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I believe there are two ways to measure Confidence Intervals of Autocorrelation one assumption is assuming the Autocorrelation is following Gaussian Distribution and assuming Lags other than Lag 0 are Equal to 0 and Bartlett’s formula which assumes Moving Average Process.

I want to measure the extent of the Autocorrelation of 10 Year Rolling Returns of Sensex. I have 404 10 Year Rolling Returns which I calculated from April 1979-Nov 2022 and I want to know what is the number of Independent Samples I am having of 10 Year Returns. The simple answer is 4 but that is assuming a very strong Autocorrelation and considering the Start Date Sensitivity of Financial Returns Data I wanted to check how the strong the Autocorrelation is.

I used the ACF function in R to see at what Lag the Autocorrelation is not significant from 0 but I am seeing different results depending on what formula I am using for calculating the ACF Confidence Intervals. If I use the first assumption, then at Lag 46 there is no Autocorrelation (meaning 10 Year Return which was calculated 46 Months Ago does not influence 10 Year Return which was calculated Today) but Bartlett's Formula is showing there is no Autocorrelation at Lag 26.

I believe the later is correct because the former assumes there is no Autocorrelation for all Lags other than Lag 0 which is incorrect in this case due to the 10 Year Rolling Returns being constructed using overlapping Monthly Periods (i.e. First 10 Year Return is calculated From Jan 1979-Jan 1989,Second From Feb 1979-Feb 1989, Third From March 1979-March 1989,etc) so the Autocorrelation of First 10 Lags will definitely not be Equal to 0 but I am not sure on this hence the question.

Thanks and Regards,

Anon9001.

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  • $\begingroup$ @RichardHardy acf(na.omit(monthlyreturns$Sensex 10 Year IRR),ci.type="ma",lag.max=26) acf(na.omit(monthlyreturns$Sensex 10 Year IRR),lag.max=46) $\endgroup$
    – Anon9001
    Commented Mar 22, 2023 at 18:14
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    $\begingroup$ Thanks for the code! I had a similar question here. $\endgroup$ Commented Mar 22, 2023 at 18:14
  • $\begingroup$ @RichardHardy I am a noob here so I don't know the terminology used in the post you linked but I looked at the R Documentaiton for plot.acf and in the Note it states the following "The confidence interval plotted in plot.acf is based on an uncorrelated series and should be treated with appropriate caution. Using ci.type = "ma" may be less potentially misleading". Considering the Rolling Returns are not uncorrelated I believe this supports calculating Confidence Intervals based on Bartlett's Formula like I assumed before. $\endgroup$
    – Anon9001
    Commented Mar 22, 2023 at 19:18
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    $\begingroup$ I am afraid this is so... $\endgroup$ Commented Mar 22, 2023 at 19:36
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    $\begingroup$ I am not sure. My gut feeling is that this might vary depending on what you are estimating, but I cannot offer any proof or even an idea of a proof... $\endgroup$ Commented Mar 22, 2023 at 20:32

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I want to know what is the number of Independent Samples I am having of 10 Year Returns.

Overlapping samples are not independent. If you have two overlapping samples that share even a single observation, they are not independent. You do not need to test this statistically, as this is an algebraic fact.

On the other hand, you could ask what your effective sample size is, as in how many independent 10-year observations your sample corresponds to. This has to be qualified by stating the target of your estimation, as I think the answer may depend on that. E.g. if you wanted to estimate the uncertainty around your estimate of the sample mean (a 10-year mean), you could obtain the effective sample size in the following way:

  1. Estimate the variance by adjusting for autocorrelation using Newey-West or some other method to obtain $\tilde\sigma^2$,
  2. Estimate the variance in the naive way to obtain $\hat\sigma^2$,
  3. Take the ratio $\frac{ \hat\sigma^2 }{ \tilde\sigma^2 }$ and multiply it by the number of 10-year periods your data covers, i.e. 404 in your case.

(The example is probably not a great one, as you would not care about the effective number of observations once you have the robust estimate for variance $\tilde\sigma^2$. The latter gives you the uncertainty you are after.)

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    $\begingroup$ I was facing a related challenge and encountered some puzzling findings; see this thread. $\endgroup$ Commented Mar 22, 2023 at 18:11

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