如何找到矩阵与循环收敛的时间 [英] How to find when a matrix converges with a loop

问题描述

我得到了一个矩阵:

P <- matrix(c(0, 0, 0, 0.5, 0, 0.5, 0.1, 0.1, 0, 0.4, 0, 0.4, 0, 0.2, 0.2, 0.3, 0, 0.3, 0, 0, 0.3, 0.5, 0, 0.2, 0, 0, 0, 0.4, 0.6, 0, 0, 0, 0, 0, 0.4, 0.6), nrow = 6, ncol = 6, byrow = TRUE)

使用功能mpow，rows_equal，matrices_equal.我想找到P^n何时收敛，换句话说n是什么，矩阵中的所有行均相等时以及P^n = P^(n+1)何时.

Using the functions, mpow, rows_equal, matrices_equal. I want to find when P^n converges, in other words what n is, when all the rows are equal in the matrix and when P^n = P^(n+1).

仅查看函数i即可推断出n=19-21周围的矩阵将收敛.

By just looking at the functions i have managed to deduce that around n=19-21 the matrix will converge.

尽管，我想使用循环找到合适的n.下面是功能mpow，rows_equal和matrices_equal.我知道它们的书写方式可能会有所不同，但请保持原样.

Although, I want to find the right n using a loop. Here under are the functions mpow, rows_equal and matrices_equal. I know they can be written differently but please keep them as they are.

mpow <- function(P, n, d=4) {
  if (n == 0) diag(nrow(P)))
  else if (n== 1) P
  else P %*% mpow(P, n - 1))
  }

rows_equal <- function(P, d = 4) {
  P_new <- trunc(P * 10^d)
  for (k in 2:nrow(P_new)) {
    if (!all(P_new[1, ] == P_new[k, ])) {
      return(FALSE)}
      }
    return(TRUE)
    }

matrices_equal <- function(A, B, d = 4) {
  A_new <- trunc(A * 10^d)
  B_new <-trunc(B * 10^d)
  if (all(A_new == B_new)) TRUE else FALSE
  }

现在，要编写循环，我们应该按照以下方式进行:

Now, to write the loop, we should do it something along the lines of:

首先创建一个像这样的函数:

First creating a function like so:

when_converged <- function(P) {...}

和

for (n in 1:50)

在t.ex n = 50时尝试.

To try for when t.ex n = 50.

尽管我不知道如何正确编写代码，但是有人可以帮助我吗?

Although i don't know how to write the code correctly to do so, can anyone help me with that?

感谢您阅读我的问题.

推荐答案

实际上，更好的方法是这样做:

Actually, a much better way is to do this:

## transition probability matrix
P <- matrix(c(0, 0, 0, 0.5, 0, 0.5, 0.1, 0.1, 0, 0.4, 0, 0.4, 0, 0.2, 0.2, 0.3, 0, 0.3, 0, 0, 0.3, 0.5, 0, 0.2, 0, 0, 0, 0.4, 0.6, 0, 0, 0, 0, 0, 0.4, 0.6), nrow = 6, ncol = 6, byrow = TRUE)

## a function to find stationary distribution
stydis <- function(P, tol = 1e-16) {
  n <- 1; e <- 1
  P0 <- P  ## transition matrix P0
  while(e > tol) {
    P <- P %*% P0  ## resulting matrix P
    e <- max(abs(sweep(P, 2, colMeans(P))))
    n <- n + 1
    }
  cat(paste("convergence after",n,"steps\n"))
  P[1, ]
  }

然后，当您调用函数时:

Then when you call the function:

stydis(P)
# convergence after 71 steps
# [1] 0.002590674 0.025906736 0.116580311 0.310880829 0.272020725 0.272020725

函数stydis本质上是连续执行的:

The function stydis, essentially continuously does:

P <- P %*% P0

直到P的收敛.收敛性由差异矩阵的L1范数确定:

until convergence of P is reached. Convergence is numerically determined by the L1 norm of discrepancy matrix:

sweep(P, 2, colMeans(P))

L1范数是所有矩阵元素的最大绝对值.当L1范数降到1e-16以下时，就会发生收敛.

The L1 norm is the maximum, absolute value of all matrix elements. When the L1 norm drops below 1e-16, convergence occurs.

如您所见，收敛需要71个步骤.现在，我们可以通过控制tol(公差)来获得更快的收敛":

As you can see, convergence takes 71 steps. Now, we can obtain faster "convergence" by controlling tol (tolerance):

stydis(P, tol = 1e-4)
# convergence after 17 steps
# [1] 0.002589361 0.025898057 0.116564506 0.310881819 0.272068444 0.271997814

但是，如果您检查:

mpow(P, 17)
#             [,1]       [,2]      [,3]      [,4]      [,5]      [,6]
# [1,] 0.002589361 0.02589806 0.1165645 0.3108818 0.2720684 0.2719978
# [2,] 0.002589415 0.02589722 0.1165599 0.3108747 0.2720749 0.2720039
# [3,] 0.002589738 0.02589714 0.1165539 0.3108615 0.2720788 0.2720189
# [4,] 0.002590797 0.02590083 0.1165520 0.3108412 0.2720638 0.2720515
# [5,] 0.002592925 0.02592074 0.1166035 0.3108739 0.2719451 0.2720638
# [6,] 0.002588814 0.02590459 0.1166029 0.3109419 0.2720166 0.2719451

只有前4位数字与您输入的tol = 1e-4相同.

Only the first 4 digits are the same, as you put tol = 1e-4.

浮点数最多可以包含16位数字，因此建议您使用tol = 1e-16进行可靠的收敛性测试.

A floating point number has a maximum of 16 digits, so I would suggest you use tol = 1e-16 for reliable convergence test.

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