# Why my proof of unbiased estimator seem fine without using expectation?

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

$$Y_{i}$$ = $$\alpha$$ + $$\betaX_{i}$$+ $$errorterm_{i}$$

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?!! 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 !!

• 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 $\hat \beta = \beta$. Jan 28, 2019 at 8:09
• 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. Jan 28, 2019 at 13:32

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