Chapter 4 — Front Matter
Purpose and Overview
Calculus was invented to reason about things that change continuously. Probability was developed to reason about things that are uncertain. Continuous random variables are where these two great mathematical traditions meet.
In Chapter 3, you worked with discrete random variables — quantities whose possible values can be listed one by one. A fair die shows 1, 2, 3, 4, 5, or 6. A coin flipped ten times produces 0, 1, 2, ..., or 10 heads. These outcomes are countable, and probability mass functions assign individual probabilities to individual values.
But many of the quantities we most want to model cannot be listed that way. How long will this lightbulb burn before it fails? What is the exact height of a randomly selected adult? How many millimeters of rain will fall tomorrow? These quantities live on continuous scales. Between any two possible values, infinitely many others exist. No list can capture them.
For continuous quantities, the machinery of Chapter 3 breaks down in a fundamental way. You cannot assign positive probability to every individual point — the solution is elegant and deep: replace probability mass with probability density. Instead of asking for the probability at a point, ask for the probability over an interval. Instead of summing probabilities, integrate them.
This chapter develops the full theory of continuous random variables and studies five essential distributions in detail: the uniform, exponential, normal, gamma, and beta. By the end, you will understand continuous probability as calculus applied to uncertainty.
Opening Question
A lightbulb manufacturer claims its bulbs last an average of 1000 hours. You install one today. What is the probability it is still burning after 1500 hours? What is the probability it fails before 200 hours? What percentage of bulbs fall within one standard deviation of the mean lifetime? By the end of this chapter, you will answer all three questions using the exponential and normal distributions — and understand why the methods work.
14Sections
11Objectives
5Key Distributions
15+Worked Examples
Learning Objectives
- 1Explain why continuous random variables require integration rather than summation for computing probabilities.
- 2Verify that a function is a valid probability density function and find normalizing constants.
- 3Interpret probability as area under a density curve and compute probabilities using integration.
- 4Derive and use the cumulative distribution function, applying the Fundamental Theorem of Calculus.
- 5Find percentiles and quantiles by solving F(xp) = p.
- 6Compute expected value as an integral and apply the Law of the Unconscious Statistician.
- 7Calculate variance and standard deviation using the shortcut formula E[X²] − (E[X])².
- 8Work with the uniform, exponential, and normal distributions: recognize, parameterize, and compute.
- 9Apply the memoryless property of the exponential distribution.
- 10Identify situations appropriate for gamma and beta distributions and apply their basic properties.
- 11Recognize and avoid common mistakes in continuous probability calculations.
Sections 4.1 – 4.3
Probability as Area: The Continuous Framework
What Continuous Means · Density Functions · Probability as Integration
Section 4.1
What Makes a Random Variable Continuous
Recall that a random variable is a function that assigns a numerical value to each outcome of a random experiment. In Chapter 3, you studied discrete random variables, whose possible values form a countable set. The word countable is precise: a set is countable if its elements can be placed in a one-to-one correspondence with the natural numbers.
Many random quantities naturally take values in an uncountable set. The time until a radioactive atom decays can be any positive real number. The proportion of defective items in a production batch can be any value in [0, 1]. For such quantities, the discrete framework fails fundamentally.
The Central Shift
To see why discrete methods fail, suppose X can take any value in [0, 1] with equal likelihood. If we tried to assign a probability p > 0 to each individual real number, the sum of all these probabilities would be infinite — not 1. The only consistent assignment is to give probability zero to every individual point.
P(X = c) = 0 for every specific value c. This is not a paradox — zero probability does not mean impossible. Probability lives in intervals, not at points.
Definition 4.1
Continuous Random Variable
A random variable X is continuous if there exists a nonnegative function f, defined on ℝ, such that for any interval [a, b]:
P(a ≤ X ≤ b) = ∫[a to b] f(x) dx
The function f is called the probability density function (PDF) of X.
Because probability at a single point requires integrating over a set of measure zero, it follows for any constant c that P(X = c) = 0. A key consequence: for continuous random variables, whether endpoints are included or excluded makes no difference.
P(a ≤ X ≤ b) = P(a < X ≤ b) = P(a ≤ X < b) = P(a < X < b)
| Feature | Discrete | Continuous |
|---|
| Possible values | Countable set | Uncountable interval |
| Probability at a point | Can be positive | Always zero |
| Probability tool | Probability mass function (PMF) | Probability density function (PDF) |
| Probability over a range | Sum of PMF values | Integral of PDF |
| CDF | Step function | Smooth, continuous function |
| Expected value | Σ x · p(x) | ∫ x · f(x) dx |
Why Calculus?
Integration — the accumulation of infinitely many infinitesimally small contributions — is precisely the tool for computing probabilities over continuous sets. When you see a probability density function, think of it as the derivative of accumulated probability. The Fundamental Theorem of Calculus will appear throughout this chapter in the relationship between f and F.
Section 4.2
Probability Density Functions
Definition 4.2
Probability Density Function
A function f : ℝ → ℝ is a probability density function (PDF) if and only if:
1. Nonnegativity: f(x) ≥ 0 for all x ∈ ℝ.
2. Normalization: ∫[−∞ to ∞] f(x) dx = 1.
If X is a continuous random variable with PDF f, then for any a ≤ b: P(a ≤ X ≤ b) = ∫[a to b] f(x) dx.
