Questions: Expectation (Measure-Theoretic)

5 questions to test your understanding

Score: 0 / 5
Question 1 Multiple Choice

Let Xₙ = n · 𝟏{0 < U < 1/n} where U ~ Uniform(0,1). Then Xₙ → 0 almost surely, yet E[Xₙ] = 1 for all n. Which theorem fails to apply here, and why?

AThe Monotone Convergence Theorem — it fails because the sequence is not monotone increasing
BThe Dominated Convergence Theorem — it fails because there is no integrable function g with |Xₙ| ≤ g for all n
CThe Law of Large Numbers — it fails because the Xₙ are not identically distributed
DBoth MCT and DCT — they fail because Xₙ does not converge in L¹
Question 2 Multiple Choice

What is the key advantage of defining expectation as E[X] = ∫_Ω X dP (a Lebesgue integral on the probability space) over the elementary definitions E[X] = Σ xᵢP(X = xᵢ) or E[X] = ∫ x f(x) dx?

AThe measure-theoretic definition is easier to compute numerically for most practical distributions
BIt provides a single unified framework that handles discrete, continuous, and mixed distributions — and distributions with no density — under one definition
CIt automatically guarantees that every random variable has a finite expectation
DIt eliminates the need to check measurability conditions for the random variable
Question 3 True / False

If Xₙ ≥ 0 for all n and Xₙ increases pointwise to X (possibly with X = ∞ on some set), the Monotone Convergence Theorem guarantees that E[Xₙ] → E[X], even when E[X] = ∞.

TTrue
FFalse
Question 4 True / False

Nearly every random variable defined on a probability space has a well-defined finite expectation, since probabilities are bounded between 0 and 1.

TTrue
FFalse
Question 5 Short Answer

Why must we verify that E[|X|] < ∞ (absolute integrability) rather than just checking that ∫_Ω X dP converges, before concluding that E[X] is well-defined?

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