EMA with a horizontal and verticle shift

• 05/11/2022 at 4:07 PM #193061
  josward1

  How would i code an EMA 3 with a 1 bar horizontal shift and a 0.034 verticle shift?

  

  josward1

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  05/11/2022 at 6:32 PM #193071
  JS

  y = ExponentialAverage[3](close[1]) + 0.034
  Return y

  

  JS

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  05/11/2022 at 8:44 PM #193074
  josward1

  thanks, and if i wanted a verticle shift down would it be

   

  ExponentialAverage[3](close[1]) – 0.034  ?

   

  also does return y needed to be coded underneith? if i was coding using ema3 as the name should it be this…

  ema3 = ExponentialAverage[3](close[1]) + 0.034

  return ema3

  

  josward1

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  05/11/2022 at 8:56 PM #193076
  JS

  True, plus (+) is a positive vertical shift and minus (-) is a negative vertical shift.

  You can use any name you want.

  

  JS

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  05/11/2022 at 9:48 PM #193079
  josward1

  thanks, is it necessary to put “return ema3” under the line?

  

  josward1

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  05/11/2022 at 9:59 PM #193080
  JS

  An indicator must be closed with RETURN

  Return EMA3 will show the values of the EMA3 in the graph.

  

  JS

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  05/12/2022 at 11:24 AM #193106
  josward1

  Thanks, I have done this:

  ema3 = ExponentialAverage[3](close[1]) + 0.034

  return ema3

   

  and the return is underlined red showing an error

  

  josward1

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  05/12/2022 at 11:57 AM #193107
  JS

  Do you use the above line in an automatic trading system?

   

  

  JS

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  05/12/2022 at 12:56 PM #193113
  JS

  Example Trading System with EMA3

  EMA3Plus = ExponentialAverage[3](close[1]) + 0.034
  EMA3Min = ExponentialAverage[3](close[1]) – 0.034

  If Close Crosses Over EMA3Plus then
  Buy 1 contract at market
  ElsIf Close Crosses Under EMA3Min then
  SellShort 1 contract at market
  EndIf

  Graph EMA3Plus as “EMA3Plus”
  Graph EMA3Min as “EMA3Min”

   

  

  JS

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  05/12/2022 at 2:36 PM #193127
  josward1

  yes its is to detect conditions for one line bing over the other by a set distance.  I have the code i’m just looking to find the correct way to set up the verticle and horizontal shift. The return was new to me but showed up as an error when i tried to implement

  

  josward1

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  05/12/2022 at 9:38 PM #193133
  JS

  Have you already optimized your system?

  

  JS

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  05/14/2022 at 1:53 PM #193179
  josward1

  Yeh I’m just experimenting with different ways of setting up the averages

  

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  05/18/2022 at 4:34 PM #193393
  MIKE PAPY

  moi j’aimerais pouvoir déplacer horizontalement un indicateur quelconque vers la gauche ou la droite

  I would like to be able to horizontally move any indicator to the left or to the right

  

  MIKE PAPY

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  05/18/2022 at 11:42 PM #193417
  robertogozzi

  @MIKE PAPY

  Only post in the language of the forum that you are posting in. For example English only in the English speaking forums and French only in the French speaking forums.

  Please abide by the rules highlighted in yellow (below).

  Thank you 🙂

   

  

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  05/18/2022 at 11:52 PM #193418
  robertogozzi

  With RETURN you can only move the indicator to the left, i.e. the past value is plotted currently (not in the future, though), such as:

  Sma = Average[20,0](close) RETURN Sma[1]

  1 2	Sma = Average[20,0](close) RETURN Sma[1]

  you cam shift both ways using graphic instructions (DrawSegment) by:
  – repainting current data into the past
  – plotting cuurrent data into the future.

  In any case each indicator must be programmed to behave this way.

  

  robertogozzi

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