Source: Introduction to Statistics, Purdue University
Tags: continuous probability distribution, PDF, probability density function, CDF, cumulative distribution function, continuous random variable, area under the curve, integration, STAT
Difficulty: Intermediate | Prerequisites: Basic integration (definite integrals), discrete probability distributions (Chapter 5).
This chapter moves from discrete random variables (where you can list every possible outcome) to continuous ones (where outcomes fall anywhere along an interval). The core shift is that probability is no longer assigned to individual points; it is the area under a curve. If you are comfortable with discrete PMFs and basic definite integrals, you have what you need. Everything here builds toward the normal distribution, which dominates the rest of the course.
A continuous random variable takes values over an interval, and its probabilities come from integrating a probability density function (PDF). The PDF must be non-negative everywhere and integrate to 1. The cumulative distribution function (CDF) gives the probability of being at or below a given value by accumulating area from left to right under the PDF.
Continuous random variable
A random variable that can take any value within an interval (or collection of intervals) on the real number line. Unlike a discrete variable, you cannot list all possible outcomes.
In simple terms: instead of counting outcomes (like dice rolls), you are measuring quantities that can land anywhere on a range, such as time, weight, or temperature.
Probability density function (PDF)
A function f(x) that describes the relative likelihood of a continuous random variable taking on a given value. The probability of the variable falling in any interval equals the area under f(x) over that interval.
Think of it as: the curve whose area gives you probabilities. The height of the curve at a point is not a probability by itself; it is a density.
Cumulative distribution function (CDF)
The function F(x) = P(X ≤ x), computed as the integral of the PDF from negative infinity up to x. It gives the total accumulated probability to the left of (and including) x.
In simple terms: "what is the probability that X ends up at or below this value?" The CDF starts at 0, increases, and ends at 1.
Probability mass function (PMF)
The discrete counterpart of the PDF. For a discrete random variable, P(x) gives the probability of each specific outcome directly.
Think of it as: the function you used in Chapter 5, where you could assign a probability to each individual value. With continuous variables, you switch from PMF to PDF.
Valid density curve
A PDF that satisfies both required properties: f(x) ≥ 0 for all x, and the total area under the curve equals 1.
A function f(x) is a valid PDF if and only if it satisfies two conditions:
f(x) ≥ 0 for all x (the curve never dips below zero)
The total area under the curve equals 1: ∫ from −∞ to ∞ of f(x) dx = 1
The probability that X falls between two values a and b is the area under the PDF between those values:
P(a < X < b) = ∫ from a to b of f(x) dx
A critical difference from discrete distributions: the probability at any single exact point is zero.
P(X = a) = ∫ from a to a of f(x) dx = 0
This means P(a < X < b) = P(a ≤ X ≤ b). Whether you include the endpoints or not makes no difference for continuous variables.
Suppose f(x) is proportional to x², defined on [0, 2], and zero elsewhere. To find the constant C:
Set up: ∫ from 0 to 2 of C·x² dx = 1
Integrate: C · (x³/3) evaluated from 0 to 2 = C · (8/3) = 1
Solve: C = 3/8
The PDF is: f(x) = (3/8)x² for 0 ≤ x ≤ 2, and 0 elsewhere
Given f(x) = x/2 for 0 ≤ x ≤ 2, and 0 elsewhere.
(a) Is this a valid density curve?
Check both conditions:
f(x) = x/2 ≥ 0 for all x in [0, 2]. Yes.
∫ from 0 to 2 of (x/2) dx = (x²/4) evaluated from 0 to 2 = 4/4 = 1. Yes.
Both conditions hold, so this is a valid PDF.
(b) P(class will end within 15 seconds of the hour)
The class runs for 2 hours (x measured in hours). 15 seconds = 0.25 minutes = 1/240 hours... but the notes model this with x in some unit where 15 seconds corresponds to the interval [1.75, 2].
P(1.75 ≤ X ≤ 2) = ∫ from 1.75 to 2 of (x/2) dx = (x²/4) evaluated from 1.75 to 2 = 4/4 − (1.75)²/4 = 1 − 0.7656 = 0.2344
The CDF of a continuous random variable X is:
F(x) = P(X ≤ x) = P(X < x) = ∫ from −∞ to x of f(s) ds
Note that P(X ≤ x) and P(X < x) are equal for continuous variables (because the probability at a single point is zero).
F(x) is non-decreasing: as x increases, accumulated probability can only grow or stay the same
As x → −∞, F(x) → 0
As x → +∞, F(x) → 1
The CDF is continuous (no jumps), unlike the step-function CDF of discrete variables
The CDF is the integral of the PDF: F(x) = ∫ from −∞ to x of f(s) ds
The PDF is the derivative of the CDF: f(x) = dF(x)/dx
To find P(a < X < b) from the CDF: P(a < X < b) = F(b) − F(a)
P(x) = Probability Mass Function (PMF), used for discrete random variables
f(x) = Probability Density Function (PDF), used for continuous random variables
F(x) = Cumulative Distribution Function (CDF), used for both (but computed differently)
Given f(x) = x/2 for 0 ≤ x ≤ 2, and 0 elsewhere.
