Consider the standard simple regression model $y= \beta_o + \beta_1 x +u$ under the Gauss-Markov Assumptions SLR.1 through SLR.5.
Let $\tilde{\beta_1}$ be the estimator for $\beta_1$ obtained by assuming that the intercept is 0. Find $E[\tilde{\beta_1}]$ in terms of the $x_i$, $\beta_0$, and $\beta_1$. Verify that $\tilde{\beta_1}$ is an unbiased estimator of $\beta_1$ obtained by assuming intercept is zero. Are there any other cases when $\tilde{\beta_1}$ is unbiased?


We need to prove that $E[\tilde{\beta_1}] = E[\beta_1]$

Using least squares, we find that $\tilde{\beta_1} = \dfrac{\sum{x_iy_i}}{\sum{(x_i)^2}}$

Then, $ \tilde{\beta_1} = \dfrac{\sum{x_i(\beta_0 +\beta_1x_i +u)}}{\sum{(x_i)^2}}$

$\implies \tilde{\beta_1} = \beta_0\dfrac{\sum{x_i}}{\sum{(x_i)^2}} +\beta_1 +\dfrac{\sum{x_iu_i}}{\sum{(x_i)^2}}$

Taking Expectayion on both sides:

$\implies E[\tilde{\beta_1}] = \beta_0E[\dfrac{\sum{x_i}}{\sum{(x_i)^2}}]+ \beta_1 +\dfrac{\sum{E(x_iu_i)}}{E[\sum{(x_i)^2}]}$ (since summation and expectation operators are interchangeable)

Then, we have that $E[x_iu_i]=0$ by assumption (results from the assumption that $E[u|x]=0$

$\implies E[\tilde{\beta_1}] = \beta_0E[\dfrac{\sum{x_i}}{\sum{(x_i)^2}}]+ \beta_1 +0$

Now, the only problem we have is with the $\beta_0$ term.

If we have that $\beta_0 =0$ or $\sum{x_i}=0$, then $\tilde{\beta_1}$ is an unbiased estimator of $\beta_1$/

Can anyone please verify this proof? Also, why don't we write $y= \beta_1x +u$ instead of $y= \beta_0 +\beta_1x +u$ if we're assuming that $\beta_0 =0$ anyway?

Please let me know if my reasoning is valid and if there are any errors.

Thank you.

  • 2
    $\begingroup$ "since summation and expectation operators are interchangeable" Yes, you are right. But division or fraction and expectation operators are NOT interchangeable. $E(\frac AB) \ne \frac{E(A)}{E(B)}$. $\endgroup$ – user158565 Oct 6 '18 at 22:23
  • $\begingroup$ Then how do I get around this problem? $\endgroup$ – A.Asad Oct 7 '18 at 19:39
  • 1
    $\begingroup$ I just found an error. I cannot understand what you want to prove. After "assuming that the intercept is 0", $\beta_0$ appears many times. In regression, generally we assume covariate $x$ is a constant. So $E(x)=x$. $\endgroup$ – user158565 Oct 7 '18 at 20:15
  • $\begingroup$ Since $x_i$'s are fixed in repeated sampling, can I take the $\dfrac{1}{\sum{x_i^2}}$ as a constant and then apply the Expectation operator on $x_iu_i$ ? $\endgroup$ – A.Asad Oct 7 '18 at 20:41
  • $\begingroup$ Like $\dfrac{1}{\sum{(x_i)^2}}\sum{E[x_iu_i]}$ $\endgroup$ – A.Asad Oct 7 '18 at 20:43

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