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I have difficulties to determine the stationary distribution of a markov model. I start to understand the theory and connections: Given a stochastic matrix, to dermine the stationary distribution we need to find the eigen vector for the largest eigenvalue (that is 1)

I started with generating a stochastic matrix

set.seed(6534)
stoma <- matrix(abs(rnorm(25)), nrow=5, ncol=5)
stoma <- (stoma)/rowSums(stoma) # that should make it a stochastic matrix rowSums(stoma) == 1

Afterward I use the R eigen function

ew <- eigen(stoma)

But I dont understand the results

> ew
$values
[1]  1.000000e+00+0.000000e+00i -6.038961e-02+0.000000e+00i -3.991160e-17+0.000000e+00i
[4] -1.900754e-17+1.345763e-17i -1.900754e-17-1.345763e-17i

$vectors
              [,1]           [,2]          [,3]                  [,4]                  [,5]
[1,] -0.4472136+0i  0.81018968+0i  0.3647755+0i -0.0112889+0.1658253i -0.0112889-0.1658253i
[2,] -0.4472136+0i  0.45927081+0i -0.7687393+0i  0.5314923-0.1790588i  0.5314923+0.1790588i
[3,] -0.4472136+0i  0.16233945+0i  0.2128250+0i -0.7093859+0.0000000i -0.7093859+0.0000000i
[4,] -0.4472136+0i -0.09217315+0i  0.4214660+0i -0.1305497-0.1261247i -0.1305497+0.1261247i
[5,] -0.4472136+0i -0.31275073+0i -0.2303272+0i  0.3197321+0.1393583i  0.3197321-0.1393583i

The vector for the largest value (1) has all the same component value "-0.4472136". Even if I change the seed, to draw different number, I get the same values again. What do i miss ? Why the components of the eigenvector are all eqaul? Why they do not sum up to 1 - since this should be a stationary distribtuion?

Thank you for your help!

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2 回答 2

5

这是左右特征向量之间的混淆。试试eigen(t(stoma))。

我发现尝试wikipedia article on stochastic matrices中的示例很有用：

p <- matrix(c(0,0,1/4,0,0,0,0,1/4,0,0,
       1/2,1,0,1/2,0,0,0,1/4,0,0,1/2,0,1/4,1/2,1),
     ncol=5)
p
     [,1] [,2] [,3] [,4] [,5]
## [1,] 0.00 0.00  0.5 0.00 0.50
## [2,] 0.00 0.00  1.0 0.00 0.00
## [3,] 0.25 0.25  0.0 0.25 0.25
## [4,] 0.00 0.00  0.5 0.00 0.50
## [5,] 0.00 0.00  0.0 0.00 1.00

检查它是一个随机矩阵：

rowSums(p)
## [1] 1 1 1 1 1

特征值：

zapsmall(eigen(p)$values)
## [1]  1.0000000  0.7071068 -0.7071068  0.0000000  0.0000000

特征向量：

print(zapsmall(eigen(p)$vectors),digits=3)

##       [,1]  [,2]   [,3]   [,4]   [,5]
## [1,] 0.447 0.354  0.354 -0.802 -0.609
## [2,] 0.447 0.707  0.707  0.535 -0.167
## [3,] 0.447 0.500 -0.500  0.000  0.000
## [4,] 0.447 0.354  0.354  0.267  0.776
## [5,] 0.447 0.000  0.000  0.000  0.000

（结果与您相同，但符号任意翻转。R 缩放特征向量，使每列的平方和为 1：sqrt(1/5)约为 0.447 ...）

您正在寻找其他（左？）特征向量：

print(zapsmall(eigen(t(p))$vectors),digits=3)
##      [,1]   [,2]    [,3] [,4]   [,5]
## [1,]    0 -0.149  0.3011 -0.5  0.707
## [2,]    0 -0.149  0.3011  0.5  0.000
## [3,]    0 -0.422 -0.8517  0.0  0.000
## [4,]    0 -0.149  0.3011 -0.5 -0.707
## [5,]    1  0.869 -0.0517  0.5  0.000
于 2014-01-17T15:48:16.600 回答
3

使用包最容易计算平稳分布（也称为稳态）markovchain。

library(markovchain)
mc <- new("markovchain", transitionMatrix = stoma)
steadyStates(mc)

这给了你相同的答案

ev <- eigen(t(stoma))
ev$vectors[, 1] / sum(ev$vectors[, 1])
于 2014-01-17T15:55:59.677 回答