Quantifying uncertain outcomes and their distributions
X: Represents the random variable
X = x: The event that random variable X takes the value x
P(X = x): Probability that X equals x
E(X): Expected value (mean) of X
Var(X): Variance of X
A probability distribution is a mathematical function that provides the probabilities of occurrence of different possible outcomes for a random variable. For a discrete random variable, it lists all possible values and their corresponding probabilities.
A probability distribution is often presented as a table:
| X | x₁ | x₂ | x₃ | ... | xₙ |
|---|---|---|---|---|---|
| P(X) | p₁ | p₂ | p₃ | ... | pₙ |
Problem: Two coins are tossed. Let X = number of heads. Construct the probability distribution.
Solution:
Sample Space: {HH, HT, TH, TT}
X can be 0, 1, or 2
P(X = 0) = P(TT) = 1/4
P(X = 1) = P(HT, TH) = 2/4 = 1/2
P(X = 2) = P(HH) = 1/4
| X | 0 | 1 | 2 |
|---|---|---|---|
| P(X) | 1/4 | 1/2 | 1/4 |
Problem: A die is rolled. Let X = outcome. Is this a valid probability distribution?
Solution:
X ∈ {1, 2, 3, 4, 5, 6}
Each outcome: P(X = i) = 1/6
Sum: 6 × (1/6) = 1 ✓
Each probability: 0 ≤ 1/6 ≤ 1 ✓
Yes, this is a valid probability distribution (uniform distribution)
Problem: For the probability distribution:
| X | 0 | 1 | 2 |
|---|---|---|---|
| P(X) | 1/4 | 1/2 | 1/4 |
Solution:
E(X) = 0 × (1/4) + 1 × (1/2) + 2 × (1/4) = 0 + 1/2 + 1/2 = 1
E(X²) = 0² × (1/4) + 1² × (1/2) + 2² × (1/4) = 0 + 1/2 + 1 = 3/2
Var(X) = E(X²) - [E(X)]² = 3/2 - 1² = 3/2 - 1 = 1/2
σ = √(1/2) ≈ 0.707
Problem: A fair die is rolled. Find E(X) and Var(X).
Solution:
E(X) = 1(1/6) + 2(1/6) + 3(1/6) + 4(1/6) + 5(1/6) + 6(1/6)
E(X) = (1+2+3+4+5+6)/6 = 21/6 = 3.5
E(X²) = 1²(1/6) + 2²(1/6) + ... + 6²(1/6) = 91/6
Var(X) = 91/6 - (3.5)² = 91/6 - 12.25 ≈ 2.917
Q: Define random variables and probability distribution
Q: Construct probability distribution from sample space
Q: Calculate E(X) and Var(X) from distributions
Common Mistake: Forgetting to verify that all probabilities sum to 1
Helpful Tip: Use the shortcut formula Var(X) = E(X²) - [E(X)]² for easier calculation
Key Insight: Understanding mean and variance of distributions is crucial for statistics
Common Pattern: Questions often ask for E(X), E(X²), Var(X) sequentially
Speed Technique: Memorize linearity properties for quick calculations
Integration: This topic connects to Chapter 14 (Mathematical Expectation) in some curricula