在计算P ^ n时，matrixpower()和markov()之间有什么区别? [英] What is the difference between matrixpower() and markov() when it comes to computing P^n?

问题描述

考虑具有状态空间S = {1, 2, 3, 4}和转移矩阵的马尔可夫链

Consider a Markov chain with state space S = {1, 2, 3, 4} and transition matrix

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 

然后，看看下面的源代码:

And, take a look at the following source code:

# 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)

输出

> 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?

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

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

推荐答案

当前，您正在比较完全不同的事物.首先，我将不专注于计算P ⁿ ，而是关注A * P ⁿ ，其中A是初始分布.在这种情况下，matrixpower可以完成工作:

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

分别是经过10个步骤(使用A进行的初始绘制之后)分别处于状态1、2、3、4的真实概率.

those are the true probabilities of being in states 1, 2, 3, 4, respectively, after 10 steps (after the initial draw made using A).

同时，markov(init = alpha, mat = tmat, n = nn)只是长度nn + 1的单个实现，并且此实现的最后一个数字与A * P ⁿ 有关.因此，为了尝试获得与理论值相似的数字，我们需要使用nn <- 10进行许多实现，如

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

我在其中模拟10000个实现，并仅获取每个实现的最后状态.

where I simulate 10000 realizations and take only the last state of each realization.

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