1
$\begingroup$

I've been looking over some regression models lately and I came across one which, although similar, differs from the "standard" simple linear model. I was hoping somebody could provide some assistance with some properties that I'm confused with.

Assuming the regression form:

$y_{i} = \beta_{0} + \beta_{1}(x_{i}-\bar{x}) + \epsilon_{i}$

with expected value:

${\bf E}[y_{i}] = \hat{\beta}_{0} + \hat{\beta}_{1}(x_{i}-\bar{x})$

where $\hat{\beta}_{0} = \bar{y}$ and $\hat{\beta}_{1} = \frac{S_{XY}}{S_{XX}}$

and, from what I've worked out:

${\bf E}[\hat{\beta}_{0}] = \beta_{0}$, ${\bf E}[\hat{\beta}_{1}] = \beta_{1}$

and:

$\text{Var}(y_{i}) = \sigma^{2}$, $\text{Var}(\hat{\beta}_{0}) = \frac{\sigma^{2}}{n^{2}}$, $\text{Var}(\hat{\beta}_{1}) = \frac{\sigma^{2}}{S_{XX}}$

How can it be shown that:

(a)

$\text{Cov}(y_{i}, \hat{\beta}_{1}) = \frac{\sigma^{2}(x_{i}-\bar{x})}{\sum (x_{i}-\bar{x})^{2}}$

I know that the covariance formula is given by:

$\text{Cov}(y_{i}, \hat{\beta_{1}}) = {\bf E}[(y_{i} - {\bf E}[y_{i}])(\hat{\beta_{1}} - {\bf E}[\hat{\beta_{1}}])]$

I'm guessing that to yield this result, the covariance formula somehow becomes of the form:

$\text{Cov}(y_{i}, \hat{\beta_{1}}) = (x_{i}-\bar{x})\text{Var}(\hat{\beta}_{1})$

since this would give:

$\text{Cov}(y_{i}, \hat{\beta}_{1}) = \sigma^{2} \frac{(x_{i}-\bar{x})}{\sum (x_{i}-\bar{x})^{2}} = \frac{\sigma^{2}(x_{i}-\bar{x})}{\sum (x_{i}-\bar{x})^{2}}$

However, although, I have tried to do this, I'm confused about how to manipulate this formula to yield the desired result.

(b)

$\text{Corr}(\hat{\beta}_{0}, \hat{\beta}_{1}) = 0$

Here, I know that if it can be shown that:

$\text{Cov}(\hat{\beta_{0}}, \hat{\beta}_{1}) = 0$

it follows that:

$\text{Corr}(\hat{\beta}_{0}, \hat{\beta}_{1}) = 0$

However, as in part (a), I'm confused about how to develop the covariance formula accordingly.

$\endgroup$
2
$\begingroup$

Here are some hints:

For (a) Try substituting $y_i = \hat{\beta}_0 + \hat{\beta}_1(x_i - \bar{x}) + \epsilon_i$. So you get: \begin{equation} Cov(y_i,\hat{\beta}_1) = Cov(\hat{\beta}_0,\hat{\beta}_1) + (x_i - \bar{x})Cov(\hat{\beta}_1,\hat{\beta}_1) + Cov(\epsilon_i,\hat{\beta}_1). \end{equation} To get this to what you want, you need to figure out part (b) to show $Cov(\hat{\beta}_0,\hat{\beta}_1) = 0$. To do this, remember that asymptotically: \begin{equation} (\hat{\beta}_0,\hat{\beta}_1)^T \sim MVN_2 ( \boldsymbol{\beta}_0, I_E(\boldsymbol{\beta}_0)^{-1}), \end{equation} where $\boldsymbol{\beta}_0 = (\beta_0,\beta_1)^T$. Here $I_E(\boldsymbol{\beta}_0)$ denotes the expected Fisher information matrix, given by the expectation of the negative second derivative of the log-likelihood (hopefully you've seen this before, if not look here: http://en.wikipedia.org/wiki/Fisher_information). If you look at the off-diagonals of $I_E^{-1}$ (or equivalently $I_E$ here), you should be able to find the covariance between the two parameters, which will mean you can answer both (a) and (b). (By the way $I_E(\boldsymbol{\beta}_0) \approx I_E(\boldsymbol{\hat{\beta}})$ provided you have a decent sample size).

Hope it helps.

$\endgroup$
  • $\begingroup$ Thanks for your reply! In your answer you state: $$(\hat{\beta}_{0}, \hat{\beta}_{1})^{T} \sim MVN_{2}(\beta_{0}, I_{E}(\beta_{0})^{-1}),$$ How does this change if the distribution is not multivariate normal, but univariate normal? Is there a more elementary way to calculate $\text{Cov}(\epsilon_{i}, \hat{\beta}_{1})$, assuming this adheres to a univariate normal distribution? Also, can the concept mentioned above be applied to yield $\text{Cov}(\epsilon_{i}, \hat{\beta}_{1})$? $\endgroup$ – user9171 Oct 31 '12 at 11:52
  • $\begingroup$ In the model the errors $\epsilon_i$ are assumed to be independent of everything else, so $Cov(\epsilon_i,\hat{\beta}_1)=0 \: \forall i$. Not sure what you're asking in the first bit, but does that help/clarify things? $\endgroup$ – Sam Livingstone Oct 31 '12 at 15:07
  • $\begingroup$ With respect to the first bit, I was intending to ask whether the approach given above used to find the covariance between the two parameters is unaltered if both parameters ($\hat{\beta}_{0}$ and $\hat{\beta}_{1}$) are univariate normal as opposed to multivariate normal (as given in the original answer). $\endgroup$ – user9171 Oct 31 '12 at 15:31
  • 1
    $\begingroup$ Erm, no I don't think it will. The marginal distributions of $\hat{\beta}_1$ and $\hat{\beta}_2$ here will be univariate Normal. If you consider them together (as I've done) you'll get a multivariate Normal, where the off-diagonals of the covariance matrix give the covariance between parameters. In other words, if you consider the parameters to be univariate Normal, and that there is some covariance between them (which may or may not be 0), then the joint distribution will be multivariate Normal here (with the appropriate covariance matrix). Hope that clears things up a bit. $\endgroup$ – Sam Livingstone Oct 31 '12 at 17:41
  • $\begingroup$ Oh, I see now. At first, I didn't realise that considering $\hat{\beta}_{0}$ and $\hat{\beta}_{1}$ would yield a multivariate normal distribution, but that makes sense. Thanks for all your assistance - it's really appreciated! $\endgroup$ – user9171 Oct 31 '12 at 23:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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