R语言实现ARMA模型的估计

1、基于R的ARMA模型的估计首先，我们给出一个ARMA模型：=一6七+匕一-8s t 1随机生成一组含200个观测值的时间序列，代码如下：#ARMA(1,1) yt=-0.6yt-1+xt-0.8xt-1set.seed(10)x-rnorm(200)y-vector(length=2)y1=x1for(i in 2:200)yi=-0.6*yi-1+xi-0.8*xi-1y事实上，在R中有更简单的语句可以生成ARIMA时间序列，以上述ARMA (1,1)模型为例: set.seed(10)y-arima.sim(list(order=c(1,0,1),ar=-0.6,ma=-0.8),n=20

2、0)在本次实验中，我们采用第一种方法生成的时间序列做估计。时间序列图如下：ts.plot(y)TimeACF和PACF图如下：acf(y,xaxp=c(0,20,20),yaxp=c(-1,1,10)pacf(y,xaxp=c(0,20,20),yaxp=c(-1,1,10)模型1:J二 ay+ 8ti t-it模型2：J二 ay+ 8 + b 8ti t-iti t-i模型3：J二 ay+a y +8ti t-i2 t-2【模型1】t下面给出三个模型的估计:a-1;b-0,c-0ARMAcall :arimaCx = y, order = c(l, 0, 0,Ii nt er ce pt -

3、0.0174 0. 0429coeffn cn ents; arl -0.8114 s. -e.0.0405esTi maxed as 1.2; log 11 kel 1 hood = -SQZ. 6, aic = 611. 2SBC准则：#SBC=-2ln(模型中极大似然函数值)+ln(n)(模型中未知参数个数)loglike-ARMA$loglikSBC-2*loglike+log(200)*1SBCA SBCEl 610.495残差平方和：residual-ARMA$residuals #残差 ssr-0for(i in 1:200)ssr=ssr+(residualiA2)ssr #残

4、差平方和ssr #残差平方和1 240.093Ljung-Box 姓验：Box.test(residual, type=Ljung,lag=8,fitdf=1)Box.test(residual, type=Ljung,lag=16,fitdf=1)Box.test(residual, type=Ljung,lag=24,fitdf=1)Eox. test (iresi dual , typ=LJung , 1 ag=B jFi tdf =1) Bow-Ljung testdata: resn dualX-squared = 34.61, df = 7, p-vjil ue = 1. 324-

5、6-05box.test(rcsldual, typ=Ljumg,1ag=16,f1tdf=1) sox-Ljung testdata: resi dualX-5quared = 42.657 , df = 15, p-value = 0.00016 57box. test (res-) dual , typ=Ljurng, 1 ag=24 ,f i tdf =1) Box-Ljung twtdata: resn dualX-squared = 55.BB2 , df = 2 3, p-value = 0.000147【模型2】(R语言代码类似模型1,以下代码省略)a-L;b-0;c-LARM

6、A-ar i maty, or der=cta, b, c) , method=rML)ARMA call :Ar 1 ria(k = y, order = c(a, b, c) , method = ML) si gma/2 esfi mated as 0. S93: log 11 kel 1 hood = -273. 45, ai c = 554. 9coeffi ci ents: arl -0.6267s. e. 0.0599-D. 71500.0604mat intercept-0.01450. 0119#SEC=- Z IriCIS中极大成菸由敬值)十1门门)但蛰中市知爹敬T教)1o

7、glIke-ARMA51ogl1ksbc-?*-!oglnke+log(200)*2SBC1 557.5011resIdual-i-ARMAjresl dual s 系残差ssr-0for(1 In 1:2003+ ssr=ssr+ (rsn dual i .ssr寺我差平方和1 17E.6069eox. test Crcsi dujt! , type=L jung, 1 Jig= tdf=23Box-Ljung TC5Tdata: residualx-squared = 2.0744, df = C, p-value = 0.9127Eox.testfresndual , typ=Ljung

8、,1ag=lt,fntdf=2)Eox-Ljung testclatA： rcsi dualX-squared = 11.407, df = 14, p-value = 0.65SSEox. Creslduiil , type=L jung, 1,f 1 tdf =2)EO?i-Ljung TC5Tdata: r*esi dualX-squared = 20.051, df = 22, p-value = 0.5799【模型3】a-Z;b-0; c-0ARMA box.Test(residual, type=Ljung,1ag=24,f1tdf=2)Bax-Ljung testdata: r

9、esi dualx-squared =43.873,肝=22,p-valU = 0.003678模型1模型2模型3J = a j +ej = a j +8 + b8j = a j + a j +8t1 t1tt1 t1t1 t1t1 t12 t2t残差平方和240.10178.61205.33a的估计值-0.8114-0.6267-1.1203（标准差）(0.0405)(0.0599)(0.0651)%的估计值-0.3771（标准差）(0.0650)b的估计值-0.7150（标准差）(0.0604)AIC； SBC611.2; 610.5554.9; 557.5582.2; 584.8Ljung-Box QQ(8)=34.610(0)Q(8)=2.074(0.9127)Q(8)=20.650(0.0021)统计量（显著Q(16)=42.857(0.0002)Q(16)=11.407(0.6538)Q(16)=31.305(0.0050)水平）Q(24)=55.882(