Stochastic down trend

Viewing 7 posts - 1 through 7 (of 7 total)

Author

Posts
03/31/2023 at 9:58 AM #212598 quote
marco7630
Participant

New
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")
```
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
03/31/2023 at 10:24 AM #212601 quote

Nicolas
Keymaster

Legend

Change C1 with:

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

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

marco7630
Participant

New

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

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

Master
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)
03/31/2023 at 9:28 PM #212632 quote

JS
Participant

Master

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…

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

JC_Bywan
Moderator

Master

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.

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

JS
Participant

Master

Hi,

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

Posts