What is this permutation?

I was recently looking for ways to resample time series, in ways that

1. Approximately preserve the auto-correlation of long memory processes.
2. Preserve the domain of the observations (for instance a resampled times series of integers is still a times series of integers).
3. May affect some scales only, if required.

I came up with the following permutation scheme for a time series of length $2^N$:

• Bin the time series by pairs of consecutive observations (there are $2^{N-1}$ such bins). Flip each of them (i.e. index from 1:2 to 2:1) independently with probability $1/2$.
• Bin the obtained time series by consecutive $4$ observations (thre are $2^{N-2}$ such bins). Reverse each of them (i.e. index from 1:2:3:4 to 4:3:2:1) independelty with probability $1/2$.
• Repeat the procedure with bins of size $8$, $16$, ..., $2^{N-1}$ always reversing the bins with probability $1/2$.

This design was purely empirical and I am looking for work that would have already been published on this kind of permutation. I am also open to suggestions for other permutations or resampling schemes.

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Your procedure is interesting but as you describe it, it appears to me that if $2^k$ is the max block size, you basically partition your data into $2^{(N-k)}$ consecutive blocks and then within each block permute pairs, each instance being equal-probable. –  muratoa Sep 11 '12 at 11:40
Instead of pairs you could define $k_{\text{min}}$ and $k_{\text{max}}$. This way you ensure at least $2^{k_{\text{min}}}$ points are preserved and can move a distance at most $2^{k_{\text{max}}}$. –  muratoa Sep 11 '12 at 11:43
@muratoa thanks for feedback. I am not sure I follow. If $2^k$ is the max block size the scheme is not like permuting pairs within blocks. For instance, for $k=2$, you can obtain the order 4:3:2:1 with probability 1/8, which is not a pair permutation. As for $k_{\min}$ and $k_{\max}$, this is what I refer to in point 3. This is the way to shuffle scales from $k_{\min}$ and $k_{\max}$. –  gui11aume Sep 11 '12 at 15:10
you're right I didn't read your first bullet correctly, I thought the min size was 2. –  muratoa Sep 11 '12 at 19:59

If you include the last bin of size $2^N$, the random permutation is uniformly chosen from the iterated wreath product of groups of order $2$, denoted $C_2 \wr C_2 \wr ... \wr C_2$. (If you leave out the last possible reversal, then you get a uniform sample from an index $2$ subgroup, the product of two iterated wreath products with $N-1$ factors.) This is also the Sylow $2$-subgroup of the symmetric group on $2^N$ elements (a largest subgroup of order a power of $2$ -- all such subgroups are conjugate). It is also the group of symmetries of a perfect binary tree with $2^N$ leaves all at level $N$ (counting the root as level $0$).

A lot of work has been done on groups like this on the mathematical side, but much of it may be irrelevant to you. I took the above image from a recent MO question on the maximal subgroups of the iterated wreath product.

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Awesome (+1)!! Thanks for the reference to the wreath product and the Sylov 2-subgroup. Forgetting the last (top) reversion was a mistake, in fact it is included in the scheme. –  gui11aume Sep 11 '12 at 12:52