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『経済・ファイナンスデータの計量時系列分析』章末問題をRで解く-第1章時系列分析の基礎概念-

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『経済・ファイナンスデータの計量時系列分析』

の章末問題で「コンピュータを用いて」とあるものをRで解いています。
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数式の問題は『経済・ファイナンスデータの計量時系列分析』章末問題を解く-第1章時系列分析の基礎概念-

1.3

wn<-function(m, s) {
  df<-rnorm(100, mean = m, sd = sqrt(s))
  plot(df, type = 'l', main = sprintf('μ=%d, σ=%d', m, s))
  abline(h = as.integer(min(df)):as.integer(max(df)), col = 'gray', lty = 2)
}

set.seed(1)
par(mfrow = c(3, 2))
par(mar = c(2, 2, 2, 0.5))
wn(0, 1)
wn(2, 1)
wn(-2, 1)
wn(0, 2)
wn(0, 3)
wn(2, 2)

1_1.3.png

1.5

data<-read.table("economicdata.txt", header=T)
data.ts<-ts(data[,2:7],start=c(1975,1),frequency=12)
  • (1)
par(mfrow=c(3,2))
par(mar = c(2, 3, 2, 0.5))
ts.plot(data.ts[,1], xlab="", ylab="", ylim=c(0,3500), main="TOPIX")
ts.plot(data.ts[,2], xlab="", ylab="", ylim=c(0,120), main="実効為替レート")
ts.plot(data.ts[,3], xlab="", ylab="", ylim=c(0,120), main="鉱工業生産指数")
ts.plot(data.ts[,4], xlab="", ylab="", ylim=c(0,120), main="CPI")
ts.plot(data.ts[,5], xlab="", ylab="", ylim=c(0,6), main="失業率")
ts.plot(data.ts[,6], xlab="", ylab="", ylim=c(0,14), main="コールレート")

1_1.5.1.png

(2)(3)

par(mfrow=c(3,1))
topix.d<-diff(log(data$topix))
topix.d.ts<-ts(topix.d*100,start=c(1975,1),frequency=12)
ts.plot(topix.d.ts, ylim=c(-20,15), main="TOPIX")
abline(h=seq(-20,15,5), col="gray", lty=2)
exrate.d<-diff(log(data$exrate))
exrate.d.ts<-ts(exrate.d*100,start=c(1975,1),frequency=12)
ts.plot(exrate.d.ts, ylim=c(-8,12), main="実効為替レート")
abline(h=seq(-8,12,2), col="gray", lty=2)
indprod.d<-diff(log(data$indprod))
indprod.d.ts<-ts(indprod.d*100,start=c(1975,1),frequency=12)
ts.plot(indprod.d.ts, ylim=c(-5,5), main="鉱工業生産指数")
abline(h=-5:5, col="gray", lty=2)

1_1.5.2_3.png

(4)

par(mfrow=c(1,1))
par(mar = c(3, 3, 3, 3))
acf(diff(log(data$indprod)), xlim=c(1,20), ylim=c(-0.5,0.5))
lag=seq(1,10)
box.t<-apply(as.matrix(lag),1,function(x) {tmp=Box.test(diff(log(data$indprod)),type="Ljung", lag=x); return(c(tmp$statistic, tmp$p.value))})
out<-data.frame(matrix(box.t[1,], nrow=1))
out<-rbind(out, box.t[2,])
colnames(out)<-lag
rownames(out)<-c('Q(m)', 'P')
out

1_1.5.4.png

                1            2            3            4            5
Q(m) 3.420025e+01 3.976335e+01 5.039731e+01 5.180377e+01 5.190793e+01
P    4.972263e-09 2.320056e-09 6.574763e-11 1.516132e-10 5.634492e-10
                6            7            8            9           10
Q(m) 5.445974e+01 5.446017e+01 5.745610e+01 5.793698e+01 5.884316e+01
P    5.958688e-10 1.907042e-09 1.466964e-09 3.342017e-09 5.995413e-09

(5)

cf(diff(log(data$topix)), xlim=c(1,20), ylim=c(-0.5,0.5))
box.t<-apply(as.matrix(lag),1,function(x) {tmp=Box.test(diff(log(data$topix)),type="Ljung", lag=x); return(c(tmp$statistic, tmp$p.value))})
out<-data.frame(matrix(box.t[1,], nrow=1))
out<-rbind(out, box.t[2,])
colnames(out)<-lag
rownames(out)<-c('Q(m)', 'P')
out

1_1.5.5.png

                1            2            3            4            5
Q(m) 3.467592e+01 3.638552e+01 3.684399e+01 3.855135e+01 3.891405e+01
P    3.894192e-09 1.255984e-08 4.964713e-08 8.622952e-08 2.471338e-07
                6            7            8            9           10
Q(m) 3.987178e+01 3.996513e+01 4.019884e+01 4.442465e+01 4.596921e+01
P    4.827198e-07 1.278282e-06 2.941722e-06 1.178549e-06 1.452777e-06

⇒ 次は 『経済・ファイナンスデータの計量時系列分析』章末問題をRで解く-第2章ARMA過程-

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