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I am trying to derive the error for the weighted arithmetic mean $$ \langle A \rangle=\frac{\sum_{i=1}^N \frac{A_i}{\sigma_{A_i}^2}}{\sum_{i=1}^N \frac{1}{\sigma_{A_i}^2}}. $$ of a Data-Set $A_i$ with the errors $\sigma_{A_i}$.

With the gaußian propagation of uncertainty for a function $f(x_1,...,x_N)$ $$ \sigma_f=\sqrt{\sum_{j=1}^{N}\left(\frac{\partial f}{\partial x_j}\right)^2\sigma_{x_j}^2}\,\,\,\,\,\,(1) $$ we get (here: $f\equiv\langle A \rangle$, $x_i\equiv A_i$) ($\sigma_i:=\sigma_{A_i}$, sums and products go from 1 to N): $$ \sigma_{\langle A \rangle}\stackrel{(1)}{=}\sqrt{\sum_j\left(\frac{1}{\sigma_j\sum_i \frac{1}{\sigma_i^2}}\right)^2}=\sqrt{\sum_j\frac{1}{\sigma_j^2\sum_i \frac{1}{\sigma_i^2}}}\sqrt{\frac{1}{\sum_i \frac{1}{\sigma_i^2}}}\stackrel{(2)}{=}\sqrt{\frac{1}{\sum_i \frac{1}{\sigma_i^2}}}. $$ This is the correct answer. So the first root must be equal to 1 $\forall \{\sigma_i|i=1,...,N\}$ (for all "error lists")(I have computed this in Python and it is correct!). I managed to bring the radicand on one denominator, but I still don't get to the conclusion I want: $$ \sum_j\frac{1}{\sigma_j^2\sum_i \frac{1}{\sigma_i^2}} =\frac{1}{\prod_k \left( \sigma_k^2 \sum_i \frac{1}{\sigma_i^2}\right)} \sum_j \frac{\prod_k \left( \sigma_k^2 \sum_i \frac{1}{\sigma_i^2}\right)}{\sigma_j^2\sum_i \frac{1}{\sigma_i^2}} =\frac{\sum_j \prod_{k\neq j} \left(\sigma_k^2 \sum_i \frac{1}{\sigma_i^2} \right)} {\prod_k \left( \sigma_k^2 \sum_i \frac{1}{\sigma_i^2}\right)} \stackrel{?}{=}1.\,\,\,\,\,\,\, (2) $$ Question: How do I show that Equation (2) is equal to 1?

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I have the answer now. I overcomplicated it by trying to put it on one denominator. $$ \sum_j\frac{1}{\sigma_j^2\sum_i \frac{1}{\sigma_i^2}}=\sum_j\frac{1}{\sigma_j^2}\left(\sum_i \frac{1}{\sigma_i^2}\right)^{-1}=1 $$

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