从vars包估计的VAR输出中提取系数和方差-协方差矩阵

3
我希望从我的VAR模型输出中提取系数和方差协方差矩阵(使用vars软件包进行估计)。有没有办法将系数保存到数组中,并将方差协方差矩阵保存到矩阵中,以便我稍后可以从中提取某些数字并用作后续函数的输入(这是我的最终目标)。
以下是一个模拟的估计和结果。
sale1_ts <- ts(sale1, frequency=12, start=c(2012,1))
sale2_ts <- ts(sale2, frequency=12, start=c(2012,1))

#VARMODEL
varendo<-data.frame(sale1_ts,sale2_ts)
names(varendo)<-c("sale1_ts","sale2_ts")
attach(varendo)

library(vars)
fitvar<- VAR(varendo, type = c("both"), season = NULL, lag.max = 12,ic = c("AIC"))
summary(fitvar)

这将导致以下输出。
Endogenous variables: sale1_ts, sale2_ts 
Deterministic variables: both 
Sample size: 21 
Log Likelihood: -43.463 
Roots of the characteristic polynomial:
1.19 1.06 1.06 0.961 0.961 0.96 0.96 0.914 0.914 0.883 0.883 0.84 0.84 0.75
Call:
VAR(y = varendo, type = c("both"), lag.max = 12, ic = c("AIC"))


Estimation results for equation sale1_ts: 
========================================= 
sale1_ts = sale1_ts.l1 + sale2_ts.l1 + sale1_ts.l2 + sale2_ts.l2 + sale1_ts.l3 + sale2_ts.l3 + sale1_ts.l4 + sale2_ts.l4 + sale1_ts.l5 + sale2_ts.l5 + sale1_ts.l6 + sale2_ts.l6 + sale1_ts.l7 + sale2_ts.l7 + const + trend 

            Estimate Std. Error t value Pr(>|t|)
sale1_ts.l1  -0.0576     0.3603   -0.16     0.88
sale2_ts.l1  -0.0384     0.3538   -0.11     0.92
sale1_ts.l2  -0.4768     0.3933   -1.21     0.28
sale2_ts.l2   0.1963     0.3137    0.63     0.56
sale1_ts.l3  -0.0837     0.3245   -0.26     0.81
sale2_ts.l3  -0.2097     0.6108   -0.34     0.75
sale1_ts.l4  -0.5665     0.3653   -1.55     0.18
sale2_ts.l4  -0.3511     0.5174   -0.68     0.53
sale1_ts.l5  -0.1479     0.4199   -0.35     0.74
sale2_ts.l5   0.1263     0.4640    0.27     0.80
sale1_ts.l6  -0.3143     0.3684   -0.85     0.43
sale2_ts.l6   0.1174     0.2700    0.43     0.68
sale1_ts.l7  -0.4727     0.4810   -0.98     0.37
sale2_ts.l7   0.3255     0.3071    1.06     0.34
const        12.5453    13.4200    0.93     0.39
trend         0.1062     0.0759    1.40     0.22

Residual standard error: 1.5 on 5 degrees of freedom
Multiple R-Squared: 0.741,  Adjusted R-squared: -0.0376 
F-statistic: 0.952 on 15 and 5 DF,  p-value: 0.575 

Estimation results for equation sale2_ts: 
========================================= 
sale2_ts = sale1_ts.l1 + sale2_ts.l1 + sale1_ts.l2 + sale2_ts.l2 + sale1_ts.l3 + sale2_ts.l3 + sale1_ts.l4 + sale2_ts.l4 + sale1_ts.l5 + sale2_ts.l5 + sale1_ts.l6 + sale2_ts.l6 + sale1_ts.l7 + sale2_ts.l7 + const + trend 

            Estimate Std. Error t value Pr(>|t|)  
sale1_ts.l1  -0.7135     0.3899   -1.83    0.127  
sale2_ts.l1  -0.0598     0.3829   -0.16    0.882  
sale1_ts.l2  -0.4054     0.4257   -0.95    0.385  
sale2_ts.l2  -0.0300     0.3395   -0.09    0.933  
sale1_ts.l3  -0.7129     0.3512   -2.03    0.098 .
sale2_ts.l3   0.6945     0.6611    1.05    0.342  
sale1_ts.l4  -0.7165     0.3954   -1.81    0.130  
sale2_ts.l4   0.4709     0.5600    0.84    0.439  
sale1_ts.l5   0.1324     0.4545    0.29    0.783  
sale2_ts.l5   0.7127     0.5022    1.42    0.215  
sale1_ts.l6   0.2318     0.3987    0.58    0.586  
sale2_ts.l6  -0.0859     0.2922   -0.29    0.781  
sale1_ts.l7   0.4826     0.5206    0.93    0.397  
sale2_ts.l7   0.5011     0.3324    1.51    0.192  
const        -0.6180    14.5252   -0.04    0.968  
trend         0.1744     0.0822    2.12    0.087 .
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 1.6 on 5 degrees of freedom
Multiple R-Squared: 0.819,  Adjusted R-squared: 0.276 
F-statistic: 1.51 on 15 and 5 DF,  p-value: 0.343 


