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](https://arxiv.org/pdf/1710.04055.pdf)).


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?



Jeffreys' priors as coefficients of a linear estimator
The coefficients $a\_i$ of the random variables $X\_i$ are not any prior probabilities at all -- because prior probabilities are coefficients, not of random variables, but of probability distributions.


The choice $a\_i\propto 1/\sigma\_i$ in your setting equalizes the variances of the random variables $a\_iX\_i$, and that is all it does.




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Even though priors have actually nothing to do with your question, for criticisms of Jeffreys' "noninformative" priors one may want to see e.g. [Section 4.1](https://projecteuclid.org/journals/statistical-science/volume-24/issue-2/Harold-Jeffreyss-Theory-of-Probability-Revisited/10.1214/09-STS284.full).


