Minimum-variance unbiased linear estimator 
Suppose that it is known that the mean of RV $X_i$ is $\mu_i\theta$, (i = 1, 2,..., n), where $\mu_i$ are known constants, whereas $\theta$ is unknown. Let $\Sigma$ be the variance matrix of the random vector $X = \left[X_1, X_2,\ldots, X_n\right]$.
Show that the minimum-variance unbiased linear estimator of $\theta$ is given by
$\hat \theta\left(x\right) = \frac{\mu^T\Sigma^{-1}x}{\mu^T\Sigma^{-1}\mu}$ where
$\mu = \left[\mu_1, \mu_2, \ldots, \mu_n \right]$.

For the unbiased part, it is easily to proved. But for the minimum-variance part, I don't have the PDF of $X_i$ so it cannot be proved using traditional way. Anyone has suggestions on this problem?
 A: Because the estimate is subjected to linear estimator, it can be written as $\tilde{\theta} = a^T x, a \in \mathbb{R}^n$. Since it is unbiased, it must satisfy $\mathbb{E}(\tilde{\theta}) = a^T\mathbb{E}(X) = \theta$, i.e., $a^T \mu = 1$. Therefore to find the minimum-variance unbiased linear estimator, it is equivalent to solve
\begin{align}
& \min_{a \in \mathbb{R}^n} \text{Var}(a^Tx) = a^T\Sigma a \\
& \text{subject to }\; a^T\mu = 1. \tag{1}
\end{align}
This optimization problem can be easily solved by Lagrange multiplier, and the result is $\hat{\theta}(x) = \frac{\mu^T\Sigma^{-1}x}{\mu^T\Sigma^{-1}\mu}$.
Details:
To solve $(1)$, first construct the Lagrangian
$$f(a) = a^T\Sigma a - \lambda (a^T\mu - 1). \tag{2}$$
Differentiate $(2)$ with respect to $a$, and set it equals to zero, we have
$$2\Sigma a - \lambda \mu = 0.$$
Solve it for $a$, we have 
$$a = \frac{1}{2}\lambda\Sigma^{-1}\mu.\tag{3}$$
Substitute this back to the constraint $\mu^T a = 1$, we can solve $\lambda$ as follows:
$$\lambda = \frac{2}{\mu^T\Sigma^{-1}\mu}.\tag{4}$$
Plug $(4)$ back to $(3)$, we obtain 
$$a = \frac{\Sigma^{-1}\mu}{\mu^T\Sigma^{-1}\mu}.$$
Therefore $\hat{\theta}(x) = a^Tx = \frac{\mu^T\Sigma^{-1} x}{\mu^T\Sigma^{-1}\mu}$.
