Difficulty: Intermediate to Advanced | Prerequisites: Expected value, basic integration, normal distribution basics
This set of notes covers three areas that regularly appear on STAT 350 Exam 1. First, variance properties and linear transformations, which underpin many computational problems across the course. Second, the exponential distribution, which models waiting times and pairs naturally with the Poisson distribution. Third, working with piecewise probability density functions, including finding normalising constants, building CDFs, and computing medians. These topics require comfort with algebra and basic integration.
Var(aX) = a²Var(X), but Var(X1 + X2 + ... + Xn) = nVar(X) for independent, identically distributed variables, so multiplying one observation by n gives a different variance than summing n observations. The exponential distribution models the time between Poisson events and has the memoryless property. Continuous PDFs must integrate to 1 over their support, and the CDF is the running integral of the PDF.
Variance of a linear transformation
If Y = aX + b, then Var(Y) = a²Var(X). The additive constant b does not affect the variance.
In simple terms, scaling X by a factor of a scales the variance by a².
Variance of a sum of independent variables
If X1, X2, ..., Xn are independent, then Var(X1 + X2 + ... + Xn) = Var(X1) + Var(X2) + ... + Var(Xn).
If they are also identically distributed (each with variance σ²), this simplifies to nσ².
Expected value of Y²
For Y = aX + b: E[Y²] = E[(aX + b)²] = E[a²X² + 2abX + b²] = a²E[X²] + 2abE[X] + b².
Since E[X²] = Var(X) + (E[X])² = σ² + μ², this gives: E[Y²] = a²(σ² + μ²) + 2abμ + b².
This holds for any random variable X with finite mean μ and variance σ², regardless of its distribution.
Exponential distribution, Exp(λ)
Models the waiting time between events in a Poisson process with rate λ.
PDF: f(x) = λe^(-λx) for x ≥ 0
CDF: F(x) = 1 - e^(-λx) for x ≥ 0
Mean: E[X] = 1/λ
Variance: Var(X) = 1/λ²
Support: [0, ∞)
In simple terms, if events happen at an average rate of λ per unit time, the exponential tells you how long you will wait for the next one.
Memoryless property
P(X > s + t | X > s) = P(X > t). The exponential is the only continuous distribution with this property.
Think of it as: the distribution "forgets" how long you have already waited.
Probability density function (PDF)
For a continuous random variable, f(x) gives the density of probability at each point. Probabilities come from integrating the PDF over an interval: P(a < X < b) = ∫ from a to b of f(x) dx.
Two requirements for a valid PDF: f(x) ≥ 0 everywhere, and the total integral over the entire support equals 1.
Cumulative distribution function (CDF)
F(x) = P(X ≤ x) = ∫ from -∞ to x of f(t) dt.
The CDF is non-decreasing, goes from 0 to 1, and its derivative (where it exists) is the PDF.
Median
The value m such that P(X ≤ m) = 0.5, equivalently F(m) = 0.5.
In simple terms, half the distribution is below the median and half is above.
This is a classic exam distinction. Let Y have variance σ².
X = 3Y → Var(X) = 9Var(Y) = 9σ²
Z = Y1 + Y2 + Y3 (independent copies) → Var(Z) = Var(Y1) + Var(Y2) + Var(Y3) = 3σ²
So Var(X) = 9σ² ≠ 3σ² = Var(Z), unless σ² = 0. Since the problem specifies nonzero variance, Var(X) = Var(Z) is false.
The intuition: multiplying one observation magnifies its randomness (variance scales by a²). Summing independent observations lets their randomness partially cancel out (variance scales by n, not n²).
For X with mean μ and variance σ², and Y = aX + b:
E[Y] = aμ + b
Var(Y) = a²σ²
E[Y²] = Var(Y) + (E[Y])² = a²σ² + (aμ + b)² = a²σ² + a²μ² + 2abμ + b² = a²(σ² + μ²) + 2abμ + b²
This identity holds for any distribution with finite mean and variance. It follows from the general fact that E[X²] = Var(X) + (E[X])², which is always true.
