Consider the following population regression model:

$$y_{i} = \beta _{1} + \beta_{2}x_{i} + \epsilon _{i},$$

where $i=1,...,n$. Assume $\epsilon \sim iid$, with the pdf in equation: $f(\epsilon ) = \alpha \epsilon$ for $0\leq \epsilon \leq 1$.

I have the following data: $X = (3,1,2,2)^{'}$, and $Y = (3,0,1,4)^{'}$.

By calculating the estimated intercept and the slope, I have $b_{1} = -1$, and $b_{2}=1.5$.

Two classical assumptions are violated in this model: one is no mean-zero errors $E(\epsilon {_i})\neq 0$, and the other one is the errors not being normal.

From this, can I infer anything about the true intercept of the model? I think we could either be overestimating or underestimating the true intercept.

Also, as long as I have an intercept in the model, then is the OLS still BLUE? I think that our estimate of the intercept is biased, but the OLS is still BLUE because we have an intercept in the model.


1 Answer 1


Your last sentence is self-conradictory: "BLUE" means "Best Linear Unbiased Estimator" - so you are saying "the OLS estimator is biased but it is still Best Linear Unbiased" -except if you meant to say "the OLS estimator of the slope", i.e. of $b_2$.

Properties like these have nothing to do with the specific sample. Also, normality is not a prerequisite for the BLUE property. Denoting $\mathbf X = [\mathbf 1 : X]$ the regressor matrix we have that

$$\mathbf b = \left(\mathbf X'\mathbf X \right)^{-1}\mathbf X'\mathbf y = \beta +\left(\mathbf X'\mathbf X \right)^{-1}\mathbf X'\mathbf \epsilon$$

and treating the $X$'s as deterministic, we have

$$E(\mathbf b) = \beta +\left(\mathbf X'\mathbf X \right)^{-1}\mathbf X'E(\mathbf \epsilon)$$

and since $E(\mathbf \epsilon) \neq 0$ neither $b_1$ nor $b_2$ are unbiased. Now it is often said that "by including an intercept in the model, we can safely assume that the error term has mean equal to zero". This is correct, but it has the following implication: that what we are estimating as an intercept, is not the theoretical "true" coefficient, but augmented by the expected value of the error. In other words, we can declare the OLS as unbiased if we accept that $b_1$ estimates $\beta_1 + E(\epsilon)$. In other words to preserve unbiasedness, we are obliged to accept that we are estimating the model

$$y_{i} = \gamma _{1} + \beta_{2}x_{i} + u_i, \\\;\;u_i = \epsilon_i - E(\epsilon_i)\\ \gamma_1 = \beta_1 + E(\epsilon_i)$$

It is for this last model that the OLS estimator (of $\gamma_1$ and $\beta_2$) is BLUE.

  • $\begingroup$ Nice explanation. Since $b_1 = \beta_1 + E(\epsilon)$, the true intercept is overestimated by $-\frac{2}{3}$. $\endgroup$
    – OGC
    Nov 6, 2014 at 2:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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