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> 1000 * exp( Cum( log( y / 1000 ) ) )
In *theory*, this would be quite useful in accumulating large values
(such as Volume), but unfortunately MetaStock cannot handle large
values for the Exp() function, so there is no advantage here.
Here is an example:
Exp(Cum(Log(V/1000)))*1000
> Exp( Sum( Log( y ), n ) / n )
> or simply
> Exp( Mov( Log( y, n, S ) ) )
Neither of these is an improvement on Mov(y,n,S).
MG, back to your modeling cubicle with you...
jose '-)
http://www.metastocktools.com
--- In equismetastock@xxxxxxxxxxxxxxx, "MG Ferreira" <quant@xxxx>
wrote:
>
> OK, here is a quick, off the cuff example. The Cum function
>
> Cum(y)
>
> gives
>
> y1 + y2 + y3 + ...
>
> What if you want
>
> y1 * y2 * y3 * ...
>
> Well, use
>
> exp( Cum( log( y ) )
>
> This will generally give an overflow quickly, unless you use
> smallish values, so you may need to rescale it to get it to work,
> ie calculate
>
> 1000 * exp( Cum( log( y / 1000 ) ) )
>
> If you think the Cum function is useful, then this must appeal to
> you as well!
>
> Another example, the geometric average. The simple moving average
> is defined as
>
> ( y1 + y2 + y3 + ... + yn ) / n
>
> Of course, in MSFL we all use
>
> Mov(y,n,S)
>
> to calculate this, but we could also use
>
> Sum(y,n) / n
>
> to get the answer the brute-force way.
>
> The geometric moving average, which may actually be more applicable
> to markets due to the exponential growth often seen in prices,
> is defined as
>
> ( y1 x y2 x y3 x ... x yn ) ^ ( 1 / n )
>
> To do this in MSFL, use
>
> Exp( Sum( Log( y ), n ) / n )
>
> or simply
>
> Exp( Mov( Log( y, n, S ) ) )
>
> Regards
> MG Ferreira
> TsaTsa EOD Programmer and trading model builder
> http://www.ferra4models.com
> http://fun.ferra4models.com
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