# Should $X$ be full column rank in normal Gauss Markov model to make $(\mathbf{y'y},\mathbf{X'y})$ be a complete statistic?

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.

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.