Week 4.2: Common Continuous Distributions

Prerequisites: Continuous RVs (Week 4.1: Continuous Random Variables & PDFs) Cross-links: BSMA3012 (Linear Stat Models) — error distributions Core question: What are the workhorse continuous distributions in statistics?

1. Uniform Distribution $\text{Uniform}(a,b)$

PDF: $f_X(x) = \frac{1}{b-a}$, $a \leq x \leq b$ CDF: $F_X(x) = \frac{x-a}{b-a}$, $a \leq x \leq b$ Mean: $\frac{a+b}{2}$ Variance: $\frac{(b-a)^2}{12}$ Uses: Random number generation, "uninformed" priors in Bayesian stats.

Example

2. Exponential Distribution $\text{Exp}(\lambda)$

PDF: $f_X(x) = \lambda e^{-\lambda x}$, $x \geq 0$ CDF: $F_X(x) = 1 - e^{-\lambda x}$, $x \geq 0$ Mean: $1/\lambda$ Variance: $1/\lambda^2$ Uses: Waiting times, inter-arrival times, service times. Memoryless property: $\mathbb{P}(X > s + t \mid X > s) = \mathbb{P}(X > t)$.

Example

3. Normal (Gaussian) Distribution $\mathcal{N}(\mu, \sigma^2)$

PDF: $f_X(x) = \frac{1}{\sqrt{2\pi\sigma^2}} \exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)$, $x \in \mathbb{R}$ Mean: $\mu$ Variance: $\sigma^2$ Standard Normal: $Z \sim \mathcal{N}(0,1)$, CDF denoted $\Phi(z)$ 68-95-99.7 Rule:

• $\mathbb{P}(|X-\mu| \leq \sigma) \approx 0.68$
• $\mathbb{P}(|X-\mu| \leq 2\sigma) \approx 0.95$
• $\mathbb{P}(|X-\mu| \leq 3\sigma) \approx 0.997$ Standardisation: $Z = \frac{X - \mu}{\sigma} \sim \mathcal{N}(0,1)$.

Example

4. Gamma Distribution $\text{Gamma}(\alpha, \beta)$

PDF: $f_X(x) = \frac{\beta^\alpha}{\Gamma(\alpha)} x^{\alpha-1} e^{-\beta x}$, $x \geq 0$ Mean: $\alpha/\beta$ Variance: $\alpha/\beta^2$ Special cases:

• $\alpha = 1$: Exponential($\beta$)
• $\alpha = n$ (integer): Sum of $n$ i.i.d. Exponential($\beta$)

5. Summary Table

6. Practice Questions

Q1 (Easy)

Full Solution

$\mathbb{P}(X > 4) = \frac{5-4}{5-2} = \frac{1}{3}$.

Q2 (Medium)

Full Solution

$Z = (65-50)/10 = 1.5$. $\mathbb{P}(Z > 1.5) = 1 - \Phi(1.5) = 1 - 0.9332 = 0.0668$.

Q3 (Hard)

Full Solution

Median $m$ satisfies $F(m) = 0.5$. $1 - e^{-0.5m} = 0.5 \implies e^{-0.5m} = 0.5 \implies -0.5m = \ln(0.5) \implies m = -2\ln(0.5) = 2\ln 2 \approx 1.386$.

Next topic: Week 4.3: Expectations for Continuous Random Variables — Expected values for continuous RVs. Join Discord PreviousWeek 4.1: Continuous Random Variables & PDFsNextWeek 4.3: Expectations for Continuous Random Variables