I asked the following question in a forum more suitable for statistics, but I didn't get any answer; I hope, someone could shed light on my question:
I have three random variables, $X_1$, $X_2$, and $X_3$, which they are distributed normally. If we consider an estimator which reads as: $$\hat{\theta} = a_1 X_1 + a_2 X_2 +a_3X_3,$$ with the constraint: $a_1 + a_2 + a_3 = 1$, we know that for $a_i \propto 1/\sigma_i^2$, where $\sigma_i$'s are the standard deviations of $X_i$'s, the estimator has the smallest variance (for proof, see Theorem 3.2., here).
Now, my question is: What is the significance of choosing the coefficients in the above sum as Jeffreys' priors, that is, $a_i \propto 1 / \sigma_i$?
Does $a_i \propto 1 / \sigma_i$ result, for example, in the largest acceptable variance of the estimator? Perhaps, one can say, in the case of complete ignorance, i.e., uninformative priors, the variance of $\hat{\theta}$ should be the largest acceptable variance. Is this argument valid?