Even / Odd decomposition

JS · Fri, 13 Aug 2021 12:42:52 +0000

I continue to promote Digital Signal Processing (DSP) on this forum because I think it has a lot to offer us… DSP is all about signals, in our case the (input)signal is the share price or the price of the Dow Jones Index, etc. An important part of DSP is the decomposition of the input signal. Decomposition is to decompose the input signal into two or more components. The purpose of decomposition is to break up a complicated input signal into several simpler components. The opposite of decomposition is superposition, where you add the different components back together and thus get the original input signal back. A decomposition is valid when the components form the original signal after the superposition (summation). There are several strategies to dissect signals (the most important are the impulse decomposition and the Fourier decomposition) and Even-Odd decomposition is one of them. When we look in the x-y plane, a signal has an “Even symmetry” when it can be mirrored around the y-axis and an “odd symmetry” when it can be mirrored around the y-axis AND its values is opposite to the original signal. You can also view this as a rotation around point zero (decomposition is part of the concept of circular symmetry). Of course most signals neither have a even- or odd symmetry but the beauty of it is that it works for all signals. Equations (discreet): Xe[n] = (x[n] + x[N-n]) / 2 (even signal) Xo[n] = (x[n] – x[N-n]) / 2 (odd signal) Xsp[n] = Xe[n] + Xo[n] (Superpositie) The most important point here is that you can dissect a signal into components and this decomposition is valid when the addition (superposition) of the components forms the original signal. N = 10000 Xe = (Close + Close[N]) / 2 Xo = (Close - Close[N]) / 2 Xsp = Xe + Xo Return Xe as "Even", Xo as "Odd", Xsp as "Superposition"

This topic has 21 replies, 4 voices, and was last updated 2 years ago by JS.

Viewing 7 posts - 16 through 22 (of 22 total)

← 1 2

08/25/2021 at 3:05 PM #176215

Most systems are not lineair, a simple example is a system based on a EMA…

Try your system with 1 contract and than with 10 contracts and you will see that your profit is also 10 times higher because it is lineair.

Or try your system in a other timeframe, for example current timeframe = 5 min and your period (x) = 1000 (5 x 1000 = 5000) then use a timeframe of 10 min and a period (x) of 500 (10 x 500 = 5000) and you will see that the characteristics of the system don’t change (because it is lineair)

Participant

Topics: 26

Replies: 1701

Been thanked: 768 times

08/25/2021 at 4:46 PM #176228

snucke

Ok, i underatand now.

Why is it better for a system to be linear?

snucke

Participant

Topics: 9

Replies: 143

Been thanked: 21 times

08/25/2021 at 5:13 PM #176230

When a system is not lineair then you are forced to examine the individual characteristics of unrelated systems.

When a system is lineair than certain (mathematical) rules can be applied.

Participant

Topics: 26

Replies: 1701

Been thanked: 768 times

08/26/2021 at 8:48 AM #176260

GraHal

JS wrote:

Try your system with 1 contract and than with 10 contracts and you will see that your profit is also 10 times higher because it is lineair.

I still don’t get what you are saying here … profit is directly (linearly) related to number of Contracts traded with all / any system.

1 Contract @ £1 per point and price moves 100 points … profit = £100.
10 Contacts @ £1 per point and price moves 100 points … profit = £1000.

GraHal

Participant

Topics: 211

Replies: 8765

Been thanked: 2179 times

08/26/2021 at 9:36 AM #176267

PeterSt

JS wrote:

Or try your system in a other timeframe, for example current timeframe = 5 min and your period (x) = 1000 (5 x 1000 = 5000) then use a timeframe of 10 min and a period (x) of 500 (10 x 500 = 5000) and you will see that the characteristics of the system don’t change (because it is lineair)

JS, I tend to say : Prove it. I mean, by means of a backtest of your choice. Just change the time frame as per your suggestion – use 5 min, 10 min and 15 min – don’t change anything else. Both are allowed to start at the same second.
I think you won’t succeed …

the characteristics of the system don’t change

In theory and with math, yes. But in practice you won’t be able to show double and triple results. It is not even so that the longer the period of sampling, the more close all will be. Try it. It could be fun.

PeterSt

Participant

Topics: 73

Replies: 2501

Been thanked: 805 times

08/26/2021 at 1:11 PM #176295

@GraHal: this is not true for all / any system but only for lineair systems (including lineair MM)

@PeterSt: see https://www.prorealcode.com/topic/digital-signal-processing-defining-using-the-difference-equation/

Participant

Topics: 26

Replies: 1701

Been thanked: 768 times

06/25/2022 at 6:39 PM #196099

The equations are correct, but after progressive insight, I should have used arrays here as well.

The most important point here is that you can dissect a signal into components, even- and odd, and this decomposition is valid when the addition (superposition) of the components forms the original signal.

When we look in the x-y plane, a signal has an “Even symmetry” when it can be mirrored around the y-axis and an “odd symmetry” when it can be mirrored around the x-axis AND its values is opposite to the original signal.
You can also view this as a rotation around point zero (decomposition is part of the concept of circular symmetry).

DefParam DrawOnLastBarOnly = true
UnSet($Xe)
UnSet($Xo)
UnSet($Xn)

DrawHLine(Close[N/2])
DrawVLine(BarIndex[N/2])

$Xe[0]=Close[0]
$Xo[0]=0
$Xn[0]=Close[0]

N=1000
For i = 1 to N-1
$Xe[LastSet($Xe)+1] = (Close[i] + Close[N-i])/2
$Xo[LastSet($Xo)+1] = (Close[i] - Close[N-i])/2
$Xn[LastSet($Xn)+1] = $Xe[i] + $Xo[i]
DrawPoint(BarIndex[i],$Xe[i],2) Coloured(0,255,0)
DrawPoint(BarIndex[i],$Xo[i]+Close[N/2],2) Coloured(255,0,0)
DrawSegment(BarIndex[i-1],$Xn[i-1],BarIndex[i],$Xn[i]) Coloured(0,0,255)
Next
Return

DefParam DrawOnLastBarOnly = true

UnSet($Xe)

UnSet($Xo)

UnSet($Xn)

DrawHLine(Close[N/2])

DrawVLine(BarIndex[N/2])

$Xe[0]=Close[0]

$Xo[0]=0

$Xn[0]=Close[0]

N=1000

For i = 1 to N-1

$Xe[LastSet($Xe)+1] = (Close[i] + Close[N-i])/2

$Xo[LastSet($Xo)+1] = (Close[i] - Close[N-i])/2

$Xn[LastSet($Xn)+1] = $Xe[i] + $Xo[i]

DrawPoint(BarIndex[i],$Xe[i],2) Coloured(0,255,0)

DrawPoint(BarIndex[i],$Xo[i]+Close[N/2],2) Coloured(255,0,0)

DrawSegment(BarIndex[i-1],$Xn[i-1],BarIndex[i],$Xn[i]) Coloured(0,0,255)

Return

Participant

Topics: 26

Replies: 1701

Been thanked: 768 times

Author

Posts

Viewing 7 posts - 16 through 22 (of 22 total)

← 1 2

Find exclusive trading pro-tools on