I would like to show that $\mathbf{\hat{Y}}^T\mathbf{e}=\mathbf{0}$. I can solve this by saying that it is equivalent to showing $\sum e_i\hat{y}_i=0$. However, I'm wondering if there is a way to employ matrix methods to show that $\mathbf{\hat{Y}}^T\mathbf{e}=\mathbf{0}$. Any ideas?
Asked
Active
Viewed 51 times
3
-
Could you provide some more context here? Are you taking about linear regression, with $\hat Y=X\hat\beta$? – PedroSebe Nov 08 '20 at 01:12
-
@PedroSebe yes, that is correct! – Jen Snow Nov 08 '20 at 01:18
1 Answers
2
Perhaps this is what you are looking for:
$$\mathbf{e}=(I-P)\mathbf{y}$$
$$\mathbf{\hat{y}}=P\mathbf{y}$$
Therefore, \begin{align} \mathbf{\hat{y}}'\mathbf{e}&=\mathbf{y}'P'(I-P)\mathbf{y} \\ &=\mathbf{y}'(P'-P'P)\mathbf{y} \\ &=0 \end{align}
Last step uses the fact that $P$ is idempotent and symmetric.
Dayne
- 2,113
- 1
- 7
- 24
-
1This is perfect. I like the trick of using $\mathbf{\hat{y}}=P\mathbf{y}$! – Jen Snow Nov 08 '20 at 16:03
-
It made sense when I first read your answer, but now I'm wondering: how do we know that $P$ is idempotent and symmetric? – Jen Snow Nov 10 '20 at 23:41
-
1@RonSnow: See the properties section of [this](https://en.m.wikipedia.org/wiki/Projection_matrix) page. – Dayne Nov 11 '20 at 07:32
-
1