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At this link I have seen the following formula whereas $$ r_k = \frac {\sum_{t=k+1}^n a_t*a_{t-k}} {\sum_{t=1}^n a_t^2}$$ $$Var(r_k) = \frac {n-k}{n*(n+2)}$$ where $r_k$ is the autocorrelation at relevant lag, $n$ is the number of points in the data set, and $a_t$ is the error.

I have searched the internet for the proof for variance equation, but I haven't found it. Could anyone help me prove the formula I mentioned above?

mertcan
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    In the provided link, it's not the variance, it's the definition of autocorrelation of model errors. – gunes Jun 20 '19 at 00:03
  • Yes sorry for mistake. I have edited my question. Could you help me about it? – mertcan Jun 20 '19 at 05:58
  • To be honest, I really would like to believe your knowledge is totally beyond my question but having not received any response from you just make me consider that you are so reluctant( except @Richard Hardy ) to spare your time, actually which is an ambivalence towards the aim of this forum. Therefore, please be more responsive and keep in mind that those questions are kind of which everybody may wonder and be keen on comprehending. I again ask : Could you help me about my question? – mertcan Jun 20 '19 at 18:51
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    Hi @mertcan, I actually did look at the result in the original paper of Box; however he also seems to give no explanation. I just don't have much time nowadays and I couldn't think of a quick way to prove it. Actually, I found the result quite interesting, and follow the question. – gunes Jun 21 '19 at 12:50
  • I have not even a clue to start. Could you give me at least the draft of proof? – mertcan Jun 21 '19 at 20:02
  • I don't have a draft or idea on a start. But, as far as I see, some similar (not the same) results about the variance of sample autocorrelation has been studied in Chapter 48 of the book "The Advanced Theory of Statistics" by Kendall, Stuart and Org's. I've seen it referenced in the article "Some Robust Exact Results on Sample Autocorrelations and Tests of Randomness" by Dufour and Roy. – gunes Jun 22 '19 at 10:12
  • The book called The Advanced Theory of Statistics you mentioned before only have 30 chapter within 2 volume @gunes? – mertcan Jun 24 '19 at 13:34
  • I’m sorry, forgot to mention. it’s in Volume 3 – gunes Jun 24 '19 at 13:36
  • In our library we lack the volume 3 and individually also @gunes. Is it possible to demonstrate how the book you mentioned deals with variance of autocorrelation? – mertcan Jun 24 '19 at 13:59
  • I can't since it's a long chapter and I haven't read it (it was a bit hard to read chapter). It's just a pointer for you. I've seen it in some website (I don't recall it now), and don't know if it is a legal one or not. I'd also ask @RichardHardy to see if he knows the proof of this. – gunes Jun 24 '19 at 14:03
  • @gunes, sorry, I do not happen to remember the proof. I would look for early papers introducing or elaborating on autocorrelation (serial correlation) for a proof. Papers on testing for autocorrelation may also have some useful information as they would develop test statistics under the null of no autocorrelation and probably employ standard errors of estimators of autocorrelation in constructing the test statistics. Perhaps papers introducing Box-Pierce and Ljung-Box tests have something in them? I have cited one of those papers [here](https://stats.stackexchange.com/questions/226334/). – Richard Hardy Jun 24 '19 at 17:47
  • Thanks for return. I have found box original paper but even there variance of autocorrelation is not derived @RichardHardy @gunes? – mertcan Jun 25 '19 at 07:01
  • Also, Box-Pierce just claims the variance of autocorrelation in the link you mentioned. No proof exists by Box-Pierce @RichardHardy? – mertcan Jun 25 '19 at 07:16
  • @mertcan, I would check other papers or some time series textbooks then. – Richard Hardy Jun 25 '19 at 07:27
  • I have been searching for a very long time that is why I would like to ask you @Richard Hardy. Just I don't understand why the proof of it does not exist? – mertcan Jun 25 '19 at 07:40
  • +1 for your question. I am sorry I cannot help more. – Richard Hardy Jun 25 '19 at 07:46

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