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Estimation of the position and value of a peak point

Posted 1 month ago
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Hi,

We know that the value of the 20th peak point is 38.60.

If I want to estimate the position and also the value of this point based on previous peak points. How do I that?

 data = {45.070711725125555`, 53.498297934722096`, 
   29.584510552747492`, 11.151690956807158`, 11.189625653415481`, 
   25.285839036082518`, 28.057908240982226`, 23.56037341335459`, 
   27.359877903104667`, 23.136393135264058`, 20.52783862787402`, 
   34.12780683174508`, 42.99141335930445`, 44.57290718131107`, 
   25.58296670347263`, 12.519024579537946`, 8.813650190806353`, 
   17.823927962891695`, 37.99345093666351`, 38.609478630177335`, 
   36.80195533249116`, 26.344280857176376`};

FindPeaks[data]

 {{2, 53.4983}, {7, 28.0579}, {9, 27.3599}, {14, 
  44.5729}, {20, 38.6095}}

ListLinePlot[{data, FindPeaks[data]}, 
 Joined -> {True, False}, 
 PlotStyle -> {Automatic, {Red, PointSize[.02]}}]

enter image description here

2 Replies

Here is a way to get something, but I wouldn't expect it to be useful or reliable.

peaks = FindPeaks[data]

Find an estimated distribution for the peaks (without the last peak):

peakDist = FindDistribution[Last /@ Most[peaks]]

Find an estimated distribution for the times between peaks (only 3 data points..)

dtDist = 
 FindDistribution[Differences[First /@ Most[peaks]], 
  TargetFunctions -> "Continuous"]

Estimate with a random sample from the distributions:

nextPeakSample = {First[Last[peaks]] + RandomVariate[dtDist], 
  RandomVariate[peakDist] }

I wouldn't bet on the result.

Posted 1 month ago

I appreciate your time, Prof.Carter.

At least you lit a lantern.

Alex

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