R中的马尔可夫链的手动模拟(2) [英] Manual simulation of Markov Chain in R (2)

问题描述

考虑状态空间为 S = {1，2} 的马尔可夫链，过渡矩阵

Consider the Markov chain with state space S = {1, 2}, transition matrix

和初始分布α=(1/2，1/2).

1. 模拟马尔可夫链的5个步骤(即模拟 X ₀ ， X ₁ 、. ...， X ₅ ).重复模拟100次.

我的解决方案:

states <- c(1, 2)
alpha <- c(1, 1)/2
mat <- matrix(c(1/2, 1/2, 0, 1), nrow = 2, ncol = 2) 

nextX <- function(X, pMat)
{
    probVec <- vector()

    if(X == states[1])
    {
        probVec <- pMat[1,]
    }
    if(X==states[2])
    {
        probVec <- pMat[2,]
    }

    return(sample(states, 1, replace=TRUE, prob=probVec))
}

steps <- function(alpha1, mat1, n1)
{
    X0 <- sample(states, 1, replace=TRUE, prob=alpha1)

    if(n1 <=0)
    {
        return (X0)
    }
    else
    {
        vec <- vector(mode="numeric", length=n1)

        for (i in 1:n1) 
        {
            X <- nextX(X0, mat1)
            vec[i] <- X
        }

        return (vec)
    }
}

# steps(alpha1=alpha, mat1=mat, n1=5)

simulate <- function(alpha1, mat1, n1)
{
    for (i in 1:n1) 
    {
        vec <- steps(alpha1, mat1, 5)
        print(vec)
    }
}

simulate(alpha, mat, 100)

输出

> simulate(alpha, mat, 100)
[1] 1 2 2 2 2
[1] 2 1 2 2 2
[1] 1 1 1 1 1
[1] 1 1 1 1 1
[1] 2 2 2 2 2
[1] 2 1 1 2 2
[1] 1 1 1 1 1
[1] 2 2 1 2 2
[1] 1 1 1 1 1
[1] 2 2 1 1 2
[1] 2 2 1 2 2
[1] 1 1 1 1 1
[1] 1 1 1 1 1
[1] 1 1 2 2 2
[1] 1 1 1 1 1
[1] 2 1 2 2 2
[1] 1 1 1 1 1
[1] 1 1 1 1 1
[1] 2 1 2 1 1
[1] 1 1 1 1 1
[1] 1 1 1 1 1
[1] 1 1 1 1 1
[1] 1 1 1 1 1
[1] 1 2 1 2 2
[1] 2 2 2 1 2
[1] 1 1 1 1 1
[1] 1 1 1 1 1
[1] 1 1 1 1 1
[1] 2 1 2 2 1
[1] 1 2 2 2 2
[1] 1 1 2 2 2
[1] 1 2 2 1 2
[1] 1 1 1 1 1
[1] 2 2 1 2 2
[1] 2 2 2 1 1
[1] 1 1 1 1 1
[1] 1 1 1 1 1
[1] 1 1 1 1 1
[1] 1 1 1 2 2
[1] 1 2 1 1 2
[1] 2 2 1 1 1
[1] 1 1 1 1 1
[1] 2 2 2 2 1
[1] 1 1 1 1 1
[1] 1 1 1 1 1
[1] 1 2 2 2 2
[1] 2 1 1 2 2
[1] 1 1 1 1 1
[1] 1 2 1 2 1
[1] 1 1 1 1 1
[1] 1 1 1 1 1
[1] 2 1 2 1 2
[1] 1 1 1 1 1
[1] 1 1 1 1 1
[1] 1 1 1 1 1
[1] 1 1 1 1 1
[1] 2 1 2 1 2
[1] 1 1 1 1 1
[1] 1 1 1 1 1
[1] 2 2 1 1 2
[1] 1 1 1 1 1
[1] 1 1 1 1 1
[1] 2 2 1 1 2
[1] 1 1 1 1 1
[1] 1 2 1 1 2
[1] 1 1 1 1 1
[1] 2 1 1 2 1
[1] 1 1 1 1 1
[1] 2 1 2 2 2
[1] 1 1 1 1 1
[1] 1 1 1 1 1
[1] 1 2 2 2 2
[1] 1 1 1 1 1
[1] 2 2 2 1 2
[1] 2 2 2 1 1
[1] 1 1 2 2 2
[1] 1 1 1 1 1
[1] 2 2 2 1 2
[1] 1 1 1 1 1
[1] 2 2 1 2 2
[1] 2 2 1 2 1
[1] 1 1 1 1 1
[1] 2 2 2 2 2
[1] 1 1 1 1 1
[1] 1 2 1 2 2
[1] 1 1 1 1 1
[1] 2 1 1 2 1
[1] 2 2 2 2 1
[1] 2 2 2 2 2
[1] 1 1 1 1 1
[1] 2 2 2 1 1
[1] 2 2 2 2 2
[1] 1 1 1 1 1
[1] 1 1 1 1 1
[1] 2 1 2 2 1
[1] 2 2 1 1 1
[1] 1 1 1 1 1
[1] 2 2 1 2 2
[1] 2 1 2 2 2
[1] 1 1 1 1 1

如您所见，我在每次迭代中都得到相同的输出.

As you can see, I am getting same output in each iteration.

如何修复我的代码?

推荐答案

有两个问题:

如果您检查输入的矩阵，它就是您想要的内容的转置:

If you check the matrix you have input, it is the transpose of what you wanted:

> mat
     [,1] [,2]
[1,]  0.5    0
[2,]  0.5    1

所以，改变它.

在step函数中，不使用返回的状态来启动后续状态.取而代之的是，X0一直不断地被传递:

In the step function, the returned state is not used to initiate the subsequent state. Instead, X0 just keeps getting passed in repeatedly:

for (i in 1:n1) 
{
    X <- nextX(X0, mat1)
    vec[i] <- X
}

老实说，您根本不需要X0.只需将step函数中的所有X0更改为X，它就可以工作.

Honestly, you don't need X0 at all. Just change all the X0s in the step function to X and it should work.

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