Continuous Random Variables — PDF, CDF, and Properties

🎯 Learning Objectives

• Distinguish continuous from discrete random variables
• Understand probability density functions (PDF) and why $P(X = a) = 0$
• Compute probabilities as areas under the PDF using integration
• Find and interpret the CDF for continuous RVs
• Compute expected value and variance for continuous RVs
• Find the median and percentiles of a continuous distribution

📋 Prerequisites

• Discrete Random Variables (15-discrete-random-variables) — PMF, expected value, variance
• Basic integration from BSMA1001-maths-1
• Understanding of area under a curve

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📖 Core Content

17.1 Intuition: From Counting to Measuring

Everyday analogy: The chance you arrive at exactly 3:00:00.000... PM is essentially zero. But the chance you arrive between 2:55 and 3:05 PM is meaningful. 🔑 Key Insight: For continuous RVs, probability is represented by area, not height. The density function gives the "concentration" of probability.

17.2 Probability Density Function (PDF)

17.2.1 Definition

1. $f(x) \geq 0$ for all x (density can't be negative)
2. $\int_{-\infty}^{\infty} f(x) \, dx = 1$ (total area under the PDF = 1)
3. $P(a \leq X \leq b) = \int_a^b f(x) \, dx$ (probability = area under curve)

17.2.2 Important: PDF ≠ Probability

17.3 Cumulative Distribution Function (CDF)

17.3.1 Definition

$$F(x) = P(X \leq x) = \int_{-\infty}^x f(t) \, dt$$

1. $F'(x) = f(x)$ (the derivative of the CDF is the PDF)
2. $\lim_{x \to -\infty} F(x) = 0$
3. $\lim_{x \to +\infty} F(x) = 1$
4. $P(a < X < b) = F(b) - F(a)$

17.4 Expected Value and Variance

Expected Value

$$E(X) = \mu = \int_{-\infty}^{\infty} x \cdot f(x) \, dx$$

Expected Value of a Function

$$E(g(X)) = \int_{-\infty}^{\infty} g(x) \cdot f(x) \, dx$$

Variance

$$Var(X) = \sigma^2 = \int_{-\infty}^{\infty} (x - \mu)^2 f(x) \, dx = E(X^2) - [E(X)]^2$$

Standard Deviation

$$\sigma_X = \sqrt{Var(X)}$$

17.5 Worked Examples

Example 1: Validating a PDF (Easy)

$$\int_0^2 kx \, dx = k \left[\frac{x^2}{2}\right]_0^2 = k(2) = 1$$

Example 2: Computing a Probability (Medium)

$$f(x) = \begin{cases} \frac{3}{4}(1 - x^2), & -1 \leq x \leq 1 \\ 0, & \text{otherwise} \end{cases}$$ $$P(0 \leq X \leq 0.5) = \int_0^{0.5} \frac{3}{4}(1 - x^2) \, dx$$ $$= \frac{3}{4} \left[x - \frac{x^3}{3}\right]_0^{0.5} = \frac{3}{4} \left[(0.5) - \frac{(0.5)^3}{3} - 0\right]$$ $$= \frac{3}{4} \left[0.5 - \frac{0.125}{3}\right] = \frac{3}{4} [0.5 - 0.0417] = \frac{3}{4} \times 0.4583 \approx 0.344$$

Example 3: Expected Value and Variance (Harder)

$$E(X) = \int_{-1}^1 x \cdot \frac{3}{2}x^2 \, dx = \frac{3}{2} \int_{-1}^1 x^3 \, dx = \frac{3}{2} \left[\frac{x^4}{4}\right]_{-1}^1 = \frac{3}{2} \left(\frac{1}{4} - \frac{1}{4}\right) = 0$$ $$E(X^2) = \int_{-1}^1 x^2 \cdot \frac{3}{2}x^2 \, dx = \frac{3}{2} \int_{-1}^1 x^4 \, dx = \frac{3}{2} \left[\frac{x^5}{5}\right]_{-1}^1 = \frac{3}{2} \left(\frac{1}{5} + \frac{1}{5}\right) = \frac{3}{2} \times \frac{2}{5} = \frac{3}{5}$$ $$Var(X) = E(X^2) - [E(X)]^2 = \frac{3}{5} - 0 = 0.6$$

