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0002ct
Posted : Wednesday, May 24, 2017 11:06:18 AM
Registered User
Joined: 11/19/2015
Posts: 459

Hello.

I'd like to expand on a calculation exampled yesterday:
CountTrue(O/O1>1.05,100)
which will give a count of the number of times the most recent open is 5% above the previous open over the last 100 bars.

I'm working with the same basic concept, except I'd like to calculate stats for:
percent of the time (as opposed to the count) over the last period that the open was X% or more above the previous open.

For example, I'd like to return a list of tickers that are calculated statistically:
opened 75% of the time higher than the previous open over the last 100 bars.

Does this make sense? Is it possible?

Thank you very much.

StockGuy
Posted : Wednesday, May 24, 2017 11:08:51 AM

Administration

Joined: 9/30/2004
Posts: 9,187

CountTrue(O/O1>1.05,100) > 75

0002ct
Posted : Wednesday, May 24, 2017 11:19:17 AM
Registered User
Joined: 11/19/2015
Posts: 459

Thanks StockGuy. The aribitrary "100 days" is convenient because it makes the above easy, however, I will be creating a bunch of these using various percents and periods. How is this calculated if, say, the period is 43 days?

Bruce_L
Posted : Wednesday, May 24, 2017 11:27:27 AM


Worden Trainer

Joined: 10/7/2004
Posts: 65,138

Well you could do the math on how many bars need to return true, but you could also just write it like this.

CountTrue(O > 1.05 * O1, 43) > 75 * 43 / 100

I generally try to simplify things down to the simplest form in hopes that it will make the formula less resource intensive however.

CountTrue(O > 1.05 * O1, 43) > 32



-Bruce
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0002ct
Posted : Wednesday, May 24, 2017 11:41:00 AM
Registered User
Joined: 11/19/2015
Posts: 459

Thanks very much, Bruce. The example provided does precisely what I'm looking for. Thanks again.

Bruce_L
Posted : Wednesday, May 24, 2017 12:02:18 PM


Worden Trainer

Joined: 10/7/2004
Posts: 65,138

You're welcome.



-Bruce
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