Density Is Not Probability
f(x) is not a probability — it is a rate (probability per unit of x). This means f(x) can be greater than 1. What cannot exceed 1 is the area under f over any region. If f(2) = 0.8, this does not mean P(X = 2) = 0.8. That probability is zero. The value f(2) means that near x = 2, probability accumulates at a rate of 0.8 per unit. To find any probability, you must integrate.
Example 4.2.1
Verifying a PDF and computing a probability
- ProblemVerify that f(x) = (3/8)x² for 0 ≤ x ≤ 2 (and 0 otherwise) is a valid PDF. Then find P(1 ≤ X ≤ 2).
- Nonneg.For 0 ≤ x ≤ 2: x² ≥ 0, so f(x) = (3/8)x² ≥ 0. ✓
- Normal.
∫[0 to 2] (3/8)x² dx = (3/8) · [x³/3] from 0 to 2
= (3/8) · (8/3) = 1 ✓
- Prob.
P(1 ≤ X ≤ 2) = ∫[1 to 2] (3/8)x² dx
= (3/8) · [x³/3] from 1 to 2
= (3/8) · (8/3 − 1/3)
= (3/8) · (7/3) = 7/8
Answer: Valid PDF. P(1 ≤ X ≤ 2) = 7/8 = 0.875.
Example 4.2.2
Finding a normalizing constant
- ProblemFind c such that f(x) = ce−3x for x ≥ 0 (and 0 for x < 0) is a valid PDF.
- Require∫[0 to ∞] ce−3x dx = 1
- Solve
c · [−(1/3)e^(−3x)] from 0 to ∞ = c · (0 − (−1/3)) = c/3 = 1
Answer: c = 3. The PDF is f(x) = 3e−3x for x ≥ 0.
Section 4.3
Probability as Area
When f(x) is graphed as a curve over the x-axis, the probability P(a ≤ X ≤ b) is the area of the region bounded above by y = f(x), below by the x-axis, and between x = a and x = b. The entire area under the curve is 1.
This geometric view lets you reason qualitatively before computing anything. If the density curve is tall and narrow, most probability is concentrated in a small region. If it is flat and wide, probability is spread out. The area to the left of any point a gives P(X ≤ a); the area to the right gives P(X > a) = 1 − P(X ≤ a).
Example 4.3.1
Computing probabilities by integration
- SetupLet X have PDF f(x) = 2e−2x for x ≥ 0. Find (a) P(X > 1), (b) P(0.5 ≤ X ≤ 1.5), (c) P(X < 0.3 or X > 1).
- Part (a)
P(X > 1) = ∫[1 to ∞] 2e^(−2x) dx = [−e^(−2x)] from 1 to ∞ = e^(−2) ≈ 0.1353
- Part (b)
P(0.5 ≤ X ≤ 1.5) = [−e^(−2x)] from 0.5 to 1.5 = −e^(−3) + e^(−1) ≈ 0.3181
- Part (c)Use the complement: "X < 0.3 or X > 1" is the complement of "0.3 ≤ X ≤ 1".
1 − P(0.3 ≤ X ≤ 1) = 1 − [−e^(−2x)] from 0.3 to 1
= 1 − (e^(−0.6) − e^(−2)) ≈ 1 − 0.4135 = 0.5865
Answers: (a) 0.1353 (b) 0.3181 (c) 0.5865
Sections 4.4 – 4.7
CDF, Percentiles, Expected Value, and Variance
Cumulative Distributions · Quantiles · Expectation as Integral · Variance
Section 4.4
Cumulative Distribution Functions
Definition 4.3
Cumulative Distribution Function
The cumulative distribution function (CDF) of a random variable X is F(x) = P(X ≤ x). For a continuous random variable with PDF f:
F(x) = ∫[−∞ to x] f(t) dt
Note the dummy variable t inside the integral, since x appears as the upper limit.
The Fundamental Theorem Connection
Since F(x) = ∫[−∞ to x] f(t) dt, the Fundamental Theorem of Calculus gives:
F′(x) = f(x)
The PDF is the derivative of the CDF. The CDF is the antiderivative of the PDF. This is the central relationship in continuous probability, and it will be used throughout the chapter.
Properties of Every CDF
| Property | Statement |
|---|
| Bounded | 0 ≤ F(x) ≤ 1 for all x |
| Nondecreasing | If x₁ < x₂, then F(x₁) ≤ F(x₂) |
| Left limit | limx→−∞ F(x) = 0 |
| Right limit | limx→+∞ F(x) = 1 |
| Continuous | For continuous X, F has no jumps (contrast with discrete CDFs) |
Example 4.4.1
Deriving a CDF from a PDF
- ProblemFind the CDF of X with PDF f(x) = (3/8)x² for 0 ≤ x ≤ 2.
- x < 0F(x) = 0 (no area to the left of the support).
- 0 ≤ x ≤ 2
F(x) = ∫[0 to x] (3/8)t² dt = (3/8) · t³/3 | from 0 to x = x³/8
- x > 2F(x) = 1 (all area is to the left).
- VerifyF(0) = 0, F(2) = 8/8 = 1. ✓ And F′(x) = 3x²/8 = f(x). ✓
CDF: F(x) = x³/8 for 0 ≤ x ≤ 2.
Example 4.4.2
Using the CDF to find probabilities
- ProblemUsing F(x) = x³/8, find P(1 ≤ X ≤ 1.5).
- Solve
P(1 ≤ X ≤ 1.5) = F(1.5) − F(1)
= (1.5)³/8 − 1³/8
= 3.375/8 − 1/8 = 2.375/8 ≈ 0.297
Answer: P(1 ≤ X ≤ 1.5) ≈ 0.297.
