Math

STAT 5160 HW4

1

(1) (1-1.05B+0.4B2)Xt=Zt => Xt=1.05 Xt-1-0.4 Xt-2+ Zt, Zt~iid.N(0,1) => AR(2) model => stationarity condition:

|φ2|<1 and φ1 + φ2 <1 and φ2 – φ1 <1 =>|-0.4|<1 and 1.05  + (-0.4) = 0.65<1 and  (-0.4) – 1.05 =-1.45<1 which satisfies all three conditions => This process is stationary

(2) (1-1.05B)Xt=Zt => Xt=1.05 Xt-1+ Zt, Zt~iid.N(0,1) => AR(1) model => stationarity condition:|φ1|<1 =>|1.05|>1 which not satisfies the condition

(3) (1+0.8B)Xt=(1-0.25B)Zt => Xt=-0.8Xt-1 + Zt – 0.25Zt-1, Zt~iid.N(0,1) => ARMA(1,1) model => stationarity condition: |φ1|<1 and invertibility contition: |θ1|<1 => stationarity condition: |-0.8|<1 and invertibility contition: |0.25|<1 => Both stationarity condition and invertibility contition satisfied => This process is stationary and invertible.

(4) Xt=(1-0.7B+0.5B2)Zt => Xt= Zt – 0.7Zt-1+0.5 Zt-2, Zt~iid.N(0,1) => MA(2) model => invertibility condition:

| θ2|<1 and θ 1 + θ 2 <1 and θ 2 – θ 1 <1 =>|0.7|<1 and 0.7+ (-0.5) = 0.2<1 and (-0.5) -0.7=-1.2<1 which satisfies all three conditions) => This process is invertible.

2.

(a) H0: not stationary vs Ha: stationary

> adf.test(x,alternative="stationary")

        Augmented Dickey-Fuller Test

data:  x

Dickey-Fuller = -3.1088, Lag order = 3, p-value = 0.1284

alternative hypothesis: stationary

From the analysis above, since p-value = 0.1284 > a=0.05, then we will conclude H0, this data is not stationary

However, from SACF and SPACF, this is obviously a AR(1) model. In this situation, we will use stationarity condition:|φ1|<1 to judge if it is stationary or not.

[pic 1] 

From the plot above we will clearly see that |φ1|<1 which satisfies stationarity condition:|φ1|<1 thus this data should be stationary.\

(b) H0: not stationary vs Ha: stationary

> x1<-diff(x,differences = 1)

> adf.test(x1,alternative="stationary")

        Augmented Dickey-Fuller Test

data:  x1

Dickey-Fuller = -4.2848, Lag order = 3, p-value = 0.01