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  3. Study Set
    Introductory Econometrics
  4. Exam
    Exam 11: Further Issues in Using Ols With Time Series Data
  5. Question
    Consider the Model: Y<sub>t</sub> = <Sub>0</sub> + <Sub>1</sub>z<sub>t</sub><sub>1</sub>
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Consider the Model: Yt = 0 + 1zt1

Question 1

Question 1

Multiple Choice

Consider the model: yt = Consider the model: y<sub>t</sub> =   <sub>0</sub> +   <sub>1</sub>z<sub>t</sub><sub>1</sub> +   <sub>2</sub>z<sub>t</sub><sub>2</sub> + u<sub>t</sub>. Under weak dependence, the condition sufficient for consistency of OLS is: A) E(z<sub>t</sub><sub>1</sub>|z<sub>t</sub><sub>2</sub>)  = 0. B) E(y<sub>t</sub> |z<sub>t</sub><sub>1</sub>, z<sub>t</sub><sub>2</sub>)  = 0. C)  E(u<sub>t</sub> |z<sub>t</sub><sub>1</sub>, z<sub>t</sub><sub>2</sub>)  = 0. D)  E(u<sub>t</sub> |z<sub>t</sub><sub>1</sub>, z<sub>t</sub><sub>2</sub>)  =   . 0 + Consider the model: y<sub>t</sub> =   <sub>0</sub> +   <sub>1</sub>z<sub>t</sub><sub>1</sub> +   <sub>2</sub>z<sub>t</sub><sub>2</sub> + u<sub>t</sub>. Under weak dependence, the condition sufficient for consistency of OLS is: A) E(z<sub>t</sub><sub>1</sub>|z<sub>t</sub><sub>2</sub>)  = 0. B) E(y<sub>t</sub> |z<sub>t</sub><sub>1</sub>, z<sub>t</sub><sub>2</sub>)  = 0. C)  E(u<sub>t</sub> |z<sub>t</sub><sub>1</sub>, z<sub>t</sub><sub>2</sub>)  = 0. D)  E(u<sub>t</sub> |z<sub>t</sub><sub>1</sub>, z<sub>t</sub><sub>2</sub>)  =   . 1zt1 + Consider the model: y<sub>t</sub> =   <sub>0</sub> +   <sub>1</sub>z<sub>t</sub><sub>1</sub> +   <sub>2</sub>z<sub>t</sub><sub>2</sub> + u<sub>t</sub>. Under weak dependence, the condition sufficient for consistency of OLS is: A) E(z<sub>t</sub><sub>1</sub>|z<sub>t</sub><sub>2</sub>)  = 0. B) E(y<sub>t</sub> |z<sub>t</sub><sub>1</sub>, z<sub>t</sub><sub>2</sub>)  = 0. C)  E(u<sub>t</sub> |z<sub>t</sub><sub>1</sub>, z<sub>t</sub><sub>2</sub>)  = 0. D)  E(u<sub>t</sub> |z<sub>t</sub><sub>1</sub>, z<sub>t</sub><sub>2</sub>)  =   . 2zt2 + ut. Under weak dependence, the condition sufficient for consistency of OLS is:


A) E(zt1|zt2) = 0.
B) E(yt |zt1, zt2) = 0.
C) E(ut |zt1, zt2) = 0.
D) E(ut |zt1, zt2) =
Consider the model: y<sub>t</sub> =   <sub>0</sub> +   <sub>1</sub>z<sub>t</sub><sub>1</sub> +   <sub>2</sub>z<sub>t</sub><sub>2</sub> + u<sub>t</sub>. Under weak dependence, the condition sufficient for consistency of OLS is: A) E(z<sub>t</sub><sub>1</sub>|z<sub>t</sub><sub>2</sub>)  = 0. B) E(y<sub>t</sub> |z<sub>t</sub><sub>1</sub>, z<sub>t</sub><sub>2</sub>)  = 0. C)  E(u<sub>t</sub> |z<sub>t</sub><sub>1</sub>, z<sub>t</sub><sub>2</sub>)  = 0. D)  E(u<sub>t</sub> |z<sub>t</sub><sub>1</sub>, z<sub>t</sub><sub>2</sub>)  =   . .

Correct Answer:

verifed

Verified

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