Section 4.5
Percentiles and Quantiles
Definition 4.4
Quantile / Percentile
The p-th quantile (or 100p-th percentile) of X is the value xp such that:
F(x_p) = p, i.e., P(X ≤ x_p) = p
The 0.5 quantile is the median. The 0.25 and 0.75 quantiles are the first and third quartiles. The interquartile range is IQR = x0.75 − x0.25.
Example 4.5.1
Finding the median
- ProblemFind the median of X with CDF F(x) = x³/8 for 0 ≤ x ≤ 2.
- Set upSolve F(x0.5) = 0.5.
- Solve
x³/8 = 1/2
x³ = 4
x = 4^(1/3) = 2^(2/3) ≈ 1.587
Median ≈ 1.587.
Example 4.5.2
General quantile formula for Exp(λ)
- SetupFor X ~ Exp(λ), the CDF is F(x) = 1 − e−λx. Solve for xp.
- Solve
1 − e^(−λ x_p) = p
e^(−λ x_p) = 1 − p
x_p = −ln(1−p) / λ
- MedianAt p = 0.5: x0.5 = ln(2)/λ.
p-th quantile: xp = −ln(1 − p) / λ.
Section 4.6
Expected Value as an Integral
Chapter 3 defined expected value for discrete distributions as the weighted average Σ x · p(x). For continuous random variables, the sum becomes an integral and the probability mass p(x) becomes the density f(x) dx. The density f(x) dx represents the infinitesimal probability of X falling near x, and x is the value at that location.
Definition 4.5
Expected Value of a Continuous Random Variable
If X is a continuous random variable with PDF f, then:
E[X] = μ = ∫[−∞ to ∞] x · f(x) dx
For a function g(X) (the Law of the Unconscious Statistician):
E[g(X)] = ∫[−∞ to ∞] g(x) · f(x) dx
Example 4.6.1
Computing E[X] and E[X²]
- SetupLet f(x) = (3/8)x² on [0, 2]. Find E[X] and E[X²].
- E[X]
E[X] = ∫[0 to 2] x · (3/8)x² dx = (3/8) · [x⁴/4] from 0 to 2
= (3/8) · 4 = 3/2 = 1.5
- E[X²]
E[X²] = ∫[0 to 2] x² · (3/8)x² dx = (3/8) · [x⁵/5] from 0 to 2
= (3/8) · (32/5) = 12/5 = 2.4
E[X] = 3/2, E[X²] = 12/5.
Linearity of Expectation
For any constants a and b: E[aX + b] = a · E[X] + b. Adding a constant b shifts the distribution but does not change spread. This property extends to sums: E[X + Y] = E[X] + E[Y] for any X and Y — independent or not. Chapter 5 develops this further.
Section 4.7
Variance and Standard Deviation
Definition 4.6
Variance
The variance of a continuous random variable X with mean μ = E[X] is:
Var(X) = σ² = E[(X − μ)²] = ∫[−∞ to ∞] (x − μ)² f(x) dx
The computational shortcut (usually easier):
Var(X) = E[X²] − (E[X])²
The standard deviation is σ = √Var(X).
Example 4.7.1
Computing Var(X) using the shortcut
- GivenFrom Example 4.6.1: E[X] = 3/2 and E[X²] = 12/5. Use f(x) = (3/8)x² on [0, 2].
- Variance
Var(X) = E[X²] − (E[X])²
= 12/5 − (3/2)²
= 12/5 − 9/4
= 48/20 − 45/20 = 3/20 = 0.15
- SDσ = √(3/20) = √0.15 ≈ 0.387
Var(X) = 3/20, σ ≈ 0.387.
Transformation Rules for Variance
For constants a and b:
Var(aX + b) = a² · Var(X)
Adding a constant b shifts the distribution without changing spread — variance is unaffected by b. Multiplying by a scales all deviations by a, so variance scales by a².
Sections 4.8 – 4.12
Five Essential Continuous Distributions
Uniform · Exponential · Normal · Gamma · Beta
Section 4.8
Uniform Distributions
The uniform distribution models a quantity equally likely to fall anywhere in a fixed interval — the continuous analog of rolling a fair die, extended to an entire interval [a, b]. Every subinterval of equal length has equal probability, regardless of its location.
Definition 4.7
Uniform Distribution U(a, b)
X ~ Uniform(a, b) if its PDF is:
f(x) = 1/(b − a), a ≤ x ≤ b (and 0 otherwise)
U(a, b) Summary
| Property | Value |
|---|
| Support | [a, b] |
| PDF | 1/(b − a) |
| CDF | (x − a)/(b − a) for a ≤ x ≤ b |
| Mean | (a + b)/2 |
| Variance | (b − a)²/12 |
Example 4.8.1
Uniform: Bus arrival problem
- SetupA bus arrives uniformly at random between 8:00 and 8:20 AM. You arrive at 8:10. What is the probability the bus has not yet come?
- ModelLet X ~ U(0, 20) be the arrival time in minutes after 8:00.
- Solve
P(X > 10) = 1 − F(10) = 1 − (10 − 0)/(20 − 0) = 1 − 1/2 = 1/2
Answer: 50% probability the bus has not yet arrived. You arrived at the midpoint of the window.
The Standard Uniform U(0, 1)
U(0, 1) is the foundation of computational probability. All simulation begins with U(0, 1) samples, then transforms them to the target distribution via the inverse CDF. If U ~ U(0, 1), then X = F−1(U) has CDF F. This is the inverse CDF method, the most important technique in Monte Carlo simulation.
