I am explaining the line of top of Page 40
and using example Q 3.17 as an aid.
Let's say we want to find the distribution of \(N_{12}\) given \(N_1\).
i.e. no. of transitions from state 1 to 2 given Number of times the process was in state 1.
Now if the process is in state 1, after a transition takes place it can either remain in state 1, or go to state 2 or 3.
Transition to state 2 can be taken as a success, and transition to state 1 and 3 a failure.
So distribution of \(N_{12}\) given \(N_1\) is \(Bin(N_1, ~ p_{12})\).
Now we want to estimate \(p_{12}\) from the data given.
11
12
12
13
13
13
These are the observations from the data set. So, 6 times the process was in state 1 and after a transition took place, 2 times it went to state 2.
So \(\widehat{p}_{12}= \frac{2}{6}\)
Similarly \(\widehat{p}_{11} = \frac{1}{6} and ~\widehat{p}_{13}= \frac{3}{6}\)
Therefore distribution of \(N_{12}|N_1\) is \(Bin(N_1,~ \frac{2}{6})\).
or in general distribution of \(N_{ij}|N_i\) is \(Bin(N_i,~ p_{ij})\).
You can apply the same logic to triplets as well, which was your doubt.
Last edited by a moderator: Feb 22, 2014