New version of ROC?

[mgj@xxxxxxxxxxxx]

From: "Kovacs, Ross R" <KOVACSRR@xxxxxxx>
>>Doesn't it make sense to use the value for a simple moving average N/2 days 
ago in the Momentum formula, rather than the single price n days ago?

PRICE today  -  N day SMA from n/2 days ago

should partially alleviate the bumpy movement in ROC or Momentum.
<<

Yes, it does.  Another way of looking at it is to visualize it as a simple 
1-bar momentum of a price series smoothed by a weighted average:  

                  price - average(price[1],N))  
              =   momentum(Waverage(price,N),1) * (N+1)/2

or its transposed form, a weighted moving average of a 1-bar momentum

                 Waverage(momentum(price,1),N) * (N+1)/2

Its exactly the same all three ways.  This is very similar to the formula 
mentioned by <rjbiiilis@xxxxxxxxxxxxx>:

     Input: AvgLen(13),ROCLen(21);
     Plot1(RateOfChange(XAverage(C, AvgLen)[2], ROCLen),"S ROC");

In this method, he replaces momentum by ROC, and Waverage by Xaverage.


- Mark Jurik
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Tools for financial data preprocessing
Jurik Research     http://www.jurikres.com/ 
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