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I have time-series data generated via Metropolis algorithm - Monte Carlo simulations. Since these data must have some correlation between them, the formula of the standard error for IIDs variable must not be true here.

What is the formula for standard error for correlated data? (I have estimated the correlation coefficients using: $r_k = \frac{c_k}{c_0}$ where $c_0$ is the varaiance and $c_k = \frac{1}{N}\sum_{i=1}^{n-i}(x_i - \mu)(x_{i+k} - \mu)$, $\mu$ is mean.

luffy
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  • Hi: if you estimated the correlation coefficients, then that means that you have made an assumption about how the data is correlated. For example, maybe correlation is induced by MA(1) noise which is $\epsilon_t + \theta \epsilon_{t-1}$. So, how to estimate the variance is dependent on what the model is for the how the correlation comes about. – mlofton Aug 16 '18 at 05:55
  • Hi @mlofton! I have used following formula for correlation: $r_k = \frac{c_k}{c_0}$ where $c_0$ is the varaiance and $c_k = \frac{1}{N}\sum_{i=1}^{n-i}(x_i - \mu)(x_{i+k} - \mu)$, ($\mu$ is mean), if this helps. – luffy Aug 16 '18 at 06:42
  • Hi: You've calculated the actual correlations using a non-parametric estimator. This is fine. I guess you knew the mean or estimated it ? if you want to assume your correlations estimates are the true correlations, then there's a closed form solution for the variance of the series. I can't remember it so I'll look it up and get back to you. – mlofton Aug 16 '18 at 11:04
  • Hi: I looked briefly and any time domain approach has its issues. you're probably better off smoothing the periodogram. See this link for why there are issues. https://stats.stackexchange.com/questions/79216/summing-variance-of-autocorrelated-timeseries. Then see this link for details on how to do it in the freqency domain using the periodogram. http://stat.rutgers.edu/home/rongchen/publications/97ISR_review.pdf – mlofton Aug 16 '18 at 11:20
  • The details on how to do it are in section 2 of the second link. Note that, if you do use the approach in the second link, then the variance estimate is the value of the smoothed periodogram estimate evaluated at the frequency $\omega = 0$ so $\tilde{f}(0)$. – mlofton Aug 16 '18 at 11:25

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