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In the paragraph shown here (from "Least Squares Estimation" by S.A. Geer) I don't understand how we convert the least square estimation problem into the equation 4.

Could someone please elaborate?

LSE problem

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  • $\begingroup$ You have to find such $\hat{\beta}$ that minimizes the square error so equation 4 is obtained by taking derivative $$\frac{\partial}{\partial \beta}\Vert Y - X\beta \Vert^2 = 0$$ $\endgroup$ Commented Feb 19, 2017 at 8:43
  • $\begingroup$ I would be surprised if this question weren't answered in the passage immediately following the one shown. $\endgroup$
    – whuber
    Commented Feb 19, 2017 at 18:19

1 Answer 1

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An informal derivation : your model is

$$\mathbf y = \mathbf X \beta +\mathbf u$$

premultiply by $\mathbf X'$ to get

$$\mathbf X'\mathbf y = \mathbf X'\mathbf X \beta +\mathbf X'\mathbf u$$

Now pre-multiply by $(\mathbf X'\mathbf X)^{-1}$ to get

$$(\mathbf X'\mathbf X)^{-1}\mathbf X'\mathbf y = (\mathbf X'\mathbf X)^{-1}\mathbf X'\mathbf X \beta +(\mathbf X'\mathbf X)^{-1}\mathbf X'\mathbf u$$

The error term is unknown so ignore the last term, and simplify to get

$$(\mathbf X'\mathbf X)^{-1}\mathbf X'\mathbf y = \hat \beta $$

and it is only estimated and not the exact $\beta$, because we have ignored the term $(\mathbf X'\mathbf X)^{-1}\mathbf X'\mathbf u$.

The formal treatment that validates the optimality of this approach under a certain criterion, is what a comment suggested.

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  • $\begingroup$ Than you very much. $\endgroup$
    – wmac
    Commented Feb 20, 2017 at 3:26

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