Applied probability (0809)
Solution to Assignment 3
Chapter 3
Problem 1.1
The state space is
{
0
,
1
,
2
,
3
,
4
,
5
}
.It is obvious that
P
00
=
P
55
= 1.
If there are
k
(
k
= 1
,
2
,
3
,
4) people who are diseased, the number of pairs of individuals
in which the disease transmission may take place(one of these persons is diseased and the
other not) is
C
k
1
C
5

k
1
=
k
(5

k
), and the number of the total pairs of individuals is
C
5
2
= 10. Since the probability that the disease is transmitted to the healthy people is
α
= 0
.
1, thus the (
k, k
+ 1)th entry of the transition matrix
P
is
P
k,k
+1
=
α
k
(5

k
)
10
,
and the (
k, k
)th entry
P
k,k
= 1

P
k,k
+1
= 1

α
k
(5

k
)
10
.
Other entries of
P
are 0. Therefore, the transition matrix is as follows:
P
=
1
0
0
0
0
0
0 0
.
96 0
.
04
0
0
0
0
0
0
.
94 0
.
06
0
0
0
0
0
0
.
94 0
.
06
0
0
0
0
0
0
.
96 0
.
04
0
0
0
0
0
1
.
Problem 1.4
The state space is
{
0
,
1
,
2
,
3
}
.
Since
X
n
+1
= max
{
ξ
1
,
· · ·
, ξ
n
+1
}
= max
{
X
n
, ξ
n
+1
}
,
thus
Pr(
X
n
+1
=
j

X
n
=
i
) =
0
,
if
j < i,
∑
i
k
=0
Pr
{
ξ
n
+1
=
k
}
,
if
j
=
i,
Pr
{
ξ
n
+1
=
j
}
,
if
j > i.
1
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 Spring '11
 afdsa
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