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I stated when I posted it that I hadn't thoroughly checked it ...
This morning I did check it in Excel ... Looks correct to me the way
it is.
--- In amibroker@xxxxxxxxxxxxxxx, "vlanschot" <ecbu@xxxx> wrote:
> Fred, just one comment. Unless you assume that you capture the
whole
> population (which I would argue you're not; you're dealing with
> samples), the formula below is biased and needs a slight adjustment:
>
> CDiff2n[i] = CDiff2[i] / (n[i]-1);
>
> This way you adjust for using the mean in your estimates.
>
> Sorry for being a purist.
>
> PRS
> --- In amibroker@xxxxxxxxxxxxxxx, "Fred" <ftonetti@xxxx> wrote:
> > That's probably correct given what you are after ... Just be
> careful
> > what you put inside the loop .vs. outside and that you only do
full
> > array calc's where necessary ...
> >
> > I changed your oscillator function ( 10 - 50 ) and I haven't
> checked
> > the rest thoroughly, but this should get you close to where you
> want
> > to go. Notice I calced n outside the loop as there's no reason
for
> > it to be inside and that some of the calcs inside the loop are
not
> > full array calcs. I did try playing with some things like Sum(X,
n
> > [i]) in place of Cum(X) but there don't appear to be any real
time
> > savings. Exacerbating the problem is that indicator code is
> > typically run several times. Try running what's below as a plot
> and
> > then as an explore and you'll see a noticable difference in how
> much
> > clock time is used. There are ways to keep this from happening
> with
> > indicators but I forget what they are off the top of my head.
> >
> > X = (sin(Cum(0.01)) + 2) * 20 - 10;;
> >
> > n = BarIndex() + 1;
> >
> > for (i = 0; i < BarCount; i++)
> > {
> > CumX = Cum(X);
> > Mean[i] = CumX[i] / n[i];
> > Diff = X - Mean[i];
> > Diff2 = Diff ^ 2;
> > CDiff2 = Cum(Diff2);
> > CDiff2n[i] = CDiff2[i] / n[i];
> > StDevX[i] = sqrt(CDiff2n[i]);
> > }
> >
> > Plot(StDevX, "StDevX", colorWhite);
> >
> > Filter = 1;
> > AddColumn(StDevX, "StDevX", 1.4);
> >
> >
> > --- In amibroker@xxxxxxxxxxxxxxx, "treliff" <treliff@xxxx> wrote:
> > > > That doesn't seem to be relevant as at any given bar a
> particular
> > > > Length would be in play for the MA calculated at that bar.
> There
> > > > wouldn't appear to be a need to fill an array with the same
> > > > number ...
> > >
> > > Perhaps not in the MA example, but in my particular code there
is
> > > substantial difference between dispersion and min/max values
(and
> > > consequently stdev's) related to Cycle 10 and Cycle 50, so yes,
if
> > > I'm
> > > currently in Cycle 10-mode, I want stdev calculated strictly
over
> > > Cycle 10's.
> > >
> > > > If you'd like to continue this please email me offline ...
> > >
> > > You helped me remove the 1st loop, I'm pretty certain I'll
> > > have to
> > > live with the 2nd one... Thanks for all your help.
> > >
> > > -treliff
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