Markov chain simulation [duplicate]

Question

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R library for discrete Markov chain simulation (3 answers)

Closed 8 years ago.

I'm wondering if there is an algorithm to simulate a discrete Markov chain with a specific number of occurrences of state knowing the transition matrix way.

For example, how to simulate in R a Markov chain of length n with p occurrences (p < n) of the state '0' for a transition matrix defined by:

TransitionMatrix<- matrix(c(0.7, 0.3, 0.4, 0.6),byrow=TRUE, nrow=2) 


colnames(TransitionMatrix) <- c('0','1')
row.names(TransitionMatrix) <- c('0','1')

Answer 1

So, for a two state matrix like yours, this would work to generate random sequences of ones and zeroes, with n total and p ones. It will also give you the probability:

sim.sequence<-function(n,p) {
  ones<-sort(sample(1:n,p,replace=FALSE))
  x<-rep(0,n)
  x[ones]<-1
  row<-as.character(x[1:(n-1)])
  col<-as.character(x[2:n])
  probs<-TransitionMatrix[cbind(row,col)]
  list(states=x,probs=probs)
}

sim.sequence(20,12)
# $states
# [1] 0 0 1 1 1 0 1 0 1 1 0 1 0 1 1 1 1 0 1 0

# $probs
# [1] 0.7 0.3 0.6 0.6 0.4 0.3 0.4 0.3 0.6 0.4 0.3 0.4 0.3 0.6 0.6 0.6
# [17] 0.4 0.3 0.4

---

Okay, based on your comment, I think I know what you want now. Randomly generated 0's and 1's, and then the proportion of transitions from each state. TransitionMatrix isn't used at all.

So, this would be do it:

sim.sequence<-function(n,p) {
  ones<-sort(sample(1:n,p,replace=FALSE))
  x<-rep(0,n)
  x[ones]<-1
  row<-as.character(x[1:(n-1)])
  col<-as.character(x[2:n])
  tab<-table(row,col)
  probs<-as.matrix(tab) / rowSums(tab)
  list(states=x,probs=probs)
}
set.seed(1)
sim.sequence(20,12)

# $states
# [1] 1 1 1 1 0 1 0 1 1 0 1 0 0 1 1 1 0 0 1 0
# 
# $probs
#    col
# row       0         1
# 0 0.2857143 0.7142857
# 1 0.5000000 0.5000000

---

I claim that what you are asking for, in the example provided in the comments, is impossible.

You state sequence must have 20 states, meaning 19 transitions. Of those transitions, there are only four possibilities:

(1,1) - m1
(1,0) - m2
(0,1) - n1
(0,0) - n2

So m1+m2+n1+n2==19 and all four are also integers. To restrain b==0.4 it is necessary that m1/(m1+m2)==0.4. Because m1 and m2 also have to be integers, this means that there are only three possibilities for the pair (m1,m2): (2,3), (4,6), (6,9). (8,12) is impossible, because that would imply more than 19 transitions.

Similarly, to restrain a==0.3, you have only one possibility for (n1,n2): (3,7). Therefore, n1+n2==10. This implies that m1+m2==9. However, the only possibilities for m1+m2 is (5,10,15). That is why there will be no string that ever meets your criteria.

Furthermore, we haven't even got into the next constraint where there must be exactly 12 ones. That would make things even more difficult.