I checked all the books and on-line materials I could find for the proof, but found all of them have a derivation problem, which I cannot understand.

To prove the least squares estimator is the $BLUE$ for the linear model $y = X*\beta + v$, one assumes $c = C*y$ is any linear unbiased estimator of $\beta$. Using the fact that $c$ is an unbiased estimator, we can easily obtain $(C*X-I)\beta = 0$.

Then all the books and on-line materials just conclude that $C*X-I$ must be 0. I don't understand this at all. If the equality $(C*X-I)\beta = 0$ should hold for any $\beta$, then certainly we would have $C*X-I = 0$. But I don't know any reason that $(C*X-I)\beta$ = 0 should hold for any $\beta$. Is there some one who can explain this to me?

  • $\begingroup$ When you are doing regression, you don't know what $\beta$ is. If you want your procedure to be BLUE, then, it must be BLUE regardless of the true value of $\beta$. It's not enough for it to be BLUE for just some $\beta$ (such as your particular one) because even when you have made your estimate, you still don't know what the true $\beta$ is. $\endgroup$
    – whuber
    Mar 29, 2017 at 16:10

1 Answer 1


The OLS estimator is $$ \hat{\boldsymbol{\beta}} = (\mathbf{X}'\mathbf{X})^{-1}\mathbf{X}'\boldsymbol{Y} $$

The class of linear unbiased estimators is $\tilde{\boldsymbol{\beta}} = \mathbf{C}\boldsymbol{Y}$, for $\mathbf{C} = \mathbf{f}(\mathbf{X})$ (that is $\mathbf{C}$ is a matrix valued function of $\mathbf{X}$), such that $$ \begin{align} \mathbb{E}(\tilde{\boldsymbol{\beta}} \mid \mathbf{X}) &= \boldsymbol{\beta} \\ \mathbf{C}\mathbb{E}(\boldsymbol{Y} \mid \mathbf{X}) &= \boldsymbol{\beta} \\ \mathbf{C}\mathbf{X}\boldsymbol{\beta} &=\boldsymbol{\beta} \end{align} $$ where the last step follows from the linear regression model $\mathbb{E}(\boldsymbol{Y} \mid \mathbf{X}) = \mathbf{X}\boldsymbol{\beta}$.

Hence, it follows that $$ \mathbf{C}\mathbf{X} = \boldsymbol{\iota}_K $$ where $\boldsymbol{\iota}_K$ is the identity matrix of size $K$.

  • $\begingroup$ Yes, CX*\beta should be equal to \beta, but why should CX be equal to I ? $\endgroup$
    – Alan
    Feb 17, 2014 at 18:48
  • $\begingroup$ @Alan From the definition of the identity matrix. $\boldsymbol{\iota}$ is the identity matrix if $\forall$ matrices $\mathbf{a}$, $\boldsymbol{\iota}\mathbf{a} = \mathbf{a}$. $\endgroup$ Feb 17, 2014 at 18:50
  • $\begingroup$ My question was why should (CX-I)*\beta=0 hold for any \beta? Can not \beta be a fixed parameter vector? $\endgroup$
    – Alan
    Feb 17, 2014 at 18:56
  • $\begingroup$ @Alan What difference does that make? $\endgroup$ Feb 17, 2014 at 19:17
  • 1
    $\begingroup$ Yes, this should hold for any unbiased estimator $\tilde{\beta}$ --- different $C$ gives different estimator. But $\beta$ in $(CX-I)\beta=0$ is the true parameter vector, not an estimator. $\endgroup$
    – Alan
    Feb 17, 2014 at 21:08

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