normal markov chain implies that the likelihood of going from one state to each other state is non - zero in some limited number of steps
so to demonstrate that a given markov chain is normal, we are going to ascertain p^2,p^3, etc until the point when we see each component of the network non zero.
p^2=p.p=[ 0.5000 0.3333 0.1667
0.3750 0.5000 0.1250
0.7500 0 0.2500]
still one term is zero
along these lines, we should attempt once more
p^3=p.p^2= [0.5000 0.3333 0.1667
0.5625 0.2500 0.1875
0.3750 0.5000 0.1250]
so each term is non zero, henceforth we can say that the markov chain is standard for 3 stages.
b.)
to figure likelihood of moving between various states state in 2 stages , we should discover the 2 stage change framework which is p^2.
so p^2 is as appeared
the likelihood of going from state 1 t 3 of every two stages will be(p^2)13 ie the component in line 1 and segment 3.
c.) the constraining vector
the constraining vector is a line vector which upon augmentation with the tansition likelihood grid gives indistinguishable vector from itself and entirety of its components ought to be equivalent to 1.
in this way, w.P=w
or then again [w1 w2 w3].[1/2 1/3 1/6
3/4 0 1/4
0 1 0]=[w1 w2 w3]
we get three direct conditions from this network condition which are homogenius
furthermore, the fourth condition is w1+w2+w3=1
after settling these fourth condition, you will get the outcome.
this is it.
I surmise you were not requesting that how unravel these direct conditions or discover framework square or shape cuz that should be possible utilizing any logical adding machine.
ematics of Discrete-Time Markov Chaill Develop a Markov chain model for each of the following situations....