Tell me more ×
Signal Processing Stack Exchange is a question and answer site for practitioners of the art and science of signal, image and video processing. It's 100% free, no registration required.
up vote 2 down vote favorite

This is a page from the book linear algebra,geodesy and gps by Gilbert Strang....

This is a page from the book linear algebra,geodesy and gps by Gilbert Strang.... the page explains about the justification of the inverse of the of the co variance matrix of measurement vector $b$ in the over determined system $Ax = b$ as the best weight matrix for best estimate of x.

  1. what does the line errors contained in the matrix mean?(underlined in blue color..)
  2. how does that substitution take place marked with a blue curve...
  3. is E{$rr^T$} = E{$bb^T$} ??

any explanation is welcome...

share|improve this question
1  
Do not cross-post the same question to multiple sites (or leave links). So far you have posted this here, on Cross Validated and Mathematica... If you keep doing this, you might end up getting suspended. – Lorem Ipsum Sep 29 '12 at 16:54

1 Answer

up vote 2 down vote

  1. The linear model is $b=Ax+r$, so you can see the errors $r$ as the noise added to the measurements $b$.

  2. The covariance $\Sigma_b=E\{(b-\bar{b})(b-\bar{b})^T\}=E\{(b-Ax)(b-Ax)^T\}=E\{rr^T\}$

  3. $E\{rr^T\} \neq E\{bb^T\}$ unless $Ax=0$

share|improve this answer
1...can you explain what is this $\hat{b}$ ? is it the estimated value like $\hat{x}$ if yes then where do you get that from ? or is it the mean of $b$ ? 2...how is $\hat{b} = Ax $?? – rotating_image Sep 29 '12 at 13:00
Sorry, what I mean is $\bar{b}$, which is the mean of $b$. Since $r$ is assumed to be zero mean, then $\bar{b}=Ax$. – chaohuang Sep 29 '12 at 13:48
$E(b) = E(Ax + r)$....$E(b) = E(Ax) + E(r)$....$\bar{b} = AE(x)+ E(r)$....$\bar{b} = A\bar{x} + 0$....is $A\bar{x} = Ax$ ?? – rotating_image Sep 29 '12 at 13:51
For BLUE, $x$ is deterministic – chaohuang Sep 29 '12 at 14:15
i think for BLUE $E(x - \hat{x}) = 0$ not $\bar{x} = x$ – rotating_image Sep 29 '12 at 14:46
show 1 more comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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