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1
1
0.8
0.8
0.6
0.6
P
P
0.4
0.4
0.2
0.2
0
0
8
16
24
32
40
48
56
64
72
8
16
24
32
40
48
56
64
72
trial
trial
Fig. 3. Learning process of the Markov model: deterministic order (left) and random
order (right). The parameters were
p suc
=
p sr
=
p or
=0 . 9. Every line shows the
probability of success P
(cf. 5) for the concerning object.
which results in ten trials on every object and 80 trials in total. Further we
distinguish a deterministic and a random order. In case of a deterministic order
the objects are presented in the same succession in every cycle. For the random
case the succession of the objects is randomised in every cycle.
3 Analysis of the Markov Model
The advantage of a Markov model is the possibility to an analytic view on the
learning process. The following analysis is limited on the explicit associative
learning task (Fig. 1) on a single object. This is the model as it is described
above until (4). The steady state and convergence of the Markov chain illustrate
the correlation of the parameters p suc , p sr and their influence during learning.
We gain no further finding on the role of the parameter p or by this analysis,
since this parameter is influential on the learning process for a series of objects.
The steady state has its relevance in the asymptotic character of the state prob-
ability vector π ( t )for t →∞ . For convenience, we shorten T
in (4).
Repetitive replacement in (4) of π ( t +1) by A · π ( t ) gives the steady state:
T
· C
=
A
t
π stat = lim
t→∞ A
π
(0)
=
π stat =
A π stat ,
(7)
leading to the linear system of equations
π j stat =
i
A ij π i stat
,
, j
S.
(8)
With the analytic solution for the steady state (8), we are able to calculate the
final state probability vector of the Markov chain. A closed solution can be found
with a computer algebra system. The final probability of success π 16 stat depends
only on p suc and p sr in a fraction which numerator is a sixth-degree polynomial
of p sr and second-degree of p suc . The denominator is a fourth-degree polynomial
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