I'm wondering why the two statements below are equal regardless of using $\vec{y}$ in deviation form/mean-deviaton/centered form or not. In other words, why isn't the result changed when you use the observation vector $ \vec{y} $ instead of the centered observation vector $ \vec{y_{d}} $.
$ \ X_d $ is the design matrix X in deviation form/mean-deviation form (containing the deviation values)/centered design matrix.
$\vec{y_{d}} $ is the observation vector $\vec{y} $ in deviation form/mean-deviation form/centered, also referred to as the centered observation vector.
$\mbox{ }\\$ $\mbox{ }\\$ $\mbox{ }\\$
$ \vec{\widehat y_d} = X_d ( X_d^{T}X_d)^{-1} X_d^{T} \vec{y_{d}}$
is equal to:
$ \vec{\widehat y_d} = X_d ( X_d^{T}X_d)^{-1} X_d^{T} \vec{y}$
$\mbox{ }\\$ $\mbox{ }\\$ $\mbox{ }\\$
Where $ X_d (X_d^{T}X_d)^{-1} X_d^{T} $ is the column space orthogonal projector P
And $\hat{\beta}=(X_d'X_d)^{-1}X_d'y_d$
Any help greatly appreciated!
To see why first lets take an ordinary regression:
$ y_i=\alpha + X_i'\beta +\epsilon_i \space \space \space \space $ (1)
Therefore we have
$\hat{y_i} = \hat{\alpha} + X_i'\hat{\beta}$
But by definition, $\hat{\alpha} = \bar{y}-\bar{X}'\hat{\beta}$ and thus
$\hat{y_i} = \bar{y} + (X_i-\bar{X})'\hat{\beta} = \bar{y} + X_d'\hat{\beta}$
Once again, take the relation: $ \bar{y}=\alpha + \bar{X}'\beta \space \space \space$ (2)
Subtract (2) from (1) to get
$ y_d= X_d'\beta +\epsilon_i $ Now note that if we run this regression, the coefficient on $\beta$ will be identical to the one above and
$\hat{y_d} = \hat{y_i} - \bar{y} = X_d'\hat{\beta}$ which should give you what you need.