Question

(a) Show that E[ε|zi-0 implies Elg(z)E-0 f this result or any function qz. Use to conclude that (eq1) and (eq2) (b) Write down the sample-analogue estimating equations based on the two moment conditions (eql) and (eq2) derived in (a), and show that where, as usual, ỹ Σί.1Vi, ž--Σί-iz, and Ζ-n Σί.i zi. (Note that Ao and β1 are not the usual least-squares estimators presented in class.)

Answer 1

$(a)~E(g(z_i)\epsilon_i )=E(g(z_i)E(\epsilon_i|z_i))=E(g(z_i)*0)=0.\\ E(\epsilon_i )=EE(\epsilon_i|z_i )=0\\ E(z_i\epsilon _i)=E(z_iE(\epsilon_i|z_i ))=0\\ (b)~Here,~model:~y_i=\beta_0+\beta_1x_i+\epsilon_i ;~i=1(1)n\\then~\epsilon_i=y_i-\beta_0-\beta_1x_i;~i=1(1)n \\ eq1\Rightarrow \frac{1}{n}\sum_{i=1}^n\epsilon_i=\frac{1}{n}\sum_{i=1}^n(y_i-\beta_0-\beta_1x_i)=0.........(1)\\ or,~\breve{\beta}_0=\bar{y}-\bar{x}\breve{\beta}_1\\ eq2\Rightarrow \frac{1}{n}\sum_{i=1}^nz_i\epsilon_i=\frac{1}{n}\sum_{i=1}^nz_i(y_i-\beta_0-\beta_1x_i)=0\\ or,~\frac{1}{n}\sum_{i=1}^n(z_i-\bar{z})(y_i-\beta_0-\beta_1x_i)=0~(using~(1),~where~\bar{z}=\frac{1}{n}\sum_{i=1}^nz_i)\\ or,~\frac{1}{n}\sum_{i=1}^n(z_i-\bar{z})(y_i-\bar{y})-\beta_1\frac{1}{n}\sum_{i=1}^n(z_i-\bar{z})(x_i-\bar{x})=0\\ or,~\breve{\beta}_1=\frac{\sum_{i=1}^n(z_i-\bar{z})(y_i-\bar{y})}{\sum_{i=1}^n(z_i-\bar{z})(x_i-\bar{x})}=\frac{\sum_{i=1}^n(z_i-\bar{z})y_i}{\sum_{i=1}^n(z_i-\bar{z})x_i},~since~\sum_{i=1}^n(z_i-\bar{z})=0$