i'm trying to compute the variance of the random variable

$$X = \frac{1}{N}\sum_{i=1}^N x_i$$

where $x_i$ are correlated identical random variables (mean and variance defined) obtained from a stationary stochastic process.

I've reached the point where, with algebra only, I obtain

$$\left\langle \left(X-\langle X\rangle\right)^{2}\right\rangle = \frac{\left\langle x^{2}\right\rangle -\left\langle x\right\rangle ^{2}}{N}+2\frac{1}{N^{2}}\sum_{\Delta=1}^{N}\sum_{i=1}^{N-\Delta-1}\left\langle \left(x_{i}-\left\langle x\right\rangle \right)\left(x_{i+\Delta}-\left\langle x\right\rangle \right)\right\rangle $$

I identify

$$R\left(\Delta\right)=\frac{1}{\left\langle x^{2}\right\rangle -\left\langle x\right\rangle ^{2}}\frac{1}{N-\Delta-1}\sum_{i=1}^{N-\Delta-1}\left\langle \left(x_{i}-\left\langle x\right\rangle \right)\left(x_{i+\Delta}-\left\langle x\right\rangle \right)\right\rangle$$

to be the autocorrelation function and thus I can write

$$\left\langle \left(X-\langle X\rangle\right)^{2}\right\rangle = \frac{\left\langle x^{2}\right\rangle -\left\langle x\right\rangle ^{2}}{N}+\frac{\left\langle x^{2}\right\rangle -\left\langle x\right\rangle ^{2}}{N}2\sum_{\Delta=1}^{N}\frac{N-\Delta-1}{N}R\left(\Delta\right)$$

Can I further simplify this? I mean, does the quantity


means anything in the context of statistics that makes sense to write

$$\left\langle \left(X-\overline{X}\right)^{2}\right\rangle = \frac{\left\langle x^{2}\right\rangle -\left\langle x\right\rangle ^{2}}{N}\left(1+2\tau\right)$$

In the (likely) case this is a standard text book calculation, can you provide a reference on a (good) book that presents it?

  • 1
    $\begingroup$ David Mitra gives the general solution in his answer here. $\endgroup$ – COOLSerdash Mar 28 '14 at 15:13
  • 1
    $\begingroup$ @COOLSerdash I saw I forgot to mention that $x_i$ are like a random walk. I.e. they are successive. Thus, my question in the context of David's solution, is if there is a name for the sum of the covariances/variance in the context of random walks. In any case, thanks a lot for the link! $\endgroup$ – Jorge Leitao Mar 28 '14 at 15:24

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