1
$\begingroup$

Suppose normal Gauss-Markov model $\mathbf{y=Xb+e}$ where $y\sim N(\mathbf{Xb},\sigma^2 \mathbf{I})$, the pdf of y in exponential family:

Set $\theta=(\mathbf{b},\sigma^2)$: $$ \begin{aligned} f _ { \mathbf { Y } } ( \mathbf { y } \mid \mathbf { \theta } ) & = ( 2 \pi ) ^ { - n / 2 } \left( \sigma ^ { 2 } \right) ^ { - n / 2 } \exp \left\{ - ( \mathbf { y } - \mathbf { X } \mathbf { b } ) ^ { \prime } ( \mathbf { y } - \mathbf { X } \mathbf { b } ) / 2 \sigma ^ { 2 } \right\} \\ & = ( 2 \pi ) ^ { - n / 2 } \left( \sigma ^ { 2 } \right) ^ { - n / 2 } \exp \left\{ - \mathbf { y } ^ { \prime } \mathbf { y } / 2 \sigma ^ { 2 } + \mathbf { y } ^ { \prime } \mathbf { X } \mathbf { b } / \sigma ^ { 2 } - ( \mathbf { X } \mathbf { b } ) ^ { \prime } \mathbf { X } \mathbf { b } / 2 \sigma ^ { 2 } \right\} \\ & = ( 2 \pi ) ^ { - n / 2 } \left( \sigma ^ { 2 } \right) ^ { - n / 2 } \exp \left\{ - ( \mathbf { X } \mathbf { b } ) ^ { \prime } \mathbf { X } \mathbf { b } / 2 \sigma ^ { 2 } \right\} \exp \left\{ - \mathbf { y } ^ { \prime } \mathbf { y } / 2 \sigma ^ { 2 } + \mathbf { b } ^ { \prime } \mathbf { X } ^ { \prime } \mathbf { y } / \sigma ^ { 2 } \right\} \\&= h(\mathbf{y})c(\theta)\exp\{\sum_{i=1}^2 w_i(\theta)t_2(\mathbf{y})\} \end{aligned} $$

where $w_1(\theta)=-\frac{1}{2\sigma^2},w_2(\theta)=\frac{\mathbf{b}}{\sigma^2}$ and $t_1(\mathbf{y})=\mathbf{y'y},t_2(\mathbf{y})=\mathbf{X'y}$.

Seems the family is full ranks and apply

In a exponential family, the statistic $$ T=T(\mathbf{X})= \left( \sum _ { j = 1 } ^ { n } t _ { 1 } \left( X _ { j } \right) , \sum _ { j = 1 } ^ { n } t _ { 2 } \left( X _ { j } \right) , \ldots , \sum _ { j = 1 } ^ { n } t _ { k } \left( X _ { j } \right) \right)$$ is complete if the parameter space $$ \left\{ \boldsymbol { \eta } = \left( \eta _ { 1 } , \eta _ { 2 } , \ldots , \eta _ { k } \right) : \eta _ { i } = w _ { i } ( \boldsymbol { \theta } ) ; \boldsymbol { \theta } \in \Theta \right\} $$ contains an open set in $\mathcal{R}^k$. For the most part, this means the dimension $d=k$.

hence $(\mathbf{y'y},\mathbf{X'y})$ will be a complete statistic and hence $\mathbf{\hat{b}}=\mathbf{(X'X)^gX'y}$ will be a UMVUE for $\mathbf{b}$. However if $\mathbf{X}$ isn't full column rank,

$$ E[\mathbf{(X'X)^gX'y}]=\mathbf{(X'X)^gX'Xb} $$

can't be $\mathbf{b}$.

I think that may because if $\mathbf{X}$ isn't full column rank, then there is some $\mathbf{a}$ such that $\mathbf{Xa=0}$, that suggests $\mathbf{b'a}=0$. Then

$$ \left\{ \boldsymbol { \eta } = \left( \eta _ { 1 } , \eta _ { 2 } , \ldots , \eta _ { k } \right) : \eta _ { i } = w _ { i } ( \boldsymbol { \theta } ) ; \boldsymbol { \theta } \in \Theta \right\} $$

does not contain an open set in $\mathcal{R} ^{p+1}$($\mathbf{X}$ is $N\times p$ )?Is this sounds reasonable? And I still curious about why $\mathbf{X}$ be not full column rank will result to the family be not full rank.

$\endgroup$
0

1 Answer 1

0
$\begingroup$

Now I have an answer, when $\mathbf{X}$ is not full column rank, the statistic is still complete, but $\mathbf{\hat{b}}$ is a UMVUE for its expectation, not equal to b.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .