# 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.

EDIT:

Here's where I got the slope estimate from:

You are to show $$E(\tilde\beta_1)=\beta_1$$ but your formula for $$\tilde\beta_1$$ is not correct. It should be: \begin{align*} \tilde\beta_1&=\frac{\sum_i(x_i-\bar{x})y_i}{\sum_i(x_i-\bar{x})^2}\quad\text{where}\quad\bar{x}=\frac{1}{n}\sum_ix_i. \end{align*} Now, first, observe $$\sum_i(x_i-\bar{x})^2=\sum_i(x_i-\bar{x})x_i$$. Second, using $$y_i=\beta_0+\beta_1x_i+u_i$$, we have: $$\sum_i(x_i-\bar{x})y_i=\sum_i(x_i-\bar{x})(\beta_0+\beta_1x_i+u_i)=\beta_1\sum_i(x_i-\bar{x})x_i+\sum_i(x_i-\bar{x})u_i.$$ Therefore, $$\tilde\beta_1=\beta_1+\frac{\sum_i(x_i-\bar{x})u_i}{\sum_i(x_i-\bar{x})^2}=\beta_1+\sum_i\lambda_iu_i,\quad\text{with}\quad\lambda_i=\frac{x_i-\bar{x}}{\sum_j(x_j-\bar{x})^2}.$$ Under the Gauss-Markov's assumptions, the $$\lambda_i$$ are not random. And so: $$E(\tilde\beta_1)=\beta_1+E\left(\sum_i\lambda_iu_i\right)=\beta_1+\sum_i\lambda_iE(u_i)=\beta_1.$$