Standard deviation for a reconciled value? I have $n$ sensors, each of them providing a value $y_i$ for a measurement with a corresponding standard deviation $\sigma_i$. I know that the most probable value $Y$ is given by
$$Y=\left(\sum_{i=1}^n \frac {y_i}{\sigma_i^2}\right)\left(\sum_{i=1}^n \frac {1}{\sigma_i^2}\right)^{-1}$$ My problem is that I do not remember how is computed $\sigma_Y$.
Could you provide me the formula ? 
 A: The maximum likelihood estimator of $\mu$ when $Y_{i}\sim N\left(\mu,\sigma_{i}^{2}\right)$
are independently sampled is $\widehat{\mu}=\left(\sum_{i=1}^{n}\frac{1}{\sigma_{i}^{2}}\right)^{-1}\sum_{i=1}^{n}\frac{Y_{i}}{\sigma_{i}^{2}}$
as claimed. This can be verified by differentiating the log-likelihood of the normal distribution. To find its variance, we will use two properties of the
variance: 
(1) that $\textrm{Var}\left(aX\right)=a^{2}\textrm{Var}\left(X\right)$
whenever $a$ is constant, (2) that $\textrm{Var}\left(X+Y\right)=\textrm{Var}\left(X\right)+\textrm{Var}\left(Y\right)$
when $X$ and $Y$ are independent, see e.g. wikipedia for a reference.
Now it is straightforward to show
\begin{eqnarray*}
\textrm{Var}\widehat{\mu} & = & \left(\sum_{i=1}^{n}\frac{1}{\sigma_{i}^{2}}\right)^{-2}\textrm{Var}\left(\sum_{i=1}^{n}\frac{Y_{i}}{\sigma_{i}^{2}}\right),\\
 & = & \left(\sum_{i=1}^{n}\frac{1}{\sigma_{i}^{2}}\right)^{-2}\sum_{i=1}^{n}\frac{\textrm{Var}\left(Y_{i}\right)}{\sigma_{i}^{4}},\\
 & = & \left(\sum_{i=1}^{n}\frac{1}{\sigma_{i}^{2}}\right)^{-2}\sum_{i=1}^{n}\frac{\sigma_{i}^{2}}{\sigma_{i}^{4}},\\
 & = & \left(\sum_{i=1}^{n}\frac{1}{\sigma_{i}^{2}}\right)^{-1}.
\end{eqnarray*}
It follows that $\widehat{\mu}\sim N\left(\mu,\sigma^{2}\right)$,
where $\sigma^{2}=\left(\sum_{i=1}^{n}\frac{1}{\sigma_{i}^{2}}\right)^{-1}.$
Note that this is consistent with the case of equal variances.
Allo of this presupposes known variances for each $Y_i$, of course.
