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Consider the Model: Yt = 0 + 1zt1

Question 13

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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>) = . .

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