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The assumption is the following. $$E[Y_i]=\beta _0 +\beta _{1}X_i, Var[Y_i]=\sigma ^2, Cov[Y_i,Y_j]=0, \forall i\ne j $$

Where $\hat \beta_0$ and $\hat \beta_1 $ are the least squares estimators.

I want to prove that $E[\hat \beta_1]=\beta_1$ and I am looking at my professor's work,

enter image description here

I get the first two rows, but I do not understand how he got the third row.

understand that

$$E[\Sigma (x_i-\bar{x})y_i]=\Sigma \left( E[(x_i-\bar{x})y_i] \right)$$

but I don't know how he got rid of the Expected value part.

May I get some help?

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  • $\begingroup$ Because $x$ is non-random and $y$ is random( the denominator in $\hat\beta_1$ is just a constant). $\endgroup$ – StubbornAtom Feb 3 at 6:26
  • $\begingroup$ Can you elaborate on that, please? How can you tell that x is non-random and y is? $\endgroup$ – hyg17 Feb 3 at 6:46
  • $\begingroup$ That part must have been mentioned right at the start, because otherwise how can you proceed? $\endgroup$ – StubbornAtom Feb 3 at 6:47
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$X$ is a variable, but not a random variable. To be more rigorous, I would prefer conditioning on $X$, i.e, \begin{align} \mathbb{E}[\hat{\beta}_1|X] &=\mathbb{E}\left(\frac{\sum(X_i - \bar{X})Y_i}{\sum (X_i - \bar{X})^2} |X=\mathrm{x} \right)\\ &= \left(\frac{\sum(X_i - \bar{X})(\beta_0 + \beta_1 X_i)}{\sum (X_i - \bar{X})^2} |X=\mathrm{x} \right)\\ &=\beta_1 \end{align}

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