I am reading through the Introduction to Time Series and Forecasting Springer series textbook by Brockwell and Davis. On page 16 they discuss a simulated sequence of 200 iid normal random variables with mean 0 and variance 1, and the autocorrelation function for that data at lags h=0,1,...,40. They then state that it can be shown that for iid noise with finite variance, the sample autocorrelations p_hat(h) for h>0 are approximately IID N(0,1/n) for large n. They continue to say that hence, approximately 95% of the sample autocorrelations should fall between the bounds of +/-1.96/sqrt(n), since 1.96 is the 97.5% quantile of the standard normal distribution.

My question is why the value of 1.96/sqrt(n), specifically the square root part. I think I understand that 97.5% of the sample autocorrelations should fall within 1.96 standard deviations of the sample mean, which in this case is 0, but I'm confused where the square root is coming from.

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    $\begingroup$ $1/n$ is the variance, so standard deviation is $1/\sqrt{n}$. $\endgroup$ – Moss Murderer Jan 13 '18 at 2:06

Why the square-root? You have a variance of $1/n$, so by definition, standard deviation is $1/\sqrt{n}$.

Why is the variance of the correlation $1/n$? Because for a sequence of IID normal random variables, the autocorrelation $\rho(h)$ at any lag $h\neq0$ is simply the mean of the product of two independent standard normal variables. If the sequence length $n$ is large, then by central limit theorem, this mean tends towards a normal distribution with mean $0$ (because the mean of the product is equal to the product of the means for independent random variables) and variance $1/n$.


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