Averaging results from different experiments where errors may be underestimated Let's say we have three different published results that measure the same physical quantity $x$ using completely different methods. Experiment 1 says $x=100\pm0.1$, Experiment 2 says $x=102\pm0.2$, and Experiment 3 says $x=99\pm0.3$. Obviously these results are inconsistent, and we suspect that all three experimenters have underestimated their errors.
Is there any rigorous way to average together these three numbers an get a sensible answer for the error of the mean? I would think the answer is no and the only way to deal with it would be to scale all three errors by some ad-hoc factor to make the results more sensible.
What I would like from a "sensible" procedure is that we acknowledge that the error in experiment 1 is likely greater than 0.1, but it is still likely that experiment 1 is the most precise experiment, and experiment 3 is the least precise.
Like I said, it seems kind of hopeless, the only reason I'm asking is because wikipedia seems to suggest that there is a solution to exactly this problem. However the wikipedia article doesn't provide any references or any justification for the formula it gives. It says you should compute the mean in the normal way (weighted by $1/\sigma^2$) and that the error on the mean is $$\sigma_{\bar{x}}^2 = \frac{ 1 }{\sum_{i=1}^n \sigma_i^{-2}} \times \frac{1}{(n-1)} \sum_{i=1}^n \frac{ (x_i - \bar{x} )^2}{ \sigma_i^2 } .$$
So my question is: what is the source of this formula? Does it apply to the situation I described above?
 A: A quantity $x$ is measured in three different ways, where $x$ is constant and independent of the measurement.  Experiments are subject to 2 sources of errors and the measurements record instead $$X_i = x + \epsilon_i + \mu_i.$$  The experimenters only know about $\epsilon$ and can come up with a corresponding error estimate $\sigma_i$.
Obviously, unless you know/assume more about $\mu$, you cannot say anything further.  But assume all $\mu_i$ are realizations of the same RV $\mu$ with mean 0, variance $s^2$ and $\mu \perp \epsilon$.  
A simple estimator for $x$ is simply an average over $X_i$: $$\hat x = \frac{1}{n} \sum_i X_i.$$  (you can come up with variance minimizing estimators as well.)  You can easily see that the value you are looking for, $var\ \hat x$ is $$var\ \hat x = \frac{1}{n^2}\left( \sum_i \sigma_i + n s^2 \right).$$ For unknown $s^2$ you can use the relationship $$var\ X = \sum_i \sigma_i^2 + ns^2.$$


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*If you think it makes sense, I can think further but it will take some time.

*Think if you are happy with the assumptions.

*Try it with Monte-Carlo simulations.  I am afraid 3 measurements is way too little to provide anything useful.

