I am working on the following problem:

In a problem of simple linear regression, $$Y = \hat\beta_0 + \hat\beta_1 x(bar),$$ show that the random variables $\hat\beta_1$ and $Y$ are un-correlated (All the betas have "hats").

I have come up with a solution, but I am not convinced whether it is correct logically. Any alternative ways/suggestions/corrections would be helpful.

My solution goes through the following steps:

  1. Express $\hat\beta_0 = Y - \hat\beta_1 x$
  2. Find the variance of the expression: $Var(\hat\beta_0) = Var(Y - \hat\beta_1 x)$
  3. Use the variance formula for $β_0$ on the left side, and split up the right side, so that we have a term that is $Cov(Y,\hat\beta_1)$
  4. Simplify and show that the covariance is $0$. Therefore the correlation will be $0$.

This question does not make sense as stated. What are $x,Y$? Vectors? Is $x$ fixed? Is there an error term?

For example, let $$\left[ \begin{array}{c} y_1 \\ y_2 \end{array} \right] = \beta_1 \left[ \begin{array}{c} 1 \\ 2 \end{array} \right] + \left[ \begin{array}{c} e_1 \\ e_2 \end{array} \right]$$

Then we know that $ \hat{\beta}_1 = (X^\top X)^{-1} X^\top Y = y_1/5 + 2y_2/5 $

can they really be uncorrelated?

  • $\begingroup$ I mean x-bar instead of x. Y is a random variable $\endgroup$ – FranXh Apr 28 '14 at 20:13
  • $\begingroup$ ok, for example $Y = 5x + E$? Where $x$ is scalar (fixed) and $E$ is random error? $\endgroup$ – PA6OTA Apr 28 '14 at 20:18
  • $\begingroup$ I understand your question, but I would guess that E is zero in that scenario? $\endgroup$ – FranXh Apr 28 '14 at 20:19
  • $\begingroup$ The error term is never 0. Unless you deal with two points or a single point (as you imply here). But if you have a single point $(x,Y)$ how are you going to fit two parameters, slope and intercept? $\endgroup$ – PA6OTA Apr 28 '14 at 20:28

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