Let ${(Y_1,X_1),...,(Y_i, X_i),...,(Y_n, X_n)}$ be a random sample (observations are independents and i.i.d) and $Y_i=X_i'b+e_i$ a linear regression. My question: Is the following statement correct? If $(Y_i,X_i)$ and $(Y_j,X_j)$ are independent, then $e_i$ and $e_j$ are independent. My intuition says me which $e_i$ and $e_j$ independent because errors are a transformation of $Y$ and $X$, however I am not sure. I would like have a proof.
$\begingroup$
$\endgroup$
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
I think so.
$$f(e_i,e_j)$$
$$=f(e_i|e_j)f(e_j)$$
$$=f(Y_i - X_i'b|Y_j-X_j'b)f(e_j)$$
$$=f(Y_i-X_i'b)f(e_j)$$
$$=f(e_i)f(e_j)$$
$\begingroup$
$\endgroup$
The result you are looking for is a consquence of the general fact that if two random vectors are independent of one another then any functions of those respective random vectors are also independent of one another. To apply this in the present case, we note that we can rearrage the regression equation to get the function:
$$e_i = h(Y_i,X_i) = Y_i - X_i^\text{T} b.$$
Within classical statistics the parameter $b$ is treated as a fixed constant, so the error term is a function of the random vector $(Y_i,X_i)$. If $(Y_i,X_i)$ and $(Y_j,X_j)$ are independent then any functions of these respective random vectors are independent, so your errors are independent.
