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Registered User
Joined: 10/7/2004
Posts: 69
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HI,
WOULD YOU WRITE A MIN(FORCE13) FORMULA (GENERIC, FOR OTHER SYSTEMS) FOR 10D BACK?
THANKS
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Worden Trainer
Joined: 10/7/2004
Posts: 65,138
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I really am not sure if this is what you are requesting, but if you want the 10-Period Minimum of the 13-Period Exponential Moving Average of (C - C1) * V, I do not know of a way to create a Personal Criteria Formula that would be short enough to be practical.
You may wish to review the following:
Min Max PCFs
Cascades of Moving Averages
How to create a Personal Criteria Forumula (PCF)
PCF Formula Descriptions
Handy PCF example formulas to help you learn the syntax of PCFs!
-Bruce
Personal Criteria Formulas
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Registered User
Joined: 10/7/2004
Posts: 69
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NEED TO STIPULATE THAT TODAY'S BAR IS THE LOW FOR 10 D. THANKS
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Registered User
Joined: 10/7/2004
Posts: 69
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BRUCE,
WOULD I GET THE RIGHT RESULT BY WRITING: (C-C13)*V13 < (C1-C14)*V14......?
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Worden Trainer
Joined: 10/7/2004
Posts: 65,138
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I'm pretty sure that would not produce the correct result... that said, I'm not even sure I understand the request.
-Bruce
Personal Criteria Formulas
TC2000 Support Articles
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Registered User
Joined: 10/7/2004
Posts: 69
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SORRY,
13D FORCE = (C-C13)*V ?
AND THEN :
(C-C13)*V < (C1-C14)*V1 AND (C-C13)*V<(C2-C15)*V2....(C-C13)*V<(C14-C15)*V13??
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Worden Trainer
Joined: 10/7/2004
Posts: 65,138
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If the 13-Period Force Index is (C - C13) * V or (C - C13) * AVGV13 then it should probably work. If on the other hand, it is the 13-Period Exponential Moving Average of (C - C1) * V (as has been mentioned in other topics), it will not work.
-Bruce
Personal Criteria Formulas
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Registered User
Joined: 10/7/2004
Posts: 69
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THANKS BRUCE. IF YOU COME UP WITH A REASONABLE FX FOR THE EXP'L FX I WOULD APPRECIATE IT AND IN THE MEANTIME I'LL GO WITH THE ABOVE.
SC
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Worden Trainer
Joined: 10/7/2004
Posts: 65,138
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You're welcome. I'm still playing with a few ideas for the Exponential version... but am not particularly hopeful at this point.
-Bruce
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Registered User
Joined: 1/1/2005
Posts: 2,645
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Bruce,
I assume the request is for a PCF equivalent to the form:
XAVGC13.0 < XAVGC13.1
AND XAVGC13.0 < XAVGC13.2
AND ...
AND XAVGC13.0 < XAVGC13.9
with C replaced by (C-C1)*V.
The direct method would require 18 series expansions of the EMAs. We will show that it can be done with 9 series expansions of the EMAs and 9 truncated series expansions of the EMAs. The number of terms in the 9 truncated series expansions is 1,2,...,9.
The series expansion for XAVGCP is:
(1-a)*(a^0*C0+a^1*C1+...+a^k*Ck+...)
where a = (P-1)/(P+1). Let XAVGCP.Q,R represent XAVGCP.Q with the series portion truncated to R terms, i.e., XAVGCP.Q,R is defined to be:
(1-a)*(a^0*CQ+a^1*C(Q+1)+...+a^(R-1)*C(Q+R-1))
Theorem: Define a = (P-1)/(P+1). Then, for N < M,
XAVGP.N - XAVGCP.M = XAVGP.N,(M-N) - (1-a^(M-N))*XAVGCP.M
As an application of the Theorem, let P = 13, N = 2. M = 7. Then,
XAVG13.2 - XAVGC13.7
is equivalent to:
(1-6/7)
*(C2+6/7
*(C3+6/7
*(C4+6/7
*(C5+6/7
*(C6)))))
-(1-(6/7)^5)*XAVGC13.7
An application of the Theorem to each of the 9 inequalities results in each of the Cs replaced by (C-C1)*V being written only once and then in a particularly simple form.
For better accuracy, I would be inclined to approximate the infinite series of only XAVGC13.9.
I do not believe we can simplify more than I have indicated.
Your comments will be appreciated.
Thanks,
Jim Murphy
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Worden Trainer
Joined: 10/7/2004
Posts: 65,138
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bustermu,
Thank you very much for the input. I think youhave provided what I need, but I'm not fully understanding your suggestion.
QUOTE (bustermu) Theorem: Define a = (P-1)/(P+1). Then, for N < M,
XAVGP.N - XAVGCP.M = XAVGP.N,(M-N) - (1-a^(M-N))*XAVGCP.M
I made it that point yesterday on my own. It's the leap that comes next that I don't get. I'm probably missing something fundamental.
QUOTE (bustermu) For better accuracy, I would be inclined to approximate the infinite series of only XAVGC13.9.
I still end up making what essentially amounts one "full" expansion for each comparison, for nine in all (which is still very, very long). Just thinking about it in the middle of working on other stuff (after thinking about it far more rigorously yesterday), it is not immediately obvious to me why the first eight comparison would not require the expansion of the last term.
-Bruce
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Registered User
Joined: 1/1/2005
Posts: 2,645
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QUOTE (Bruce_L) It's the leap that comes next that I don't get. I'm probably missing something fundamental.
As i'm sure you suspect, the reason you don't understand the "leap" that comes next is that there is no "leap". We have only reduced the original 18 full expansions to 9.
QUOTE (Bruce_L) I still end up making what essentially amounts one "full" expansion for each comparison, for nine in all (which is still very, very long).
The expression:
XAVGC13.0 < XAVGC13.1
AND XAVGC13.0 < XAVGC13.2
AND ...
AND XAVGC13.0 < XAVGC13.9
can be written in the form:
F(1) < XAVGC13.9
AND F(2) < XAVGC13.9
AND ...
AND F(9) < XAVGC13.9
where each F is a linear combination of the C0,C1,...,C8. Now we have:
max(F(1),F(2),...,F(8)) < XAVGC13.9
which requires only one full expansion at the expense of writing the maximum 2^3 expressions.
Still too long for me, how about you?
Thanks,
Jim Murphy
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Worden Trainer
Joined: 10/7/2004
Posts: 65,138
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bustermu,
I had not thought of the second form and finding the maxmimum. I'm suspecting the length will still stretch the definition of practical, but I might have to plug it into a spreadsheet to at least satisfy my curiosity.
-Bruce
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