1
$\begingroup$

can u guys give some hint on how to prove that tilde beta is a linear estimator and that it is unbiased? $$\tilde\beta=\frac1n\sum_{i=1}^n\frac{y_i-\bar{y}}{x_i-\bar{x}}$$

i have attempted to transform the mean of y part to express it in terms of yi by separating yi and y bar, so that it can be expressed in the summation of wi and yi format $$\sum_{i=1}^nw_iy_i$$

$$\tilde\beta=\frac1n\sum_{i=1}^n\frac{y_i}{x_i-\bar{x}} - \frac1n\sum_{i=1}^n\frac{\bar{y}}{x_i-\bar{x}}$$

$$=\frac1n\sum_{i=1}^n\frac{y_i}{x_i-\bar{x}} - \frac{\bar{y}}n\sum_{i=1}^n\frac{1}{x_i-\bar{x}}$$

This $$\sum_{i=1}^n\frac{1}{x_i-\bar{x}}$$ should not be equal to zero and so i dont know how to move forward

$\endgroup$
3
  • $\begingroup$ This is elemental and covered by all introductory textbooks on linear model. There are proofs all over the Internet. Google! $\endgroup$
    – SmallChess
    Mar 29, 2017 at 15:38
  • $\begingroup$ @MichaelChernick i have make the self-study tag, i have attempted to do the questions but i dont know how to make it in the summation of wi and yi format $\endgroup$
    – Lida
    Mar 29, 2017 at 16:00
  • $\begingroup$ We appreciate your efforts to clarify the question. In order for it to be re-opened, it has to stand by itself without requiring us to visit another Web site to understand it. But before you go to the effort of providing those details, why not explore your questions about regression and regression formulas? I'm sure many of them already answer your question. Try our site search. $\endgroup$
    – whuber
    Mar 29, 2017 at 16:04

1 Answer 1

4
$\begingroup$

For clarity, when we say linear estimator in the regression context we mean linear with respect to $\mathbf{y}$, the response vector. The fact that

$$\tilde\beta=\frac1n\sum_{i=1}^n\frac{y_i-\bar{y}}{x_i-\bar{x}}$$

is a linear estimator then is easily seen if we switch to matrix notation. Define $\mathbf{w} = \left[ \left(x_1-\bar{x}\right)^{-1} \ldots \left(x_n-\bar{x}\right)^{-1} \right]^{T}$, the vector of weights, so that

$$\tilde\beta =\sum_{i=1}^n w_i \left(y_i-\bar y \right) $$

Finally, to get a cleaner expression we may use a centering matrix, which subtacts the mean from each component of the vector. This is of course a projection matrix onto the orthogonal complement of the vector $ \frac{1}{n} \mathbf{1}$. Hence we can write the whole expression as

$$\tilde \beta = \mathbf{w}^{T} \left( \mathbf{I} -\frac{1}{n} \mathbf{1} \mathbf{1}^{T} \right) \mathbf{y}$$

and this is now seen to be linear in $\mathbf{y}$, hence a linear estimator.

$\endgroup$
2
  • $\begingroup$ Thanks a lot for answering, but i was studying in an elementary course where i havent learnt how to use matrix operation in linear regression yet. Could you kindly explain in an easier way. Thank you $\endgroup$
    – Lida
    Mar 30, 2017 at 2:19
  • $\begingroup$ It should be possible to prove this without reference to vector notation, which is not used at all in most introductory courses in, say, econometrics. In fact, the methods therein get quite sophisticted, as do the proofs, with not a mention of vectors and scalars, only summation and expetation operators. $\endgroup$
    – Hexatonic
    Mar 10, 2021 at 2:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.