# What is the formula to compute the height of significance line on autocorrelation plot?

Autocorrelation of a time series can be plotted in R with use of acf function. For example:

acf(ldeaths) # built-in series


I couldn't find any information about the blue dashed line in function's docs but after watching some videos about time series I came to conclussion that it must be the cut-off point of critical region for testing the hypothesis

$$H_0: \forall_{h \in \{1...T-1\}} \rho(h) = 0, \text{ vs } H_1: \exists h \in \{1...T-1\} \quad \rho(h) \neq 0,$$

i.e. autocorrelation is zero for all lags $$h$$ vs for some lag the autocorrelation is non-zero. $$T$$ denotes the length of the series.

I don't remember where I found it but the formula $$2/\sqrt{T}$$ popped up in my mind and I tried to test it:

t = length(mdeaths)
acf(mdeaths)
abline(h = 2/sqrt(t), col = "red")


It seems to be right, but I'm not sure how to derive it. I guess it has something to do with the distribution of $$\hat \rho$$ estimator. If it was normal with zero mean and $$\sigma = 1$$ and we took $$z_{\alpha} \approx 2$$ and $$n = T$$, then the term $$\frac{2}{\sqrt(T)}$$ would be $$z_{\alpha}\frac{\sigma}{\sqrt{n}}$$ used in construction of 95% confidence interval for a normally distributed variable. But the information I found about the distribution of $$\hat \rho$$ is inconsistent with that https://en.wikipedia.org/wiki/Pearson_correlation_coefficient#Standard_error. Can anyone help?

Edit: I've changed the value of 2 in the formula to 1.96 and the new line fits even better to the dashed one, which suits my theory.

Edit 2: I've found this: https://otexts.com/fpp2/wn.html#wn

For white noise series, we expect each autocorrelation to be close to zero. Of course, they will not be exactly equal to zero as there is some random variation. For a white noise series, we expect 95% of the spikes in the ACF to lie within ±2/√Twhere T is the length of the time series.

• In any useful time series, $n$ is sufficiently large that one needn't fuss about the difference between $n$ and $n-2$ -- especially because the effective value of $n$ decreases as the lag increases, anyway. You don't have to reverse-engineer R code: simply inspect stats:::plot.acf (which is written in R itself), where you can verify your theory.
– whuber
Apr 11 at 17:01
• There have been other questions (at least one) about the same thing here on Cross Validated. You may try to find them and see if they contain any useful information. Apr 11 at 19:00