The exponential distribution models the time between events, not the count of events.
The Poisson distribution counts how many events occur in a fixed interval.
The exponential distribution measures how long you wait until the next event.
Exam trap: a statement claiming the exponential models "the total number of events in a fixed interval" is false. That describes the Poisson.
For Exp(λ):
Mean = 1/λ
Variance = 1/λ²
Note that the variance equals the square of the mean: Var(X) = (E[X])², which is the same numerical relationship as saying the standard deviation equals the mean. The mean and variance are not the same number (variance = 1/λ², mean = 1/λ), but the standard deviation (1/λ) does equal the mean.
The exponential is supported on [0, ∞). The normal is supported on (-∞, +∞).
For any y ≥ 0 in the support of Exp(λ), there exists an x in the support of N(μ, σ) such that x = y, because the normal's support includes all non-negative real numbers. This makes the statement "for any y in the support of Y, there exists an x in the support of X such that x = y" true when Y ~ Exp(λ) and X ~ N(μ, σ).
When given a piecewise function and told it is a valid PDF, integrate over the entire support and set the result equal to 1, then solve for k.
Worked example from the exam:
f(x) = k(6.25 - (x - 7.5)²) for 5 ≤ x < 7.5, and f(x) = (25/4)k for 7.5 ≤ x < 12.5, and 0 otherwise.
Step 1: Integrate the first piece from 5 to 7.5.
Let u = x - 7.5, so du = dx. When x = 5, u = -2.5; when x = 7.5, u = 0.
∫ from 5 to 7.5 of k(6.25 - (x - 7.5)²) dx = k ∫ from -2.5 to 0 of (6.25 - u²) du
= k [6.25u - u³/3] from -2.5 to 0
= k [0 - (6.25(-2.5) - (-2.5)³/3)]
= k [0 - (-15.625 + 15.625/3)]
= k [0 - (-15.625 + 5.2083)]
= k [0 - (-10.4167)]
= k * 10.4167
= k * 125/12
Step 2: Integrate the second piece from 7.5 to 12.5.
∫ from 7.5 to 12.5 of (25/4)k dx = (25/4)k (12.5 - 7.5) = (25/4)k 5 = (125/4)k
Step 3: Set the total equal to 1.
k 125/12 + k 125/4 = 1
k 125 (1/12 + 1/4) = 1
k 125 (1/12 + 3/12) = 1
k 125 4/12 = 1
k * 125/3 = 1
k = 3/125
Integrate the PDF from the left boundary of the support up to x.
For the taco truck problem, the CDF is:
F(x) = 0 for x < 5
F(x) = ∫ from 5 to x of k(6.25 - (t - 7.5)²) dt for 5 ≤ x < 7.5
F(x) = F(7.5) + ∫ from 7.5 to x of (25/4)k dt for 7.5 ≤ x < 12.5
F(x) = 1 for x ≥ 12.5
The exam provides the CDF:
F(x) = -(1/125)(x - 7.5)³ + (3/20)x - 7/8, for 5 ≤ x < 7.5
F(x) = (3/20)x - 7/8, for 7.5 ≤ x < 12.5
Set F(m) = 0.5 and solve for m.
Worked example: Check which piece of the CDF contains 0.5.
First, evaluate F(7.5) = (3/20)(7.5) - 7/8 = 22.5/20 - 7/8 = 9/8 - 7/8 = 2/8 = 0.25.
Since F(7.5) = 0.25 < 0.5, the median falls in the second piece (7.5 ≤ x < 12.5).
Set (3/20)m - 7/8 = 0.5:
(3/20)m = 0.5 + 7/8 = 4/8 + 7/8 = 11/8
m = (11/8) * (20/3) = 220/24 = 55/6 ≈ 9.1667 minutes.
