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Given the standard simple regression model: $y_i = β_0 + β_1 x_i + u_i$

What is the expected value of the estimator $\hat\beta_1$in terms of $x_i, \beta_0$ and $\beta_1$ when $\hat\beta_1=\sum x_i y_i/\sum x_i^2$?

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  • $\begingroup$ What are your ideas? $\endgroup$ – João Ramos Jul 15 '15 at 18:12
  • $\begingroup$ You must substitute y_i into the expression to get $β̂_1=∑x_i(β_0 + β_1 x_i +u_i )/∑x^2_1$ and then split up the equation into separate summations, but I'm unclear how to separate treat the expected value of $∑x_i u_i / ∑x_i^2$ $\endgroup$ – Jan Aneill Jul 15 '15 at 18:19
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\begin{align} \text{E} \left ( \frac{\sum_{i=1}^{n} x_i Y_i}{\sum_{j=1}^{n} x_j^2} \right ) &= \left ( \sum_{j=1}^{n} x_j^2 \right )^{-1} \sum_{i=1}^{n} x_i \text{E}(Y_i) \\ &= \left ( \sum_{j=1}^{n} x_j^2 \right )^{-1} \sum_{i=1}^{n} x_i (\beta_0 + \beta_1 x_i) \\ &= \beta_0 n\bar{x} \left ( \sum_{j=1}^{n} x_j^2 \right )^{-1} + \beta_1 . \end{align}

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  • $\begingroup$ You used the right formula for $\hat \beta_1$, if there is no intercept. But here we have $\beta_0$. $\endgroup$ – callculus Jul 15 '15 at 18:24
  • $\begingroup$ Correct, which is what the poster asked about. $\beta_0$ is an assumed part of the true model, not necessarily the model being estimated. The apparent point is that the estimator is unbiased if $\beta_0 = 0$ or $\bar{x} = 0$. $\endgroup$ – dsaxton Jul 15 '15 at 18:26
  • $\begingroup$ If the estimators of the parameters are unbiased, then it doesn´t imply that $\beta_0=0$-only $E(\hat \beta_0)=\beta_0$ $\endgroup$ – callculus Jul 15 '15 at 18:37
  • $\begingroup$ I don´t understand your comment. $\endgroup$ – callculus Jul 15 '15 at 18:44
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Denote $x=(x_1,\dots,x_n)$. Assuming that $\mathbb{E}[u_i|x]=0$

$$\mathbb{E}[\hat\beta_1|x]=\mathbb{E}\left[\frac{\sum x_i y_i}{\sum x_i^2}\mid x\right]=\mathbb{E}\left[\frac{\sum x_i (\beta_0 + \beta_1 x_i + u_i)}{\sum x_i^2}\mid x\right]=\frac{\beta_0\sum x_i+\beta_1\sum x_i^2+\sum x_i\mathbb{E}[u_i|x]}{\sum x_i^2}=\beta_1+\beta_0\frac{\sum x_i}{\sum x_i^2}$$

and

$$\mathbb{E}[\hat\beta_1]=\beta_1+\beta_0\mathbb{E}\left[\frac{\sum x_i}{\sum x_i^2}\right]$$

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  • $\begingroup$ I want to clear my concept in this topic. If $$\mathbb{E}[\hat{\beta_1}]=\beta_1$$ is true, then from the result (that you have proved nicely), how can I derive $\mathbb{E}\hat{\beta_1}=\beta_1$? $\endgroup$ – vbm Jun 26 '18 at 15:33
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    $\begingroup$ @thevbm $\hat{\beta}_1$ is biased in this case (i.e. it's expectation is not $\beta_1$). $\endgroup$ – d.k.o. Jun 27 '18 at 3:35

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