Price to reverse trend of SMA

• 07/06/2021 at 5:46 PM #173237
  Khaled

  Hello Community,

  I’m looking for the simplest way to estimate the price at which the SMA20 would reverse from downtrend to uptrend or the opposite.

  I tried this code. Can you please review and comment?

  Thank you for your time.

  Price to reverse trend of SMA

  DEFPARAM DRAWONLASTBARONLY=TRUE N=20 SMA7= Average[N](close) ATR=AverageTrueRange[14](close) IF SMA7[1]<SMA7[2] THEN //currently in downtrend and waiting for a reverse x =round((1 + SMA7 - SMA7[1] * (1 - 2 / (N+1)))*(N+1)/2) ENDIF IF SMA7[1]>SMA7[2] THEN //currently in uptrend and waiting for a reverse x =round((-1 + SMA7 - SMA7[1] * (1 - 2 / (N+1)))*(N+1)/2) ENDIF drawtext(" SMA7 #x# ",barindex,0.5*ATR+highest[20](BollingerUp[20](close)),SansSerif,Bold,16)coloured(0,0,255) Return x

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  DEFPARAM DRAWONLASTBARONLY=TRUE   N=20 SMA7= Average[N](close) ATR=AverageTrueRange[14](close)   IF SMA7[1]<SMA7[2] THEN //currently in downtrend and waiting for a reverse x =round((1 + SMA7 - SMA7[1] * (1 - 2 / (N+1)))*(N+1)/2) ENDIF   IF SMA7[1]>SMA7[2] THEN //currently in uptrend and waiting for a reverse x =round((-1 + SMA7 - SMA7[1] * (1 - 2 / (N+1)))*(N+1)/2) ENDIF       drawtext("                                  SMA7 #x# ",barindex,0.5*ATR+highest[20](BollingerUp[20](close)),SansSerif,Bold,16)coloured(0,0,255)   Return x

   

   

  

  Khaled

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  07/06/2021 at 5:59 PM #173238
  Khaled

  I realize a made a few mistakes. In addition, my code tries to find what price is required to move the EMA by “1”, which is not what I’m trying to acheive.

  I’d like to find the price that reverses the EMA trend.

  Thanks for your help!

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  DEFPARAM DRAWONLASTBARONLY=TRUE N=20 EMA20= ExponentialAverage[N](close) ATR=AverageTrueRange[14](close) IF EMA7[1]<EMA7[2] THEN //currently in downtrend and waiting for a reverse x =round((1 + EMA7 – EMA7[1] * (1 – 2 / (N+1)))*(N+1)/2) ENDIF IF EMA7[1]>EMA7[2] THEN //currently in uptrend and waiting for a reverse x =round((–1 + EMA7 – EMA7[1] * (1 – 2 / (N+1)))*(N+1)/2) ENDIF drawtext(”                                  SMA7 #x# “,barindex,0.5*ATR+highest[20](BollingerUp[20](close)),SansSerif,Bold,16)coloured(0,0,255) Return x

   

   

   

  

  Khaled

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  07/06/2021 at 8:10 PM #173252
  JC_Bywan

  Hi, here is post #96165 from french forum with answer to same question: https://www.prorealcode.com/topic/formulation-ema/page/2/#post-96165

  

  JC_Bywan

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  07/06/2021 at 8:24 PM #173256
  Khaled

  Thank you for your feeback.

  Doyou mean this code to find the “Close” at which EMA reverses?

  defparam drawonlastbaronly=true

  once customEMA=close

  once prevEMA=close

  EMALength = 10

  alpha = 2/(EMALength+1)

  once irange = 200 //ticks

  mincalc = prevEMA*100

  lasttest=mincalc

  if barindex>EMALength then

  CustomEMA= alpha*Close + (1–alpha)*CustomEMA[1]

  for i = –irange to irange do

    pricetest=(Close+i*ticksize)

    calc = alpha*pricetest + (1–alpha)*CustomEMA[1]

    mincalc = min(mincalc,abs(calc–CustomEMA[1]))

    if mincalc<=ticksize and mincalc<lasttest then

     lasttest=mincalc

     result=pricetest

     //break

    endif

  next

  DRAWTEXT(”                                            ———-    #result#  EMA”,barindex,result,dialog,standard,20) coloured(200,0,204)

  endif

  return result as “Close result”,mincalc as “min distance found”,customEMA coloured(200,0,0) style(line,2) as “real EMA”,CustomEMA[1] coloured(255,0,255) as “prevema”

  

  Khaled

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  07/06/2021 at 8:41 PM #173259
  Khaled

  Noobywan, As you mentionned yourself in post #96165, the code only provides the EMA[1], but not the EMA at which the trend reverses.

  

  Khaled

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  07/07/2021 at 7:42 AM #173269
  JC_Bywan

  Hi, for ema case it’s both:

  • the ema at which it reverses is ema[1] (“by definition” because it is the limit for ema between going up or down), so that’s what was used in the post to create the equation to solve,
  • and the solution to the equation ema=ema[1] in the post was close=ema[1] (if not mistaken at the time I did it)

   

  For comparison, if I also come back to first post and look at an sma of N periods case, then:

  • similarily the level where sma would reverse is sma=sma[1],
  • that would be for close=sma[N]

  1 user thanked author for this post.

  Khaled

  

  JC_Bywan

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