Section 4.9
Exponential Distributions
The exponential distribution models waiting times between events in a process where events occur at a constant average rate. How long until the next customer arrives? How long until a machine fails? These waiting times are exponential when the rate of occurrence is constant over time.
The exponential and Poisson distributions are closely linked: if events occur as a Poisson process with rate λ (expected events per unit time), then the waiting time between consecutive events is Exponential(λ).
Definition 4.8
Exponential Distribution Exp(λ)
X ~ Exp(λ) with rate parameter λ > 0 if its PDF is:
f(x) = λe^(−λx), x ≥ 0 (and 0 for x < 0)
CDF: F(x) = 1 − e−λx for x ≥ 0.
Exp(λ) Summary
| Property | Value |
|---|
| Support | [0, ∞) |
| PDF | λe−λx |
| CDF | 1 − e−λx |
| Mean | 1/λ |
| Variance | 1/λ² |
| SD = Mean? | Yes. CV = σ/μ = 1 always. |
Theorem 4.1 — Memoryless Property
If X ~ Exp(λ), then for any s, t ≥ 0:
P(X > s + t | X > s) = P(X > t)
Interpretation. Given that you have already waited s units of time, the remaining wait time has the same exponential distribution. The process has no memory of elapsed time. The exponential is the only continuous distribution with this property.
Proof
P(X > s+t | X > s) = P(X > s+t) / P(X > s) = e−λ(s+t) / e−λs = e−λt = P(X > t). ∎
Example 4.9.1
Exponential: Customer service line
- SetupA customer service line receives 4 calls per hour. Call interarrival times follow Exp(4). Find: (a) P(next call within 10 min), (b) P(next call after 20 min), (c) given no call in 10 min, P(call in next 10 min).
- Part (a)
P(X ≤ 1/6) = 1 − e^(−4/6) = 1 − e^(−2/3) ≈ 0.487
- Part (b)
P(X > 1/3) = e^(−4/3) ≈ 0.264
- Part (c)By the memoryless property: P(X > 1/6 + 1/6 | X > 1/6) = P(X > 1/6) ≈ 0.487. The same as (a).
Answers: (a) 0.487 (b) 0.264 (c) 0.487
Section 4.10
Normal Distributions
The normal distribution is the most important distribution in all of probability and statistics. It appears in natural phenomena (measurement errors, heights, test scores), it is the limiting distribution for sums of many independent random variables (the Central Limit Theorem, Chapter 7), and it underlies virtually every classical inferential procedure.
Definition 4.9
Normal Distribution N(μ, σ²)
X ~ N(μ, σ²) with mean μ and variance σ² (σ > 0) if its PDF is:
f(x) = (1 / (σ√(2π))) · exp(−(x − μ)² / (2σ²)), −∞ < x < ∞
E[X] = μ and Var(X) = σ² exactly — the parameters are the mean and variance.
Definition 4.10
Standard Normal N(0, 1)
The standard normal Z ~ N(0, 1) has PDF:
φ(z) = (1/√(2π)) · e^(−z²/2)
Its CDF Φ(z) = ∫[−∞ to z] φ(t) dt has no closed form and is evaluated from tables or software. Key symmetry: Φ(−z) = 1 − Φ(z).
Standardization
Any X ~ N(μ, σ²) converts to standard normal via:
Z = (X − μ) / σ ~ N(0, 1)
Therefore: P(X ≤ x) = Φ((x − μ)/σ). This reduces all normal probability calculations to the standard normal table.
The 68 – 95 – 99.7 Rule
For any X ~ N(μ, σ²), these probabilities are exact and should be memorized:
P(μ − σ < X < μ + σ) ≈ 0.6827 (68%)
P(μ − 2σ < X < μ + 2σ) ≈ 0.9545 (95%)
P(μ − 3σ < X < μ + 3σ) ≈ 0.9973 (99.7%)
Example 4.10.1
Normal: Heights of adult men
- SetupHeights of adult men: X ~ N(70, 9) in inches (μ = 70, σ = 3). Find (a) P(X < 73), (b) P(64 < X < 76), (c) the 90th percentile.
- Part (a)z = (73 − 70)/3 = 1.00. P(X < 73) = Φ(1.00) ≈ 0.8413.
- Part (b)
z₁ = (64−70)/3 = −2, z₂ = (76−70)/3 = 2
P(−2 < Z < 2) = Φ(2) − Φ(−2) = 2Φ(2)−1 ≈ 2(0.9772)−1 = 0.9544
- Part (c)Find z0.90 with Φ(z0.90) = 0.90. From table: z0.90 ≈ 1.282. Then x = 70 + 3(1.282) ≈ 73.85 in.
Answers: (a) 0.8413 (b) 0.9544 (c) 73.85 inches
Linear Transformations of Normal Variables
If X ~ N(μ, σ²) and Y = aX + b (a ≠ 0), then Y ~ N(aμ + b, a²σ²). Linear transformations of normal random variables are normal. This closure property makes the normal distribution especially tractable in theory and applications.
Section 4.11
Gamma Distributions: An Introduction
The exponential distribution models waiting for one event. What about waiting for the k-th event? If a server receives calls at rate λ and you want the time until the 3rd call, the gamma distribution is the right model. It generalizes the exponential to multiple events and uses the Gamma function to achieve this.
