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and identify the states with their indices. To this end, we consider the following
3 sessions with altogether 3 products (for simplicity, we forgo the absorbing state):
•1
!
3
!
2
!
3
•2
!
1
!
2
!
3
!
1
•1
!
1
!
2
!
2
!
3
!
3
!
1
!
1
This yields the following global matrix of transition probabilities for
k ¼
1:
2
4
3
5
2
5
2
5
1
5
1
5
1
5
3
5
p ¼
1
2
1
4
1
4
Subsequently, we consider
k ¼ 2
, i.e., the transition probability does not only
depend on the current product, but also on its predecessor. Then we obtain the
following tensor of transition probabilities
P
:
2
3
2
4
3
5
,
2
4
3
5
000
001
1
2
010
001
010
100
001
001
4
5
p
ðÞ
¼
^
p
ðÞ
¼
^
,
^
p
ðÞ
¼
1
2
0
s
,
s
0
∈
S
denotes the matrix of transition probabilities if
i
has
been visited previously. Let us, e.g., consider the entry with index (3,1) of
p
ðÞ
¼ p
ðÞss
0
where
p
ðÞ
, i.e.,
^
1
p
ðÞ
3
,
1
¼
2
: after 2 had been visited, a transition to 3 occurred altogether twice.
Then, from there, there was one transition to 1 (and one transition to 3). Thus, the
probability of a transition from 3 to 1 given that 2 has been considered previously is
precisely 0.5.
We now choose the following partition of the state space
S ¼
{1,2,3},
m ¼
2:
^
G
1
¼
f
,
G
2
¼
1
;
fg
and thus obtain the corresponding aggregation prolongator and its pseudo-inverse
2
4
3
5
,
U
þ
¼
2
4
3
5
1
2
1
2
10
10
01
0
U ¼
001
We now determine the core tensor
C
, whereat we matricify
P:
