Updating mean and variance estimates an improved method 00 dating site

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Algorithms for calculating variance play a major role in computational statistics.

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To create a marker, choose the ‘Create Project Location’ tab from the top set of tabs: There are several options for creating a project marker name.Using: $$\text(X):=\frac\sum_i \omega_i x_i$$ The "naive", non-corrected variance I'm using is this: $$\text(X):=\frac\sum_i\omega_i(x_i - \text(X))^2$$ So I'm wondering whether the correct way of correcting bias is A) $$\text(X):=\frac\sum_i\omega_i(x_i - \text(X))^2$$ or B) $$\text(X):=\frac\frac\sum_i\omega_i(x_i - \text(X))^2$$ or C) $$\text(X):=\frac\sum_i\omega_i(x_i - \text(X))^2$$ A) does not make sense to me when the weights are small. The third, C) is my interpretation of the answer to this question: https://mathoverflow.net/questions/22203/unbiased-estimate-of-the-variance-of-an-unnormalised-weighted-mean For C) I have just realized that the denominator looks a lot like $\text(\Omega)$. I think it does not entirely align; and obviously there is the connection that we are trying to compute the variance...The normalization value could be 0 or even negative. All three of them seem to "survive" the sanity check of setting all $\omega_i=1$. ''Update:'' whuber suggested to also do the sanity check with $\omega_1=\omega_2=.5$ and all remaining $\omega_i=\epsilon$ tiny. When you consider cases where the two largest weights are equal and all the rest become vanishingly small, both (A) and (B) drop from contention (because they disagree with the known results for $n=2$).The end result is that environmental data from multiple sources is sorted into a single table.This user guide will step through the process of obtaining data, and describe available transforms and processing available for the data.

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