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ad the slope should be "k3"
btw folks i can't try it now cuz i don't have MS on this pc
any1 would care to let me know if it works ?
--- In equismetastock@xxxxxxxxxxxxxxx, "lecorbeauxmasque7"
<lecorbeauxmasque7@xxxx> wrote:
> ok ,this is it
>
>
> q1:=Input("set lenght of sample",1,10000,20);
> k0:=(q1+1)/2;
> k1:=Sum(1,q1);
> k2:=Mov(C,q1,S);
> k3:=(1/q1)*Sum((k1-k0)*(C-k2),q1)/Power(k1-k0,2);
> k4:=k2-(k3*k0);
> k5:=k4+(k3*k1);
> k5;
>
> --- In equismetastock@xxxxxxxxxxxxxxx, "Lionel Issen" <lissen@xxxx>
wrote:
> > What is pds?
> >
> > -----Original Message-----
> > From: Jose [mailto:josesilva22@x...]
> > Sent: Saturday, August 14, 2004 11:15 PM
> > To: equismetastock@xxxxxxxxxxxxxxx
> > Subject: Re: [EquisMetaStock Group] Subject: Linear Regression /
Linear
> > Regression Slope Relationship
> >
> >
> > Roy,
> >
> > > b=the slope (Linear Regression Slope)
> > > and
> > > b=[(x1-xbar)(y1-ybar)+.......+(xn-xbar)(yn-ybar)]/
> > > [(x1-xbar)sq +......+(xn-xbar)sq]
> >
> > Which is almost the same as:
> >
> > startBar:=Cum(1)-pds;
> > y:=pds*Sum(startBar*x,pds)
> > -Sum(startBar,pds)*Sum(x,pds);
> > z:=pds*Sum(Pwr(startBar,2),pds)
> > -Pwr(Sum(startBar,pds),2);
> > LRS:=y/z;
> >
> > There must be an easy way to get rid of the Cum(1).
> >
> > jose '-)
> >
> >
> >
> > --- In equismetastock@xxxxxxxxxxxxxxx, "Roy Larsen" <rlarsen@xxxx>
> > wrote:
> > > Hi Harry
> > >
> > > I'm totally lost with this mathematical jargon (which I admit is
> > what I asked for), but there a
> > > couple of lines of your post that make sense to me. They are...
> > >
> > > b=the slope (Linear Regression Slope)
> > >
> > > and
> > >
> > > b=[(x1-xbar)(y1-ybar)+.......+(xn-xbar)(yn-ybar)]/
> > > [(x1-xbar)sq +......+(xn-xbar)sq]
> > >
> > > Putting these two statements, can I assume that converting b
> > correctly into MetaStock code will give
> > > me an exact replica of the MFL Linear Regression Slope function?
> > >
> > > If that's the case then I am indebted to you.
> > >
> > > Regards
> > >
> > > Roy
> > >
> > > > On 14 Aug 2004 13:57:04 -0000, "Roy Larsen" <rlarsen@xxxx>
> > > > wrote:
> > > >
> > > > >>
> > > > >>Can anyone help with the mathematical relationship between
> > Linear Regression and Linear
> > > Regression
> > > > >>Slope.
> > > >
> > > > The formula for linear regression is:
> > > > y=a+bx+E
> > > >
> > > > a=the intercept
> > > > b=the slope (Linear Regression Slope)
> > > > E=the error
> > > >
> > > > normally you already have x and y. They are your paired data
> > points.
> > > > You calculate b, then you estimate a (i.e., assume E=0).
> > > >
> > > > Since I don't have symbols on my E-mail program, my definitions of
> > > > terms are as follows:
> > > > n=number of pair values (x and y)
> > > > x1=first value of x
> > > > xn=nth value of x
> > > > xbar=sample mean of the x values
> > > > sq=term squared
> > > > sqrt=square root of term
> > > > *=multiply
> > > >
> > > > b=[(x1-xbar)(y1-ybar)+.......+(xn-xbar)(yn-ybar)]/
> > > > [(x1-xbar)sq +......+(xn-xbar)sq]
> > > >
> > > > a=(ybar)-(b)(xbar)
> > > >
> > > > You can calculate the linear regression with the trend ()
function
> > in
> > > > Excel.
> > > >
> > > > My statistics book reminded me that the slope by itself cannot
> > tell
> > > > you how strongly correlated x and y are. For that you use the
> > Pearson
> > > > correlation. In Excel, it is Pearson().
> > > >
> > > > The formula for the Pearson correlation is:
> > > >
> > > > r=[(x1-xbar)(y1-ybar)+.......+(xn-xbar)(yn-ybar)]/
> > > > [{(x1-xbar)sq +......+(xn-xbar)sq}sqrt * {(y1-ybar)sq
> > > > +......+(yn-ybar)sq}sqrt]
> > > >
> > > > That is the easy part, Roy. I'll leave the MS coding to you!
> > > >
> > > > Harry
> > > >
> > > >
> > > >
> > > >
> > > >
> > > > Yahoo! Groups Links
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> >
> >
> >
> >
> >
> > Yahoo! Groups Links
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