# Calculation of the error for the weighted arithmetic mean

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?

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