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13: Time-Series Analysis-LOS l习题精选

Session 3: Quantitative Methods: Quantitative
Methods for Valuation
Reading 13: Time-Series Analysis

LOS l: Explain autoregressive conditional heteroskedasticity (ARCH) and discuss how ARCH models can be applied to predict the variance of a time series.

 

 

 

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

re

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

εt2 = 0.4 + 0.80εt-12 + μt, where ε = ε^

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

A)

3.6.

B)

3.2.

C)

2.0.

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

εt2 = 0.4 + 0.80εt-12 + μt, where ε = ε^

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

A)

3.6.

B)

3.2.

C)

2.0.




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

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



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.

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

εt2 = 0.25 + 0.6ε2t-1 + μt, where ε = ε^

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

A)

0.790.

B)

0.736.

C)

0.850.

TOP

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

εt2 = 0.25 + 0.6ε2t-1 + μt, where ε = ε^

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

A)

0.790.

B)

0.736.

C)

0.850.




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

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