在计算 P^n 时，matrixpower() 和 markov() 有什么区别？

考虑具有状态空间的马尔可夫链S = {1, 2, 3, 4}和转移矩阵

P =  0.1  0.2  0.4  0.3
     0.4  0.0  0.4  0.2
     0.3  0.3  0.0  0.4
     0.2  0.1  0.4  0.3 

并且，看一下以下源代码：

# markov function
markov <- function(init,mat,n,labels) 
{ 
    if (missing(labels)) 
    {
        labels <- 1:length(init) 
    }

    simlist <- numeric(n+1) 
    states <- 1:length(init) 
    simlist[1] <- sample(states,1,prob=init)  

    for (i in 2:(n+1))
    { 
        simlist[i] <- sample(states, 1, prob = mat[simlist[i-1],])  
    }

    labels[simlist] 
}

# matrixpower function
matrixpower <- function(mat,k) 
{
    if (k == 0) return (diag(dim(mat)[1])) 
    if (k == 1) return(mat)
    if (k > 1) return( mat %*% matrixpower(mat, k-1))
}

tmat = matrix(c(0.1, 0.2, 0.4, 0.3,
                0.4, 0.0, 0.4, 0.2,
                0.3, 0.3, 0.0, 0.4,
                0.2, 0.1, 0.4, 0.3), nrow=4, ncol=4, byrow=TRUE)

p10 = matrixpower(mat = tmat, k=10)  
rowMeans(p10)

nn <- 10 
alpha <- c(0.25, 0.25, 0.25, 0.25)

set.seed(1)

steps <- markov(init=alpha, mat=tmat, n=nn)
table(steps)/(nn + 1)

Output

> rowMeans(p10)
[1] 0.25 0.25 0.25 0.25
> 
.
.
.
> table(steps)/(nn + 1)
steps
         1          2          3          4 
0.09090909 0.18181818 0.18181818 0.54545455 
> ?rowMeans

为什么结果如此不同？

What is the difference between using matrixpower() and markov() when it come to compute Pⁿ?

Currently you are comparing completely different things. First, I'll focus not on computing Pⁿ, but rather A*Pⁿ, where A is the initial distribution. In that case matrixpower does the job:

p10 <- matrixpower(mat = tmat, k = 10)  
alpha <- c(0.25, 0.25, 0.25, 0.25)
alpha %*% p10
#           [,1]      [,2]      [,3]      [,4]
# [1,] 0.2376945 0.1644685 0.2857105 0.3121265

这些分别是 10 个步骤后（使用 A 进行初始抽签后）处于状态 1、2、3、4 的真实概率。

Meanwhile, markov(init = alpha, mat = tmat, n = nn) is only a single realization of length nn + 1 and only the last number of this realization is relevant for A*Pⁿ. So, as to try to get similar numbers to the theoretical ones, we need many realizations with nn <- 10, as in

table(replicate(markov(init = alpha, mat = tmat, n = nn)[nn + 1], n = 10000)) / 10000
#
#      1      2      3      4 
# 0.2346 0.1663 0.2814 0.3177

我模拟了 10000 个实现，并且仅获取每个实现的最后状态。