Determine the value of $α$ for which the $MSE(T)$ is minimal.

Question

Let $X_1$ be an estimator for the probability $θ$ of unauthorized access.
Let $X_2$ be another estimator for $θ$.
Assume that $X_1$ and $X_2$ are independent, unbiased estimators for $θ$.
Furthermore, let $σ_1^2$ be the variance of $X_1$ and $σ_2$ the variance of $X_2$.
Finally, define a new estimator $T$ for $θ$ by $T = αX_1+(1−α)X_2$.
Determine the value of $α$ for which the $MSE(T)$ is minimal.

$$MSE(T)=Var(T)+(E[T]-\Theta)^2$$ $$E[T]=\alpha E[X_1]+(1-\alpha)E[X_2]=\alpha\cdot\Theta+(1-\alpha)\cdot\Theta$$ $$MSE(T)=Var(T)=\alpha^2\cdot Var(X_1)+(1-\alpha)^2\cdot Var(X_2)=\alpha^2\cdot \sigma_1^2+(1-\alpha)^2\cdot \sigma_2^2$$

Now from this point how can I calculate the minimum value of $\alpha$ that minimize the $MSE(T)$, in my opinion I should put $MSE(T)=0$, then differentiate respect to $\alpha$ and solve for $\alpha$, but if I do it, then $\alpha$ will be equal to $0$, and I think is incorrect and I should find a finite value for it, where I'm wrong?

Answer 1

First differentiate, then set equal to zero: \begin{align*} \frac{d}{d\alpha}\text{MSE}(\alpha) &= 2\alpha \sigma_1^2 - 2(1 - \alpha)\sigma_2^2 = 0\\ & \alpha (\sigma_1^2 +\sigma_2^2) - \sigma_2^2 = 0\\ & \alpha = \frac{\sigma_2^2}{\sigma_1^2 + \sigma_2^2} \end{align*}