Q1. Which of the following is least likely a consequence of a model containing ARCH(1) errors? The:

A)   variance of the errors can be predicted.

B)   regression parameters will be incorrect.

C)   model's specification can be corrected by adding an additional lag variable.

Q2. Suppose you estimate the following model of residuals from an autoregressive model:

ε_t² = 0.25 + 0.6ε²_t-1 + µ_t, where ε = ε^{^}

If the residual at time t is 0.9, the forecasted variance for time t+1 is:

A)         0.736.

B)        0.790.

C)        0.850.

Q3. Suppose you estimate the following model of residuals from an autoregressive model:

ε_t² = 0.4 + 0.80ε_t-1² + µ_t, where ε = ε^{^}

If the residual at time t is 2.0, the forecasted variance for time t+1 is:

A)         3.2.

B)        2.0.

C)        3.6.

答案和详解如下：

Q1. Which of the following is least likely a consequence of a model containing ARCH(1) errors? The:

A)   variance of the errors can be predicted.

B)   regression parameters will be incorrect.

C)   model's specification can be corrected by adding an additional lag variable.

Correct answer is C)

The presence of autoregressive conditional heteroskedasticity (ARCH) indicates that the variance of the error terms is not constant. This is a violation of the regression assumptions upon which time series models are based. The addition of another lag variable to a model is not a means for correcting for ARCH (1) errors.

Q2. Suppose you estimate the following model of residuals from an autoregressive model:

ε_t² = 0.25 + 0.6ε²_t-1 + µ_t, where ε = ε^{^}

If the residual at time t is 0.9, the forecasted variance for time t+1 is:

A)         0.736.

B)        0.790.

C)        0.850.

Correct answer is A)

The variance at t=t+1 is 0.25 + [0.60 (0.81)] = 0.25 + 0.486 = 0.736.

Q3. Suppose you estimate the following model of residuals from an autoregressive model:

ε_t² = 0.4 + 0.80ε_t-1² + µ_t, where ε = ε^{^}

If the residual at time t is 2.0, the forecasted variance for time t+1 is:

A)         3.2.

B)        2.0.

C)        3.6.

Correct answer is C)

The variance at t=t+1 is 0.4 + [0.80 (4.0)] = 0.4 + 3.2. = 3.6.

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