Definition 4.11
Gamma Function
For α > 0:
Γ(α) = ∫[0 to ∞] t^(α−1) e^(−t) dt
Key properties: Γ(1) = 1 · Γ(n) = (n−1)! for positive integers · Γ(α+1) = αΓ(α) · Γ(1/2) = √π
Definition 4.12
Gamma Distribution Gamma(α, λ)
X ~ Gamma(α, λ) with shape α > 0 and rate λ > 0 has PDF:
f(x) = (λ^α / Γ(α)) · x^(α−1) · e^(−λx), x > 0
Gamma(α, λ) Summary
| Property | Value |
|---|
| Support | (0, ∞) |
| PDF | (λ^α / Γ(α)) x^(α−1) e^(−λx) |
| Mean | α/λ |
| Variance | α/λ² |
| Special case α = 1 | Exponential(λ) |
| Sum of exponentials | X₁ + ··· + X_k ~ Gamma(k, λ) when X_i ~ Exp(λ) iid |
Example 4.11.1
Gamma: Machine shutdown time
- SetupA machine shuts down when 3 independent components each fail. Component lifetimes are Exp(2) (rate 2 per year). Find E[shutdown time] and Var(shutdown time).
- ModelTotal time X = X₁ + X₂ + X₃ where X_i ~ Exp(2) independently. So X ~ Gamma(3, 2).
- Solve
E[X] = α/λ = 3/2 = 1.5 years
Var(X) = α/λ² = 3/4 = 0.75 years²
Expected shutdown: 1.5 years. Variance: 0.75 years².
Section 4.12
Beta Distributions: An Introduction
The uniform, exponential, normal, and gamma distributions have unbounded support or support on [0, ∞). Many important quantities are bounded: proportions, probabilities, rates, and fractions live on [0, 1]. The beta distribution is a flexible two-parameter family on (0, 1) and the natural distribution for modeling probabilities and proportions.
The beta distribution is also foundational in Bayesian inference: when modeling an unknown probability, the beta distribution is the conjugate prior for the binomial likelihood (Chapter 10).
Definition 4.13
Beta Distribution Beta(α, β)
X ~ Beta(α, β) with shape parameters α, β > 0 has PDF:
f(x) = (Γ(α+β) / (Γ(α)Γ(β))) · x^(α−1) · (1−x)^(β−1), 0 < x < 1
where B(α, β) = Γ(α)Γ(β)/Γ(α+β) is the Beta function.
Beta(α, β) Summary
| Property | Value |
|---|
| Support | (0, 1) |
| Mean | α/(α + β) |
| Variance | αβ / [(α+β)²(α+β+1)] |
| Special case α = β = 1 | Uniform(0, 1) |
| Parameter Values | Shape | When Useful |
|---|
| α = β = 1 | Flat (uniform) | No prior information about a probability |
| α > 1, β > 1 | Unimodal, bell-shaped | Most mass in the interior of [0,1] |
| α < 1, β < 1 | U-shaped | Mass concentrated near 0 and 1 |
| α > β | Left-skewed | Probability likely high |
| α < β | Right-skewed | Probability likely low |
Example 4.12.1
Beta: Defective item proportion
- SetupA quality control engineer models the proportion p of defective items as Beta(2, 8). Find E[p] and Var[p].
- Solve
E[p] = 2/(2+8) = 2/10 = 0.20
Var(p) = (2·8) / (10² · 11) = 16/1100 ≈ 0.0145
The model expects 20% defective items, with standard deviation ≈ √0.0145 ≈ 12% — moderate uncertainty around that value.
Sections 4.13 – 4.14
Applications, Mistakes, and the Road Ahead
Science & Engineering · Eight Common Errors · Preparing for Joint Distributions
Section 4.13
Continuous Models in Science and Engineering
The distributions studied in this chapter are not merely mathematical exercises. They are working models used daily in science, engineering, finance, medicine, and data science. The ability to recognize which distribution fits which situation — and why — is a core statistical skill.
| Distribution | Applied Context | Why It Fits |
|---|
| Normal | Measurement error; heights; quality control; financial returns | Many independent additive factors; Central Limit Theorem (Ch. 7) |
| Exponential | Component lifetime; interarrival times; radioactive decay | Constant hazard rate; memoryless failure mechanism |
| Gamma | Waiting for k-th event; rainfall amount; aggregate insurance claims | Sum of k independent exponential stages |
| Beta | Proportion defective; click rates; Bayesian prior for a probability | Natural support on [0,1]; flexible shape; conjugate to binomial |
| Uniform | Random number generation; roundoff error; randomized algorithms | Complete indifference across an interval; simulation foundation |
Example 4.13.1
Multi-distribution model: Hospital emergency room
- ContextA hospital ER uses several distributions simultaneously to model different aspects of patient flow.
- ArrivalsInterarrival time between patients: Exp(λ = 4/hour). Expected gap between patients: 15 minutes.
- MinorTreatment time for minor cases: U(15, 45) minutes. Expected treatment: 30 minutes.
- MajorTreatment time for major cases: N(90, 400) minutes (mean 90, SD 20).
- Prop.Proportion of visits that are major: Beta(2, 18). E[proportion] = 2/20 = 10%.
Each distribution is chosen to match the nature of the quantity: bounded or not, rate-driven or additive, with or without natural support constraints.
Section 4.14
Common Continuous Random Variable Mistakes
Mistake 1 — Treating f(x) as a probability
"f(0.7) = 1.4, so P(X = 0.7) = 1.4"
Correct: f(x) is a density. P(X = 0.7) = 0 for any continuous X.
