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In OLS we assume that Given the model : $Y|X = F(X) + U|X $ Where U is the residuals ,we then ASSUME $E(U|X) = 0$ in order to have $prediction = F(X) = E(Y|X)$ . So the $E(U|X) = 0$ is an assumption and not a consequence of using OLS , wich means that we could have wrong parameters estimates if the assumption does not hold ( including bias in the parameters )

However It is known that minimizing the MSE in linear regression gives the $E(Y|X)$ as a minima , So we could also say that Given the model : $Y|X = F(X) + U|X $ Where U is the residuals . We search for $F(X)$ such that it minimizes the MSE . and that gives $prediction = F(X) = E(Y|X)$ and so now it becomes a CONSEQUENCE that $E(U|X) = 0$ because of $Y|X = F(X) + U|X $

So the question is this statement : $E(U|X) = 0$ an assumption that we need to guarantee before using OLS or a consequence that OLS guarantees when we use it on the residual ?

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  • $\begingroup$ $E(U \mid X) = 0$ is an assumption that is part of the model, since you are modeling the mean. Minimizing MSE in linear regression does not yield $E(Y \mid X)$. In fact, $E(Y \mid X)$ is what you get when you minimize MSE over all predictors, both linear and nonlinear. $\endgroup$
    – Mason
    Commented Dec 2, 2023 at 16:27
  • $\begingroup$ ' Minimizing MSE in linear regression does not yield E(Y∣X)' , How is this true ? $\endgroup$
    – dbzadnen khiari
    Commented Dec 6, 2023 at 10:29
  • $\begingroup$ In linear regression, the estimate is $X^T\hat{\beta}$, whereas $E(Y \mid X) = X^T\beta$. $\endgroup$
    – Mason
    Commented Dec 6, 2023 at 16:41
  • $\begingroup$ Still Beta hat will converge to B assuming a good sample so i do not think it is the reason, i think it comes more from the fact that minimizing the MSE when U(x) is the unknown yields E(Y | X ) , here we are minimizing U(x) such that is depends on the parameters B ( linearly ) so we put contraints on U(x) , so the minima of the MSE cannot always be satisfied and therefore not always E(Y | X) , However under the assumption that E( U | X ) = 0 , it becomes the case even when constraining the function. Correct me if i'm wrong $\endgroup$
    – dbzadnen khiari
    Commented Dec 7, 2023 at 14:45

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