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I have $n$ successive observation $A_\mu $ of a quantity $A$ and I need to understand how the expectation values of the square of the statistical error depends from the autocorrelation time but a single step in the demonstration is giving me trouble. the author of the book (Monte Carlo simulations in statistical physics -Binder) wrote :

\begin{align*} \langle {(\delta A)^2} \rangle &= \bigg\langle \bigg[\frac{1}{n} \sum_{\mu =1}^n (A_\mu - \langle A \rangle) \bigg]^2 \bigg\rangle \\ &= \frac{1}{n^2} \sum_{\mu=1}^n \langle(A_\mu -\langle A \rangle)^2 \rangle + \frac{2}{n^2}\sum_{ \mu_1= 1}^n \sum_{\mu_2 = \mu_1 +1}^n \bigg(\langle A_{\mu_1} A_{\mu_2} \rangle- \langle A\rangle \bigg) \end{align*}

changing the summation index $\mu_2 $ to $\mu_1 +\mu $ with $ \mu = \mu_2 - \mu_1$this equation can be rewritten as :

$$ \langle {(\delta A)^2} \rangle = \frac{1}{n} \bigg[\langle A^2\rangle - \langle A \rangle ^2 + 2 \sum_{\mu =1}^n \bigg( 1-\frac{\mu}{n} \bigg) \bigg(\langle A_0 A_\mu \rangle - \langle A \rangle^2 \bigg) \bigg] $$

where <> mean the expectation value.

how does this change in the summation index works? i originally posted this question on mathematics.stackexchange but users there advised me to try here.

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Stationarity (time origin invariance) of the correlation function means that $$ \langle A_{\mu_1} A_{\mu_2}\rangle = \langle A_{0} A_{\mu}\rangle $$ and the double sum of interest may be expressed as $$ \sum_{\mu_1=1}^n \sum_{\mu=1}^{n-\mu_1} \left( \langle A_{0} A_{\mu}\rangle - \langle A\rangle^2 \right) . $$ By the way, I think there's a typo in your question, the very last term should be $\langle A\rangle^2$ not $\langle A\rangle$, so that's what I'm writing.

Now the summand is independent of $\mu_1$, and the idea is to exchange the order of the two sums. This can be illustrated by drawing a 2D grid, with $\mu$ and $\mu_1$ as the axes, and shading in the relevant terms. For any given value of $\mu$, the sum over $\mu_1$ ranges from $1$ to $n-\mu$. $$ \sum_{\mu_1=1}^n \sum_{\mu=1}^{n-\mu_1} = \sum_{\mu=1}^{n}\sum_{\mu_1=1}^{n-\mu} $$ But since the summand is independent of $\mu_1$, there are $(n-\mu)$ identical terms in the inner sum and we can do it immediately. We are left with a single sum, over $\mu$, and a factor $(n-\mu)$ inside. In your answer, a factor $1/n$ is taken inside too, producing the overall $(1-\frac{\mu}{n})$ factor.

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