To get probability, integrate: P(a ≤ X ≤ b) = ∫[a to b] f(x) dx.
Mistake 2 — Confusing PDF and CDF
"F(x) = f(x)"
Correct: F(x) = ∫[−∞ to x] f(t) dt. And F′(x) = f(x).
F is the integral of f, not f itself.
Mistake 3 — Exponential parameterization confusion
"X ~ Exp(4), so E[X] = 4"
Correct: If the rate is λ = 4, then E[X] = 1/λ = 1/4.
Mean = 1/rate. If mean = 4, then λ = 1/4. Always check which parameterization is in use.
Mistake 4 — Integrating outside the support
"P(−1 ≤ X ≤ 1) = ∫[−1 to 1] f(x) dx" when support is [0, 2]
Correct: f(x) = 0 on [−1, 0), so P(−1 ≤ X ≤ 1) = P(0 ≤ X ≤ 1) = ∫[0 to 1] f(x) dx.
Always identify the support before integrating.
Mistake 5 — Misreading the normal table
"Φ(1.5) = P(Z > 1.5)"
Correct: Φ(z) = P(Z ≤ z) — the standard normal table gives left-tail probability.
P(Z > 1.5) = 1 − Φ(1.5). For P(a < Z < b): use Φ(b) − Φ(a).
Mistake 6 — Not standardizing before using the normal table
"X ~ N(70, 9), P(X < 73): look up 73 directly in the Z table"
Correct: First standardize. z = (73 − 70)/3 = 1.00.
Then P(X < 73) = Φ(1.00) ≈ 0.8413. The Z table only applies to N(0,1).
Mistake 7 — Choosing the wrong distribution
"I'll model this proportion with a normal distribution"
Correct: Proportions live in [0, 1] — a bounded support. The normal distribution has infinite support.
Use Beta(α, β) for proportions and probabilities on [0, 1].
Mistake 8 — Ignoring the condition for E[X] to exist
"E[X] always exists for any continuous distribution"
Correct: E[X] requires ∫|x|f(x)dx < ∞. Some heavy-tailed distributions
(e.g., Cauchy) have undefined means. For distributions in this chapter, means always exist,
but verify this in advanced work with non-standard distributions.
Chapter 4 — Back Matter
Chapter Summary
- Sec 4.1Continuous random variables require integration because individual points have probability zero. Probability lives in intervals. Calculus is the right tool for the uncountable.
- Sec 4.2A PDF must be nonneg. and integrate to 1. f(x) is a density — not a probability. It can exceed 1. Probability always requires integration over an interval.
- Sec 4.3Probability is area under the density curve. All interval probabilities are integrals. Endpoints make no difference for continuous distributions.
- Sec 4.4CDF: F(x) = P(X ≤ x) = ∫ f. Key relation: F′(x) = f(x) by the Fundamental Theorem. Interval probabilities: P(a ≤ X ≤ b) = F(b) − F(a).
- Sec 4.5Quantile x_p satisfies F(x_p) = p (the inverse CDF). Median = x_{0.5}. IQR = x_{0.75} − x_{0.25}.
- Sec 4.6E[X] = ∫ x f(x) dx. Law of Unconscious Statistician: E[g(X)] = ∫ g(x)f(x)dx. Linearity: E[aX + b] = aE[X] + b.
- Sec 4.7Var(X) = E[X²] − (E[X])². Transformation: Var(aX + b) = a²Var(X). Standard deviation σ = √Var(X).
- Sec 4.8U(a, b): constant PDF 1/(b−a), mean (a+b)/2, var (b−a)²/12. Standard uniform U(0,1) is the foundation of all simulation.
- Sec 4.9Exp(λ): PDF λe^{-λx}, mean 1/λ, var 1/λ². Unique memoryless property: past waiting time is irrelevant to future waiting.
- Sec 4.10N(μ, σ²): symmetric bell curve, mean μ, var σ². Standardize: Z = (X−μ)/σ ~ N(0,1). Use Φ for all normal probabilities. 68-95-99.7 rule.
- Sec 4.11Gamma(α, λ): generalization of Exp to k events. Uses Γ(α) = ∫ t^{α−1}e^{−t}dt. Mean α/λ, var α/λ². Gamma(1, λ) = Exp(λ).
- Sec 4.12Beta(α, β) on (0,1): flexible shape for proportions and probabilities. Mean α/(α+β). Foundation for Bayesian inference with binomial models.
- Sec 4.13Each distribution has a natural domain of application. Distribution choice should match the nature of the process: support, shape, generating mechanism.
- Sec 4.14Eight errors: density-as-probability, PDF/CDF confusion, exponential parameterization, support boundaries, normal table direction, failing to standardize, wrong distribution, mean existence.
