I have the following regression
$$y = X\beta +u$$
where $y$ and $u$ are $(n\times 1)$ and $X$ is a fixed $(n \times k)$ matrix with full column rank and $\beta$ is an unknown $(k\times 1)$ vector of parameters.
$E(u)=0$ and $var(u)=\sigma^2 V$ with $(n\times n) $ symmetric and positive definite non-diagonal matrix $V$.
It is defined as $s^2= e_{GLS}' V^{-1} e_{GLS} /(n-k)$ with $e_{GLS}=y-X\hat{\beta}_{GLS}$
I want to show that $E(s^2)= \sigma^2$
What I did
By Cholesky decomposition, there is a $(n\times n) $ non-singular matrix $P$ such that $V^{-1} = P'P$
I transform the regression $y = X\beta +u$ as
$$Py = PX\beta +Pu$$
$$y^* = X^* \beta +u^*$$
In this transformed model, there is no heteroskedasticity problem since
$$var(u^*)=var(Pu)=\sigma^2$$
Next, I re-consider the transformed model $y^* = X^* \beta +u^*$
by the OLS estimation, I will obtain that
$y^* = X^* \hat{\beta_{GLS}} +e^*$
where $e^*$ is OLS residuals from this transformed model.
$$e^* =y^* - X^* \hat{\beta_{GLS}} = Py - PX \hat{\beta_{GLS}}$$
$$e^* = P(y - X \hat{\beta_{GLS}})= P e_{GLS}$$
Then,
$$s^2= \frac{e_{GLS}' V^{-1} e_{GLS}}{(n-k)}=\frac{{e^* }' e^*}{(n-k)}$$
I re-consider the regression
$y^* = X^* \hat{\beta_{GLS}} +e^*$
I multiply with the projection matrix $M_{X^* } = I - X^* ({X^* }' X^* ){X^* }'$
$$M_{X^* } y^* = M_{X^* } X^* \hat{\beta_{GLS}} + M_{X^* } e^* $$
$$e^* = M_{X^* } y^* = M_{X^* } [X^* \beta + u^*] = M_{X^* }u^* $$
So, ${e^*}' e^* = {u^*}' M_{X^* } u^* $
Then,
$$E(s^* ) = E(\frac{{u^* }' M_{X^* } u^* }{n-k}) = \frac{E({u^* }' M_{X^* } u^* ) }{n-k}$$
Rules:
(i) tr(A) = A for scalar. (tr = trace)
(ii) E(tr(A)) = tr(E(A))
(iii) tr(ABC) = tr(ACB)= ...
By using these rules, and since ${u^* }' M_{X^* } u^* $ is scalar,
$$E({u^* }' M_{X^* } u^* ) = tr ( E({u^* }' M_{X^* } u^* )) = E(tr({u^* }' M_{X^* } u^* )) = E(tr({u^* }'u^* M_{X^* } )) $$ $$= tr(E({u^* }'u^* ) M_{X^* } ) = tr(\sigma^2 M_{X^* } ) = \sigma^2 tr (M_{X^* }) = \sigma^2 (n-k) $$
So,
$$E(s^* ) = \frac{E({u^* }' M_{X^* } u^* ) }{n-k} = \frac{\sigma^2 (n-k) }{n-k}= \sigma^2$$
What I did is that. But, I am not sure because it seems very complicated and a bit fabrication and incorrect. Please discuss how can I show that $E(s^2)=\sigma^2$? How can I correct and simplify my solution?
How can you solve this question in a shortest and logical way?
