# Proof Verification: $\tilde{\beta_1}$ is an unbiased estimator of $\beta_1$ obtained by assuming intercept is zero

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?

Proof:

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.

• "since summation and expectation operators are interchangeable" Yes, you are right. But division or fraction and expectation operators are NOT interchangeable. $E(\frac AB) \ne \frac{E(A)}{E(B)}$. – user158565 Oct 6 '18 at 22:23
• Then how do I get around this problem? – A.Asad Oct 7 '18 at 19:39
• 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 $E(x)=x$. – user158565 Oct 7 '18 at 20:15
• 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 $x_iu_i$ ? – A.Asad Oct 7 '18 at 20:41
• Like $\dfrac{1}{\sum{(x_i)^2}}\sum{E[x_iu_i]}$ – A.Asad Oct 7 '18 at 20:43