17.6 Percentiles and Median

$$F(x) = \int_0^x \frac{t}{2} \, dt = \left[\frac{t^2}{4}\right]_0^x = \frac{x^2}{4}$$

17.7 Discrete vs. Continuous — Summary Table

17.8 Why This Matters

• Physical measurements (height, weight, time)
• Financial variables (stock returns, interest rates)
• Natural phenomena (temperature, rainfall)
• They lead directly to Weeks 12-13 distributions (Uniform, Exponential, Normal)

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📐 Key Formulas / Concepts

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⚠️ Common Pitfalls

Pitfall 1: Thinking $f(x)$ Is a Probability

Pitfall 2: Forgetting to Check PDF Validity

Pitfall 3: Using Summation Instead of Integration

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📝 Practice Questions

Q1: PDF Validation

</strong>

Is $f(x) = 2x$ for $0 \leq x \leq 1$ a valid PDF?

<details> <strong>Solution</strong>

$f(x) \geq 0$ ✓ for $x \in [0,1]$

$\int_0^1 2x \, dx = [x^2]_0^1 = 1 - 0 = 1$ ✓

Yes, it's a valid PDF.

</details> > **Q2: Finding a Constant** > > </strong> > > $f(x) = k$ for $0 \leq x \leq 5$. Find k. > > <details> <strong>Solution</strong> > > $\int_0^5 k \, dx = k \times 5 = 1$, so $k = 1/5$. > > This is the **Uniform(0,5)** distribution. </details> > **Q3: Computing Probability** > > </strong> > > $f(x) = 3x^2$ for $0 \leq x \leq 1$. Find $P(0.2 < X < 0.5)$. > > <details> <strong>Solution</strong> > > $P(0.2 < X < 0.5) = \int_{0.2}^{0.5} 3x^2 \, dx = [x^3]_{0.2}^{0.5} = 0.125 - 0.008 = 0.117$ > > $\boxed{0.117}$ </details> > **Q4: Expected Value** > > </strong> > > $f(x) = 2(1-x)$ for $0 \leq x \leq 1$. Find $E(X)$. > > <details> <strong>Solution</strong> > > $E(X) = \int_0^1 x \cdot 2(1-x) \, dx = 2 \int_0^1 (x - x^2) \, dx = 2 \left[\frac{x^2}{2} - \frac{x^3}{3}\right]_0^1$ > > $= 2 \left(\frac{1}{2} - \frac{1}{3}\right) = 2 \times \frac{1}{6} = \frac{1}{3}$ > > $\boxed{E(X) = 1/3}$ </details> > **Q5: CDF from PDF** > > </strong> > > $f(x) = \frac{1}{2}x$ for $0 \leq x \leq 2$. Find $F(x)$ and $P(X > 1)$. > > <details> <strong>Solution</strong> > > **CDF:** $F(x) = \int_0^x \frac{1}{2}t \, dt = \left[\frac{t^2}{4}\right]_0^x = \frac{x^2}{4}$ for $0 \leq x \leq 2$ > > $F(x) = 0$ for $x < 0$, $F(x) = 1$ for $x > 2$ > > **$P(X > 1)$:** $= 1 - P(X \leq 1) = 1 - F(1) = 1 - \frac{1}{4} = \frac{3}{4}$ > > Or directly: $\int_1^2 \frac{1}{2}x \, dx = [\frac{x^2}{4}]_1^2 = \frac{4}{4} - \frac{1}{4} = \frac{3}{4}$ ✓ > > $\boxed{F(x) = x^2/4,\ P(X > 1) = 3/4}$ </details> > **Q6: Variance** > > </strong> > > $f(x) = \frac{3}{2}\sqrt{x}$ for $0 \leq x \leq 1$. Find $Var(X)$. > > <details> <strong>Solution</strong> > > $E(X) = \int_0^1 x \cdot \frac{3}{2}\sqrt{x} \, dx = \frac{3}{2} \int_0^1 x^{3/2} \, dx = \frac{3}{2} \left[\frac{2}{5}x^{5/2}\right]_0^1 = \frac{3}{2} \times \frac{2}{5} = \frac{3}{5}$ > > $E(X^2) = \int_0^1 x^2 \cdot \frac{3}{2}\sqrt{x} \, dx = \frac{3}{2} \int_0^1 x^{5/2} \, dx = \frac{3}{2} \left[\frac{2}{7}x^{7/2}\right]_0^1 = \frac{3}{2} \times \frac{2}{7} = \frac{3}{7}$ > > $Var(X) = \frac{3}{7} - (\frac{3}{5})^2 = \frac{3}{7} - \frac{9}{25} = \frac{75}{175} - \frac{63}{175} = \frac{12}{175}$ > > $\boxed{Var(X) = 12/175}$ </details> > **Q7: Median** > > </strong> > > $f(x) = \frac{1}{2}$ for $2 \leq x \leq 4$. Find the median. > > <details> <strong>Solution</strong> > > CDF: $F(x) = \int_2^x \frac{1}{2} \, dt = \frac{x-2}{2}$ for $2 \leq x \leq 4$ > > Set $F(x) = 0.5$: $\frac{x-2}{2} = 0.5$ > > $x - 2 = 1$, $x = 3$ > > The median is 3. (For a symmetric uniform distribution, median = mean.) > > $\boxed{\text{Median} = 3}$ </details> > **Q8: Application** > > </strong> > > The lifetime of a battery (in hours) has PDF $f(x) = \frac{1}{100}e^{-x/100}$ for $x \geq 0$. > > Find the probability a battery lasts more than 100 hours. > > <details> <strong>Solution</strong> > > $P(X > 100) = \int_{100}^\infty \frac{1}{100} e^{-x/100} \, dx = \left[-e^{-x/100}\right]_{100}^\infty$ > > $= 0 - (-e^{-1}) = e^{-1} \approx 0.368$ > > About 36.8% of batteries last more than 100 hours. This is an **Exponential** distribution. > > $\boxed{P(X > 100) = e^{-1} \approx 0.368}$ </details> > **Q9: Expected Value of Exponential** > > </strong> > > Using the PDF from Q8, find $E(X)$. > > <details> <strong>Solution</strong> > > $E(X) = \int_0^\infty x \cdot \frac{1}{100} e^{-x/100} \, dx$ > > Using integration by parts: $\int x e^{-x/100} dx = -100x e^{-x/100} - 10000 e^{-x/100}$ > > $E(X) = \frac{1}{100} \left[-100x e^{-x/100} - 10000 e^{-x/100}\right]_0^\infty$ > > As $x \to \infty$, $e^{-x/100} \to 0$, so the upper limit = 0. > > At $x = 0$: $0 - 10000(1) = -10000$ > > $E(X) = \frac{1}{100}[0 - (-10000)] = \frac{10000}{100} = 100$ > > The expected lifetime is 100 hours. This matches the parameter of the exponential distribution. > > $\boxed{E(X) = 100}$ </details> > **Q10: PDF from CDF** > > </strong> > > Given > > $$> F(x) = \begin{cases} 0, & x < 0 \\ x^3, & 0 \leq x \leq 1 \\ 1, & x > 1 \end{cases} >$$

