Manual simulation of Markov Chain in R (3) - r

Question

I have tried to improve my previous code so that I can incorporate conditional probability.
Source Code
states <- c(1, 2)
alpha <- c(1, 1)/2
mat <- matrix(c(0.5, 0.5,
0, 1), nrow = 2, ncol = 2, byrow = TRUE)
# this function calculates the next state, if present state is given.
# X = present states
# pMat = probability matrix
nextX <- function(X, pMat)
{
#set.seed(1)
probVec <- vector() # initialize vector
if(X == states[1]) # if the present state is 1
{
probVec <- pMat[1,] # take the 1st row
}
if(X==states[2]) # if the prsent state is 2
{
probVec <- pMat[2,] # take the 2nd row
}
return(sample(states, 1, replace=TRUE, prob=probVec)) # calculate the next state
}
# this function simulates 5 steps
steps <- function(alpha1, mat1, n1)
{
vec <- vector(mode="numeric", length = n1+1) # initialize an empty vector
X <- sample(states, 1, replace=TRUE, prob=alpha1) # initial state
vec[1] <- X
for (i in 2:(n1+1))
{
X <- nextX(X, mat1)
vec[i] <- X
}
return (vec)
}
# this function repeats the simulation n1 times.
# steps(alpha1=alpha, mat1=mat, n1=5)
simulate <- function(alpha1, mat1, n1)
{
mattt <- matrix(nrow=n1, ncol=6, byrow=T);
for (i in 1:(n1))
{
temp <- steps(alpha1, mat1, 5)
mattt[i,] <- temp
}
return (mattt)
}
Execution
I created this function so that it can handle any conditional probability:
prob <- function(simMat, fromStep, toStep, fromState, toState)
{
mean(simMat[toStep+1, simMat[fromStep+1, ]==fromState]==toState)
}
sim <- simulate(alpha, mat, 10)
p <- prob(sim, 0,1,1,1) # P(X1=1|X0=1)
p
Output
NaN
Why is this source code giving NaN?
How can I correct it?

Answer 1

I didn't inspect the rest of your code, but it seems that only prob has a mistake; you are mixing up rows with columns and instead it should be
prob <- function(simMat, fromStep, toStep, fromState, toState)
mean(simMat[simMat[, fromStep + 1] == fromState, toStep + 1] == toState)
Then NaN still remains a valid possibility for the following reason. We are looking at a conditional probability P(X1=1|X0=1) which, by definition, is well defined only when P(X0=1)>0. The same holds with sample estimates: if there are no cases where X0=1, then the "denominator" in the mean inside of prob is zero. Thus, it cannot and should not be fixed (i.e., returning 0 in those cases would be wrong).