Noting that zn is the sum of zn 1 independent random

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Unformatted text preview: ✓µ) If we pick ✓ = 2µ/ 2 2 ✓)2 ◆ dx ✓ and variance (5.15) , then ✓ /2 + ✓µ = 2µ2 / 22 So Example 5.6 implies exp( 2µSn / 0}. Theorems 5.13 and 5.11 gives exp( 2µS0 / 2 2 2 2µ2 / 2 =0 ) is a martingale. Let T = min{n : Sn ) = E exp( 2µST ^n ) P (T n) since exp( 2µST / 2 ) 1 and the contribution to the expected value from {T > n} is 0. Letting n ! 1 now and noticing P (T n) ! P (B ) gives the desired result. 172 5.5 CHAPTER 5. MARTINGALES Convergence This section is devoted to the proof of the following remarkable result and some of its applications. Theorem 5.17. If Xn and EX1 EX0 . 0 is a supermartingale then X1 = limn!1 Xn exists The bad martingale in Example 5.9 shows that we can have X0 = 1 and X1 = 0. The key to the proof of this is the following maximal inequality. Lemma 5.18. Let Xn 0 be a supermartingale and ✓ ◆ P max Xn > EX0 / > 0. n0 Proof. Let T = min{n 0 : Xn > }. Theorem 5.13 implies that EX0 E (XT ^n ) P (T n) i.e., P (T n) EX0 / . Since this holds for all n the desired r...
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