I have the following problem:

  • a measurement produces a certain value $x_i$ and I have an associated error $\Delta x_i$ for every measurement;
  • the $i$ measurements are not independent and there are many of them $i \in [0, N]$ with $N \approx 2000$;

I would like to calculate the standard error of the (weighted) mean $_w\mu(x_i, w_i)$, defined as:

$$_w\mu(x_i, w_i) = \frac{\sum_i w_i x_i}{\sum_i w_i}$$

where the weights are given by $w_i = 1 / \Delta x_i^2$.

Having defined the weighted variance $_w\sigma(x_i, w_i)^2$ as:

$$_w\sigma(x_i, w_i)^2 = \frac{\sum_i w_i^2 (x_i - _w\mu)^2}{\sum_i w_i^2}$$

I assume that the standard error cannot be simply obtained dividing $_w\sigma(x_i, w_i)$ by $\sqrt{N}$. To the very least I should calculate the correction for the correlation in the sample and Wikipedia suggests to estimate/calculate a correction factor $f$ defined as:

$$f = \sqrt{\frac{1+\rho}{1-\rho}}$$

where $\rho$ is an estimate of the autocorrelation coefficient.

The standard error on the weighted mean $\Delta _w\mu(x_i, w_i)$ would then become:

$$\Delta _w\mu(x_i, w_i) = \frac{_w\sigma(x_i, w_i)}{\sqrt{N}} f$$

However, I have trouble estimating $\rho$.

What I have tried was to calculate the autocorrelation function $R(\tau)$ (normalized to the variance), but I do not understand how could I compute $\rho$ from it, and what modification would be required to account for the different weights.

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