MVUE is unique - wrong proof? Here is the proof of "MVUE is unique" that my lecturer gave:


Now I understand the following:


*

*The first expansion is done using the formula for the sum of correlated random variables (https://en.wikipedia.org/wiki/Variance#Sum_of_correlated_variables).

*I understand that, having arrived at the result $Var(T+\lambda(T'-T)) = Var(T) - \frac{\rho^2}{\gamma}$ we require $\rho=0$ because otherwise $Var(T) - \frac{\rho^2}{\gamma} < Var(T)$ which is not possible, since $T$ is MVUE.


However, it seems to me that the lecturer made a mistake in the statement just after this. Indeed, if we require $\rho=0$ then surely $\lambda:=\frac{\rho}{\gamma} = \frac{0}{\gamma}=0$ right? 
And, consequently, we don't have $Var(T+\lambda(T'-T)) = Var(T)+\lambda^2 Var(T'-T)$ but rather $Var(T+0(T'-T))= Var(T)$ which is just an identity obviously.
What's more is that we cannot choose $\lambda$ to be $1$ because clearly, it turned out to be $0$.
Am I reading the proof correctly?
 A: Your lecturer intends for $\lambda$ to be a free variable in the variance equation.  He initially uses a specific value $\lambda=\rho/\gamma$
 to show that $\rho=0$.  Having done this, he reverts to the general form of the variance equation and shows the rest of the proof using the new specific value $\lambda=1$.  Perhaps he could have been a bit clearer about this in his proof.  Here is a clearer version of the same proof.

Result: The MVUE is (almost surely) unique.
Proof: Suppose that $T$ and $T'$ are both MVUE for $g(\theta)$.  If $\mathbb{V}(T'-T)=0$ then $T' \overset{\text{a.s}}{=} T$ which completes the proof.  We now proceed for the non-trivial case where $\mathbb{V}(T'-T)>0$.
For any scalar $\lambda \in \mathbb{R}$ the estimator $T + \lambda (T'-T)$ is an unbiased estimator with variance:
  $$\mathbb{V}(T + \lambda (T'-T)) = \mathbb{V}(T) + \lambda^2 \mathbb{V}(T'-T) + 2 \lambda \mathbb{C}(T,T'-T).$$
  Now, using a specific value of our scalar we have:
  $$\lambda = - \frac{\mathbb{C}(T,T'-T)}{\mathbb{V}(T'-T)} \quad \quad \implies \quad \quad \mathbb{V}(T + \lambda (T'-T)) = \mathbb{V}(T) - \frac{\mathbb{C}(T,T'-T)^2}{\mathbb{V}(T'-T)}.$$
  Since $T + \lambda (T'-T)$ is an unbiased estimator, it cannot have a variance lower than $\mathbb{V}(T)$ (since $T$ is MVUE).  This means that $\mathbb{C}(T,T'-T)=0$ (i.e., $T$ and $T'-T$ are uncorrelated), which allows us to simplify the variance equation down to:
  $$\mathbb{V}(T + \lambda (T'-T)) = \mathbb{V}(T) + \lambda^2 \mathbb{V}(T'-T).$$
  Using a new value for our scalar we then obtain:
  $$\lambda = 1 \quad \quad \implies \quad \quad \mathbb{V}(T') = \mathbb{V}(T) + \mathbb{V}(T'-T).$$
  Since $T'$ and $T$ are both MVUE, they must have the same variance as each other, and so we must have $\mathbb{V}(T'-T)=0$.  This gives $T' \overset{\text{a.s}}{=} T$, which completes the proof. $\text{ } \blacksquare$

