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?
Best Answer
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
$$e_z = e_x \sqrt{a^2 +(1-a)^2}$$