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FoldList for decaying local maximum?

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  • Subject: [mg113849] FoldList for decaying local maximum?
  • From: Garapata <warsaw95826 at>
  • Date: Mon, 15 Nov 2010 05:52:24 -0500 (EST)

Use the following to represent a list of 500 data points:

data = RandomReal[ParetoDistribution[1, 4], 500];

Now, I need to find the decay from a local maximum (typically as time^.

The following illustrates my latest attempt to do this:

data = RandomReal[ParetoDistribution[1, 4], 500];
=E2=80=A8maxData = Rest[FoldList[Max, 0, data]];
=E2=80=A8t = Flatten[Range[1, {#[[2]]}] & /@ Tally[maxData]];
maxDecayed = maxData - (t^.5)*.1;
ListLinePlot[{data, maxData, maxDecayed}, PlotRange -> All]

But this only shows the decay from the maximum data values.  I need to
set a local maximum when:

maxData > data >=  maxDecayed

then begin the decay again from that point.

Note: you may need to run the code a couple of times to see the
condition where a data value exceeds the decay line, but has not
produced a new maximum.

I think I need a different approach to do this, which calculates what
I need in one pass through the list, "data" rather than calculating
maxData and t once then trying to calculate the decay.

So, I think I need a function, which would include the conditions that

1. resets the local maximum and restarts the t count or
2. calculates the decays and increments the t count.

It would include something like:

data > decayedLocalMaximum, {data, 1},
 (* new local maximum, restarts t at 1 *)

data <= decayedLocalMaximum, {decayedLocalMax - (t + 1)^.05*.1, t +
(* local maximum still in effect so it decays and t increments *)

This should give me what I'd need for each successive data point's

Then I could use that function, call it "localMaxDecay", in something

FoldList[localMaxDecay, 0, data];

But I'm stumped.

- How do I set up the initial decayedLocalMaximum value (it should
equal data[[1]])?
- How do I keep track of t, the intervals since a local maximum
- Will FoldList do this?

 I'm not clear how to approach this.  Any suggestions appreciated.


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