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Unformatted text preview: ) = 1. To calculate h(x)
for 0 < x < N , we set q = 1 p to simplify the formulas, and consider what
happens on the ﬁrst step to arrive at
h(x) = ph(x + 1) + qh(x 1) (1.19) To solve this we rearrange to get p(h(x + 1) h(x)) = q (h(x)
conclude
q
h(x + 1) h(x) = · (h(x) h(x 1))
p
If we set c = h(1) h(0) then (1.20) implies that for x
✓ ◆x 1
q
h(x) h(x 1) = c
p h(x 1)) and
(1.20) 1 Summing from x = 1 to N , we have
1 = h(N ) h(0) = N
X h(x) h(x x=1 1) = c N
X ✓ q ◆x x=1 1 p Now for ✓ 6= 1 the partial sum of the geometric series is
N1
X
j =0 ✓j = 1 ✓N
1✓ (1.21) 48 CHAPTER 1. MARKOV CHAINS To check this note that
(1 ✓)(1 + ✓ + · · · ✓N 1 ) = (1 + ✓ + · · · ✓N ) (✓ + ✓ + · · · ✓ ) = 1
2 Using (1.21) we see that c = (1 ✓)/(1
using the fact that h(0) = 0, we have
h(x) = h(x) 1 h(0) = c x1
X
i=0 N ✓N ✓N ) with ✓ = q/p. Summing and ✓i = c · 1
1 ✓x
=
1✓
1 ✓x
✓N Recalling the deﬁnition of h(x) and rearranging the fraction we have
Px (VN < V0 ) = ✓x
✓N 1
1 where ✓ = 1p
p (1.22) To see what (1.22) says in a concrete exam...
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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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