Covariance matrix of residuals:
         sale1_ts sale2_ts
sale1_ts     2.14    -1.25
sale2_ts    -1.25     2.51

Correlation matrix of residuals:
         sale1_ts sale2_ts
sale1_ts    1.000   -0.541
sale2_ts   -0.541    1.000

我尝试过print()和coefficients(),但是得到的结果很奇怪。例如:

test<-array(coefficients(fitvar), c(16,4,2))

> test
, , 1

      [,1]       [,2]       [,3]       [,4]      
 [1,] Numeric,64 Numeric,64 Numeric,64 Numeric,64
 [2,] Numeric,64 Numeric,64 Numeric,64 Numeric,64
 [3,] Numeric,64 Numeric,64 Numeric,64 Numeric,64
 [4,] Numeric,64 Numeric,64 Numeric,64 Numeric,64
 [5,] Numeric,64 Numeric,64 Numeric,64 Numeric,64
 [6,] Numeric,64 Numeric,64 Numeric,64 Numeric,64
 [7,] Numeric,64 Numeric,64 Numeric,64 Numeric,64
 [8,] Numeric,64 Numeric,64 Numeric,64 Numeric,64
 [9,] Numeric,64 Numeric,64 Numeric,64 Numeric,64
 [10,] Numeric,64 Numeric,64 Numeric,64 Numeric,64
 [11,] Numeric,64 Numeric,64 Numeric,64 Numeric,64
 [12,] Numeric,64 Numeric,64 Numeric,64 Numeric,64
 [13,] Numeric,64 Numeric,64 Numeric,64 Numeric,64
 [14,] Numeric,64 Numeric,64 Numeric,64 Numeric,64
 [15,] Numeric,64 Numeric,64 Numeric,64 Numeric,64
 [16,] Numeric,64 Numeric,64 Numeric,64 Numeric,64

, , 2

      [,1]       [,2]       [,3]       [,4]      
 [1,] Numeric,64 Numeric,64 Numeric,64 Numeric,64
 [2,] Numeric,64 Numeric,64 Numeric,64 Numeric,64
 [3,] Numeric,64 Numeric,64 Numeric,64 Numeric,64
 [4,] Numeric,64 Numeric,64 Numeric,64 Numeric,64
 [5,] Numeric,64 Numeric,64 Numeric,64 Numeric,64
 [6,] Numeric,64 Numeric,64 Numeric,64 Numeric,64
 [7,] Numeric,64 Numeric,64 Numeric,64 Numeric,64
 [8,] Numeric,64 Numeric,64 Numeric,64 Numeric,64
 [9,] Numeric,64 Numeric,64 Numeric,64 Numeric,64
 [10,] Numeric,64 Numeric,64 Numeric,64 Numeric,64   
 [11,] Numeric,64 Numeric,64 Numeric,64 Numeric,64
 [12,] Numeric,64 Numeric,64 Numeric,64 Numeric,64
 [13,] Numeric,64 Numeric,64 Numeric,64 Numeric,64
 [14,] Numeric,64 Numeric,64 Numeric,64 Numeric,64
 [15,] Numeric,64 Numeric,64 Numeric,64 Numeric,64
 [16,] Numeric,64 Numeric,64 Numeric,64 Numeric,64
1个回答

2

要访问系数,您需要运行

fitvar$varresult$sale1_ts$coefficients

这将提供命名向量,例如:

> fitvar$varresult$sale1_ts$coefficients
sale1_ts.l1 sale2_ts.l1       const       trend 
 -0.1799802  -0.1048722  -0.3722163   1.2865624 

要访问协方差/相关性,请使用

summary(fitvar)$covres
summary(fitvar)$corres

提供 matrix 对象:

> summary(fitvar)$covres
           sale1_ts   sale2_ts
sale1_ts  0.7955669 -0.1140167
sale2_ts -0.1140167  0.7013895
> summary(fitvar)$corres
           sale1_ts   sale2_ts
sale1_ts  1.0000000 -0.1526337
sale2_ts -0.1526337  1.0000000

我有类似的问题,我想在某个文件夹中同时访问var result和“Pr>|t|”的系数,请帮忙,非常感谢! - Stackuser

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