6 k1 proof expanding out the square of the sum e n x

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Unformatted text preview: . E.g., in Example 5.4 Mn = Xn n. To explain the reason for our interest in martingales, we will now give a number of examples. In what follows we will often be forced to write the conditioning event, so we introduce the short hand Av = {Xn = xn , Xn 1 = xn 1 , . . . , X0 where v is short for the vector (xn , . . . , x0 , m0 ) = x0 , M0 = m0 } (5.5) 162 CHAPTER 5. MARTINGALES Example 5.2. Random walks. Let X1 , X2 , . . . be i.i.d. with EXi = µ. Let Sn = S0 + X1 + · · · + Xn be a random walk. Mn = Sn nµ is a martingale with respect to Xn . Proof. To check this, note that Mn+1 Mn = Xn+1 µ is independent of Xn , . . . , X0 , M0 , so the conditional mean of the di↵erence is just the mean: E (Mn+1 Mn |Av ) = EXn+1 µ=0 In most cases, casino games are not fair but biased against the player. We say that Mn is a supermartingale with respect to Xn if a gambler’s expected winnings on one play are negative: E (Mn+1 Mn |Av ) 0 To help remember the direction of the inequality, note that there is nothing “super” about a supermartingal...
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This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell University (Engineering School).

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