Find the PDF $f(x)$ and $P(0.2 \leq X \leq 0.5)$.

<details> <strong>Solution</strong>

PDF:

$$> f(x) = F'(x) = \begin{cases} 3x^2, & 0 \leq x \leq 1 \\ 0, & \text{otherwise} \end{cases} >$$

$P(0.2 \leq X \leq 0.5)$: $= F(0.5) - F(0.2) = (0.5)^3 - (0.2)^3 = 0.125 - 0.008 = 0.117$

$\boxed{f(x) = 3x^2,\ P = 0.117}$

</details> * * * ## 🔗 Cross-References - **Next topic:** [Expectation](/courses/bsma1002/notes/17-expectation) — deeper dive into expected value properties - **Previous:** [Discrete Random Variables](/notes/01-foundation-bsma1002-stats-1-week09-15-discrete-random-variables) — the discrete analogue - **Week 12 (Distributions):** Uniform, Exponential, Normal — specific continuous distributions - **BSMA1001-maths-1:** Integration techniques (substitution, by parts) - **BSMA1004 (Stats 2):** Continuous distributions in statistical inference [Join Discord](https://discord.gg/gE2m4Qrdqv) [Previous**Discrete Random Variables**](/notes/01-foundation-bsma1002-stats-1-week09-15-discrete-random-variables)[Next**Expectation & Variance**](/notes/01-foundation-bsma1002-stats-1-week10-17-expectation-variance)