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https://math.stackexchange.com/questions/787939/show-that-the-least-squares-estimator-of-the-slope-is-an-unbiased-estimator-of-t

The above link is what I used as the related reference that I got the same answer as mine

$Y_{i}$ = $\alpha$ + $\beta$$X_{i}$+ $errorterm_{i}$

Prove that the estimator of $\beta$ is unbiased What I did is proving this thing without using any expectation at all, so I wonder why it works?!!enter image description here

Basically, what unbiasedness is the expectation of the estimator equal to the true parameter we interest in which is beta here, but what I did I did not use any expectation so What I got still the true parameter?

Can I conclude that the expectation of what I got the constant is that true parameter?

My feeling is that I might get something wrong...

P.S.I wish to type all of those Math formattings but maybe next time and I wish to post this in mathematics but unfortunately, I do not have more than 10 reputations to post the picture !!

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    $\begingroup$ The point where you eliminate $\sum \epsilon_i x_i + \sum \epsilon_i \bar {x} = 0+0$ you are using expectation. This equation is not true for specific cases. Therefore your conclusion at the end that $\hat \beta = \beta $ should have been $E (\hat \beta) = E (\beta) = \beta $ which makes more sense. You can not have (in general, in all cases) that $\hat \beta = \beta $. $\endgroup$ Jan 28 '19 at 8:09
  • $\begingroup$ To be specific, assume that errors are from N(0,1), and that $n=5$. Then, mean(rnorm(5)) will generally not be equal to zero. $\endgroup$ Jan 28 '19 at 13:32
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The point where you eliminate $\sum \epsilon_i x_i + \sum \epsilon_i \bar {x} = 0+0$ you are using expectation. This equation is not true for specific cases. Therefore your conclusion at the end that $\hat \beta = \beta$ should have been $E (\hat \beta) = E (\beta) = \beta$ which makes more sense. You can not have (in general, in all cases) that $\hat \beta = \beta$.

Typically, $\hat\beta$ will be like a random variable. This is because it is a linear combination of the $y_i$ and will depend on the sampled $y_i$.

$$\hat\beta = \sum_{i=1}^n c_i y_i \quad\quad \text{with} \quad\quad c_i = \frac{x_i-\bar{x}}{\sum_{j=1}^n(x_j-\bar{x})^2}$$

If the $y_i$ are i.i.d normal distributed and have mean $E(y_i) = \alpha_i + \beta x_i$ then this elimination of the error terms (like you do with $\sum \epsilon_i x_i + \sum \epsilon_i \bar {x} = 0+0$) can be done earlier

$$E(\hat\beta) = \sum_{i=1}^n c_i E(y_i) = \frac{ \sum_{i=1}^n (x_i-\bar{x})(\alpha + \beta x_i)}{\sum_{j=1}^n(x_j-\bar{x})^2} $$

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