Regression model + expected value, variance and autocorrelation of the error term

Question

Consider this regression model $$Y_t=X_t\beta+\epsilon_t, ~~~~~~~~~~\epsilon_t \sim WN(0, \sigma^2_{\epsilon})$$

with 3 different specifications of the error term:

1. $\epsilon_t=\alpha_1\epsilon_{t-1}+\alpha_2\epsilon_{t-2}+\eta_t,~~~~~~~~~~~~~~~~~~~~~~~~~~\eta_t \sim WN(0, \sigma^2_{\eta}) $
2. $\epsilon_t=\eta_t+\theta_1\eta_{t-1}+\theta_2\eta_{t-2},~~~~~~~~~~~~~~~~~~~~~~~~~~\eta_t \sim WN(0, \sigma^2_{\eta})$
3. $\epsilon_t=\alpha_1\epsilon_{t-1}+\eta_t~~~~~~~~~~$ with $|\alpha|=1, ~~~~~~~~~~\eta_t \sim WN(0, \sigma^2_{\eta})$

a) Compute the expected value, the variance and the autocorrelation coefficients of $\epsilon_t$ for each specification

b) Transform the model $Y_t=X_t\beta+\epsilon_t$ in first differences and let $\varepsilon_t$ be the transformed error. Show that $\varepsilon_t$ is not white noise.

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My attempt was the following:

POINT A) - (1) FIRST SPECIFICATION

$E(\epsilon_t) =E(\alpha_1\epsilon_{t-1}+\alpha_2\epsilon_{t-2}+\eta_t)$

$~~~~~~~~~~~~~~~~=\alpha_1E(\epsilon_{t-1})+\alpha_2E(\epsilon_{t-2})+E(\eta_t)$

$~~~~~~~~~~~~~~~~=0$

$\operatorname{Var}(\epsilon_t)=\operatorname{Var}(\alpha_1\epsilon_{t-1}+\alpha_2\epsilon_{t-2}+\eta_t)$

$~~~~~~~~~~~~~=\alpha_1^2 \operatorname{Var}(\epsilon_{t-1})+\alpha_2^2\operatorname{Var}(\epsilon_{t-2})+\operatorname{Var}(\eta_t)$

$~~~~~~~~~~~~~=\alpha_1^2\cdot\sigma^2_{\epsilon}+\alpha_2^2\cdot\sigma^2_{\epsilon}+\sigma^2_{\eta}$

$\rho(h)=\frac{\operatorname{Cov}(\epsilon_t,\epsilon_{t-h})}{\operatorname{Var}(\epsilon_t)}$ From this point, how can I compute $\operatorname{Cov}(\epsilon_t,\epsilon_{t-h})$?

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For the other ones

Basically, I have to follow the same reasoning (hoping that the previous calculus are correct)

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POINT B) - SHOW THAT $\varepsilon_t$ IS NOT WHITE NOISE

$$Y_t-Y_{t-1}=(X_t-X_{t-1})\beta+\varepsilon_t,~~~~~~~~~~\varepsilon_t=\epsilon_t-\epsilon_{t-1}$$

How can I prove that the transformed error term is not white noise?

Any help would be appreciated

Answer 1

Your variance is wrong: $\epsilon_{t-1}$ and $\epsilon_{t-2}$ are not independent!

Hint: $$ \mathbf{E}\big[\eta_t \epsilon_t] = \mathbf{E}\big[\eta_t(\alpha_1\epsilon_{t-1} + \alpha_2\epsilon_{t-2} + \eta_t)] = \sigma_\eta^2; $$ because $\epsilon_k$ is independent of $\eta_j$ if $j>k$ (think about why this is the case!); similarly, $$\begin{aligned} \mathbf{E}\big[\eta_{t-1} \epsilon_t] &= \mathbf{E}\big[\eta_{t-1}(\alpha_1\epsilon_{t-1} + \alpha_2\epsilon_{t-2} + \eta_t)] = \alpha_1\sigma_\eta^2;\\ \mathbf{E}\big[\eta_{t-2} \epsilon_t] &= \mathbf{E}\big[\eta_{t-2}(\alpha_1\epsilon_{t-1} + \alpha_2\epsilon_{t-2} + \eta_t)] = \alpha_2\sigma_\eta^2;\\ \end{aligned}$$ Hence, $$\begin{aligned} \mathrm{Var}(\epsilon_t) &= \mathbf{E}\big[\epsilon_t \epsilon_t\big]\\ &= \mathbf{E}\big[(\alpha_1 \epsilon_{t-1} + \alpha_2\epsilon_{t-2} + \eta_t) \epsilon_t\big]\\ &= \mathbf{E}\big[(\alpha_1 (\alpha_1 \epsilon_{t-2} + \alpha_2 \epsilon_{t-3} + \eta_{t-1}) + \alpha_2(\alpha_1\epsilon_{t-3} + \alpha_2\epsilon_{t-4} + \eta_{t-2}) +\eta_t\big) \epsilon_t\big] \\\ &=\cdots \end{aligned} $$