Uniform distribution

Welcome to the Discrete Uniform Distribution!

Hello future Further Mathematician! In the FS1 statistics module of OxfordAQA Further Mathematics, your study of discrete probability models begins with the Discrete Uniform Distribution. It is one of the cleanest and most intuitive distributions you will encounter!

We call it "uniform" because every possible discrete outcome is equally likely. In section FS1.2, you will learn the conditions for using this model, how to compute probabilities, and how to carry out the formal algebraic derivations for its mean and variance using standard summation identities.

Key Takeaway from the Introduction

The Discrete Uniform Distribution models a scenario where a finite number of discrete outcomes all have the exact same probability.

1. Understanding the Discrete Uniform Distribution (FS1.2: Conditions for application)

A random variable \(X\) follows a discrete uniform distribution when it takes a finite set of discrete values with equal probability.

What are the conditions for using it?

A discrete random variable \(X\) follows a discrete uniform distribution over the integers \(1, 2, \dots, n\) if:

1. \(X\) is a discrete random variable with a finite number of possible values \(n\).
2. Each outcome is mutually exclusive and has the same probability of occurring, namely \(P(X = x) = \frac{1}{n}\).

Standard Model: In its standard formulation, \(X\) takes values \(x \in \{1, 2, 3, \dots, n\}\) such that:

\(P(X = x) = \frac{1}{n}\quad \text{for } x = 1, 2, \dots, n\)

Classic Example: An Ordinary Fair Die

When rolling a fair 6-sided die, the score \(X\) can be \(1, 2, 3, 4, 5,\) or \(6\). Each outcome has a probability of \(\frac{1}{6}\). Here \(n = 6\), so \(P(X = x) = \frac{1}{6}\) for \(x = 1, 2, \dots, 6\).

Key Takeaway from Section 1

For \(n\) equally likely consecutive integers from \(1\) to \(n\), the probability of each single outcome is simply \(\frac{1}{n}\).

2. Calculating Probabilities (FS1.2: Calculation of probabilities)

Because the outcomes are discrete and mutually exclusive, finding the probability that \(X\) falls in any subset of values is found by summing the individual probabilities.

General Probability Rule

For any set of favorable outcomes \(A \subseteq \{1, 2, \dots, n\}\):

\(P(X \in A) = \sum_{x \in A} P(X = x) = \frac{\text{number of favorable outcomes}}{n}\)

For example, if \(X\) is uniformly distributed on \(\{1, 2, \dots, n\}\), the probability of obtaining a value between \(a\) and \(b\) inclusive (where \(1 \le a \le b \le n\)) is:

\(P(a \le X \le b) = \frac{b - a + 1}{n}\)

Step-by-Step Example

A fair 10-sided spinner has sections numbered \(1\) to \(10\). Let \(X\) be the number spun.

1. State the distribution: \(X\) is discrete uniform on \(\{1, 2, \dots, 10\}\), so \(n = 10\) and \(P(X = x) = \frac{1}{10}\).
2. Find \(P(X \ge 7)\): The outcomes are \(7, 8, 9, 10\) (4 outcomes).
   \(P(X \ge 7) = \frac{4}{10} = \frac{2}{5}\)
3. Find \(P(3 \le X \le 8)\): The number of outcomes is \(8 - 3 + 1 = 6\).
   \(P(3 \le X \le 8) = \frac{6}{10} = \frac{3}{5}\)

Key Takeaway from Section 2

For discrete uniform variables, probabilities are calculated directly by dividing the count of favorable integers by the total number of outcomes \(n\).

3. Mean and Variance (FS1.2: Mean and variance - Knowledge and derivations expected)

The OxfordAQA specification states that knowledge and derivations will be expected for the mean and variance of the discrete uniform distribution. You must be confident proving these results using standard summation formulas.

Required Standard Summation Formulas

• Sum of first \(n\) natural numbers: \(\sum_{r=1}^n r = \frac{n(n+1)}{2}\)
• Sum of squares of first \(n\) natural numbers: \(\sum_{r=1}^n r^2 = \frac{n(n+1)(2n+1)}{6}\)

3.1 Derivation of the Mean \(E(X)\)

By definition of the expected value of a discrete random variable:

\(E(X) = \sum_{x=1}^n x P(X=x) = \sum_{x=1}^n x \left(\frac{1}{n}\right) = \frac{1}{n} \sum_{x=1}^n x\)

Substitute the summation identity \(\sum_{x=1}^n x = \frac{n(n+1)}{2}\):

\(E(X) = \frac{1}{n} \left( \frac{n(n+1)}{2} \right) = \frac{n+1}{2}\)

Required Result: \(E(X) = \frac{n+1}{2}\)

3.2 Derivation of the Variance \(Var(X)\)

To find \(Var(X)\), we use the identity \(Var(X) = E(X^2) - [E(X)]^2\).

First, derive \(E(X^2)\):

\(E(X^2) = \sum_{x=1}^n x^2 P(X=x) = \frac{1}{n} \sum_{x=1}^n x^2\)

Substitute the identity \(\sum_{x=1}^n x^2 = \frac{n(n+1)(2n+1)}{6}\):

\(E(X^2) = \frac{1}{n} \left( \frac{n(n+1)(2n+1)}{6} \right) = \frac{(n+1)(2n+1)}{6}\)

Now substitute \(E(X^2)\) and \(E(X)\) into the variance formula:

\(Var(X) = E(X^2) - [E(X)]^2 = \frac{(n+1)(2n+1)}{6} - \left(\frac{n+1}{2}\right)^2\)

\(Var(X) = \frac{(n+1)(2n+1)}{6} - \frac{(n+1)^2}{4}\)

Factor out the common term \(\frac{n+1}{12}\):

\(Var(X) = \frac{n+1}{12} \left[ 2(2n+1) - 3(n+1) \right]\)

Expand and simplify the expression inside the square brackets:

\(2(2n+1) - 3(n+1) = 4n + 2 - 3n - 3 = n - 1\)

Multiply the factored terms:

\(Var(X) = \frac{(n+1)(n-1)}{12} = \frac{n^2 - 1}{12}\)

Required Result: \(Var(X) = \frac{n^2 - 1}{12}\)

Quick Review: Discrete Uniform Distribution Formulas

4. Worked Example and Standard Deviation

Let's apply these results to a standard exam-style problem.

Example

A fair 8-sided die has faces numbered \(1, 2, 3, 4, 5, 6, 7, 8\). The score after one roll is denoted by \(X\).

1. Calculate the expectation \(E(X)\):
   Here \(n = 8\).
   \(E(X) = \frac{n+1}{2} = \frac{8+1}{2} = \frac{9}{2} = 4.5\)
2. Calculate the variance \(Var(X)\):
   \(Var(X) = \frac{n^2-1}{12} = \frac{8^2 - 1}{12} = \frac{64 - 1}{12} = \frac{63}{12} = \frac{21}{4} = 5.25\)
3. Find the standard deviation:
   \(\sigma = \sqrt{Var(X)} = \sqrt{5.25} \approx 2.29\) (3 s.f.)

You now have a complete understanding of FS1.2 for the Discrete Uniform Distribution, including the conditions for application, probability calculations, and the essential algebraic derivations for its mean and variance.