我试图通过马尔可夫链模拟一个步骤，目的是通过过程多次循环，直到满足条件。(即，找出平均需要多少步才能达到特定状态)。
在这种情况下，状态只能走一条路。例如，状态4可以向前过渡到状态5，但不能向后过渡到状态3。这意味着转换矩阵的左下半部被设置为零。这也是为什么下面的方法在"先验"状态中放置任意大的值。我试图通过检查转移矩阵的指定行中哪个概率最接近随机数来找到正确的新状态。
get_new_state <- function(current_state, trans_matrix)
{
# generate a random number between 0-1 to compare to the transition matrix probabilities
rand <- runif(1)
# transition to where the
# random number falls within the transition matrix
row <- current_state # initial condition determines the row of the trans_matrix
col = current_state # start in the column
# loop thru all columns and find the correct state
potential_states <- rep(0, each=ncol(trans_matrix)) # holds the value of the potential state it transitions to
# in this case, we can't transition to a previous state so we set the previous state values arbitrarily high
# so they don't get selected in the which.min() function later
potential_states[1:col] <- 999
for(k in row:ncol(trans_matrix)) # loop thru non-zero matrix values
{
if(trans_matrix[row,k] > rand)
{
potential_states[k] <- trans_matrix[row,k] / rand
potential_states[k] <- 1 - potential_states[k]
}
}
# new state is equal to the index of the lowest value
# lowest value = closest to random number
new_state = which.min(potential_states)
return(as.numeric(new_state))
}
我不确定这种方法是否合理。我假设有一种更好的方法来模拟，而不需要在potential_states[]中放置任意大的值。
这样做会更好吗(它是一行马尔可夫转换):
> new_state <- sample(1:number_states, size=1,
prob=transition_matrix[old_state,])
并将其放入(例如)带有计数器的while()循环中