Re: Find a period in multiple delayed time series

Martin.Camitz@xxxxxxxxx wrote:
> Hi!
>
> I'm a beginnner to time series analysis and I was wondering what
> techniques there are for analysing multiple time-series with the same
> period. In particular, if the time span of the data is shorter than the
> expected period, is there a way to retrieve the period if you have
> several time series and there is a delay (unknown or known) between
> them?
>
> Any input greatly appreciated.
>
> Martin

I don't know of any standard technique, which doesn't mean there
isn't one. Do you have some specific application or is this a
general interest question? How do you know these individual time
series all have one dominant period? Is than an assumption? How
much noise is there? What is the form of the full series expected
to be (pure sine, saw tooth, ... ?). How much of this analysis do
you have to do? If you just have one set of low noise segments,
I'd start (and maybe end) by just plotting up the segments and
laying them out on a template of what you expect the total series
to look like, and moving the individual segments around until they
fit the underlying template, or just move plots of the individual
segments around until they seem to compose a nice curve, and then
find the period from that curve. Can the individual segments be
allowed to overlap? It is often a good idea to start with simply
looking at the data in various ways in any data analysis situation.
Of course, there is no way (that I know of) to avoid ambiguities
introduced by the fact that there may be an unknown number of
multiple missing cycles of data between your segments, unless you
know some additonal information. Tree ring analysts deal with
this sort of thing all the time. They have what they call (IIRC)
floating series for which they have no absolute dates until they
can find other series which overlap the the floating ones and are
tied to the present through the set of overlapping tree rings. Of
course, that data has multiple periods and quite a bit of noise.
HTH.

Cheers,
Russell

.

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