Let's carefully address each part of the stochastic process question:

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Given Process:

$p_t = 2 + 0.2p_{t-1} + \varepsilon_t - 0.2\varepsilon_{t-1}, \quad \varepsilon_t \sim i.i.d. \ N(0,1)$

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(a) Is the process stationary? Explain.

A process is stationary if its mean, variance, and autocovariance do not change over time.

• Mean and variance: The autoregressive term $0.2p_{t-1}$ and the moving average component $-0.2\varepsilon_{t-1}$ have constant coefficients.
• Autocovariance: The dependence on past $\varepsilon$ terms and the autoregressive coefficient ensure that autocovariances stabilize over time.

Conclusion: The process is stationary because the coefficients are stable, and the white noise $\varepsilon_t$ has constant properties over time.

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(b) Can the process be represented by an AR($\infty$) process? Explain.

An AR($\infty$) representation is possible for MA processes or mixed ARMA models.

Here, the process is an ARMA(1,1) (due to $0.2p_{t-1}$ and $-0.2\varepsilon_{t-1}$):

• AR terms can theoretically extend indefinitely, making it representable as AR($\infty$).

Conclusion: Yes, the process can be represented by AR($\infty$).

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(c) Does the unconditional mean of $\{p_t\}$ exist? If yes, what is it?

The unconditional mean of a stationary process is: $\mu = \frac{\text{Constant term}}{1 - \phi}, \quad \phi = \text{autoregressive coefficient}.$

Here, the constant term is $2$ and $\phi = 0.2$: $\mu = \frac{2}{1 - 0.2} = 2.5.$

Conclusion: The unconditional mean exists, and it is 2.5.

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(d) Does the unconditional variance of $\{p_t\}$ exist? If yes, what is it?

The unconditional variance for an ARMA process is given by: $\text{Var}(p_t) = \frac{\sigma^2}{1 - \phi^2}, \quad \sigma^2 = \text{variance of }\varepsilon_t.$

Here, $\sigma^2 = 1$ and $\phi = 0.2$: $\text{Var}(p_t) = \frac{1}{1 - 0.2^2} = \frac{1}{1 - 0.04} = \frac{1}{0.96} \approx 1.04.$

Conclusion: The unconditional variance exists, and it is approximately 1.04.

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(e) Can the process be used for forecasting the long-run value of $p_{t+\infty}$?

For a stationary ARMA process, the long-run forecast converges to the unconditional mean: $p_{t+\infty} = \mu = 2.5.$

Conclusion: Yes, the process can forecast the long-run value, and it is 2.5.

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(f) Forecast for $p_{t+2}$ and forecast variance $\sigma^2_{t+2|t}$:

Given:

• $p_t = 2$,
• $p_{t-1} = 1$,
• $p_{t-2} = -1$.

The forecast equations follow: [ f_{t,1} = E[p_{t+1}|t] = 2 + 0.2p_t + E[\varepsilon_{t+1}] = 2 + 0.2(2) = 2.4, ] with $\sigma^2_{t+1|t} = \text{Var}(\varepsilon_{t+1}) = 1$.

For $t+2$: $f_{t,2} = E[p_{t+2}|t] = 2 + 0.2f_{t,1} = 2 + 0.2(2.4) = 2.48,$ and $\sigma^2_{t+2|t} = \text{Var}(\varepsilon_{t+2}) = 1$.

Conclusion:

• Forecast for $p_{t+2}$: 2.48.
• Forecast variance: 1.

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Additional Queries and Tip:

1. Why does stationarity depend on the stability of the coefficients?
2. How does an ARMA(1,1) process compare to a pure AR(1) process in terms of variance?
3. Why does the long-run forecast converge to the mean for stationary processes?
4. Could this process exhibit non-stationarity under different parameter values?
5. How do the autoregressive and moving average terms interact in an ARMA process?

Tip: For ARMA processes, ensure the stationarity condition ($|\phi| < 1$) is satisfied before computing long-run forecasts.