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Linear model without covariates

$$y_i=a+\epsilon_i$$

I'm trying to calculate the $\hat{a}$ using OLS by $min\sum(y_i-a)^2$. $$\frac{d}{d(a)}\sum(y_i-a)^2=0$$ $$\sum y_i=n \hat{a}$$ $$ \hat{a}=\frac{\sum y_i}{n} =\bar{y}.$$

Now for $E(\hat{a})$:

$$\hat{a}=\frac{\sum (a+\epsilon_i)}{n}=a+\frac{\sum \epsilon_i}{n}$$

But if I well understood the idea of $\epsilon$, $\sum \epsilon_i=0$.

So, $$\hat{a}=\bar{y}=a ???$$

Where was the problem ?

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  • $\begingroup$ With $i=1,2,3,...n$ $\endgroup$ – Math Dec 13 '16 at 18:03
  • $\begingroup$ You forgot to take the expectation. $\endgroup$ – whuber Dec 13 '16 at 18:53
  • $\begingroup$ You took $E(\hat{a})$ and showed that was $a$ ... but then just started writing $\hat a$ instead of $E(\hat{a})$... why? $\endgroup$ – Glen_b Dec 14 '16 at 4:47
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It is not the case that $n^{-1} \sum \epsilon = 0$. However, it is the case that $n^{-1} \sum \epsilon \to \mathbb{E}(\epsilon) $ (due to the law of large numbers), and the assumption about the error term is that $\mathbb{E}(\epsilon) = 0$.

In other words, $\hat{a} \to a$, meaning that the sample average is a consistent estimator for $a$.

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  • $\begingroup$ This doesn't answer the question, which concerns computing $\mathbb{E}(\hat a)$. $\endgroup$ – whuber Dec 13 '16 at 18:53
  • $\begingroup$ OP finds that $\hat{a} = a$ and asks what goes wrong. I show what goes wrong (sample average confused with expectation). $\endgroup$ – Superpronker Dec 13 '16 at 18:55
  • $\begingroup$ The only thing that went wrong is the OP did not take the expectation. Convergence with $n$ is not part of the question, nor is it part of its solution. $\endgroup$ – whuber Dec 13 '16 at 18:56
  • $\begingroup$ OP explicitly writes "But if I well understood the idea of $\epsilon$, $\sum \epsilon = 0$", and this is not true. $\endgroup$ – Superpronker Dec 13 '16 at 18:59
  • $\begingroup$ Your are correct. The solution is to point out that $\mathbb{E}\left(\sum\epsilon\right)=0$ is true. $\endgroup$ – whuber Dec 13 '16 at 19:01

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