Let's work through each of the two questions based on the ARMA(1,1) model:

ARMA(1,1) Model:

$Y_t = 2 + 0.5 Y_{t-1} + \epsilon_t + 0.5 \epsilon_{t-1}$ Where $\epsilon_t \sim i.i.d.$ (independent and identically distributed) with mean 0 and variance $\sigma_\epsilon^2 = 1$.

Question 8: Unconditional Mean of $Y_t$

The general form of the ARMA(1,1) model is:

$Y_t = \mu + \phi Y_{t-1} + \theta \epsilon_t + \theta_1 \epsilon_{t-1}$

Where $\mu$ is a constant, $\phi$ is the autoregressive coefficient, and $\theta$ is the moving average coefficient.

For the unconditional mean of $Y_t$, we take the expected value of both sides:

$E(Y_t) = 2 + 0.5 E(Y_{t-1}) + E(\epsilon_t) + 0.5 E(\epsilon_{t-1})$

Given that $\epsilon_t$ has a mean of 0 ($E(\epsilon_t) = 0$):

$E(Y_t) = 2 + 0.5 E(Y_{t-1})$

Since the expected value of $Y_t$ is constant over time (stationarity assumption), we can set $E(Y_t) = E(Y_{t-1}) = \mu$. Thus, the equation becomes:

$\mu = 2 + 0.5 \mu$

Solving for $\mu$:

$\mu - 0.5 \mu = 2$ $0.5 \mu = 2$ $\mu = 4$

So, the unconditional mean of $Y_t$ is 4.0.

Thus, the correct answer is (c) 4.0.

Question 9: Unconditional Variance of $Y_t$

Now, let's find the unconditional variance of $Y_t$.

We start by calculating the variance of the ARMA model. The variance of $Y_t$ is given by:

$\text{Var}(Y_t) = \text{Var}(2 + 0.5 Y_{t-1} + \epsilon_t + 0.5 \epsilon_{t-1})$

The variance of a constant is zero, so we have:

$\text{Var}(Y_t) = \text{Var}(0.5 Y_{t-1} + \epsilon_t + 0.5 \epsilon_{t-1})$

We can use the following properties of variance:

$\text{Var}(aX + bY) = a^2 \text{Var}(X) + b^2 \text{Var}(Y) + 2ab \, \text{Cov}(X,Y)$

Now, considering the components:

1. Autoregressive part: $\text{Var}(Y_{t-1}) = \text{Var}(Y_t)$
2. Moving average part: Since $\epsilon_t$ are i.i.d., $\text{Var}(\epsilon_t) = 1$

Substituting these into the variance formula:

$\text{Var}(Y_t) = 0.5^2 \text{Var}(Y_t) + 1 \cdot \text{Var}(\epsilon_t) + 0.5^2 \text{Var}(\epsilon_{t-1})$ $\text{Var}(Y_t) = 0.25 \text{Var}(Y_t) + 1 + 0.25$ $\text{Var}(Y_t) = 0.25 \text{Var}(Y_t) + 1.25$

Let $\text{Var}(Y_t) = \sigma_Y^2$. Solving for $\sigma_Y^2$:

$\sigma_Y^2 - 0.25 \sigma_Y^2 = 1.25$ $0.75 \sigma_Y^2 = 1.25$ $\sigma_Y^2 = \frac{1.25}{0.75} = 1.6667$

Thus, the unconditional variance of $Y_t$ is approximately 1.667.

So, the correct answer is (d) 1.667.

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Final Answers:

• Unconditional Mean of $Y_t$: (c) 4.0
• Unconditional Variance of $Y_t$: (d) 1.667

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Would you like more detailed explanations or clarifications on any steps? Here's a set of related questions you may find interesting:

1. What is the interpretation of the coefficient $\phi = 0.5$ in the ARMA(1,1) model?
2. How would the unconditional mean and variance change if the autoregressive parameter $\phi$ was increased?
3. How does the variance of the errors $\epsilon_t$ affect the variance of $Y_t$?
4. What is the difference between an ARMA model and an ARIMA model?
5. How do you test for stationarity in an ARMA model?

Tip: When solving for variances in time series models like ARMA, it's useful to remember that the process should be stationary, which means the mean and variance do not change over time, simplifying the calculation.