Continuous Random Variables

Mathematics· 9709 S1 Chapter 6, S2 Chapter 3· 12 min read

1. Core Properties of Probability Density Functions (PDF)★★☆☆☆⏱ 15 min

Unlike discrete random variables, continuous random variables cannot have their probability represented at single points, as the total number of possible outcomes is uncountably infinite. The probability density function f(x) must satisfy two non-negotiable conditions for all values of x in its domain.

f(x) \geq 0 \text{ for all } x, \quad \int_{-\infty}^{\infty} f(x) dx = 1

📘 Definition

Valid Probability Density Function

A function that is non-negative across its entire domain, and the total area under the curve over the full range of the random variable equals exactly 1

📐 Worked Example

Verify that the function f(x) = k(4 - x^2) defined for 0 ≤ x ≤ 2, and 0 otherwise, is a valid PDF, and find the value of constant k

1
First confirm f(x) is non-negative over 0 ≤ x ≤ 2: 4 - x² ranges from 4 to 0, so f(x) ≥ 0 for all x in the domain
2
Set the total definite integral over the domain equal to 1 to solve for k:
$\int_{0}^{2} k(4 - x^2) dx = 1$
3
Evaluate the integral:
$k \left[ 4x - \frac{x^3}{3} \right]_0^2 = k \left( 8 - \frac{8}{3} \right) = \frac{16k}{3} = 1$
4
Rearrange to find k:
$k = \frac{3}{16}$

✓ Quick check

Test your understanding of PDF validity rules:

Which of the following functions meets the basic requirements for a valid PDF over 1 ≤ x ≤ 3?
- f(x) = -0.5x + 2
- f(x) = 0.5x - 1
- f(x) = 2x - 1
- f(x) = x²
Reveal answer
f(x) = -0.5x + 2 —
This function is non-negative across 1 ≤ x ≤3, and the total integral equals 1

2. Cumulative Distribution Functions (CDF)★★★☆☆⏱ 18 min

The cumulative distribution function F(x) gives the probability that the random variable X takes a value less than or equal to x, calculated as the definite integral of the PDF from the lower bound of the domain up to x. For piecewise PDFs, you will derive piecewise CDFs for each interval.

F(x) = P(X \leq x) = \int_{a}^{x} f(t) dt, \quad \text{for domain } a \leq X \leq b

🔬 Derivation

Goal:

Derive CDF from the PDF f(x) = 3/16 (4 - x²) for 0 ≤ x ≤ 2

1
Integrate the PDF from 0 to x:
$F(x) = \int_{0}^{x} \frac{3}{16}(4 - t^2) dt$
2
Evaluate the antiderivative:
$F(x) = \frac{3}{16} \left[ 4t - \frac{t^3}{3} \right]_0^x = \frac{12x - x^3}{16}$
3
Add boundary conditions: F(x) = 0 for x < 0, F(x) = 1 for x > 2

Result:

The full CDF is a piecewise function that increases monotonically from 0 to 1 across the domain

📐 Worked Example

Given F(x) = \frac{x^2}{9} for 0 ≤ x ≤ 3, find P(1 ≤ X ≤ 2)

1
Recall that for continuous variables, P(a ≤ X ≤ b) = F(b) - F(a)
2
Substitute b=2 and a=1:
$P(1 \leq X \leq 2) = F(2) - F(1) = \frac{4}{9} - \frac{1}{9} = \frac{3}{9} = \frac{1}{3}$

3. Expectation and Variance for Continuous Random Variables★★★☆☆⏱ 15 min

The expected value E[X] (mean) of a continuous random variable is calculated by integrating x multiplied by the PDF across the full domain. Variance is calculated as E[X²] - (E[X])², the same identity used for discrete variables.

