Stochastic down trend

• 03/31/2023 at 9:58 AM #212598
  marco7630

  Hello,

  I have this code:

  TimeFrame(Daily) S1=Average[20](Close) C1=Summation[10](S1<S1[1])=10 lengthRSI = 14 //RSI period lengthStoch = 14 //Stochastic period myRSI = RSI[lengthRSI](close) MinRSI = lowest[lengthStoch](myrsi) MaxRSI = highest[lengthStoch](myrsi) StochRSI = ((myRSI-MinRSI) / (MaxRSI-MinRSI))*100 C2 = StochRSI[1] > 90 C3 = StochRSI < 85 C4 = Close > 5 C5 = Average[7](Volume) > 500000 C6 = S1 < 0.94 * S1[10] SCREENER[C1 and C2 and C3 and C4 and C5 and C6](Volume as "Volume")

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  TimeFrame(Daily) S1=Average[20](Close) C1=Summation[10](S1<S1[1])=10 lengthRSI = 14 //RSI period lengthStoch = 14 //Stochastic period myRSI = RSI[lengthRSI](close) MinRSI = lowest[lengthStoch](myrsi) MaxRSI = highest[lengthStoch](myrsi) StochRSI = ((myRSI-MinRSI) / (MaxRSI-MinRSI))*100 C2 = StochRSI[1] > 90 C3 = StochRSI < 85 C4 = Close > 5 C5 = Average[7](Volume) > 500000 C6 = S1 < 0.94 * S1[10] SCREENER[C1 and C2 and C3 and C4 and C5 and C6](Volume as "Volume")

   

   

  I would like to see only the assets that have a declining SMA20 between 5 and 10 days instead of the requirement that the SMA20 needs to be negative for at least 10 days.

  Can you adjust the code for me? Thanks

   

  gr Marco

  

  marco7630

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  03/31/2023 at 10:24 AM #212601
  Nicolas

  Change C1 with:

  C1=Summation[5](S1<S1[1])=5

  

  Nicolas

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  03/31/2023 at 7:03 PM #212627
  marco7630

  Hi Nicolas,

   

  Thanks, but i need it to be  between 5 and 10 days. Now i also get results that are already longer then 10 days in downtrend.

   

  gr Marco

  

  marco7630

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  03/31/2023 at 7:40 PM #212628
  JC_Bywan

  C1= summation[5](S1<S1[1])=5 AND summation[11](S1<S1[1])<=10

  1. says S1<S1[1] happened 5 times most recently,
  2. and didn’t happen at least once:

  • • either the 11th time (so… ok 10 days max)
    • or at least once before (so… ok less than 10 consecutive times)

  

  JC_Bywan

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  03/31/2023 at 9:28 PM #212632
  JS

  Hi,

  I think a summation only says something about how often something occurs and nothing about WHERE it occurs…

  Suppose I have 11 elements and 10 meet a certain condition (and 1 does not), then you cannot say with certainty where that 1 element will occur, it can be the last element but also the first… In both cases, the summation is true…

  

  JS

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  04/04/2023 at 9:16 AM #212764
  JC_Bywan

  Hi,

  agreed if summation[11](S1<S1[1])<=10 is on its own, then it doesn’t say where is at least one missing occurrence, my bullet point (2.) above says the same thing.

  When I combine it with “summation[5](S1<S1[1])=5 AND …”, then to have C1 true we need both summation subconditions to be true, and for this summation[5] subcondition to be true, it means none is missing in the latest 5 occurrences, so when C1 is true, both subconditions are true, necessarily the missing occurrence is between 6th and 11th. If it was in the latest 5, then the summation[5] subcondition would be false and C1 false with it.

  

  JC_Bywan

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  04/04/2023 at 8:52 PM #212806
  JS

  Hi,

  Actually, it wasn’t a response to your correct code but more of a personal brain wave that I wanted to share… 😉

  

  JS

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