STAT
A Probability Path.pdf

Theorem 10101 krickeberg decomposition if xn bn n 0

Info icon This preview shows pages 398–401. Sign up to view the full content.

Theorem 10.10.1 (Krickeberg Decomposition) If {(Xn. Bn). n :::: 0} is a sub- martingale such that supE(X:) < oo , n then there exists a positive martingale {(Mn, Bn), n :::: 0} and a positive super- martingale {(Yn, Bn), n 2:: 0} and Xn = Mn- Yn. Proof. If {Xn} is a submartingale, then also {X:} is a submartingale. (See Exam- ple 10 .6.1.) Additionally, {E(XtiBn), p :::: n} is monotone non-decreasing in p. To check this, note that by smoothing, E(X;+ 1 ll3n) = E(E(X;+1 l l3p)ll3n) 2:: E(XtlBn) where the last inequality follows from the submartingale property . Monotonicity in p implies exists. We claim that {(Mn, Bn), n 2:: 0} is a positive martingale. To see this, observe that (a) Mn E Bn, and Mn 2:: 0. (b) The expectation of Mn is finite and constant inn since E(Mn) = E( lim t E(X;ll3n)) p-+00 = lim t £(£(X;Il3n)) p-+00 = lim t Ex+ p-+00 p =sup Ex;< oo, p?:O (monotone convergence) since expectations of submartingales increase. Thus E(Mn) < oo.
Image of page 398

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

10.10 Martingale and Submartingale Convergence 387 (c) The martingale property holds since E(Mn+IIBn) = E{ lim t E(XtiBn+I)iBn) p-+00 = lim t E(E<XtiBn+I)iBn) (monotone convergence) p-+00 = lim t E(XtiBn) = Mn· p-+00 (smoothing) We now show that is a positive supermartingale. Obviously, Yn E Bn. Why is Yn 2: 0? Since Mn = limp-+oo t E(XtiBn). if we take p = n, we get Mn ::: E(X:IBn) = x: ::: x: - x; = Xn . To verify the supermartingale property note that E(Yn+IIBn) = E(Mn+IIBn)- E(Xn+tiBn) :5 Mn- Xn = Yn 0 10.10.2 Doob's (Sub)martingale Convergence Theorem K.rickeberg's decomposition leads to the Doob submartingale convergence theo- rem. Theorem 10.10.2 (Submartingale Convergence) lf{(Xn. Bn). n ::: 0} is a (sub)- martingale satisfying supE(X:) < oo, neN then there exists X 00 e L 1 such that X a.s.x n oo· Remark. If {Xn} is a martingale supE(X:) < oo iff supE(IXnD < oo neN neN in which case the martingale is called L 1-bounded. To see this equivalence, ob- serve that if {(Xn, Bn). n eN} is a martingale then E<IXnD = E(X:) + E(X;) = 2E(X:)- E(Xn) = x: - const.
Image of page 399
388 10. Martingales Proof. From the Krickberg decomposition, there exist a positive martingale {Mn} and a positive supermartingale {Yn} such that From Theorem 10 .8. 5, the following are true: M a.s.M n 00• y; a.s . y; n oo so and M 00 and Y 00 are integrable. Hence M 00 and Y 00 are finite almost surely, X 00 =Moo- Y 00 exists, and Xn X 00 0 10.11 Regularity and Closure We begin this section with two reminders and a recalled fact. Reminder 1. (See Subsection 10.9.2.) Let {Zn} be a simple branching process with P( extinction)= 1 =: q. Then with Zo = 1, E(Z 1) = m So the martingale { Wn} satisfies E(Wn) = 1 £(0) = 0 so { Wn} does NOT converge in L Also, there does NOT exist a random variable W 00 such that Wn = E(W 00 1Bn) and {Wn} is NOT uniformly integrable (ui). Reminder 2. Recall the definition of uniform integrability and its character- izations from Subsection 6.5.1 of Chapter 6. A family of random variables (X, , t e I} is ui if X, e L 1 for all t e I and lim sup] IX 1 IdP = 0. b-+OO IE/ IXri>b Review Subsection 6.5.1 for full discussion and characterizations and also re- view Theorem 6.6.1 on page 191 for the following FACT: If {Xn} converges a. s.
Image of page 400

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

Image of page 401
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern