Error propagation on weighted mean I understand that, if errors are random and independent, the addition (or difference) of two measured quantities, say $x$ and $y$, is equal to the quadratic sum of the two errors. In other words, the error of $x + y$ is given by $\sqrt{e_1^2 + e_2^2}$, where $e_1$ and $e_2$ and the errors of $x$ and $y$, respectively.
However, I have not yet been able to find how to calculate the error of both the arithmetic mean and the weighted mean of the two measured quantities. How do errors propagate in these cases?
 A: The first assertion deals wtih mean squared errors, which in probabilistic terms translates into standard deviations. 
Now, probability says that the variance of the sum of two independent variables is the sum of the variances. That is, if $z = x + y$ ($x$ and $y$ indep), then $\sigma_z^2 =  \sigma_x^2  + \sigma_y^2 $ and  $$e_z = \sigma_z =  \sqrt{\sigma_x^2  + \sigma_y^2} = \sqrt{e_x^2  + e_y^2}  $$
Knowing this, and knowing that $Var(a X) = a^2 Var(X)$, if $z = a x + (1-a) y$ (weighted mean, if $ 0\le a \le1$) we get:
$$\sigma_z^2 =  a^2\sigma_x^2  + (1-a)^2\sigma_y^2 $$
$$e_z =  \sqrt{a^2 e_x^2  + (1-a)^2 e_y^2} = a \sqrt{ e_x^2  + \left(\frac{1-a}{a}\right)^2 e_y^2}  $$ 
In particular, if $a=1/2$ , then $e_z =  \frac{1}{2}\sqrt{ e_x^2  + e_y^2} $
Another particular case: if $e_x = e_y$ then 
$$e_z =  e_x \sqrt{a^2 +(1-a)^2}$$
A: The arithmetic mean is just a scaled version of the sum, so you just need to know that the error scales as the quantity itself under scaling; thus the error in the arithmetic mean is $\sqrt{e_1^2+e_2^2}/2$. (You need to enclose the argument of the root in curly braces instead of parentheses to have it displayed under the square root.)
For more general error propagation, you need to multiply the errors with the partial derivatives with respect to the individual quantities. In the case of the geometric mean, $g(x,y)=\sqrt{xy}$, these are
$$\frac{\partial g}{\partial x}=\frac12\sqrt{\frac yx}\;,\quad\frac{\partial g}{\partial y}=\frac12\sqrt{\frac xy}\;,$$
so the error $e$ is
$$
\begin{eqnarray}
e
&=&
\sqrt{\left(\frac{\partial g}{\partial x}e_x\right)^2+\left(\frac{\partial g}{\partial y}e_y\right)^2}\\
&=&
\frac12\sqrt{\frac yxe_x^2+\frac xye_y^2}
\\
&=&
\frac1{2g}\sqrt{(e_xy)^2+(e_yx)^2}\;.
\end{eqnarray}
$$
