1
$\begingroup$

$Y \ = \ X' \beta \ + \ e $

Where $Y = (y_1, ..., y_n)$ and $\beta = (\beta_0,..., \beta_k)$.

Why would Weak Exogenity under random sampling produce Strong Exogenity?

I know that weak exogenity is defined as $ \ E(e_i | x_i) = 0 \hspace{1em} \forall \ i \ $ and strong exogenity is defined as $E(e_i | x_j) = 0 \hspace{1em} \forall \ i,j \in \left\lbrace 1, \cdots, n \right\rbrace $.

And I didn't understand the claim that $ \ \ E(e_i | X) = E(e_i | x_1, \cdots, x_n) = E(e_i | x_i) = 0 \ \ $

I know $ \ \ E(e_i| x_i, x_j) = E(e_i| x_i) \ \ $ if $e_i$ is independent of $x_j$.

And I thought that random sampling is about $e_i$ being independent of $e_j$ for $i \neq j$. (or is it about $x_i$ and $x_j$? I'm confused...)

But why would $e_i$ be independent of $x_j$s due to random sampling?

Thank you so much for your help in advance!

$\endgroup$

1 Answer 1

2
$\begingroup$

Random sampling is about the multivariate realization $(y_i,x_{1i},x_{2i},...x_{ki})$ and imply that realization $i$ is independent of any others. $i=1,...,n$

Moreover the condition $E[e_i|x_{1i},x_{2i},...x_{ki}]=0$ imply $E[e_i|x_{ji}]=0$ fon any $j=1,...,k$. In word, and in the sense that you intend, strict exogeneity imply the weak.

So the claim that you do not understand is correct.

I see that you looking for other related statistical aspects here: Exogenity: What does E(eX) really mean and why is it used?

All of them are relevant. Maybe you are student and I do not want to put you in the confusion. However I have to warning you that the proper meaning of exogeneity not stay in detail about usual statistical dependence measures (mean independance, scorrelation, orthogonality, indipendence, conditional independence, ecc): Exogeneity have, or should have, a clear causal role. This role must be enclosed in causal models, then the role of structural equations, not only regression equations, is crucial. Read here can help:

What is the actual definition of endogeneity?

How are standard exogeneity assumptions and indepent of potential outcomes concepts linked?

Zero conditional expectation of error in OLS regression

Endogeneity testing using correlation test

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.