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Jose,
I just did: (365/12)*(5/7)*3
Of course I assume no-holaidays and an "median-month"
(30.4167 days)
Thank you for your correction :)
Best regards,
Pablo
JS> I would use 63 periods for 3 months (21 periods/month x 3).
JS> This assumes that there is no security data missing or holidays
JS> during that time.
JS> jose '-)
JS> http://www.metastocktools.com
JS> --- In equismetastock@xxxxxxxxxxxxxxx, Paolo <italoarg76@xxx> wrote:
>>
>> Lionel,
>>
>> LI> Just check if the close of is above a moving average at an
>> LI> earlier time and the current time. Not perfect but it may do
>> LI> the job.
>>
>> It was my answer too :)
>>
>> ColA
>> ====
>>
>> pds:=65; { 3 months in days }
>> EMA:= Mov(C,pds,E);
>> cond:=(C>EMA) and (C>REF(C,-pds));
>> cond;
>>
>> Filter
>> ======
>>
>> ColA<>0
JS> Yahoo! Groups Links
Pablo Bozzolo
---
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