bootstrap time series using the rank

Question

Don Walpola wrote a good overview of the bootstrap techniques used for time series (https://stats.stackexchange.com/a/450894).

I searched through the literature, and did not find a single method taking advantage of the rank of a time series to create bootstraps with similar properties.

The idea is simple. Let's consider a time series $y_{n}$ with n $\in [1,N]$. Create $z_{n}$ a bootstrap with replacement of $y_{n}$ so that both time series share same ranks. That's it.

A short example in R with 5 bootstraps:

x <- 1:120 # months.
y <- cos(2*pi*(x-1)/12)+runif(x)+0.03*x # your favorite non-stationary time series
rank.y <- rank(y) # rank of your time series

col5 <- c('red','blue','green','yellow','pink')
plot(x,y,type='l')
for(i in 1:5)
{
  y.boot <- sample(y,replace=T)
  y.boot <- y.boot[order(y.boot)]
  y.boot <- y.boot[rank.y]
  lines(x,y.boot,type='l',col=col5[i])
}

This method looks more simple and efficient than a block bootstrap. Am I missing something?

Thanks !

it would help to explain the logic with formulas, not everybody knows R, otherwise this looks like an off-topic — Aksakal, Dec 01 '21 at 14:30
something is missing. if the samples have the same ranks wouldnt they be essentially the same every time? — Aksakal, Dec 01 '21 at 14:49
y = (1,2,3,4,5), z = (1,1,2,5,5). both have the same rank but they have not the same values. — François Ritter, Dec 01 '21 at 15:24
Right but it appears that the shape of the time series will always be the same — Aksakal, Dec 01 '21 at 15:30

score 0 · Answer 1 · answered Dec 08 '21 at 13:57

I found the terminology for what I am trying to do, it is very close to "maximum entropy bootstrap" (https://www.jstatsoft.org/article/view/v029i05).

One difference is that the bootstrap is performed on a larger sample that includes nearby values of the original time series, which solves the problem of repeating adjacent values.

This method is actually powerful for bootstrapping time series.

bootstrap time series using the rank

1 Answers1