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Unformatted text preview: 1) = p and P (Xi = 1) = 1 p where p 2 (0, 1) and p 6= 1/2. Let Sn = S0 + X1 + · · · + Xn . Mn =
a martingale with respect to Xn . ⇣ 1p
p ⌘Sn is 163 5.2. EXAMPLES, BASIC PROPERTIES Proof. Using Theorem 5.5 with h(x) = ((1 p)/p)x , we need only check that
P
h(x) = y p(x, y )h(y ). To do this we note that
X
y p(x, y )h(y ) = p ·
= (1 ✓ 1 p p
p) · which proves the desired result. ◆x+1 ✓ 1 p
p + (1
◆x p) · +p· ✓ ✓ 1 1
p p ◆x p
◆x
p = 1 ✓ 1 p
p ◆x Example 5.4. Symmetric simple random walk. Let Y1 , Y2 , . . . be independent with
P (Yi = 1) = P (Yi = 1) = 1/2
2
and let Xn = X0 + Y1 + · · · + Yn . Then Mn = Xn n is a martingale with
2
n it is enough to show that
respect to Xn . By Theorem 5.5 with f (x, n) = x 1
1
(x + 1)2 + (x
2
2 1)2 1 = x2 To do this we work out the squares to conclude the lefthand side is
12
[x + 2x + 1 + x2
2 2x + 1] 1 = x2 Example 5.5. Products of independent random variables. To build a
discrete time model of the stock market we let X1 , X2 , . . . be independent 0
with EXi = 1. Then Mn = M0 X1 · · · Xn is a martingale with respect to Xn .
To prove this we note...
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This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell University (Engineering School).
 Spring '10
 DURRETT
 The Land

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