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[amibroker] Re: Weighted Average



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Ken,

I use a simple 4 period weighted ROC formula which gives 
more "weight" to the near term than to the longer term. The key is to 
divide by the sum of your weight factors. Here it is:

weighted_roc =(ROC(C,5)*4) + (ROC(C,15)*3) + (ROC(C,25)*2) + ROC
(C,35)) / 10);
 
regards,
Tom
--- In amibroker@xxxxxxxxxxxxxxx, Ken Close <ken45140@xxx> wrote:
>
> I found the following in some old code written by someone else.
> This is not AFL but you can see the way the weighted average is 
created.
> Is this the mathematically correct way?
>  
> 
> f1 = fselx  (In AFL this would be f1 = Foreign("FSELX","C")  )
> 
> w1 = 60
> 
> f2 = fsvlx
> 
> w2 = 30
> 
> f3 = fsrbx
> 
> w3 = 10
> 
>  
> 
> //Sum their weighted exponential growth rates.
> 
> RateSum = w1 * Log(f1 / First(f1)) + 
> 
>           w2 * Log(f2 / First(f2)) + 
> 
>           w3 * Log(f3 / First(f3))
> 
>  
> 
> //Average the weighted exponential growth rates.
> 
> RateAvg = RateSum / (w1 + w2 + w3)
> 
>  
> 
> //Create a price series using the new growth rates.
> 
> WeightedAvg = Pow(10, RateAvg)
> 
> 
>   _____  
> 
> From: Mr. Valley [mailto:valleymj@...] 
> Sent: Tuesday, January 23, 2007 12:32 PM
> To: ken45140@xxx
> Subject: RE: [AmiBroker] Weighted Average
> 
> 
> If the total of the four symbols = 100%
> Then each are a portion of 100%, then what portion is each to the 
total?
>  
> 
> -----Original Message-----
> From: amibroker@xxxxxxxxxxxxxxx [mailto:amibroker@xxxxxxxxxxxxxxx]
On Behalf
> Of Ken Close
> Sent: Tuesday, January 23, 2007 10:25 AM
> To: amibroker@xxxxxxxxxxxxxxx
> Subject: [amibroker] Weighted Average
> 
> 
> 
> 
> 
> What would be the correct way to get a weighted average of three or 
four
> tickers, with the weightings unequal?
>  
> WtAvg = Wt1 * Price1 + Wt2 * Price2 + Wt3 * Price3.   ??????
>  
> I am asking because I thought I remember seeing here or on another 
list that
> doing what I wrote above will give an incorrect result.  But I do 
not
> remember what alternative was suggested.
>  
> Any help?
>  
> Thanks,
> Ken
>



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