My teacher gave me a problem, but he only give me the $(\pmb{X}'\pmb{X})^{-1}$ matrix.
If I have only $(\pmb{X}'\pmb{X})^{-1}$, how can I find $\pmb{X}$ (the design matrix)?
I think this is an algebra problem, but now I can't think a solution.
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Frans Rodenburg
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igorkf
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6$\frac 1 {x^2} = 1$, $x$ can be 1 or -1. – user158565 Jul 13 '19 at 12:09
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I didn't understand why you simply sayed this. Can you be more specific, please? – igorkf Jul 13 '19 at 12:13
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Think $x$ is 1 $\times$ 1 matrix. – user158565 Jul 13 '19 at 12:19
1 Answers
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It is in general impossible. $(X^TX)^{-1} \in \mathbb{R}^{k\times k}$ and $X \in \mathbb{R}^{n\times k}$ with $n > k$. There are many matrices $K \in \mathbb{R}^{n\times k}$ such that $(K^TK) = (X^TX)$. Do you know something else about X?
user158565
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Grada Gukovic
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My teacher gave us the SSE (say 200) and the number of observartions (say 15). He want the unbiased $\sigma^2$. I thought use the fact of $\frac{SSE}{n-k} = \hat{\sigma}^2$, but he wants this in matrix form! – igorkf Jul 13 '19 at 12:01
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Both SSE and the variance are numbers, not matrices. You need $\frac{SSE}{n-k}$. That part with the matrix form is probably just a confusion. – Grada Gukovic Jul 13 '19 at 12:07
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Yes, I used this in my exam, but he sayed we should have used a matrix way. – igorkf Jul 13 '19 at 12:11
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Did the guy ask for a value, or for the derivation of the estimator? – Grada Gukovic Jul 13 '19 at 12:27
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See here https://stats.stackexchange.com/questions/24921/distribution-of-sum-of-squares-error-for-linear-regression – Grada Gukovic Jul 13 '19 at 13:04
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I know that $\hat{\sigma}^2 = \frac{\pmb{y}^T(\pmb{I} - \pmb{H})\pmb{y}}{n-p}$, but I can't figure out how to achieve this from $(\pmb{X}^T\pmb{X})^{-1}$ – igorkf Jul 13 '19 at 13:31
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What else do you know about the components of $X and about (X^TX)^{-1} ? – Grada Gukovic Jul 13 '19 at 16:32
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3@igor given your comments your question seems to omit relevant information. Please edit to describe the circumstances. – Glen_b Jul 14 '19 at 00:58