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Dimitri,
Thank you ....yes it is clear..... I will revert to using either Ref(value
-1) or Value....
Thank you
Anthony
-------Original Message-------
From: amibroker@xxxxxxxxxxxxxxx
Date: Friday, April 04, 2003 01:30:55
To: amibroker@xxxxxxxxxxxxxxx
Subject: [amibroker] Re: finding the value of a cross condition
Anthony,
there are many ways to interpolate [linear or non linear]. Any
interpolation would not correspond to reality. We use EOD dots,
nothing else. The continuous lines we see do not exist, they are just
for visual purposes. As you understand cum(1)==815 and cum(1)==816
correspond to EOD reality, but, cum(1)==815.32 does not make sense in
an EOD database. There is in amibroker, as in other T/A softwares, an
internal method to "smooth" the curves, a crooked line would be a bit
ugly. Any function we use is just an aggregate of discontinuous dots.
Our StochD() goes from 80.32 to 84.47 WITHOUT passing from the
intermediate values !!
I hope it is more clear now.
DT
--- In amibroker@xxxxxxxxxxxxxxx, "Anthony Faragasso" <ajf1111@xxxx>
wrote:
> Ken,
>
> How would you interpolate a value.....the day before the
cross=987.86 and
> the day after =995.61....
>
> Anthony
> -------Original Message-------
>
> From: amibroker@xxxxxxxxxxxxxxx
> Date: Thursday, April 03, 2003 17:40:34
> To: amibroker@xxxxxxxxxxxxxxx
> Subject: RE: [amibroker] Re: finding the value of a cross condition
>
> Anthony: I suppose you could interpolate algebraically from the
values
> the day after the cross and the day before the cross. What would
you do
> with the value?
>
> Ken
>
> -----Original Message-----
> From: Anthony Faragasso [mailto:ajf1111@x...]
> Sent: Thursday, April 03, 2003 3:48 PM
> To: amibroker@xxxxxxxxxxxxxxx
> Subject: Re: [amibroker] Re: finding the value of a cross condition
>
> Thank you Dimitri....I was just going to post the following:
>
>
>
> I was debugging in AA and came up with the following solution;
>
>
>
> AddColumn(ValueWhen(Cross(fast,slow),fast),"");
>
> AddColumn(ValueWhen(Cross(fast,slow),slow),"");
>
> My next question was going to be....is it possible to find the
value of
> the
> exact cross value...I assume not....so I must decide to use either
the
> fast
> or slow value......
>
> Again, thank you ....BTW...did you receive my post today.......is it
> helpful
> ? I have another Idea.....
>
>
>
> Anthony
>
>
>
>
>
>
>
>
>
> -------Original Message-------
>
>
>
> From: amibroker@xxxxxxxxxxxxxxx
>
> Date: Thursday, April 03, 2003 15:40:54
>
> To: amibroker@xxxxxxxxxxxxxxx
>
> Subject: [amibroker] Re: finding the value of a cross condition
>
>
>
> Anthony ,
>
> In AA
>
> fast=EMA(C,10);
>
> slow=EMA(C,20); cond=Cross(fast,slow);
>
> Filter=Cond;
>
> AddColumn(fast,"fast");
>
> AddColumn(slow,"slow");
>
> or, in IB
>
> fast=EMA(C,10);
>
> slow=EMA(C,20);
>
> Plot(fast,"",1,1);
>
> Plot(slow,"",9,1);
>
> Plot(Cross(fast,slow)*fast,"",5,2);
>
> Plot(Cross(fast,slow)*slow,"",4,8);
>
> Remember that Cross(fast,slow) is binary, 1 or 0.
>
> Note also that you read the values AFTER the cross. [If you want
>
> values BEFORE the cross, use Ref() function]
>
> DT
>
> --- In amibroker@xxxxxxxxxxxxxxx, "Anthony Faragasso"
<ajf1111@xxxx>
>
> wrote:
>
> > How can I find the value of a cross condition. I do not what the
H,
>
> l, C, or
>
> > Open...but the value when the 2 lines cross....is it possible...
>
> >
>
> > example:
>
> >
>
> > fast=ema(c,10);
>
> > slow=ema(c,20);
>
> >
>
> > cond=cross(fast,slow);
>
>
>
>
>
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