Find $P(X > 4.5)$

$P(X > 4.5) = \frac{5 - 4.5}{5 - 1} = \frac{0.5}{4} = 0.125$

1.2 Cumulative Distribution Function - The Linear Climb

Building the CDF step by step:

For $X \sim \text{Uniform}(a, b)$, we calculate $F(x) = P(X \leq x)$:

Case 1: $x < a$

$F(x) = 0 \text{ (No probability accumulated yet)}$

Case 2: $a \leq x \leq b$

$F(x) = \int_a^x \frac{1}{b-a} dt = \frac{x-a}{b-a} \text{ (Linear accumulation)}$

Case 3: $x > b$

$F(x) = 1 \text{ (All probability accumulated)}$

The Uniform CDF

For $X \sim \text{Uniform}(a, b)$:

$F(x) = \begin{cases} 0 & \text{if } x < a \\ \frac{x - a}{b - a} & \text{if } a \leq x \leq b \\ 1 & \text{if } x > b \end{cases}$

Key Properties:

• $F(\frac{a+b}{2}) = 0.5$ (median always at center)

Example 2: CDF Applications

For $X \sim \text{Uniform}(-2, 6)$:

(a) Write down the CDF

$F(x) = \begin{cases} 0 & \text{if } x < -2 \\ \frac{x - (-2)}{6 - (-2)} = \frac{x + 2}{8} & \text{if } -2 \leq x \leq 6 \\ 1 & \text{if } x > 6 \end{cases}$

(b) Find $P(0 < X \leq 4)$

$P(0 < X \leq 4) = F(4) - F(0) = \frac{4 + 2}{8} - \frac{0 + 2}{8} = \frac{6}{8} - \frac{2}{8} = \frac{1}{2}$

(c) Find the first quartile

$Q_1$ satisfies $F(Q_1) = 0.25$:

$\frac{Q_1 + 2}{8} = 0.25 \Rightarrow Q_1 + 2 = 2 \Rightarrow Q_1 = 0$

Example 3: Airport Waiting Times

The waiting times, in minutes, between flight take-offs at an airport are modelled by the continuous random variable $X$ with probability density function

$f(x) = \begin{cases} \frac{1}{5} & 2 \leq x \leq 7 \\ 0 & \text{otherwise} \end{cases}$

1. Write down the name of this distribution.
2. A randomly selected flight takes off at 9 am. Find the probability that the next flight takes off before 9:05 am.
3. Find the probability that at least 1 of the next 5 flights has a waiting time of more than 6 minutes.
4. Find the cumulative distribution function of $X$, for all $x$.
5. Sketch the cumulative distribution function of $X$ for $2 \leq x \leq 7$.

1.3 Numerical Characteristics - Mean, Variance, and Transformations

From Rectangle to Balance Point: The Natural Center

For continuous random variables, we learned that expected value is the “center of gravity.” For uniform distributions, where does this balance point lie?

For $X \sim \text{Uniform}(a, b)$:

$\begin{aligned} E(X) &= \int_a^b x \cdot \frac{1}{b - a} \, dx = \frac{1}{b - a} \int_a^b x \, dx = \frac{1}{b - a} \cdot \frac{x^2}{2}\Big|_a^b = \frac{b^2 - a^2}{2(b - a)} = \frac{(b - a)(b + a)}{2(b - a)} = \frac{a + b}{2} \end{aligned}$

Beautiful symmetry: The expected value is exactly the midpoint of the interval!

Variance: How Wide is “Wide”?

Using $\text{Var}(X) = E(X^2) - [E(X)]^2$:

Step 1: Calculate $E(X^2)$

$E(X^2) = \int_a^b x^2 \cdot \frac{1}{b - a} \, dx = \frac{1}{b - a} \cdot \frac{x^3}{3}\Big|_a^b = \frac{b^3 - a^3}{3(b - a)} = \frac{a^2 + ab + b^2}{3}$

Step 2: Apply the variance formula

$\text{Var}(X) = \frac{a^2 + ab + b^2}{3} - \left(\frac{a + b}{2}\right)^2 = \frac{(b - a)^2}{12}$

Uniform Distribution: Key Properties

For $X \sim \text{Uniform}(a, b)$:

$E(X) = \frac{a + b}{2} \quad \text{(always the midpoint)}$

$\text{Var}(X) = \frac{(b - a)^2}{12} \quad \text{(depends only on interval width)}$

Key Insights:

• Mean = Median = Midpoint (perfect symmetry)
• Variance depends only on interval width, not location

Example 4: Parameter Relationships

The continuous random variable $X$ is uniformly distributed over the interval $[a, \beta]$ where $\beta > a$.

