MathIsimple

Stochastic Processes – Problem 6: Prove that the sequence of random variables converges in the mean-square sense

Question

Let {Yn,Fn,n1}\{Y_{n},\mathcal{F}_{n},n\geq1\} be a martingale with EYn2K,n1E Y_{n}^{2}\leq K,n\geq1, where KK is a constant. Prove that the sequence of random variables {Yn}\{Y_{n}\} converges in the mean-square sense.

Step-by-step solution

Step 1. Given that {Yn,Fn}\{Y_n, \mathcal{F}_n\} is a martingale with supnE[Yn2]K<\sup_n \mathbb{E}[Y_n^2] \le K < \infty. By Doob's martingale convergence theorem, there exists a random variable YL1Y_\infty \in L^1 such that YnYY_n \to Y_\infty almost surely and in L1L^1. However, here we need to verify mean-square convergence (L2L^2 convergence), which requires showing that an L2L^2-bounded martingale also converges in L2L^2.

Step 2. For an L2L^2-bounded martingale, {Yn}\{Y_n\} is a uniformly bounded set in L2L^2 and is a martingale in L2L^2: E[YnYm]=E[Ym2],mn\mathbb{E}[Y_n Y_m] = \mathbb{E}[Y_m^2], \quad m \le n (since by the martingale property E[YnFm]=Ym\mathbb{E}[Y_n \mid \mathcal{F}_m] = Y_m, and taking square-integrability into account the covariance is as above). More directly: for mnm \le n, E[(YnYm)2]=E[Yn2]E[Ym2].\mathbb{E}[(Y_n - Y_m)^2] = \mathbb{E}[Y_n^2] - \mathbb{E}[Y_m^2]. Since Ym=E[YnFm]Y_m = \mathbb{E}[Y_n \mid \mathcal{F}_m], by conditional Jensen's inequality (or orthogonality): E[Yn2]=E[Ym2]+E[(YnYm)2]E[Ym2].\mathbb{E}[Y_n^2] = \mathbb{E}[Y_m^2] + \mathbb{E}[(Y_n - Y_m)^2] \ge \mathbb{E}[Y_m^2]. Therefore E[Yn2]\mathbb{E}[Y_n^2] is nondecreasing in nn and bounded above by KK, hence converges to some finite limit LL.

Step 3. For m<nm < n, E[(YnYm)2]=E[Yn2]E[Ym2]0as m,n\mathbb{E}[(Y_n - Y_m)^2] = \mathbb{E}[Y_n^2] - \mathbb{E}[Y_m^2] \to 0 \quad \text{as } m,n \to \infty since E[Yn2]\mathbb{E}[Y_n^2] converges. Therefore {Yn}\{Y_n\} is a Cauchy sequence in L2L^2.

Step 4. Since L2(Ω,F,P)L^2(\Omega, \mathcal{F}, P) is complete, there exists YL2Y \in L^2 such that E[(YnY)2]0.\mathbb{E}[(Y_n - Y)^2] \to 0. Moreover, this YY coincides with the almost sure limit YY_\infty (since mean-square convergence implies convergence in probability, and the almost sure limit is unique). Therefore YnY_n converges in mean square to YY.

Final answer

QED.

Marking scheme

This rubric is based on the official solution and is designed to assess the student's mastery of the proof of L2L^2 convergence (mean-square convergence) of martingales.

1. Checkpoints (Total: 7 pts)

Score exactly one chain (Chain A or Chain B). If a student mixes methods, grade the most complete chain.

Chain A: Based on Cauchy Sequence [Official Solution Path]

  • Key identity and orthogonality (3 pts): Derive or use the martingale property to prove the L2L^2 orthogonality relation: for mnm \le n, prove E[YnYm]=E[Ym2]E[Y_n Y_m] = E[Y_m^2] OR directly derive E[(YnYm)2]=E[Yn2]E[Ym2]E[(Y_n - Y_m)^2] = E[Y_n^2] - E[Y_m^2]. (Note: If using the orthogonal increment method Yn=DkY_n = \sum D_k and noting E[DiDj]=0E[D_i D_j]=0 for iji \ne j, thereby obtaining E[Yn2]=E[Dk2]E[Y_n^2] = \sum E[D_k^2], this item receives full marks.) [additive]
  • Monotonicity and convergence of the norm (2 pts): State that E[Yn2]E[Y_n^2] is a nondecreasing sequence in nn (or that Yn2Y_n^2 is a submartingale), combined with the upper bound KK from the hypothesis, conclude that the sequence {E[Yn2]}\{E[Y_n^2]\} converges. [additive]
  • Cauchy sequence determination (1 pt): Combine the conclusions from the previous two steps to state that limn,mE[(YnYm)2]=0\lim_{n,m \to \infty} E[(Y_n - Y_m)^2] = 0, thereby asserting that {Yn}\{Y_n\} is a Cauchy sequence in L2L^2. [additive]
  • Completeness and conclusion (1 pt): Invoke the completeness of L2L^2 (or Hilbert space property) to conclude that the Cauchy sequence must converge to some random variable YY. (Note: If the student only writes "since it is a Cauchy sequence it converges," this implicitly uses completeness and should receive credit.) [additive]

Chain B: Based on Maximal Inequality and Dominated Convergence (Doob's Maximal Inequality & DCT)

  • Almost sure convergence (1 pt): Cite Doob's Martingale Convergence Theorem to state that YnYY_n \to Y_\infty almost surely. [additive]
  • Maximal inequality and integrable domination (3 pts): Use Doob's L2L^2 maximal inequality to prove supnYnL2\sup_{n} |Y_n| \in L^2 (i.e., E[(supYn)2]4supE[Yn2]4KE[(\sup |Y_n|)^2] \le 4 \sup E[Y_n^2] \le 4K). [additive]
  • Application of the Dominated Convergence Theorem (3 pts): Use the above integrable dominating function (supYn)2(\sup |Y_n|)^2 and apply the Dominated Convergence Theorem to prove E[(YnY)2]0E[(Y_n - Y_\infty)^2] \to 0. [additive]

2. Zero-credit items

  • Only copying the problem conditions: e.g., only writing that {Yn}\{Y_n\} is a martingale or EYn2KE Y_n^2 \le K, with no subsequent derivation.
  • Circular reasoning: Assuming mean-square convergence and then deriving properties from it.
  • Incorrect logic: Asserting convergence solely from "boundedness" (e.g., Yn|Y_n| bounded     \implies convergence), ignoring that this does not hold in infinite-dimensional spaces (Cauchy property or monotonicity is needed).
  • Only citing the theorem name: For a proof problem, if only writing "by Doob's L2L^2 convergence theorem the conclusion holds" with no derivation, this is treated as an incomplete proof (unless the exam explicitly allows direct citation of this theorem; otherwise typically receives 0 or very low marks).

3. Deductions

  • Logic gap (Max -2): In Chain A, if the specific expression relating E[(YnYm)2]E[(Y_n - Y_m)^2] to E[Yn2]E[Y_n^2] is not established, and convergence of the norm is directly asserted to imply sequence convergence (confusing convergence of real sequences with convergence of random variable sequences).
  • Notational confusion (Max -1): Confusing E[Yn2]E[Y_n^2] as a random variable with a constant, or failing to specify whether the limit symbol lim\lim is in the L2L^2 sense or the real number sense, causing logical ambiguity.
  • Limit object undefined (No penalty): If the Cauchy sequence property is proved but the convergence target "YY" is not explicitly written, as long as completeness is implicitly used, no deduction.
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