Questions: Martingale Convergence Theorems

Question 1

Doob's forward convergence theorem states that if {M_n} is a martingale with sup_n E[|M_n|] < ∞, then:

• A. M_n converges in L¹ to a limit M_∞
• B. M_n converges almost surely to a finite limit M_∞, but M_∞ may not satisfy E[M_∞] = E[M_0]
• C. M_n converges in probability but not almost surely
• D. M_n converges in distribution to a normal random variable

Answer: (B)

Doob's theorem guarantees almost sure convergence under the L¹-boundedness condition sup_n E[|M_n|] < ∞. However, the limit M_∞ may not preserve the martingale property — we might have E[M_∞] < E[M_0] due to 'mass escaping to infinity.' The classic counterexample is a simple random walk stopped at 0: it converges a.s. to 0, but starts at a positive value. L¹ convergence (which would give E[M_∞] = E[M_0]) requires the stronger condition of uniform integrability.

Question 2

A uniformly integrable martingale {M_n} converges in L¹ to M_∞, and M_n = E[M_∞ | ℱ_n] for all n.

• T. True
• F. False

Answer: True

Uniform integrability is the exact condition that upgrades a.s. convergence to L¹ convergence and preserves the martingale property in the limit. If {M_n} is uniformly integrable, then M_n → M_∞ in L¹, and the conditional expectation identity M_n = E[M_∞ | ℱ_n] holds for every n. Conversely, any martingale of the form M_n = E[X | ℱ_n] for some integrable X is automatically uniformly integrable. This equivalence — UI martingales are exactly those 'closed' by a terminal random variable — is one of the most useful characterizations in probability.

Question 3

Give an example of a martingale that converges almost surely but NOT in L¹, and explain why uniform integrability fails.

Model answer: Let M_n = 2^n · 1_{A_n} where A_n is an event with P(A_n) = 2^{-n} (e.g., a fair coin shows heads n times in a row). Then E[M_n] = 1 for all n, so {M_n} is L¹-bounded. M_n → 0 a.s. since P(A_n) → 0, but E[M_∞] = 0 ≠ 1 = E[M_0], so L¹ convergence fails. Uniform integrability fails because M_n concentrates large mass on a set of vanishing probability — sup_n E[|M_n| · 1_{|M_n|>K}] does not go to 0 as K → ∞.

This is the prototypical 'mass escaping to infinity' example. The martingale property E[M_n] = 1 holds for each n, but the mass concentrates on increasingly improbable events with increasingly large values. The a.s. limit is 0, losing the mass entirely. Uniform integrability is precisely the condition that prevents this: it requires the tails of the M_n to be uniformly controlled.

Question 4

Which condition is sufficient for both a.s. and L² convergence of a martingale {M_n}?

• A. E[M_n] = 0 for all n
• B. sup_n E[M_n²] < ∞ (the martingale is L²-bounded)
• C. The increments M_{n+1} - M_n are bounded
• D. The filtration ℱ_n is generated by i.i.d. random variables

Answer: (B)

The L² martingale convergence theorem: if sup_n E[M_n²] < ∞, then M_n converges a.s. and in L² to a limit M_∞ with E[M_∞²] ≤ sup_n E[M_n²]. L²-boundedness implies L¹-boundedness (by Jensen), so Doob's theorem gives a.s. convergence. L²-boundedness also implies uniform integrability (since L² domination gives uniform integrability), upgrading to L¹ convergence. The L² convergence follows from the orthogonality of martingale increments: Var(M_n) = Σ_{k=1}^n E[(M_k - M_{k-1})²], and the bound sup E[M_n²] < ∞ forces this series to converge, giving L² Cauchy convergence.

Question 5

Doob's upcrossing inequality is the key technical tool behind the forward convergence theorem. What does it bound?

Model answer: The upcrossing inequality bounds the expected number of times a submartingale {X_n} crosses upward through an interval [a,b]. If U_N([a,b]) counts the number of complete upcrossings of [a,b] by X_0, ..., X_N, then E[U_N([a,b])] ≤ E[(X_N - a)⁺] / (b - a). Since the right side is bounded when sup E[|X_n|] < ∞, the number of upcrossings is a.s. finite for every interval [a,b], which forces convergence.

The upcrossing inequality converts the analytic condition sup E[|M_n|] < ∞ into a pathwise geometric statement: the process cannot oscillate infinitely often between any two levels a < b. If M_n failed to converge on a set of positive probability, there would exist a < b and a positive-probability set on which M_n crosses [a,b] infinitely often — contradicting the finite expected upcrossing count. This is one of the most elegant proofs in probability theory.