If there are two sets of measurement results on the same object but coming from different methods, from each results we can calculate the averages and variances of them, how can I moderate two sets of results and lower the variance in general?
Quantitively speaking, if from set one we get average $\mu_1$ and variance $\sigma_1^2$, and likewise for set 2, what is the general variance $\sigma^2$?
Obviously, $σ<σ_i$ (i=1,2), and when $σ_1=σ_2$, $σ=\frac{σ_1}{\sqrt{2}}$, so I can infer that: $$\sigma=\frac{σ_1σ_2}{\sqrt{σ_1^2+σ_2^2}}$$ Is there any way to prove the relation? I feel that it may need to use the property of χ^2 distribution, but how to do it exactly?
Ps: By applying the equation above, it’s easy to get the situation when there are n measurements by induction method. Here’s the result:
When n=3, $$\sigma=\frac{\sigma_1\sigma_2\sigma_3}{\sqrt{\sigma_1^2\sigma_2^2+\sigma_2^2\sigma_3^2+\sigma_3^2\sigma_1^2}}$$
When n is arbitrary positive integer, $$\sigma=\frac{\prod_{i=1}^n \sigma_i}{\sqrt{\sum_{i=1}^n {\frac{\prod_{j=1}^n \sigma_j^2}{\sigma_i^2}}}}$$
Equivalently, we can rewrite the equation as: $$\frac{1}{\sigma^2}=\sum_{i=1}^n \frac{1}{\sigma_i^2}$$
Update: I’ve got some reasoning by thinking of two dimensional normal distribution. Since two measurements discussed above aren’t correlated, the combined distribution of the stochastic vectors X and Y is independent 2d normal distribution, with the probability density function as: $$f(x,y)=\frac{1}{2\pi\sigma_1\sigma_2}\exp \left(-\frac{(x-\mu_1)^2}{2\sigma_1^2}-\frac{(y-\mu_2)^2}{2\sigma_2^2}\right)$$ To make things easier, here we suppose $\mu_1=\mu_2=0$ (since it can be proved that the final result of variance is irrelevant to the average). The next step is to calculate the distribution function of $Z=(X+Y)/2$. To do this,$\newcommand{\dx}{\; \mathrm{d}}x\newcommand{\dy}{\; \mathrm{d}}y$ $$F(Z)=\int_{-\infty}^Z \dx \int_{-\infty}^{\frac{y-x}{2}} f(x,y)\dy$$
This is data validation or data reconciliation.
If you have a look here, we will see that I addressed basically the same problem.