OLS results
Problem
\(e_i\) are the OLS residuals and \({\hat{y}}_i\) are the OLS fitted values.
a) Show that
\[\sum_{i=1}^{n}{ x_{i,j}e_i }=0\]
b) Show that if \(x_{i,1}=1\)
\[\sum_{i=1}^{n}{ e_i }=0\]
c) Show that
\[\bar{y}=\bar{\hat{y}}\]
d) Show that
\[\sum_{i=1}^{n}{ e_i{\hat{y}}_i }=0\]
Solution
a) From the first order condition of the least squares problem we have
\[X'e=0\]
Both sides are \(k×1\) . If you do the matrix multiplication of \(X'e\) then the first row of this equation is
\[\sum_{i=1}^{n}{ x_{i,1}e_i }=0\]
Similarly, row \(j\) , where \(1≤j≤k\) , is
\[\sum_{i=1}^{n}{ x_{i,j}e_i }=0\]
b) Follows immediately from a)
c) \(y=\hat{y}+e\) . Therefore, \(\bar{y}=\bar{\hat{y}}+\bar{e}=\bar{\hat{y}}\) since \(\bar{e}=0\) .
d) \({\hat{y}}_i=x'_ib\) so
\[\sum_{i=1}^{n}{ e_i {\hat{y}}_i }=\sum_{i=1}^{n}{ e_ix'_ib }=\left( \sum_{i=1}^{n}{ e_ix'_i } \right)b=0\]
since
\[\sum_{i=1}^{n}{ e_ix'_i }=\sum_{i=1}^{n}{ e_i\left( x_{i,1}, \ldots ,x_{i,k} \right) }=\left( \sum_{i=1}^{n}{ e_ix_{i,1} }, \ldots ,\sum_{i=1}^{n}{ e_ix_{i,k} } \right)=(0, \ldots ,0)\]