= Solution
{c}
The function below counts the initial uniformly chosen state as the first of the $n$ visited states, and then generates $n-1$ transitions using the appropriate row of $P$:
``
simchain <- function(n, P) {
if (!is.numeric(n) || length(n) != 1L || !is.finite(n) ||
n < 1 || n != floor(n)) stop("n must be a positive integer")
if (!is.matrix(P) || !is.numeric(P) || nrow(P) == 0L ||
nrow(P) != ncol(P) || any(!is.finite(P)) ||
any(P < 0) || any(abs(rowSums(P) - 1) > 1e-12)) {
stop("P must be a finite square stochastic matrix")
}
k <- nrow(P)
state <- sim1(rep(1/k, k))
counts <- numeric(k)
counts[state] <- 1
if (n > 1) {
for (step in seq_len(n - 1)) {
state <- sim1(P[state, ])
counts[state] <- counts[state] + 1
}
}
counts / n
}
``
Conditional on the current state $i$, `sim1(P[i, ])` selects $j$ with <probability> $P_{ij}$, proving the <Markov property> with the desired <transition matrix>. Initialization is uniform by the same sampler. Dividing the visit counts by $n$ gives the requested <Empirical occupation proportions of a finite Markov chain>, including a valid result when $n=1$.
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