For x < 0: F(x) = 0 (no area accumulated yet).
For 0 ≤ x ≤ 2:
F(x) = ∫ from 0 to x of (s/2) ds
= (1/4) · s² evaluated from 0 to x
= x²/4
For x > 2: F(x) = 1 (all area accumulated).
So the complete CDF is: F(x) = 0 for x < 0; x²/4 for 0 ≤ x ≤ 2; 1 for x > 2.
PDF conditions:
f(x) \geq 0 \quad \text{for all } x\int_{-\infty}^{\infty} f(x)\, dx = 1Probability from a PDF:
P(a < X < b) = \int_a^b f(x)\, dxPoint probability (always zero for continuous):
P(X = a) = \int_a^a f(x)\, dx = 0CDF definition:
F(x) = P(X \leq x) = \int_{-\infty}^{x} f(s)\, dsPDF from CDF:
f(x) = \frac{dF(x)}{dx}Interval probability from CDF:
P(a < X < b) = F(b) - F(a)Continuous distributions model any measurement that varies smoothly: how long a phone call lasts, the exact weight of a cereal box off the production line, or the time between arrivals at a hospital A&E. Engineers use CDFs to answer questions like "what percentage of parts will measure below the tolerance limit?" Quality control, finance (stock returns), and biostatistics all rely on continuous PDFs daily.
Students often think that f(x) gives the probability of X equalling x. It does not. f(x) is a density, not a probability. Only the area under the curve over an interval is a probability.
Students sometimes believe that f(x) must be ≤ 1 because probabilities cannot exceed 1. The density itself can exceed 1 at certain points; what matters is that the total area under the curve equals 1.
Confusing PDF and CDF is common. Remember: f(x) is the curve you integrate (density), and F(x) is the running total of area (cumulative probability).
Students forget that for continuous variables, P(X = a) = 0, so P(X < 5) and P(X ≤ 5) are the same. This is not true for discrete variables.
⚠️ Expect to be asked to verify whether a given function is a valid PDF. Check both conditions every time: non-negativity and total area = 1.
⚠️ Setting up the correct integration limits is where most marks are lost. Read the support of the PDF carefully (the interval where f(x) is non-zero) and match your limits to the question.
⚠️ Deriving a CDF from a PDF (and vice versa) is a standard exam question. Know the integral relationship and be ready to differentiate a CDF to recover the PDF.
⚠️ Finding a normalising constant C so that a function integrates to 1 appears frequently. Set up ∫ C·g(x) dx = 1, integrate, solve for C.
True or False: For a continuous random variable, P(X = 3) can be greater than zero.
Answer: False. For continuous variables, the probability at any exact point is always zero.
Fill in the blank: The total area under any valid PDF equals ______.
Answer: 1.
True or False: If f(x) = 1.5 at some point, the function cannot be a valid PDF.
Answer: False. The density can exceed 1; only the total area must equal 1.
Fill in the blank: To get from a PDF to a CDF, you ______. To get from a CDF to a PDF, you ______.
Answer: Integrate; differentiate.
True or False: For a continuous random variable, P(X < 5) = P(X ≤ 5).
Answer: True. Since P(X = 5) = 0, including or excluding the point makes no difference.
Q: Given f(x) = 3x² for 0 ≤ x ≤ 1 and 0 elsewhere, verify that this is a valid PDF.
A: Check f(x) ≥ 0: since x² ≥ 0 and 3 > 0, yes. Check total area: ∫ from 0 to 1 of 3x² dx = 3 · (x³/3) from 0 to 1 = 1. Both conditions met; it is a valid PDF.
Q: Using the PDF f(x) = x/2 for 0 ≤ x ≤ 2, find P(0.5 < X < 1.5).
A: ∫ from 0.5 to 1.5 of (x/2) dx = (x²/4) from 0.5 to 1.5 = (2.25/4) − (0.25/4) = 0.5625 − 0.0625 = 0.5.
Q: A continuous random variable has CDF F(x) = x²/4 for 0 ≤ x ≤ 2. What is P(X > 1)?
A: P(X > 1) = 1 − F(1) = 1 − (1/4) = 0.75.
Q: Derive the PDF from the CDF F(x) = x²/4 for 0 ≤ x ≤ 2.
A: f(x) = dF/dx = d(x²/4)/dx = 2x/4 = x/2, which matches the PDF from Example 6.1.
Q: If f(x) = C·x³ for 0 ≤ x ≤ 1 and 0 elsewhere, find C.
A: ∫ from 0 to 1 of C·x³ dx = C·(x⁴/4) from 0 to 1 = C/4 = 1, so C = 4.
This material connects directly to expected value and variance of continuous random variables (the next section of Chapter 6), where you use the PDF inside an integral to compute E(X) and Var(X). It also lays the groundwork for the normal distribution (Chapter 7), which is a specific continuous PDF with its own CDF table (the Z-table). If you are studying for applied work, the CDF concept returns in hypothesis testing and confidence intervals later in the course.
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