Questions: Relationships Between Modes of Convergence

Question 1

A sequence of random variables satisfies Xₙ → X in distribution. Under which additional condition does this imply convergence in probability?

• A. When the Xₙ are mutually independent
• B. When E[|Xₙ|] is uniformly bounded
• C. When X is a constant (the limit distribution is degenerate)
• D. Convergence in distribution always implies convergence in probability

Answer: (C)

Convergence in distribution generally does NOT imply convergence in probability. Canonical counterexample: let Xₙ = X for all n and Y be an independent copy of X ~ N(0,1). Then Xₙ → Y in distribution but P(|Xₙ − Y| > ε) = P(|X − Y| > ε) > 0 for all n — no convergence in probability. The exception is when the limit is a constant c: P(|Xₙ − c| > ε) = Fₙ(c−ε) + (1−Fₙ(c+ε)) → 0 + 0 = 0 by convergence of CDFs at continuity points.

Question 2

You know that E[|Xₙ − X|²] → 0 (convergence in L²). Which conclusions are guaranteed?

• A. Xₙ → X almost surely
• B. Xₙ → X in probability and in L¹
• C. Xₙ → X almost surely and in distribution
• D. Only convergence in distribution is guaranteed

Answer: (B)

L² convergence implies: (1) convergence in probability, via Markov's inequality P(|Xₙ−X| > ε) ≤ E[|Xₙ−X|²]/ε² → 0; and (2) L¹ convergence, since Lᵖ ⟹ Lq for p > q by Hölder's inequality. But L² convergence does NOT imply almost sure convergence — the typewriter sequence converges to 0 in all Lᵖ but fails to converge a.s. at any point.

Question 3

Almost sure convergence implies convergence in probability, but convergence in probability does not imply almost sure convergence.

• T. True
• F. False

Answer: True

The implication a.s. ⟹ in probability is standard: if Xₙ → X on a probability-1 set of paths, then P(|Xₙ−X| > ε) → 0. The reverse fails: the 'typewriter sequence' on [0,1] — indicators of sliding intervals whose lengths shrink to zero — converges to 0 in probability but not a.s., since every ω is hit by infinitely many indicators and thus Xₙ(ω) oscillates between 0 and 1 for every ω.

Question 4

If Xₙ → X almost surely, then Xₙ → X in L¹.

• T. True
• F. False

Answer: False

Almost sure convergence and L¹ convergence are incomparable — neither implies the other in general. Counterexample for a.s. ⟹ L¹ failing: Xₙ = n·𝟏_{[0,1/n]} on [0,1]. Then Xₙ(ω) → 0 for all ω > 0 (a.s. convergence), but E[|Xₙ|] = n·(1/n) = 1 for all n — no L¹ convergence. The bridge from a.s. to L¹ requires the additional condition of uniform integrability (Vitali's theorem).

Question 5

Explain why convergence in distribution is strictly weaker than convergence in probability, and what conceptual difference accounts for this.

Model answer: Convergence in distribution (Xₙ →_d X) only requires that CDFs converge: Fₙ(t) → F(t) at continuity points of F. This is a statement about probability laws — not about how Xₙ and X are related as functions on a common probability space. Xₙ and X don't even need to be defined on the same space. Convergence in probability (Xₙ →_P X) requires both to live on the same space with P(|Xₙ−X| > ε) → 0. Two sequences can converge to the same distributional limit while the random variables themselves remain far apart in probability — as illustrated by independent copies of the same distribution.

The practical consequence: the CLT gives convergence in distribution. Strengthening to almost sure convergence (as in the strong LLN) requires different techniques, and the distinction matters when you need pathwise arguments rather than distributional ones.