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I am preparing for my final in Econometrics but I am confused over a new problem I encountered. I think I have solved it but I am unsure whether I am not making any gross mistakes.

This is the study of a simple macroeconomic model that is defined as:

$$C_t=\alpha_0+\alpha_1Y_t+u_t$$ and $$Y_t=C_T+I_t$$ We further assume that $Cov(I,u)=0$ and denote $\sigma_I^2$ and $\sigma_u^2$ the variance of $I$ and $u$.

Questions are as follows:

a) Express $Y_t$ in terms of $I_t$, $u_t$, and $\alpha's$

$Y_t=\dfrac{\alpha_0+u_t+I_t}{1-\alpha_1} $

b) Express $Cov(Y,u)$ in terms of $\sigma_I^2$, $\sigma_u^2$, and $\alpha's$

I attempt the following:

$Cov(Y,u)=Cov(C+I,u)=Cov(C,u)+Cov(I,u)=Cov(C,u)$ as we know $Cov(I,u)=0$

$Cov(C,u)=Cov(\alpha_0+\alpha_1Y+u,u)=\alpha_1Cov(Y,u)+Var(u)$

And this is where I get confused as I loop back to the beginning. Should I approach it using $Y_t$ derived in question 1?

$Cov(Y,u)=Cov(\dfrac{\alpha_0+u+I}{1-\alpha_1},u)=\dfrac{1}{1-\alpha_1}Cov(u+I,u)=\dfrac{1}{1-\alpha_1}\sigma_u^2$

c) Express $Cov(Y,I)$ in terms of $\sigma_I^2$, $\sigma_u^2$, and $\alpha's$

$Cov(Y,I)=Cov(\dfrac{\alpha_0+u+I}{1-\alpha_1},I)=\dfrac{1}{1-\alpha_1}Cov(u+I,I)=\dfrac{1}{1-\alpha_1}\sigma_I^2$

d) Express $Cov(C,I)$ in terms of $\sigma_I^2$, $\sigma_u^2$, and $\alpha's$

$Cov(C,I)=Cov(Y-I,I)=\dfrac{1}{1-\alpha_1}\sigma_I^2-\sigma_I^2=\sigma_I^2\left[\dfrac{\alpha_1}{1-\alpha_1}\right]$

e) Express $Var(Y)$ in terms of $\sigma_I^2$, $\sigma_u^2$, and $\alpha's$

$Var(Y)=Var(\dfrac{\alpha_0+u+I}{1-\alpha_1})=\left[\dfrac{1}{1-\alpha_1}\right]^2Var(u+I)=\left[\dfrac{1}{1-\alpha_1}\right]^2(\sigma_I^2+\sigma_u^2)$

f) What is $plim(\hat{\alpha_{1_{OLS}}})$

$plim(\hat{\alpha_{1_{OLS}}})=\alpha_1+\dfrac{Cov(Y,u)}{Var(Y)}=\alpha_1+\dfrac{\dfrac{1}{1-\alpha_1}\sigma_u^2}{\left[\dfrac{1}{1-\alpha_1}\right]^2(\sigma_I^2+\sigma_u^2)}$

$=\alpha_1+\dfrac{(1-\alpha_1)\sigma_u^2}{(\sigma_I^2+\sigma_u^2)}$

f) What is $plim(\hat{\alpha_{1_{IV}}})$ using $I$ as an instrument for GNP (Instrumental Variable approach)

$plim(\hat{\alpha_{1_{IV}}})=\alpha_1+\dfrac{Cov(I,u)}{Cov(Y,I)}=\alpha_1$

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Yes. Given that you have solved (a), it is easiest to solve (b) by using the result in (a). It gives you $Cov(Y,u)(1-\alpha_1)=Cov(\alpha_0,u)+Cov(u,u)+Cov(I,u)$. Since $Cov(\alpha_0,u)=Cov(I,u)=0$ and $Cov(u,u)=Var(u)$, $Cov(Y,u)(1-\alpha_1)=Var(u)$. Hence, $Cov(Y,u)=Cov(u)/(1-\alpha_1)$.

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