Random variables X and Y satisfy E[f(X) | σ(Y)] = E[f(X)] for every bounded measurable function f. What can you conclude?
AX and Y have the same distribution
BX and Y are uncorrelated but may not be independent
CX and Y are independent
DThe sigma-algebras σ(X) and σ(Y) overlap but their events are uncorrelated
This is the measure-theoretic characterization of independence: X and Y are independent if and only if conditioning on σ(Y) does not change the expectation of any function of X. This links independence to information: 'σ(Y) gives no information about X.' Option B (uncorrelated) is strictly weaker — uncorrelated variables can still be dependent (e.g., X uniform on [-1,1] and Y = X²).
Question 2 Multiple Choice
A researcher wants to define independence for a mixed pair (X continuous, Y discrete). Which approach works without case-splitting?
ARequire the joint CDF to factor: F_{X,Y}(x,y) = F_X(x)F_Y(y)
BRequire the joint PDF to factor, treating Y's distribution as a limiting case
CRequire P(A ∩ B) = P(A)P(B) for all A ∈ σ(X), B ∈ σ(Y)
DRequire E[XY] = E[X]E[Y], which is necessary and sufficient for all distribution types
The sigma-algebra definition P(A ∩ B) = P(A)P(B) for all measurable events A ∈ σ(X), B ∈ σ(Y) works universally — for discrete, continuous, singular, and mixed types — because it makes no assumptions about the form of the distribution. Option A covers only CDF-level events, not all sigma-algebra events. Option D (E[XY] = E[X]E[Y]) is merely uncorrelation, which is necessary but not sufficient for independence.
Question 3 True / False
If σ(X) and σ(Y) are independent sigma-algebras, then knowing any event about Y gives no probabilistic information about any event about X.
TTrue
FFalse
Answer: True
This is the informational meaning of sigma-algebra independence. The formal condition P(A ∩ B) = P(A)P(B) for all A ∈ σ(X), B ∈ σ(Y) means conditional on any event in σ(Y), the probability of any event in σ(X) is unchanged. Equivalently, E[1_A | σ(Y)] = P(A) for all A ∈ σ(X) — no observation about Y updates probabilities of events about X.
Question 4 True / False
Two random variables X and Y are independent if and mainly if their joint probability distribution factors as the product of their marginals, regardless of distribution type.
TTrue
FFalse
Answer: False
This factoring statement is correct for continuous (joint PDF = product of marginals) and discrete (joint PMF = product of marginals) distributions, but 'joint distribution factors' has no clean formulation for singular or mixed-type distributions without the sigma-algebra framework. The sigma-algebra definition P(A ∩ B) = P(A)P(B) for all A ∈ σ(X), B ∈ σ(Y) is universally correct; the density-factoring approach only works in specific cases.
Question 5 Short Answer
Why is the sigma-algebra definition of independence more powerful than simply requiring P(A ∩ B) = P(A)P(B) for two specific events A and B?
Think about your answer, then reveal below.
Model answer: The elementary definition covers only two specific events, while sigma-algebra independence requires the product rule to hold for *all* pairs of events drawn from both sigma-algebras. A sigma-algebra σ(X) contains every observable event about X — {X > 3}, {X ∈ [1,2]}, any Borel event. Independence of σ(X) and σ(Y) means no question about Y helps predict any question about X. Two variables can have some independent event pairs while being statistically dependent overall.
Pairwise independence of specific events is strictly weaker than sigma-algebra independence. The sigma-algebra definition also works across all distribution types and connects directly to conditional expectation: independence is equivalent to E[f(X)|σ(Y)] = E[f(X)], which is the right framing for the law of large numbers and martingale theory.