Suppose my true Model is:

$$y = Xb + u \tag 1$$

But I am estimating:

$$y = Xb + Zd + u \tag 2$$

I can get the estimate of $b$ from $(2)$ by using $Mz$ operator as:

$$\hat{b} = (X'MzX)^{-1}(X'MzY)$$

Why is an estimate from the first equation is better than the second?

  • $\begingroup$ Is this a question from a course or textbook? If so, please add the [self-study] tag & read its wiki. $\endgroup$ Nov 27, 2016 at 23:12
  • $\begingroup$ @gung done. Thanks any help though? $\endgroup$
    – A-dude
    Nov 28, 2016 at 1:45
  • 1
    $\begingroup$ I am sure you know the variance covariance matrix for $b$ when estimating model (1)? There likewise is an expression for the variance of the coefficients corresponding to $b$ when estimating (2). Can you derive that one as well? $\endgroup$ Nov 28, 2016 at 5:00


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.