STA3007
Applied probability (0809)
Solution to Assignment 6
Chapter 4
Problem 3.1
(a) By deﬁnition,
f
(0)
00
= 0, and
f
(1)
00
= Pr(
X
1
= 0

X
0
= 0) =
P
00
= 1

a,
f
(
n
)
00
= Pr
{
X
1
6
= 0
,
···
, X
n

1
6
= 0
, X
n
= 0

X
0
= 0
}
= Pr
{
X
1
= 1
,
···
, X
n

1
= 1
, X
n
= 0

X
0
= 0
}
= Pr(
X
1
= 1

X
0
= 0)Pr(
X
n
= 0

X
n

1
= 1)
n

1
Y
k
=2
Pr(
X
k
= 1

X
k

1
= 1)
=
ab
(1

b
)
n

2
,
(
n
≥
2)
.
where
Q
1
k
=2
Pr(
X
k
= 1

X
k

1
= 1) = 1 is deﬁned as 1.
(b) We are asked to show
b
a
+
b
+
a
a
+
b
(1

a

b
)
n
=
P
(
n
)
00
=
n
X
k
=0
f
(
k
)
00
P
(
n

k
)
00
for
n
≥
1. Since
b
a
+
b
n
X
k
=2
f
(
k
)
00
=
b
a
+
b
n
X
k
=2
ab
(1

b
)
k

2
=
ab
2
a
+
b
1

(1

b
)
n

1
1

(1

b
)
=
ab
a
+
b
(1

(1

b
)
n

1
)
.
and
a
a
+
b
n
X
k
=2
f
(
k
)
00
(1

a

b
)
n

k
=
a
2
b
(1

a

b
)
n
(
a
+
b
)(1

b
)
2
n
X
k
=2
(
1

b
1

a

b
)
k
=
a
2
b
(1

a

b
)
n
(
a
+
b
)(1

b
)
2
(
1

b
1

a

b
)
2

(
1

b
1

a

b
)
n
+1
1

1

b
1

a

b
=
ab
a
+
b
[(1

b
)
n

1

(1

a

b
)
n

1
]
.
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThus,
n
X
k
=0
f
(
k
)
00
P
(
n

k
)
00
=
f
(1)
00
P
(
n

1)
00
+
n
X
k
=2
f
(
k
)
00
P
(
n

k
)
00
= (1

a
)[
b
a
+
b
+
a
a
+
b
(1

a

b
)
n

1
] +
ab
a
+
b
[1

(1

a

b
)
n

1
]
=
(1

a
)
b
+
ab
a
+
b
+
a
a
+
b
(1

a

b
)
n

1
(1

a

b
)
=
b
a
+
b
+
a
a
+
b
(1

a

b
)
n
=
P
(
n
)
00
as was to be shown.
Problem 3.2
This is the end of the preview. Sign up
to
access the rest of the document.
 Spring '11
 afdsa
 Markov chain, αk, A+B A+B A+B, a+b a+b

Click to edit the document details