Given that $E(X) = 8$:

1. Write down an equation involving $a$ and $\beta$.

Given also that $P(X \leq 13) = 0.7$:

2. Find the value of $a$ and the value of $\beta$.
3. Find $\text{Var}(X)$.
4. Find $P(5 \leq X \leq 35)$.

Example 5: Kitchen Weighing Scales

Albert uses scales in his kitchen to weigh some fruit. The random variable $D$ represents, in grams, the weight of the fruit given by the scales minus the true weight of the fruit. The random variable $D$ is uniformly distributed over the interval $[-2.5, 2.5]$.

1. Specify the probability density function of $D$.
2. Find the standard deviation of $D$.
3. Albert weighs a banana on the scales. Write down the probability that the weight given by the scales equals the true weight of the banana.
4. Find the probability that the weight given by the scales is within 1 gram of the banana’s true weight.
5. Albert weighs 10 bananas on the scales, one at a time. Find the probability that the weight given by the scales is within 1 gram of the true weight for at least 6 of the bananas.

Homework Exercises

Exercise 1: WST02/01/June17/7

The continuous random variable $X$ is uniformly distributed over the interval $[a, b]$

1. Find an expression, in terms of $a$ and $b$, for $E(3 - 2X)$ (2)
2. Find $P(X > \frac{1}{3}b + \frac{2}{3}a)$ (2)

Given that $E(X) = 0$

3. find an expression, in terms of $b$ only, for $E(3X^2)$ (3)

Given also that the range of $X$ is $18$

4. find $\text{Var}(X)$ (2)

Exercise 2: WST02/01/June15/3

A piece of spaghetti has length $2c$, where $c$ is a positive constant. It is cut into two pieces at a random point. The continuous random variable $X$ represents the length of the longer piece and is uniformly distributed over the interval $[c, 2c]$.

1. Sketch the graph of the probability density function of $X$ (2)
2. Use integration to prove that $\text{Var}(X) = \frac{c^2}{12}$ (6)
3. Find the probability that the longer piece is more than twice the length of the shorter piece. (3)

Exercise 3: 6684/01/June15/4

The continuous random variable $L$ represents the error, in metres, made when a machine cuts poles to a target length. The distribution of $L$ is a continuous uniform distribution over the interval $[0, 0.5]$

1. Find $P(L < 0.4)$. (1)
2. Write down $E(L)$. (1)
3. Calculate $\text{Var}(L)$. (2)

A random sample of $30$ poles cut by this machine is taken.

4. Find the probability that fewer than $4$ poles have an error of more than $0.4$ metres from the target length. (3)

When a new machine cuts poles to a target length, the error, $X$ metres, is modelled by the cumulative distribution function $F(x)$ where

$F(x) = \begin{cases} 0 & x < 0 \\ 4x - 4x^2 & 0 \leq x \leq 0.5 \\ 1 & \text{otherwise} \end{cases}$

5. Using this model, find $P(X > 0.4)$ (2)

A random sample of $100$ poles cut by this new machine is taken.

6. Using a suitable approximation, find the probability that at least $8$ of these poles have an error of more than $0.4$ metres. (3)

2 Challenge: Buffon’s Needle Experiment

The Ultimate Test of Uniform Distribution Theory

Now we’ll tackle one of the most famous problems in probability theory - one that beautifully demonstrates how uniform distributions can unlock the secrets of $\pi$ itself.

In 1777, French mathematician Georges-Louis Leclerc, Comte de Buffon, posed this challenge: “What’s the probability that a randomly dropped needle will cross a line on lined paper?”

Buffon’s Needle Setup

Materials:

• Paper with parallel lines spaced distance $1$ apart
• Needle of length $1$

Experiment: Drop the needle randomly onto the paper.

Question: What’s $P(\text{needle crosses a line})$?

Amazing Discovery: The answer involves $\pi$!

Step 1: Identify the Random Variables

Your Task

Question: What random variables should we use to model this problem?

Think about:

• What determines whether the needle crosses a line?
• What aspects of the needle’s position are random?
• What distributions would naturally model these random aspects?

Hint 1: Consider the needle’s position relative to the lines.