Key Terms Review
continuous random variable
A random variable whose probabilities are computed by integrating a PDF. P(X = c) = 0 for every constant c.
probability density function
A nonneg. function f with ∫f = 1. Not a probability; a rate. Probability requires integrating f over an interval.
cumulative distribution function
F(x) = P(X ≤ x) = ∫[−∞ to x] f(t)dt. Nondecreasing, continuous, with limits 0 and 1.
expected value (mean)
E[X] = ∫ x f(x) dx. The center of mass of the distribution. Denoted μ.
variance
Var(X) = E[(X−μ)²] = E[X²] − μ². Average squared deviation from the mean. Standard deviation σ = √Var(X).
support
The set of x where f(x) > 0. Outside the support, f = 0 and those values are impossible.
quantile / percentile
x_p satisfies F(x_p) = p. The inverse CDF. Median = x_{0.5}, quartiles = x_{0.25} and x_{0.75}.
uniform distribution
U(a, b): constant density 1/(b−a). Every equal-length subinterval has equal probability.
exponential distribution
Exp(λ): models waiting times. PDF λe^{−λx}, mean 1/λ. Characterized by the memoryless property.
memoryless property
P(X > s+t | X > s) = P(X > t). The exponential is the only continuous distribution with this property.
normal distribution
N(μ, σ²): symmetric bell curve, the most important distribution in statistics. Underlies the Central Limit Theorem.
standard normal
N(0, 1). CDF denoted Φ(z). Symmetry: Φ(−z) = 1 − Φ(z). All normal computations reduce to Φ via standardization.
standardization
Z = (X − μ)/σ transforms X ~ N(μ, σ²) to Z ~ N(0, 1). Required before using the standard normal table.
Gamma function
Γ(α) = ∫[0 to ∞] t^{α−1}e^{−t}dt. Extends factorial: Γ(n) = (n−1)!. Used to normalize the gamma and beta PDFs.
gamma distribution
Gamma(α, λ): shape α, rate λ. Generalizes Exp to waiting for k events. Mean α/λ, var α/λ².
beta distribution
Beta(α, β) on (0, 1). Flexible shape for proportions and probabilities. Mean α/(α+β). Conjugate prior for binomial.
PDF Practice
- Determine whether each is a valid PDF. If not, explain why.
(a) f(x) = 2x for 0 ≤ x ≤ 1
(b) f(x) = 3x² − 1 for 0 ≤ x ≤ 1
(c) f(x) = (1/2)sin(x) for 0 ≤ x ≤ π
(d) f(x) = ex for x ≤ 0
- Find the constant c such that each is a valid PDF.
(a) f(x) = cx³ for 0 ≤ x ≤ 2
(b) f(x) = c/x² for x ≥ 1
(c) f(x) = c(1 − x²) for −1 ≤ x ≤ 1
(d) f(x) = ce−x/3 for x ≥ 0
- Let f(x) = (3/2)(1 − x²) for −1 ≤ x ≤ 1. Verify it is a valid PDF. Find P(0 ≤ X ≤ 0.5), P(|X| > 0.5), and explain why P(X = 0.3) = 0.
- Let f(x) = 2e−2x for x ≥ 0. Find P(X > 2), P(1 ≤ X ≤ 3), and P(X < 0.5).
CDF Practice
- Find the CDF F(x) for each PDF:
(a) f(x) = 4x³ for 0 ≤ x ≤ 1
(b) f(x) = (1/2)e−x/2 for x ≥ 0
(c) f(x) = 1/x² for x ≥ 1
(d) f(x) = 6x(1−x) for 0 ≤ x ≤ 1
- Given F(x) = x² for 0 ≤ x ≤ 1: (a) Find P(0.4 ≤ X ≤ 0.7). (b) Find the PDF f(x). (c) Find P(X > 0.8). (d) Find the median.
- Given F(x) = 1 − 1/x² for x ≥ 1: (a) Verify this is a valid CDF. (b) Find the PDF. (c) Find P(2 ≤ X ≤ 4). (d) Find P(X > 3).
- A random variable X has CDF F(x) = x²/4 for 0 ≤ x ≤ 2 (and 0 or 1 outside). (a) Find and verify the PDF. (b) Compute P(0.5 < X < 1.5) both by CDF and direct integration. (c) Find the 75th percentile.
Expected Value and Variance Practice
- Find E[X] and Var(X) for each:
(a) f(x) = 3x² on [0, 1]
(b) f(x) = (1/2)e−x/2 for x ≥ 0
(c) f(x) = 6x(1 − x) on [0, 1]
- For X ~ U(a, b): (a) Derive E[X] = (a+b)/2 using integration. (b) Derive Var(X) = (b−a)²/12. (c) Find E[X] and Var(X) for U(0, 10).
- Let X have PDF f(x) = (3/8)x² on [0, 2]. Find E[X], E[X²], Var(X), E[2X + 5], and Var(2X + 5).
- A random variable X has E[X] = 4 and E[X²] = 20. (a) Find Var(X) and SD(X). (b) Find E[3X − 2] and Var(3X − 2). (c) Can you determine the distribution from this information alone?
Distribution Applications
- (Uniform) A commuter train departs every 20 minutes. A passenger arrives uniformly at random within the window. (a) P(wait > 15 min)? (b) Expected wait? (c) 80th percentile of wait time?
- (Exponential) A hard drive has exponentially distributed lifetime with mean 5 years. (a) P(lasts more than 8 years). (b) P(fails within 2 years). (c) Given it has lasted 3 years, P(it lasts 3 more years). (d) Median lifetime.
- (Normal) Birth weights of full-term newborns: X ~ N(3400, 250000) grams (σ = 500 g). (a) Fraction between 2900 and 3900 g. (b) Fraction classified low birth weight (under 2500 g). (c) Weight separating the top 5%.
- (Gamma) A server processes requests at rate λ = 3/min. Set up (do not evaluate) the probability that the first 5 requests are all processed within 2 minutes total. Use Gamma(5, 3).
- (Beta) Market researchers model proportion p of buyers as Beta(3, 12). (a) Find E[p] and Var(p). (b) Interpret E[p]. (c) Set up ∫ for P(p > 0.25).