Variance of a linear transformation: Var(aX + b) = a²Var(X)
Variance of a sum of independent RVs: Var(X1 + X2 + ... + Xn) = Var(X1) + Var(X2) + ... + Var(Xn)
Second moment identity: E[X²] = Var(X) + (E[X])²
Exponential PDF: f(x) = λe^(-λx), x ≥ 0
Exponential CDF: F(x) = 1 - e^(-λx), x ≥ 0
Valid PDF condition: ∫ over all x of f(x) dx = 1, and f(x) ≥ 0 everywhere
Median: Solve F(m) = 0.5
Students often assume Var(3Y) = Var(Y1 + Y2 + Y3) for independent copies Yi. Scaling by 3 gives Var(3Y) = 9Var(Y), while the sum gives 3Var(Y). These are equal only when Var(Y) = 0.
The exponential distribution models waiting times (time between events), not the count of events. The count is Poisson. Mixing these up is a common true/false error.
When finding k for a piecewise PDF, students sometimes forget to integrate over all pieces of the support. Every region where f(x) > 0 must contribute to the integral.
When finding the median, students sometimes plug 0.5 into the wrong piece of the CDF. Always check which piece contains the value 0.5 by evaluating the CDF at the boundary points first.
⚠️ The Var(aX) vs Var(ΣXi) distinction is a classic true/false question. Know the formulas cold.
⚠️ The E[Y²] = a²(σ² + μ²) + 2abμ + b² identity is tested in true/false. Derive it from E[X²] = Var(X) + (E[X])².
⚠️ "The exponential models the number of events" is a false statement that appears regularly.
⚠️ Finding k for a piecewise PDF is worth 18 points on this exam. Show every integration step.
⚠️ The median-from-CDF problem requires identifying the correct piece before solving. Check boundary values.
True or false: Var(5X) = 5Var(X).
Fill in the blank: The exponential distribution models the ______ between events, not the ______ of events.
True or false: E[X²] = Var(X) + (E[X])² holds for any random variable with finite mean and variance.
Fill in the blank: For a valid PDF, the integral over the entire support must equal ______.
True or false: To find the median, solve F(m) = 0.5.
Q: Y has variance 4. Let X = 5Y and let Z = Y1 + Y2 + Y3 + Y4 + Y5 (independent copies). Compare Var(X) and Var(Z).
A: Var(X) = 25 4 = 100. Var(Z) = 5 4 = 20. They are not equal. Var(X) > Var(Z).
Q: X ~ Exp(λ = 2). What are E[X] and Var(X)?
A: E[X] = 1/2 = 0.5. Var(X) = 1/4 = 0.25.
Q: A PDF is given by f(x) = cx² for 0 ≤ x ≤ 3, and 0 otherwise. Find c.
A: ∫ from 0 to 3 of cx² dx = c[x³/3] from 0 to 3 = c(27/3) = 9c = 1. So c = 1/9.
Q: The CDF is F(x) = x²/9 for 0 ≤ x ≤ 3. Find the median.
A: Set F(m) = 0.5: m²/9 = 0.5, m² = 4.5, m = √4.5 ≈ 2.1213.
Q: X is a continuous random variable with E[X] = 3 and Var(X) = 2. If Y = 4X - 1, what is E[Y²]?
A: E[Y] = 4(3) - 1 = 11. Var(Y) = 16(2) = 32. E[Y²] = Var(Y) + (E[Y])² = 32 + 121 = 153. Alternatively: 4²(2 + 9) + 2(4)(-1)(3) + (-1)² = 16(11) - 24 + 1 = 176 - 24 + 1 = 153.
Variance properties connect to everything in the course. Any time you combine, scale, or shift random variables, these rules apply. The distinction between Var(aX) and Var(ΣXi) reappears in sampling distributions and confidence intervals later in STAT 350.
The exponential and Poisson distributions are two sides of the same coin: Poisson counts events in a fixed interval; exponential measures the gap between events. If you understand one, the other follows.
Piecewise PDFs and CDFs are the foundation for working with any non-standard continuous distribution. The same integration techniques appear in expected value calculations (E[X] = ∫ xf(x) dx) and probability computations for later topics.
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