Time Series Analysis, Exercise 5.4

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(a): Is $\{Y_t\}$ stationary when $A$ and $B$ are constants?

Suatu time series $\{Y_t\}$ dikatakan stasioner, apabila mean, variansi, dan autokovariansi independen terhadap $t$.

Karena $\{X_t\}$ random walk, maka $E(X_t)=0$.

Sehingga, $E(Y_t)=E(A+Bt+X_t)=A+Bt$.

Karena mean ($E(Y_t)$) bergantung kepada $t$, maka $\{Y_t\}$ tidak stasioner.

(b): Is $\{\nabla Y_t\}$ stationary when $A$ and $B$ are constants?

Hitung $\Delta Y_t$:

\begin{align} \nabla Y_t &= Y_t - Y_{t-1} \\ &= (A + Bt + X_t) - (A + B(t-1) + X_{t-1})\\ &= A + Bt + X_t - A - Bt + B - X_{t-1} \\ &= B + (X_t - X_{t-1})\\ &= B + e_t \quad (\text{Karena }X_t = X_{t-1} + e_{t}) \end{align}$$ Didapati: - $E(\nabla Y_t) = E(B + e_t) = B + 0 = B$. - $\operatorname{Var}(\nabla Y_t) = \operatorname{Var}(B + e_t) = \operatorname{Var}(e_t) = \sigma_e^2$. - $\operatorname{Cov}(\nabla Y_t, \nabla Y_{t-k}) = \operatorname{Cov}(B + e_t, B + e_{t-k}) = \operatorname{Cov}(e_t, e_{t-k})$. Karena $\{e_t\}$ adalah white noise, $\operatorname{Cov}(e_{t}, e_{t-k}) = 0$ saat $k\neq 0$, dan $\operatorname{Var}(e_{t}) = \sigma_{e}^2$ saat $k=0$. Karena mean, variansi, dan autokovariansi konstan, maka $\{\Delta Y_{t}\}$ stasioner. ## (c): Is $\{Y_t\}$ stationary when $A$ and $B$ are random variables independent of $\{X_t\}$? Misal $E(A) = \mu_A$, $E(B) = \mu_B$, $\operatorname{Var}(A) = \sigma_A^2$, $\operatorname{Var}(B) = \sigma_B^2$, dan $\operatorname{Cov}(A, B) = \sigma_{AB}$. Didapati, $E(Y_t) = E(A + Bt + X_t) = E(A) + E(B)t + E(X_t) = \mu_A + \mu_B t + X_0$. Karena mean konstan, maka $\{Y_t\}$ stasioner. ## (d): Is $\{\nabla Y_t\}$ stationary when $A$ and $B$ are random variables independent of $\{X_t\}$? $$\begin{align} \nabla Y_t &= Y_t - Y_{t-1} \\ &= (A + Bt + X_t) - (A + B(t-1) + X_{t-1}) \\ &= A + Bt + X_t - A - Bt + B - X_{t-1} \\ &= B + (X_t - X_{t-1}) \\ &= B + e_t \\ \end{align}$$ Didapati: - $E(\nabla Y_t) = E(B + e_t) = E(B) + E(e_t) = \mu_B$ - $\operatorname{Var}(\nabla Y_t) = \operatorname{Var}(B + e_t) = \operatorname{Var}(B) + \operatorname{Var}(e_t) = \sigma_B^2 + \sigma_e^2$. - $\operatorname{Cov}(\nabla Y_t, \nabla Y_{t-k}) = \operatorname{Cov}(B + e_t, B + e_{t-k}) = \operatorname{Cov}(B, B) + \operatorname{Cov}(e_t, e_{t-k})$. Karena $$\begin{align} \operatorname{Cov}(B, B) &= \sigma_B^2\\ \operatorname{Cov}(e_t, e_{t-k}) &= \begin{cases} 0 &, k \neq 0\\ \sigma_e^2 &, k = 0 \end{cases} \end{align}$$ maka,

\operatorname{Cov}(\nabla Y_t, \nabla Y_{t-k}) = \begin{cases} \sigma_B^2 &, k = 0\ 0 &, k \neq 0 \end{cases}

Karena mean, variansi, dan autokovariansi konstan, maka $\{\nabla Y_t\}$ stasioner.