Let $X_1,...,X_n$ be a sample, all with mean $\mu$ and $Var(X_i)<\infty$. Let $T(X_1,...,X_n)=\sum_{i=1}^na_iX_i$. If T is the UMVUE of $\mu$ and T' is another linear unbiased estimate of $\mu$, then show that
$$Cov(T,T')=Var(T)$$
So I know that $E(T)=E(T')=\mu$ since T and T' are unbiased.
Then I also know that $Cov(T,T')=E(TT')-E(T)E(T')=E(TT')-\mu^2$
Then $Var(T)=E(T^2)-E(T)^2=E(T^2)-\mu^2$
The goal then becomes to show that $E(TT')=E(T^2)$, or to show that $T=T'$.
This is where I get stuck. I want to use uniqueness of the UMVUE, so I want to show that T' is as UMVUE of $\mu$ giving the desired result. I assume that since T' is another linear unbiased estimator, it is of the form $\sum_{j=1}^nb_jX_j$.
I suppose one viable approach is to show that $Var(T')\le Var(T)$, where then using that T is the UMVUE we could conclude that T=T' and the problem is done.
Any help is appreciated, thanks in advance
I don't think your proposed approach will work - if it did, you would also be proving that when a UMVUE exists, there are no other linear unbiased estimators.
What you can do instead is to say given the UMVUE $T$ and another linear unbiased estimator $T^\prime$, consider the estimator $\lambda T + (1 - \lambda)T^\prime$. You can prove that this is still unbiased, and therfore we have
$\mathrm{Var}(\lambda T + (1 - \lambda) T^\prime) \geq \mathrm{Var}(T)$
for $\lambda \in \mathbb{R}$. This gives a quadratic in $\lambda$ which is non-negative everywhere, and takes the value $0$ when $\lambda = 1$. This means the discriminant of the quadratic must be 0, and following the consequences of this conclusion algebraically will give the result.