Uniform Distribution

Context: FIT1058_MOC, FIT2086_MOC · equal probability to every outcome in a bounded range · models fairness or pure ignorance · the simplest named parametric family — discrete on $\{a,\dots ,b\}$, continuous on $[a,b]$

Quick Revision

• 🎯 Objective: every outcome in the range equally likely ➔ discrete $\mathrm{Pr}(X=x)=\frac{1}{b-a+1}$; continuous $p(x)=\frac{1}{b-a}$ on $[a,b]$.
• 📦 Core Components: both give $E=\frac{a+b}{2}$ ➔ variance differs: discrete $\frac{(b-a+1{)}^2-1}{12}$ vs continuous $\frac{(b-a{)}^2}{12}$.
• ⚡ Key Constraint: models genuine fairness or pure ignorance (maximally uninformative); the discrete and continuous variance formulas are not interchangeable.

📝 Core

1. The Discrete Distribution

• Definition ➔ $X\sim \mathrm{Unif}(a,b)$: $\mathrm{Pr}(X=x)=\frac{1}{b-a+1}$ for $a\le x\le b$, else 0.
• Count ➔ $b-a+1$ integers in $[a,b]$.
• Generalised support ➔ for any finite $\mathcal{X}\subseteq \mathbb{Z}$, $P(X=k\mid \mathcal{X})=\frac{1}{|\mathcal{X}|}$ — the set need not be contiguous.

2. Moments

• Mean ➔ $E(X)=\frac{a+b}{2}$ (midpoint, by symmetry) — need not be an integer.
• Variance ➔ $\mathrm{Var}(X)=\frac{(b-a+1{)}^2-1}{12}$.
• Check ➔ $\mathrm{Unif}(1,6)$ = fair die: $E=3.5$, $\mathrm{Var}=\frac{35}{12}$.
• Alternative parameterisation ➔ on $\{1,2,\dots ,k\}$ written $X\sim U(k)$: $P(X=x)=\frac{1}{k}$, $E=\frac{k+1}{2}$, $\mathrm{Var}=\frac{k^2-1}{12}$ (the $a=1,b=k$ case).

3. Two Motivations

• Fairness ➔ die, coin (genuine symmetry).
• Ignorance ➔ only the range known ⟹ assume nothing more.

4. Continuous Uniform (FIT2086)

• Density ➔ $X\sim U(a,b)$ with $p(x\mid a,b)=\{\begin{array}{ll}0& x<a\\ \frac{1}{b-a}& a\le x\le b\\ 0& x>b\end{array}$
• Parameters read geometrically ➔ $a$ sets the start, $w=b-a$ the width; height $=1/w$, so a narrow interval has density $>1$ (a density is not a probability — Random Variables and Probability Distributions (FIT2086)).
• Mean ➔ $E[X]=\frac{a+b}{2}=a+\frac{w}{2}$ — the midpoint, as in the discrete case.
• Variance ➔ $V[X]=\frac{(b-a{)}^2}{12}=\frac{w^2}{12}$ ➔ no $+1$, unlike the discrete formula.
• Probabilities are length ratios ➔ $P(X<c)=\frac{c-a}{b-a}$ for $a\le c\le b$: integrating a constant density just measures a sub-interval.
• cdf derivation ➔ $P(X<x)={\int }_a^x\frac{1}{b-a}d{x}^{\prime }=\frac{1}{b-a}{\left[x^{\prime }\right]}_a^x=\frac{x-a}{b-a}$ ➔ linear in $x$, from $0$ at $a$ to $1$ at $b$.
• “Fraction of the way along” shortcut ➔ any bound expressed as a weighted midpoint reads off instantly: $\frac{a+b}{2}$ is halfway ⟹ $P=\frac{1}{2}$; $\frac{a+3b}{4}$ is three-quarters along ⟹ $P(X\in [\frac{a+3b}{4},b])=\frac{1}{4}$. No integration required.
• Outside the support the probability is exactly $0$ ➔ for $X\sim U(a,b)$ with $a,b>0$: $P(X>2b)=0$ (since $2b>b$) and $P(X\in [0,a])=0$ (the interval ends where the support starts) ➔ check the support before computing anything.

Key identities:

$\text{discrete: }\mathrm{Pr}(X=x)=\frac{1}{b-a+1},E=\frac{a+b}{2},\mathrm{Var}=\frac{(b-a+1{)}^2-1}{12}$ $\text{continuous: }p(x)=\frac{1}{b-a}(a\le x\le b),E=\frac{a+b}{2},V=\frac{(b-a{)}^2}{12}$

⚖️ Core Decision Matrix

Case	$E$	$\mathrm{Var}$
discrete $\mathrm{Unif}(0,1)$	0.5	$\frac{3}{12}=\frac{1}{4}$
discrete $\mathrm{Unif}(1,6)$	3.5	$\frac{35}{12}$
discrete general	$\frac{a+b}{2}$	$\frac{(b-a+1{)}^2-1}{12}$
continuous $U(a,b)$	$\frac{a+b}{2}$	$\frac{(b-a{)}^2}{12}$

When It Flips: uniform over a sample space = Equally Likely Outcomes; $\mathrm{Unif}(0,1)$ is a fair Bernoulli Trial. It is the natural prior when the distribution's shape is entirely unknown. Reach for the continuous form when the variable is measured rather than counted (weight, waiting time, position).

