# 9 shows that we can have x0 1 and x1 0 the key to the

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Unformatted text preview: &gt; n) (q/p)a Px (⌧ &gt; n) Letting n ! 1 we have established (5.11). We can save ourselves some work by abstracting the last argument. Theorem 5.14. Suppose Mn is a martingale and T a stopping time with P (T &lt; 1) = 1 and |MT ^n | K for some constant K . Then EMT = EM0 . Proof. Theorem 5.13 implies EM0 = EMT ^n = E (MT ; T n) + E (Mn ; T &gt; n). The second term KP (T &gt; n) and |E (MT ; T n) E (MT )| KP (T &gt; n) Since P (T &gt; n) ! 0 as n ! 1 the desired result follows. Example 5.11. Duration of fair games. Let Sn = S0 + X1 + · · · + Xn where X1 , X2 , . . . are independent with P (Xi = 1) = P (Xi = 1) = 1/2. Let ⌧ = min{n : Sn 62 (a, b)} where a &lt; 0 &lt; b. Our goal here is to prove a close relative of (1.26): E0 ⌧ = ab 2 Example 5.4 implies that Sn n is a martingale. Let ⌧ = min{n : Sn 62 (a, b)}. From the previous example we have that ⌧ is a stopping time with P (⌧ &lt; 1) = 2 1. Again if we argue casually 0 = E0 (S⌧ ⌧ ) so using (5.10) 2 E0 (⌧ ) = E0 (S⌧ ) = a2 P0 (S⌧ = a) + b2 P0 (S⌧ = b) b a ab = a2 + b2 = ab...
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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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