E[X] = \int_{-\infty}^{\infty} x f(x) dx, \quad E[g(X)] = \int_{-\infty}^{\infty} g(x) f(x) dx

📐 Worked Example

Calculate E[X] and Var(X) for the PDF f(x) = 3/16 (4 - x²) defined over 0 ≤ x ≤ 2

1
Compute E[X] first:
$E[X] = \int_{0}^{2} x \cdot \frac{3}{16}(4 - x^2) dx = \frac{3}{16} \int_{0}^{2} 4x - x^3 dx$
2
Evaluate the integral:
$E[X] = \frac{3}{16} \left[ 2x^2 - \frac{x^4}{4} \right]_0^2 = \frac{3}{16} (8 - 4) = \frac{12}{16} = 0.75$
3
Compute E[X²] for variance:
$E[X^2] = \int_{0}^{2} x^2 \cdot \frac{3}{16}(4 - x^2) dx = \frac{3}{16} \int_{0}^{2} 4x^2 - x^4 dx$
4
Simplify to find variance:
$E[X^2] = \frac{3}{16} \left[ \frac{4x^3}{3} - \frac{x^5}{5} \right]_0^2 = \frac{3}{16} \left( \frac{32}{3} - \frac{32}{5} \right) = 0.8$
5
Apply the variance identity:
$Var(X) = 0.8 - (0.75)^2 = 0.8 - 0.5625 = 0.2375$

4. Median, Mode and Percentiles★★★★☆⏱ 12 min

The median m of a continuous distribution is the value where F(m) = 0.5, the point where half the total area under the PDF lies to the left. The mode is the value of x where f(x) reaches its maximum, found via differentiation for differentiable functions.

📐 Worked Example

Find the median of the distribution with CDF F(x) = (12x - x³)/16 for 0 ≤ x ≤ 2

1
Set F(m) = 0.5:
$\frac{12m - m^3}{16} = 0.5$
2
Rearrange to form a cubic equation:
$m^3 - 12m + 8 = 0$
3
Solve the cubic numerically (as expected in CIE exams) to find m ≈ 0.694, which lies inside the domain 0 ≤ m ≤ 2

5. Common Pitfalls

Wrong move:

Treating P(X = a) as a non-zero value for continuous variables

Why:

Single points have zero width under the PDF curve, so their probability is always 0

Correct move:

Always calculate probability over an interval [a, b] instead of a single point

Wrong move:

Forgetting to add boundary conditions when deriving CDF

Why:

CIE exam markers deduct 1 mark for missing F(x)=0 for x < lower bound and F(x)=1 for x > upper bound

Correct move:

Explicitly state all piecewise intervals for the full CDF in your solution

Wrong move:

Calculating variance as E[X²] without subtracting (E[X])²

Why:

This gives you the second moment, not the variance, leading to an incorrect value

Correct move:

Write the variance identity at the top of your working before you start calculations

Wrong move:

Finding the median by locating the maximum of the PDF

Why:

That gives you the mode, not the median, which is based on cumulative probability

Correct move:

Set the CDF equal to 0.5 and solve for the median value

Wrong move:

Integrating the PDF over the entire real line for piecewise functions

Why:

This wastes time and can lead to arithmetic errors for functions zero outside a narrow interval

Correct move:

Only integrate over the non-zero domain of the PDF

6. Quick Reference Cheatsheet

Measure	Formula for continuous X
PDF Validity	$f (x) \geq 0$ , $\int_{d o m} f (x) d x = 1$
CDF	$F (x) = \int_{a}^{x} f (t) d t$
Expectation $E [X]$	$\int_{d o m} x f (x) d x$
Variance $V a r (X)$	$E [X^{2}] - (E [X])^{2}$
Median $m$	$F (m) = 0.5$
Mode	Value of $x$ that maximises $f (x)$

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · 72
PDF validity and expectation calculation
2022 · 73
Derive CDF from piecewise PDF
2021 · 71
Find median and interquartile range
2020 · 62
Combined probability with continuous variables

What's Next

Mastering continuous random variables is the critical foundation for all subsequent continuous distribution topics in your CIE 9709 exam, including the normal distribution, uniform distribution, and continuous sampling concepts. You will regularly combine these core PDF and CDF skills with integration techniques you learned in Pure Mathematics to solve 8-12 mark extended response questions that appear on almost every Statistics paper. These problems are high yield, as they often allow partial marks even if you make an error solving for the initial constant k, as long as your subsequent working follows correct rules. Move on to practice targeted problem sets to reinforce your understanding, then progress to the next related sub-topics to build your full statistics toolkit.