Markov chain:
Example:
Suppose there are 2 competing products, E and N, in the market, for example,
Explorer and Netscape. After one year, some customers will keep using the same
product but some will switch to using the other product. The proportion of the
customers who keep using the same product or switch to use the other product after
one year is summarized in the following table:
Keep
Switch
Suppose
E
N
1/4
3/4
1/3
2/3
3
2
of the customers use product E and the other
of customers use
5
5
product N in the beginning. Then,
1. What is the distribution after one year?
2. What is the distribution after two years?
3.
What is the distribution as the market is said to be stable (i.e., the distribution of
the market would be constant forever).
[solution:]
Let the matrix
E
A
N
E 1 / 4 2 / 3  E  E N  E 



N 3 / 4 1 / 3   E  N N  N  ,
The first row contains the proportions of which the customers keep using product E
and the ones switch to using product E. The second row contains the proportions of
which the customers switch to using product N and the ones keep using product N.
Let
x0 
E
N
3 / 5 
2 / 5


be the vector of distribution in the beginning.
1
1.
1 / 4 2 / 3 3 / 5 E [(1 / 4)  (3 / 5)]  [( 2 / 3)  (2 / 5)]
Ax0  
 2 / 5  N [(3 / 4)  (3 / 5)]  [(1 / 3)  (2 / 5)]
3
/
4
1
/
3





E  [ E  E ]  [ N  E ]  E 25 / 60
 

N [ E  N ]  [ N  N ] N 35 / 60
After one year,
25
60
of customers use product E while
35
60
of customers use
product N.
2.
1 / 4 2 / 3 25 / 60 E [(1 / 4)  (25 / 60)]  [( 2 / 3)  (35 / 60)]
A( Ax0 )  
 35 / 60  N [(3 / 4)  (25 / 60)]  [(1 / 3)  (35 / 60)]
3
/
4
1
/
3





E  [ E  E ]  [ N  E ]  E 355 / 720
 

N [ E  N ]  [ N  N ] N 365 / 720
After two year,
355
720
of customers use product E while
365
720
of customers
use product N.
3.
Suppose
x 
xs   1 
 x2 
is the distribution as the market is stable. Thus,
1 / 4 2 / 3  x1   x1 
Axs  
  x    x   xs
3
/
4
1
/
3

 2   2 

Axs  xs  Axs  I 2 xs  ( A  I 2 ) xs
 3 / 4

 3/ 4
2
2 / 3   x1 
 0,
 2 / 3  x2 
where I 2 is a 2  2 identity matrix. The solutions for the homogeneous system are
 x  8 / 9
xs   1   
t, t  R .

 x2   1 
Since
x1  x 2  1  x1 
8
9
, x2 
.
17
17
General case:
Suppose there are n competing products, C1, C2,…,Cn in the market. A general
Markov chain model can be employed. The proportion of the customers who
keep using the same product or switch to use the other product after one year is
summarized in the following table:
C1
C2
…
Cn
C1 (after one
p11
p12
…
p1n
year)
C2 (after one
year)
p21
p22
…
p2n

p n1

pn2

…

pnn

Cn (after one
year)
Also,
p1 j  p2 j    pnj  1, j  1,2,, n.
Let
C1
C1  p11
C 2  p21
A
  

Cn  pn1
C2
…
Cn
p12
p22



pn 2


p1n 
p2 n 
 ,

pnn 
be the matrix representing the proportions of which the customers keep using
3
the original product and those switch to using the other products. Suppose, in
the beginning, the proportions of the customers for using C1, C2,…, Cn are
 1 ,  2 , ,  n
and the vector of distribution is
C1  1 


C2 
2

,         1
x0 
1
2
n
   


Cn  n 
.
Then,
 The vector of distribution after k years is
A k x0  A  A    Ax0 , k  1,2, 
k factors

Suppose
C1  x1 

x
C2 
2


xs 
   


Cn  xn 
is the vector of distribution as the market is stable. Then,
xs
can be found by
solving the following homogeneous linear system,
 p11  1 p12
 p
p 22  1
Axs  x s  ( A  I n ) x s   21
 


pn 2
 p n1
4
p1n   x1 
p 2 n   x2 
0
.

   
 
 p nn  1  xn 

