Tugas 3

• Versi web: https://notes.fazuh.com/4-Projects/tugas-3_202603130940

Question

Suppose that $Y_t=A+Bt+X_t$, where $\{X_t\}$ is a random walk. First suppose that $A$ and $B$ are constants.

• a. Is $\{Y_t\}$ stationary?
• b. Is $\{\nabla Y_t\}$ stationary?

Now suppose that $A$ and $B$ are random variables that are independent of the random walk $\{X_t\}$.

• c. Is $\{Y_t\}$ stationary?
• d. Is $\{\nabla Y_t\}$ stationary?

Answer

a.

NOTE

Quote from 1_Pertemuan 1a.pdf page 35:

“Dalam kuliah ini, istilah stasioner jika digunakan sendiri akan selalu mengacu pada bentuk stasioneritas yang lebih lemah”

Suatu stochastic process $\{Z_t\}$ dikatakan weakly stationary apabila:

1. $E[Z_t]=\mu$ (mean konstan)
2. $\text{Var}(Z_t)<\infty ,\forall t$ (variansi ada)
3. $\text{Cov}(Z_t,Z_{t-k})={\gamma }_k$ (cov independen dari $t$)

Akan dicari apakah $Y_t=A+Bt+X_t$ dimana $\{X_t\}$ adalah random walk dan $A$ dan $B$ konstan.

Diketahui dari sifat random walk¹ bahwa $E[X_t]=0$.

Sehingga,

$$\begin{array}{rl}{Y}_t& =A+Bt+X_t\\ & \\ E[Y_t]& =E\left[A+Bt+X_t\right]\\ & =A+Bt+E[X_t]\\ & =A+Bt\end{array}$$

$\therefore$ Karena $E[Y_t]$ berbeda untuk setiap $t$, maka $Y_t$ tidak stasioner

b.

Akan dicari apakah $\nabla Y_t=Y_t-Y_{t-1}$ stasioner:

$$\begin{array}{rl}\nabla Y_t& =Y_t-Y_{t-1}\\ & =(A+Bt+X_t)-(A+B(t-1)+X_{t-1})\\ & =B+e_t\\ & \\ E[\nabla Y_t]& =E[B+e_t]\\ & =B(\text{konstan }\forall t)\\ & \\ \text{Var}(\nabla Y_t)& =\text{Var}(B+e_t)\\ & =\text{Var}(e_t)\\ & ={\sigma }_e^2<\infty ,\forall t\\ & \\ \text{Cov}(\nabla Y_t,\nabla Y_{t-k})& =\text{Cov}(B+e_t,B+e_{t-k})\\ & =\text{Cov}(e_t,e_{t-k})\\ & =0\end{array}$$

$\therefore$ Terbukti $\nabla Y_t$ adalah stasioner.

c.

Andaikan $A$ dan $B$ variabel acak yang independen terhadap $\{X_t\}$

Dapat diperoleh:

$$\begin{array}{rl}{Y}_t& =A+Bt+X_t\\ & \\ E[Y_t]& =E\left[A+Bt+X_t\right]\\ & =E[A]+tE[B]\\ & \\ \text{Var}(Y_t)& =\text{Var}(A)+\text{Var}(Bt)+\text{Var}(X_t)+2\text{Cov}(A,Bt)+2\text{Cov}(A,X_t)+2\text{Cov}(Bt,X_t)\\ & =\text{Var}(A)+t^2\text{Var}(B)+t{\sigma }_e^2+2t\text{Cov}(A,B)+0+0\\ & \\ \text{Cov}(Y_t,Y_{t-k})& =E[Y_t{Y}_{t-k}]-E[Y_t]E[Y_{t-k}]\\ & =E[(A+Bt+X_t)(A+B(t-k)+X_{t-k})]-(E[A]+tE[B])(E[A]+(t-k)E[B])\\ & =E[A^2]+E[AB](t-k)+E[AB]t+E[B^2]t(t-k)+E[X_t{X}_{t-k}]\\ & -(E[A]{)}^2-E[A]E[B](t-k)-E[A]E[B]t-(E[B]{)}^{2}t(t-k)\\ & =\text{Var}(A)+(2t-k)\text{Cov}(A,B)+t(t-k)\text{Var}(B)+(t-k){\sigma }_e^2\end{array}$$

Kondisi 1 tidak terpenuhi karena $E[Y_t]$ bergantung pada $t$ (kecuali $E[B]=0$).

Kondisi 3 tidak terpenuhi karena $\text{Cov}(Y_t,Y_{t-k})$ bergantung pada $t$ untuk semua nilai $A$, $B$, ${\sigma }_e^2$ nontrivial.

$\therefore$ $\{Y_t\}$ tidak stasioner.

d.

Dapat diperoleh:

$$\begin{array}{rl}\nabla Y_t& =Y_t-Y_{t-1}\\ & =(A+Bt+X_t)-(A+B(t-1)+X_{t-1})\\ & =B+e_t\\ & \\ E[\nabla Y_t]& =E[B+e_t]\\ & =E[B](\text{konstan }\forall t)\\ & \\ \text{Var}(\nabla Y_t)& =\text{Var}(B+e_t)\\ & =\text{Var}(B)+\text{Var}(e_t)\\ & =\text{Var}(B)+{\sigma }_e^2<\infty \forall t\\ & \\ \text{Cov}(\nabla Y_t,\nabla Y_{t-k})& =\text{Cov}(B+e_t,B+e_{t-k})\\ & =\text{Cov}(B,B)+\text{Cov}(B,e_{t-k})+\text{Cov}(e_t,B)+\text{Cov}(e_t,e_{t-k})\\ & =\text{Var}(B)+\text{Cov}(e_t,e_{t-k})\\ & =\text{Var}(B)(k\ne 0)\end{array}$$

$\therefore$ Ketiga syarat terpenuhi: mean $E[B]$ konstan, variansi $\text{Var}(B)+{\sigma }_e^2<\infty$, dan kovarians hanya bergantung pada $k$ . Maka $\{\nabla Y_t\}$ stasioner.

Footnotes

1. Random walk ↩