📊 Exam Execution Trace & Applied Exercises

Manual Execution Trace

$\mathrm{Unif}(1,6)$:

Step / State	Quantity	Value
0 (Init)	$b-a+1$	6
1	$\mathrm{Pr}(X=4)$	$\frac{1}{6}$
2	$E$	3.5
3	$\mathrm{Var}$	$\frac{35}{12}$

Applied Exercise 1 — fair die events (discrete)

Problem: $X$ = face shown by a fair six-sided die. (i) $P(X\text{ odd})$; (ii) $P(X<5)$.

$$\begin{array}{rl}P(X\text{ odd})& =P(X=1)+P(X=3)+P(X=5)=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=\frac{3}{6}=\frac{1}{2}\\ P(X<5)& =4\times \frac{1}{6}=\frac{4}{6}=\frac{2}{3}\end{array}$$

Key move: under a uniform pmf every event probability collapses to favourable count $÷|\mathcal{X}|$.

Applied Exercise 2 — car weight (continuous)

Problem: weight of a randomly selected car $\sim U(800,1500)$ kg. Find (i) $E[X]$, (ii) $V[X]$, (iii) $P(X<1000)$.

$$\begin{array}{rl}E[X]& =\frac{a+b}{2}=\frac{800+1500}{2}=1150\\ V[X]& =\frac{(b-a{)}^2}{12}=\frac{700^2}{12}=\frac{490000}{12}=40833.33\\ P(X<1000)& ={\int }_{800}^{1000}\frac{dx}{700}={\left[\frac{x}{700}\right]}_{800}^{1000}=\frac{1000-800}{700}=\frac{200}{700}=0.29\end{array}$$

Final Extracted Output: $1150$ kg, $40833.33$, $P\approx 0.29$. Key move: the integral of a constant density is a length ratio — no antiderivative work needed.

Applied Exercise 3 — deriving $E[X]$ from the definition (Studio 2)

Problem: $X\sim U(0,b)$. Derive $E[X]$ by integration rather than quoting $\frac{a+b}{2}$.

$$\begin{array}{rl}E[X]& ={\int }_0^{b}xp(x\mid 0,b)dx={\int }_0^b\frac{x}{b}dx=\frac{1}{b}{\int }_0^{b}xdx\\ & =\frac{1}{b}{\left[\frac{x^2}{2}\right]}_0^b=\frac{b^2}{2b}=\frac{b}{2}\end{array}$$

Final Extracted Output: $E[X]=\frac{b}{2}$, consistent with $\frac{a+b}{2}$ at $a=0$. Key move: the constant $\frac{1}{b}$ pulls outside the integral, leaving the elementary $\int xdx$ — the same trick works for $E[X^2]$ if the question wants $V[X]$ from scratch.

Applied Exercise 4 — two dice by linearity (Studio 2)

Problem: $X$ = one roll of a fair six-sided die, $Y$ = an independent second roll. Give $E[X]$ and $E[X+Y]$.

$$\begin{array}{rl}E[X]& =\sum _{x}xP(X=x)=\frac{1}{6}(1)+\frac{1}{6}(2)+\frac{1}{6}(3)+\frac{1}{6}(4)+\frac{1}{6}(5)+\frac{1}{6}(6)=3.5\\ E[X+Y]& =E[X]+E[Y]=3.5+3.5=7\end{array}$$

Final Extracted Output: $3.5$ and $7$. Key move: $E[X+Y]=E[X]+E[Y]$ is linearity — it needs no independence at all; independence would only be required for $V[X+Y]=V[X]+V[Y]$ (see Expectations and Covariance (FIT2086)).

⚠️ Common Mistakes

• 💡 Count is $b-a+1$, not $b-a$ ➔ inclusive endpoints; the uniform is the maximally non-committal choice given only the range.
• 💡 Discrete variance formula on a continuous uniform ➔ $\frac{(b-a+1{)}^2-1}{12}$ counts lattice points and is meaningless on an interval, where $\frac{(b-a{)}^2}{12}$ applies.
• 💡 Continuous density can exceed 1 ➔ $U(3,3.5)$ has $p(x)=2$; only the area under it is a probability.
• 💡 Computing a probability for a region outside $[a,b]$ ➔ $P(X>2b)$ and $P(X\in [0,a])$ are zero, not small; students reflexively apply the length ratio and produce a negative or $>1$ answer. Sketch the support first.
• 💡 Assuming any bounded variable is uniform ➔ daily temperature and adult heights are bounded but centre-heavy ➔ normal. Uniform requires genuine symmetry (fair die/coin) or deliberate ignorance. Selection drill ➔ Parametric Probability Distributions.

🧠 Active Recall

Give the pmf, mean, and variance of $\mathrm{Unif}(a,b)$ and verify on a fair die.

• Hint: Inclusive count.

Answer

• Short answer: $\frac{1}{b-a+1}$; $E=\frac{a+b}{2}$; $\mathrm{Var}=\frac{(b-a+1{)}^2-1}{12}$; die → $3.5,\frac{35}{12}$.
• Why: Symmetry ➔ mean at the midpoint.

What two situations call for a uniform distribution?

• Hint: Fairness or ignorance.

Answer

• Short answer: Known symmetry (die/coin), or modelling ignorance (only the range known).
• Why: Non-committal ➔ assumes nothing beyond $[a,b]$.

How do the discrete and continuous uniform differ in density, variance, and how probabilities are computed?

• Hint: Lattice points vs interval length.

Answer

• Short answer: discrete puts mass $\frac{1}{b-a+1}$ on each of $b-a+1$ integers, $\mathrm{Var}=\frac{(b-a+1{)}^2-1}{12}$; continuous spreads density $\frac{1}{b-a}$ across the interval, $V=\frac{(b-a{)}^2}{12}$. Both share $E=\frac{a+b}{2}$.
• Why: Count ratio vs length ratio ➔ discretely $P(X\in A)$ counts favourable integers over $|\mathcal{X}|$; continuously $P(a\le X\le c)=\frac{c-a}{b-a}$ is the sub-interval’s share of the width, and $P(X=x)=0$ exactly.