1
$\begingroup$

The error sum of squares is defined as:

$$ SSe(\beta)=(y-X\beta)'(y-X\beta)\tag{1}$$

I want to show that for the OLS estimator $\hat\beta$,

$$SSe(\beta)=SSres+(\beta-\hat\beta)'X'X(\beta-\hat\beta)\tag{2}$$

I know that $SSres=SSe(\hat\beta)$, but I don't understand how to reach equation $(2)$ from $(1)$.

$\endgroup$
  • 1
    $\begingroup$ Can you check your question? In line 2, you end up with $(\beta-\beta)$ twice, which is identically zero. I think I se what you want to do, but can you clarify? $\endgroup$ – eSurfsnake Jun 1 '19 at 23:05
  • $\begingroup$ It looks right to me, should be (beta - beta hat) $\endgroup$ – Oscar Jonsson Jun 3 '19 at 9:07
  • $\begingroup$ math.meta.stackexchange.com/questions/5020/…. $\endgroup$ – StubbornAtom Jun 20 '19 at 15:37
0
$\begingroup$

You can rewrite $y$ as $y=X\hat{\beta}+\epsilon$. Note that, here $\hat{\beta}$ represents estimated parameters and $\epsilon$ is the residual term.

Then, Eq. (1) becomes: \begin{equation} SSE(\beta)=(X\hat{\beta}+\epsilon-X\beta)^T(X\hat{\beta}+\epsilon-X\beta)\\ =(X(\hat{\beta}-\beta)+\epsilon)^T (X(\hat{\beta}-\beta)+\epsilon)\\ =((\hat{\beta}-\beta)^TX^T+\epsilon^T)(X(\hat{\beta}-\beta)+\epsilon)\\ =(\hat{\beta}-\beta)^TX^TX(\hat{\beta}-\beta)+\epsilon^T\epsilon\\ =(\hat{\beta}-\beta)^TX^TX(\hat{\beta}-\beta)+SSE(\hat{\beta}) \end{equation}

$\endgroup$

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.