Error Propagation Weighted Mean
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Error Propagation Average
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Error Propagation Definition
The best answers are voted up and rise to the top Error propagation on weighted mean up vote 1 down vote favorite 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 $\sqrt{e_1^2 + e_2^2}$, statistics weighted mean 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? statistics error-propagation share|cite|improve this question edited Mar 22 '12 at 17:02 Michael Hardy 158k15145350 asked Mar 22 '12 at 13:46 plok 10815 add a comment| 2 Answers 2 active oldest votes up vote 3 down vote accepted The first assertion assumes one takes mean squared errors, which in probabilistic terms translates into standard deviations. Now, probability says that the variance of two independent variables is the sum of the variances. Hence, if $z = x + y$ , $\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}
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