Let $X$ be the perpendicular distance from the needle’s center to the nearest line.

Question 1a: The lines are spaced 1 unit apart. What is the range of possible values for $X$ (the distance from needle center to nearest line)?

Answer: $X$ can range from ____ to ____

Question 1b: If the needle is dropped “randomly,” what distribution should $X$ follow?

Hint: Since the needle is dropped randomly, it’s equally likely to land anywhere. This means $X$ should be ____ distributed over its range.

Question 1c: Now consider the needle’s orientation. Let $\Theta$ be the acute angle between the needle and the parallel lines. What is the range of possible values for $\Theta$?

Answer: $\Theta$ can range from ____ to ____

Question 1d: What distribution should $\Theta$ follow?

Hint: Due to rotational symmetry, the needle is equally likely to point in any direction. So $\Theta$ should be ____ distributed over its range.

Question 1e: Are $X$ and $\Theta$ independent?

Think about it: Does the position where the needle lands affect its angle? Does the angle affect where it lands? Your answer: ____

Step 2: Determine the Crossing Condition

Geometric Analysis: When does the needle cross a line?

Question 2: Look at the diagram. The needle extends $\frac{1}{2}$ unit from its center in each direction. When does the needle cross the nearest line? Find the function $f(\Theta)$ such that the needle crosses a line if and only if $X \leq f(\Theta)$.

Step 3: Calculate the Probability Using “Averaging”

Since we have two random aspects (position $X$ and angle $\Theta$), we need a clever approach. Let’s think step by step:

The Key Insight: For each specific angle $\theta$, we can easily find the probability that the needle crosses a line. Then we “average” over all possible angles.

Question 3a: Suppose the needle is oriented at a specific angle $\theta$ (where $0 \leq \theta \leq \frac{\pi}{2}$).

• From Step 2, the needle crosses a line when $X \leq$ ____
• Since $X$ is uniformly distributed on $[0, \frac{1}{2}]$, what is $P(\text{needle crosses} \mid \text{angle} = \theta)$?

Let’s work through an example: If $\theta = 30^{\circ} = \frac{\pi}{6}$, then:

• The needle crosses when $X \leq \frac{1}{2}\sin(\frac{\pi}{6}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$
• Since $X$ is uniform on $[0, \frac{1}{2}]$: $P(\text{needle crosses} \mid \theta = 30^{\circ}) = \frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{2}$

General formula: $P(\text{needle crosses} \mid \text{angle} = \theta) =$ ____

Question 3b: Now we need to find the overall probability. Since the angle $\Theta$ is also random and uniform on $[0, \frac{\pi}{2}]$, we need to “average” the conditional probabilities from Question 3a over all possible angles.

Mathematical Averaging: When we have a continuous range of values, “averaging” becomes integration:

$P(\text{needle crosses}) = \frac{1}{\frac{\pi}{2} - 0} \int_0^{\pi/2} P(\text{needle crosses} \mid \text{angle} = \theta) \, d\theta$

Substitute your answer from Question 3a into this formula and evaluate the integral.

Step-by-step guidance:

$\begin{aligned} P(\text{needle crosses}) &= \frac{1}{\frac{\pi}{2}} \int_0^{\pi/2} \_\_\_ \, d\theta \\ &= \frac{2}{\pi} \int_0^{\pi/2} \_\_\_ \, d\theta \\ &= \_\_\_ \end{aligned}$

Step 4: Discover the Amazing Result

Question 4a: Complete the calculation from Step 3. You should get:

$P(\text{needle crosses a line}) = \_\_\_$

Question 4b: Look at your answer! What famous mathematical constant appears? This is the amazing discovery that made Buffon’s needle experiment so famous.

Question 4c: Rearrange your probability formula to express $\pi$ in terms of the crossing probability:

$\pi = \_\_\_$

Question 4d: How could you use this experiment to estimate $\pi$? If you performed the experiment 1000 times and observed 637 crossings, what would be your estimate for $\pi$?

$\text{Your calculation: } \_\_\_.$

Extension Challenge:

1. What happens if the needle length is not equal to the line spacing? Derive the general formula for probability when needle length is $\ell$ and line spacing is $d$.
2. If $\ell > d$ (needle longer than line spacing), how would you modify the analysis?
3. Design a classroom experiment using readily available materials (toothpicks, paper, etc.) to estimate $\pi$.