# How is the confidence interval calculated for the ACF function?

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? • FWIW, this isn't really about R. Commented May 9, 2016 at 15:33 ## 1 Answer 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). • 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) Commented May 11, 2016 at 2:35
• Hi Glen_b, I have tried to update my answer accordingly - I have also borrowed some of your words, which I hope is okay. Commented May 13, 2016 at 11:39