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For example, in R if you call the acf() function it plots a correlogram by default, and draws a 95% confidence interval. Looking at the code, if you call plot(acf_object, ci.type="white"), you see:

qnorm((1 + ci)/2)/sqrt(x$n.used)

as upper limit for type white-noise. Can some one explain theory behind this method? Why do we get the qnorm of 1+0.95 and then divide by 2 and after that, divide by the number of observations?

Glen_b
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Nick Nikolaev
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1 Answers1

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In Chatfield's Analysis of Time Series (1980), he gives a number of methods of estimating the autocovariance function, including the jack-knife method. He also notes that it can be shown that the variance of the autocorrelation coefficient at lag k, $r_k$, is normally distributed at the limit, and that $\operatorname{Var}(r_k) \sim 1/N$ (where $N$ is the number of observations). These two observations are pretty much the core of the issue. He doesn't give a derivation for the first observation, but references Kendall & Stuart, The Advanced Theory of Statistics (1966).

Now, we want $\alpha/2$ in both tails, for the two tail test, so we want the $1−\alpha/2$ quantile.

Then see that $(1+1−\alpha)/2=1−\alpha/2 $ and multiply through by the standard deviation (i.e. square root of the variance as found above).

User1865345
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Robert de Graaf
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    Good answer but this doesn't discuss the part of the question about why (0.95+1)/2 (or whatever other value ci takes); that's not really enough for a separate answer, I think, so I'll mention it here: That's simply because we want $\alpha/2$ in both tails, so we want the $1-\alpha/2$ quantile. and then see that $(1+1-\alpha)/2=1-\alpha/2$. (Robert: if you want to incorporate something along these lines into your answer please go ahead) – Glen_b May 11 '16 at 02:35
  • Hi Glen_b, I have tried to update my answer accordingly - I have also borrowed some of your words, which I hope is okay. – Robert de Graaf May 13 '16 at 11:39