Proof we begin by computing e expxi to do this we

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Unformatted text preview: Theorems 5.13 and 5.11 to conclude EM0 = EMT ^n then we let n ! 1 and argue that EMT ^n ! EMT . Example 5.10. Gambler’s ruin. Let X1 , X2 , . . . Xn be independent with P (Xi = 1) = p and P (Xi = 1) = q = 1 p Suppose 1/2 < p < 1 and let h(x) = (q/p)x . Example 5.3 implies that Mn = h(Sn ) is a martingale. Let ⌧ = min{n : Sn 62 (a, b)}. It is easy to see that ⌧ is a stopping time. Lemma 1.3 implies that P (⌧ < 1) = 1. Again if we argue casually then (q/p)x = Ex (q/x)S (⌧ ) = (q/p)a P (S⌧ = a) + (q/p)b [1 P (S⌧ = a)] (5.11) Solving gives Px (S⌧ = a) = (q/p)b (q/p)b (q/p)x (q/p)a (5.12) generalizing (1.22). To provide a proof for (5.11), we use Theorems 5.13 and 5.11, to conclude that (q/p)x = Ex M⌧ ^n = (q/p)a P (⌧ n, S⌧ = a) + (q/p)b P (⌧ n, S⌧ = b) + E ((q/p)Sn ; ⌧ > n) 169 5.4. APPLICATIONS Since P (⌧ < 1) = 1 we have P (⌧ n, S⌧ = c) ! P (S⌧ = c) for c = a, b. To handle the third term, we note that since p > 1/2, (q/p)x (q/p)a for a < x < b and hence E ((q/p)Sn ; ⌧...
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