next up previous contents
Next: First passage times Up: Applicability of Markov chains Previous: Applicability of Markov chains   Contents

Example

Let's find out what is the probability of our subject to be in a certain stress state $ s_{i}$. We will use her transitional probability matrix#3480#> shown in table 3.3. What is the initial probability vector? Let's pretend we do not know in what state she is now. The initial probability vector is then given by the "static" probability distribution (table 3.4).


Table 3.4: Initial probability vector (stress example)
State 1 2 3 4 5 6
Probability 0.13 0.33 0.32 0.13 0.05 0.03


What is then the probability to find the subject in a given stress state three time period ahead we observed her? We simply multiply the initial probability vector with the transitional probability matrix#3481#> multiplied by itself three time (third power); If we compute the transitional probability matrix#3482#> at a certain power $ k$, it yields the probabilities for transiting from state $ i$ to state $ j$ to $ k$ times ahead. For a power of three, the transitional probability matrix#3483#> is given in table 3.5:


Table 3.5: Transitional probability matrix at third power (stress variable)
  $ STRE(t+3)$
  1 2 3 4 5 6
$ STRE(t)$ = 1 0.13 0.34 0.32 0.13 0.05 0.03
2 0.13 0.34 0.32 0.13 0.05 0.03
3 0.13 0.33 0.33 0.13 0.05 0.03
4 0.13 0.33 0.32 0.13 0.05 0.03
5 0.13 0.34 0.32 0.13 0.06 0.03
6 0.13 0.33 0.32 0.13 0.06 0.03


So the probabilities to find the subject in a given state without knowing in what state she is now is (cf. table 3.6). The probabilities to find her in a given state are the same as with the initial probability vector! It is not necessary so, but it happens when transitions converge rapidly to their long-term behavior.


Table 3.6: Probabilities to find the subject in a stress state, three times ahead
State 1 2 3 4 5 6
Probability 0.13 0.33 0.32 0.13 0.05 0.03


Strangely enough the transitional probabilities in table 3.5 are almost identical for each column. Why is it so? If we let the transitional probability matrix#3484#> at a constantly greater power, the resulting matrix converges to its long-term behavior (the limit matrix , denoted by letter $ A$). That is, if we let this person experience stress following a dynamics given by her transitional probabilities, the best guess about the state she is would be this limit matrix. In our example the limit matrix is (table 3.7).


Table 3.7: Limit matrix for the stress variable
  $ STRE(t+\infty)$
  1 2 3 4 5 6
$ STRE(t)$ = 1 0.13 0.33 0.32 0.13 0.05 0.03
2 0.13 0.33 0.32 0.13 0.05 0.03
3 0.13 0.33 0.32 0.13 0.05 0.03
4 0.13 0.33 0.32 0.13 0.05 0.03
5 0.13 0.33 0.32 0.13 0.05 0.03
6 0.13 0.33 0.32 0.13 0.05 0.03


The result implies that if we observe her for a large number of days we can expect her to feel 13% of very low stress, 33% of low stress, 32% of moderate stress, 13% of medium stress, 5% of high stress and 3% of very high stress.

Since each row of the limit matrix $ A$ converges to identical probabilities, it may be represented by a vector $ \alpha$. It may be algebraically computed by solving the equation: $ \alpha P = \alpha$.


next up previous contents
Next: First passage times Up: Applicability of Markov chains Previous: Applicability of Markov chains   Contents
Philippe Lemay
1999-09-14