Stochastic

# Then n x wn w0 hm mm mm 1 is a supermartingale m1 we

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Unformatted text preview: ent, the submartingale property of Mn and the fact that is nondecreasing imply that the = for martingales is now . 164 CHAPTER 5. MARTINGALES 2 Since x2 is convex this implies that if Mn is a martingale then Mn is a submartingale. The next result gives another proof of this, and provides a useful formula that is the analogue of E (Y 2 ) (EY )2 = var (Y ) for martingales. Lemma 5.7. If Mn is a martingale then 2 E (Mn+1 |Av ) 2 Mn = E ((Mn+1 Mn )2 |Av ) Proof. Expanding out the square on the right-hand side, then using (5.3) and Lemma 5.3 gives 2 E (Mn+1 2 2Mn+1 Mn + Mn |Av ) 2 = E (Mn+1 |Av ) 2 2Mn E (Mn+1 |Av ) + Mn 2 = E (Mn+1 |Av ) 2 Mn since E (Mn+1 |Av ) = Mn . Using ideas from the last proof we get Lemma 5.8. Othogonality of martingale increments. If Mn is a martingale and 0 i j k < n then E [(Mn and E [(Mn Mk )(Mj Mk )Mj ] = 0 Mi )] = 0. Proof. The second result follows by subtracting the result for j = i from the one for j . Let Av = {Xk = xk , . . . , X0 = x0 , M0 = m}. Using Lemma 5.4 then Lemma 5.1 and the martingale property E [(Mn Mk )Mj ] = X x = M...
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## This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell.

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