R中马尔可夫链的手动模拟 [英] Manual simulation of Markov Chain in R

问题描述

考虑具有状态空间S = {1, 2}、转移矩阵的马尔可夫链

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X₀、X₁, . . . , X₅).重复模拟 100次.使用您的模拟结果解决以下问题.

1. Simulate 5 steps of the Markov chain (that is, simulate X₀, X₁, . . . , X₅). Repeat the simulation 100 times. Use the results of your simulations to solve the following problems.

• 估计P(X₁ = 1|X₀ = 1).将您的结果与准确概率进行比较.

• Estimate P(X₁ = 1|X₀ = 1). Compare your result with the exact probability.

我的解决方案:

# returns Xn 
func2 <- function(alpha1, mat1, n1) 
{
  xn <- alpha1 %*% matrixpower(mat1, n1+1)

  return (xn)
}

alpha <- c(0.5, 0.5)
mat <- matrix(c(0.5, 0.5, 0, 1), nrow=2, ncol=2)
n <- 10


for (variable in 1:100) 
{
   print(func2(alpha, mat, n))
}

<小时>

如果我运行此代码一次或 100 次(如问题陈述中所述)有什么区别?

---

What is the difference if I run this code once or 100 times (as is said in the problem-statement)?

如何从这里找到条件概率?

How can I find the conditional probability from here on?

推荐答案

让

alpha <- c(1, 1) / 2
mat <- matrix(c(1 / 2, 0, 1 / 2, 1), nrow = 2, ncol = 2) # Different than yours

是初始分布和转移矩阵.您的 func2 仅找到不需要的第 n 步分布，并且不会模拟任何内容.相反，我们可以使用

be the initial distribution and the transition matrix. Your func2 only finds n-th step distribution, which isn't needed, and doesn't simulate anything. Instead we may use

chainSim <- function(alpha, mat, n) {
  out <- numeric(n)
  out[1] <- sample(1:2, 1, prob = alpha)
  for(i in 2:n)
    out[i] <- sample(1:2, 1, prob = mat[out[i - 1], ])
  out
}

其中 out[1] 仅使用初始分布生成，然后对于后续项，我们使用转移矩阵.

where out[1] is generated using only the initial distribution and then for subsequent terms we use the transition matrix.

然后我们有

set.seed(1)
# Doing once
chainSim(alpha, mat, 1 + 5)
# [1] 2 2 2 2 2 2

所以链从 2 开始并由于指定的转移概率而卡在那里.

so that the chain initiated at 2 and got stuck there due to the specified transition probabilities.

重复 100 次

# Doing 100 times
sim <- replicate(chainSim(alpha, mat, 1 + 5), n = 100)
rowMeans(sim - 1)
# [1] 0.52 0.78 0.87 0.94 0.99 1.00

最后一行显示了我们进入状态 2 而不是状态 1 的频率.这给出了为什么 100 次重复提供更多信息的一个(非常多的)原因:我们陷入状态 2 只做一次模拟，同时重复100 次我们探索了更多可能的路径.

where the last line shows how often we ended up in state 2 rather than 1. That gives one (out of many) reasons why 100 repetitions are more informative: we got stuck at state 2 doing just a single simulation, while repeating it for 100 times we explored more possible paths.

然后可以找到条件概率

mean(sim[2, sim[1, ] == 1] == 1)
# [1] 0.4583333

而真实概率为 0.5(由转移矩阵的左上角条目给出).

while the true probability is 0.5 (given by the upper left entry of the transition matrix).

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