1
$\begingroup$

I want to estimate $n+m$ parameters with the equation: $$\hat\theta=\left[\frac{1}{N}\sum_{t=1}^N \varphi(t)\varphi^T(t)\right]^{-1}\left[\frac{1}{N}\sum_{t=1}^N\varphi(t)y(t)\right]$$ where $\varphi$ is a vector of regressors and $\theta$ the vector of parameters. Then my book derive an expression for the estimation error: $$\hat\theta-\theta_0 =\left[\frac{1}{N}\sum_{t=1}^N\varphi(t)\varphi^T(t)\right]^{-1} \left[\frac{1}{N}\sum_{t=1}^N\varphi(t)y(t)-\left\{\frac{1}{N}\sum_{t=1}^N\varphi(t)\varphi^T(t)\right\}\theta_0\right]$$ this makes no sense to me! Why the term in curly braces? And why inside the second factor? Please help.

$\endgroup$
2
  • $\begingroup$ whoa... I guess I really have some sort of phobia for big formulae. Thanks! $\endgroup$
    – gurghet
    Commented Jun 18, 2011 at 18:11
  • $\begingroup$ @Bogdan, please convert your comment to answer. $\endgroup$
    – mpiktas
    Commented Jun 20, 2011 at 6:57

1 Answer 1

1
$\begingroup$

Let $A=\frac{1}{N}\sum_{t=1}^N \varphi(t)\varphi^T(t)$ and $B=\frac{1}{N}\sum_{t=1}^N\varphi(t)y(t)$
Then by distributivity $A^{-1}[B-A\theta_0]=A^{-1}B-A^{-1}A\theta_0=\hat{\theta}-\theta_0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.