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 righthand 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.
 Spring '10
 DURRETT
 The Land

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