Suppose I have a set of independent measurements $x_1,\ldots, x_n$ such that each $x_i$ is normally distributed i.e. $x_i \sim p(x_i) = \mathcal{N}(\mu_i, \sigma_i^2)$. I then define a variable $X$ as the empirical average of these random variables. \begin{equation} X = \frac{1}{n} \sum_{i=1}^n x_i \end{equation} With mean $\mu = \frac{1}{n} \sum \mu_i$ and variance $\sigma^2 = \frac{1}{n^2} \sum \sigma_i^2$
I want to reduce the variance of this estimate. I can use Inverse-variance weighting (https://en.wikipedia.org/wiki/Inverse-variance_weighting) which implies that I introduce weights $w_i$ with $w_i = \frac{1}{\sigma_i^2}\left ( \sum \frac{1}{\sigma_i^2} \right)^{-1}$. This would reduce the total variance of $X$ to $\frac{1}{\sum \frac{1}{\sigma_i^2}}$, which is the minimal variance. Introducing the weights will also shift the mean to $\mu =\left( \sum \frac{1}{\sigma_i^2} \right)^{-1} \sum_{i=1}^n \frac{\mu_i}{\sigma_i^2}$
An alternative approach What if I instead consider $X$ as being a sample from the joint distribution $X \sim \prod_{i=1}^n p(x_i)$. From the product of gaussian densities being a gaussian, this will have the same distribution as the variance reduced estimate i.e. $\mu =\left( \sum \frac{1}{\sigma_i^2} \right)^{-1} \sum_{i=1}^n \frac{\mu_i}{\sigma_i^2}$ and variance $\sigma^2 = \frac{1}{\sum \frac{1}{\sigma_i^2}}$
My question is how these relate. In the first, I assume the distribution of $X$ follows the sum of random variables while in the second I assume it's the product. But assuming it's a sum and then variance reducing yields the second one. Is it possible to explain what's going on in the first and second with bayesian probability theory because the two approaches relate and they all become equivalent when the variance is equal. In fact, for my experiments, the variance reduced/joint distribution yields a better fit, but I can't explain why simply taking the average is wrong (what assumptions am I making in that case and why are they wrong?).