How to find the Markov Chain Probability? - r

Question

I am trying to find the probability that the chain jumps from state k-1 to state 1 before it hits state k.
Can anyone spot my mistake?
I tried to simulate the markov chain, but i want to make a code that allows me to find probability of k ={1, 2, 3, ........17}. But I can really not get the code.
This is the error message I always get
Error in while (X[i] > 1 && X[i] < k) { :
missing value where TRUE/FALSE needed
Here is my code:
k <- 17
{ p <- 0.5
q <- 0.1
P <- matrix (0, nrow = k, ncol = k, byrow = TRUE)
for (i in 1:k)
{ for (j in 1:k)
{ if (i == 1 && i == j)
{ P[i,j] <- 1
}
else if (i == k && i == j)
{ P[i,j] <- 1
}
else if (i == j)
{ P[i,j] <- p*(1-q)
}
else if (j == k && i != 1)
{ P[i,j] <- q
}
else if (i == j+1 && i != k)
{ P[i,j] <- (1-p)*(1-q)
}
}
}
P
X <- (k-1)
trials <- 1000
hits <- 0 #counter for no. of hits
for (i in 1:trials)
{ i <- 1 #no. of steps
while(X[i] > 1 && X[i] < k)
{ Y <- runif(1) #uniform samples
p1 <- P[X[i],] #calculating the p-value
p1 <- cumsum(p1)
# changes in the chain
if(Y <= p1[1])
{ X[i+1] = 1}
else if(Y <= p1[2])
{ X[i+1] = 2}
else if(Y <= p1[3])
{ X[i+1] = 3}
else if(Y <= p1[4])
{ X[i+1] = 4}
else if(Y <= p1[5])
{ X[i+1] = 5}
else if(Y <= p1[6])
{ X[i+1] = 6}
else if(Y <= p1[7])
{ X[i+1] = 7}
else if(Y <= p1[8])
{ X[i+1] = 8}
else if(Y <= p1[9])
{ X[i+1] = 9}
else if(Y <= p1[10])
{ X[i+1] = 10}
else if(Y <= p1[11])
{ X[i+1] = 11}
else if(Y <= p1[12])
{ X[i+1] = 12}
else if(Y <= p1[13])
{ X[i+1] = 13}
else if(Y <= p1[14])
{ X[i+1] = 14}
else if(Y <= p1[15])
{ X[i+1] = 15}
else if(Y <= p1[16])
{ X[i+1] = 16}
else if(Y <= p1[17])
{ X[i+1] <= 17}
i <- i+1
}
if(X[i]==1)
{ hits <- hits+1}
else
{ hits <- hits+0}
}
Probability <- hits/trials
Probability
}

Answer 1

I think the line
i <- 1 #no. of steps
should not be there. Try this:
k <- 17
{ p <- 0.5
q <- 0.1
P <- matrix (0, nrow = k, ncol = k, byrow = TRUE)
for (i in 1:k)
{ for (j in 1:k)
{ if (i == 1 && i == j)
{ P[i,j] <- 1
}
else if (i == k && i == j)
{ P[i,j] <- 1
}
else if (i == j)
{ P[i,j] <- p*(1-q)
}
else if (j == k && i != 1)
{ P[i,j] <- q
}
else if (i == j+1 && i != k)
{ P[i,j] <- (1-p)*(1-q)
}
}
}
P
X <- (k-1)
trials <- 1000
hits <- 0 #counter for no. of hits
for (i in 1:trials)
{
while(X[i] > 1 && X[i] < k)
{ Y <- runif(1) #uniform samples
p1 <- P[X[i],] #calculating the p-value
p1 <- cumsum(p1)
# changes in the chain
if(Y <= p1[1])
{ X[i+1] = 1}
else if(Y <= p1[2])
{ X[i+1] = 2}
else if(Y <= p1[3])
{ X[i+1] = 3}
else if(Y <= p1[4])
{ X[i+1] = 4}
else if(Y <= p1[5])
{ X[i+1] = 5}
else if(Y <= p1[6])
{ X[i+1] = 6}
else if(Y <= p1[7])
{ X[i+1] = 7}
else if(Y <= p1[8])
{ X[i+1] = 8}
else if(Y <= p1[9])
{ X[i+1] = 9}
else if(Y <= p1[10])
{ X[i+1] = 10}
else if(Y <= p1[11])
{ X[i+1] = 11}
else if(Y <= p1[12])
{ X[i+1] = 12}
else if(Y <= p1[13])
{ X[i+1] = 13}
else if(Y <= p1[14])
{ X[i+1] = 14}
else if(Y <= p1[15])
{ X[i+1] = 15}
else if(Y <= p1[16])
{ X[i+1] = 16}
else if(Y <= p1[17])
{ X[i+1] <= 17}
i <- i+1
}
if(X[i]==1)
{ hits <- hits+1}
else
{ hits <- hits+0}
}
Probability <- hits/trials
Probability
}

Answer 2

You're setting X to k-1. In R, that's treated as a vector of length 1. As soon as i reaches 2, X[i] return an index error, because X does not have a second element.
Further notes: using the same index in two different nesting levels is very bad form. Also, when you start having a massive list of if-then-else statements, it's time to rethink your code. In this case, you could just subset 1:17 on p1[i] >=Y, take the minimum value, and then set X to that.