- (Multi-distribution) A hospital ER models interarrival times as Exp(4/hr), minor treatment as U(15, 45) min, major treatment as N(90, 400) min, and proportion major as Beta(2, 18). Find expected values for each quantity and P(major case > 2 hours).
Error Analysis
Each problem shows a student's work containing an error. Identify it and provide the correct solution.
Problem EA-1
X has PDF f(x) = 2x for 0 ≤ x ≤ 1.
Student: "P(X = 0.7) = f(0.7) = 2(0.7) = 1.4"
What is wrong? What is the correct interpretation of f(0.7) = 1.4?
Problem EA-2
X ~ Exp(2). Student wants E[X].
Student: "λ = 2, so E[X] = λ = 2"
What is wrong? What is the correct E[X]?
Problem EA-3
X ~ N(80, 16) (mean 80, σ = 4). Student wants P(X < 86).
Student: "Look up z = 86 in the normal table. Φ(86) ≈ 1."
What is wrong? Provide the complete correct solution.
Problem EA-4
X has CDF F(x) = x⁴ for 0 ≤ x ≤ 1.
Student: "The PDF is f(x) = F(x) = x⁴"
What is wrong? What is the correct PDF?
Problem EA-5
X has PDF f(x) = 3x² on [0, 1]. Student computes P(−0.5 ≤ X ≤ 0.5).
Student: "∫[−0.5 to 0.5] 3x² dx = [x³] from −0.5 to 0.5 = 0.125 − (−0.125) = 0.25"
The numerical answer 0.25 is correct — but is the reasoning complete? Identify any issue with the setup.
Chapter 4 Checkpoint
Chapter 4 Checkpoint
Complete without reference to the chapter — show all integration steps
Part A — Concepts
- Explain in your own words why P(X = c) = 0 for any continuous random variable X and any constant c.
- What two conditions must a function f satisfy to be a valid probability density function?
- State the relationship between the PDF f and the CDF F using calculus notation.
- What does the memoryless property of the exponential distribution mean? Give a real-world example.
- Why is the normal distribution important in statistics? Name two contexts where it arises naturally.
Part B — Computation
- Let f(x) = cx(2 − x) for 0 ≤ x ≤ 2. (a) Find c. (b) Find F(x). (c) Find E[X]. (d) Find Var(X). (e) Find the median.
- Let X ~ Exp(0.5). (a) Write the PDF and CDF. (b) Find P(X > 3). (c) Find the 25th percentile. (d) Find E[X] and Var(X).
- Suppose X ~ N(100, 225) (μ = 100, σ = 15). (a) Find P(X < 115). (b) Find P(85 < X < 130). (c) Find x such that P(X > x) = 0.05. (d) At what percentile does a score of 85 fall?
Part C — Applications
- A chemical reaction has normally distributed yield with μ = 82% and σ = 4%. (a) What proportion exceed 90% yield? (b) What is the 1st percentile? (c) Reactions below 75% are failures — what percentage fail?
- An electrical component has exponentially distributed lifetime with mean 200 hours. (a) Find λ. (b) P(lasts > 300 hours). (c) Median lifetime. (d) Given survival to 100 hours, P(survives 100 more hours).
- A PDF is given by: f(x) = kx for 0 ≤ x ≤ 1, f(x) = k(2−x) for 1 < x ≤ 2, and 0 otherwise. (a) Find k. (b) Find F(x) for all x. (c) Find E[X] (use symmetry or integration). (d) Find P(0.5 ≤ X ≤ 1.5).
Part D — Analysis
- A student claims: "Since the mean of X is 5, the probability X equals 5 must be higher than any other value." Is this correct for continuous random variables? Explain carefully.
- Compare Exp(λ) and Gamma(k, λ) for k = 1, 2, 3. How are they related? What physical process does each model?
- N(0, 1) and Beta(α, β) with α = β can both be symmetric. What fundamental difference makes the beta appropriate for some applications and the normal for others?
Looking Ahead
Chapter 5: Joint Distributions and Multiple Random Variables
Chapter 4 developed the complete theory for a single continuous random variable. You can now describe any continuous distribution with a PDF, compute probabilities by integration, find the CDF and use it for interval probabilities and quantiles, calculate expected values and variances, and recognize and work with five major distribution families.
But many real problems involve more than one random variable at once. A patient's blood pressure and cholesterol level are both random. The height and weight of a person are both random. Understanding how multiple random variables relate — whether independent, correlated, or causally linked — requires extending the framework developed here. Chapter 5 introduces joint PDFs for two continuous variables simultaneously, marginal distributions obtained by integrating out one variable, conditional distributions, and covariance and correlation as measures of linear dependence. The tools carry over directly: double integrals replace single integrals, and the conceptual architecture of Chapter 4 is the foundation for everything that follows.
MGU Library Connection
- Integration Techniques for Probability — Appendix B of this textbook
- Common Continuous Distributions Reference — Appendix G of this textbook
- Expected Value and Variance Formula Sheet — Appendix H of this textbook
- Calculus, Chapters 5–6: Definite Integrals, Integration Techniques, Improper Integrals
- Chapter 3 of this book: Discrete Random Variables — the discrete analogs of every concept in Chapter 4
- Chapter 7: Limit Theorems — the Central Limit Theorem explains why the normal distribution is ubiquitous
- MGU Dictionary entries: density function, expected value, variance, normal distribution, exponential distribution