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0.05
F
U
0.9
0.95
0.1
FIGURE 9.5
Directed graph representation of a discrete Markov chain with a state space Q
.
The vertices of the digraph correspond to the states of the process and the labels on the
directed edges stand for the transition probabilities. The beginning and ending states are
not pictured.
={
f,u
}
=
P
{ π l | π l − 1 }
P
{ π l − 1 | π l − 2 }
P
{ π l − 2 | π l − 3 }···
P
{ π 1 | π 0 }
=
a π l − 1 π l a π l − 2 π l − 1 ···
a π 0 π 1
l
=
a π i − 1 π i .
i
=
1
Example 9.1. Assume that a game is played by successively rolling a die. Two dice
are available, one fair and one unfair. Either die is equally likely to be chosen for the
first game. After that, the process of switching between the dice can be depicted by
the transition diagram in Figure 9.5 : if the fair die is used in the current game, it will
be retained for the next game with probability 0.95 and switched with the unfair die
with probability 0.05. If the unfair die is used for the current game, we continue to
use it for the next game with probability 0.9 and switch to the fair die with probability
0.1. The process
of switching between the dice can be described by a Markov
chain with a state space of two elements Q
π
={
F
,
U
}
, with transition probabilities
a FF =
P
(π t + 1
=
F
| π t
=
F
) =
0
.
95
,
a FU =
P
(π t + 1
=
F
| π t
=
U
) =
0
.
05,
a UF = P (π t + 1
=
F
| π t
=
U
) =
0
.
10, and a UU =
P
(π t + 1
=
U
| π t
=
U
) =
0
.
90.
a FF a FU
a UF a UU
0
and the initial
.
.
95 0
05
The transition matrix is then P
=
=
.
.
0
10
0
9
distribution is a 0 F =
5.
We can noweasily compute the probability of any path of this process. For example,
0
.
5 and a 0 U =
0
.
π =
if
π =
UFUF and
UUFF , we obtain P
( π) =
a 0 U a UF a FU a UF = (
0
.
5
) ∗
( π =
(
0
.
1
) ∗ (
0
.
05
) ∗ (
0
.
1
) =
0
.
0025 and P
a 0 U a UU a UF a FF = (
0
.
5
) ∗ (
0
.
9
) ∗ (
0
.
1
) ∗
(
0428.
Exercise 9.1. For the Markov chain from Example 9.1 , compute the probabili-
ties of the following paths: (a)
0
.
95
) =
0
.
FUUUU ;(b) π =
π =
UUFFUF ;(c)
π =
FFFFFU .
 
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