Week 4.1: Continuous Random Variables & PDFs

Prerequisites: BSMA1002 (Stats 1) — discrete PMF, CDF Cross-links: BSMA1001 (Maths 1) — integration Core question: How do we handle random variables with uncountably infinite ranges?

1. Intuition: From Sums to Integrals

Key analogy: PMF is like mass at discrete points; PDF is like density (mass per unit length).

2. Probability Density Function (PDF)

Definition (PDF) $X$ is continuous if there exists a function $f_X(x) \geq 0$ such that:

1. $f_X(x) \geq 0$ for all $x$
2. $\int_{-\infty}^{\infty} f_X(x) \, dx = 1$
3. For any interval $(a,b)$: $\mathbb{P}(a < X < b) = \int_a^b f_X(x) \, dx$ Unlike PMF: $f_X(x)$ is not a probability. It is a density. Only integrals over intervals give probabilities.

3. Cumulative Distribution Function (CDF)

Definition (CDF) For any RV $X$ (discrete or continuous):

• $F_X(x) = \int_{-\infty}^x f_X(t) \, dt$
• $f_X(x) = F_X'(x)$ (where differentiable) CDF properties:

1. $F$ is non-decreasing: $x_1 < x_2 \implies F(x_1) \leq F(x_2)$
2. $\lim_{x \to -\infty} F(x) = 0$, $\lim_{x \to \infty} F(x) = 1$
3. $F$ is right-continuous

4. Working with PDFs: Examples

Example 1: Verify PDF

Example 2: Find CDF from PDF

Example 3: Find the constant $c$

5. Common Pitfalls

6. Mermaid: Discrete vs Continuous

7. Practice Questions

Q1 (Easy)

Full Solution

$\int_{-1}^1 \frac{3}{4}(1-x^2) dx = \frac{3}{4}[x - x^3/3]_{-1}^1 = \frac{3}{4}[(1-1/3) - (-1+1/3)] = \frac{3}{4}[2/3 + 2/3] = \frac{3}{4} \cdot \frac{4}{3} = 1$. ✓

Q2 (Medium)

Full Solution

For $x < 0$: $F(x) = \int_{-\infty}^x \frac{1}{2}e^{t} dt = \frac{1}{2}[e^{t}]_{-\infty}^x = \frac{1}{2}e^x$.

For $x \geq 0$: $F(x) = \int_{-\infty}^0 \frac{1}{2}e^{t} dt + \int_0^x \frac{1}{2}e^{-t} dt = \frac{1}{2} + \frac{1}{2}(1-e^{-x}) = 1 - \frac{1}{2}e^{-x}$.

So

Next topic: Week 4.2: Common Continuous Distributions — Uniform, Exponential, Normal distributions. Join Discord PreviousWeek 3.4: Markov & Chebyshev InequalitiesNextWeek 4.2: Common Continuous Distributions