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kram
Posted : Thursday, September 20, 2018 5:41:50 PM
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Joined: 10/7/2004
Posts: 80

Bruce, is it possible to replicate this approach?

The model is built using the “Microsoft Excel” spread sheet application.  Based on this data the model calculates the 10 day moving average of the PRICE. As the actual PRICE should closely resemble the value of the moving average the difference between the actual PRICE and the moving average is calculated to arrive at the ‘Difference’. Also moving 10 day standard deviation is calculated for the PRICE This standard deviation is multiplied with the ‘K-Value’ to obtain the ‘K times standard deviation’. Thus the aberration in the  price indicated by the difference between actual data and moving average is divided by the volatility of the PRICE represented by the standard deviation multiplied by optimal K-Value to obtain the ‘Price Difference by K times standard deviation’. By using these data the signals signifying the market movements are derived. If the ‘Price Difference by K times standard deviation’ is greater than ‘1’ then prices are expected to increase and the signal is ‘+’ indicating an upward trend. If the ‘Price Difference by K times standard deviation’ is lesser than ‘-1’ then the prices are expected to fall and hence the signal is ‘-‘. If the ‘Price Difference by K times standard deviation’ is not very significant, the price movement cannot be predicted with certainty and hence the signal is ‘0’.

 

K values can be  .25   .50  .75   1.0  1.25   1.50 for example

 

thank you for the consideration. 

Bruce_L
Posted : Friday, September 21, 2018 11:06:11 AM


Worden Trainer

Joined: 10/7/2004
Posts: 65,138

Um, maybe it is the big block of text, but I am not actually understanding what is being described. My best guess would be something like the following.

(C - AVGC10) / STDDEV10

This could be returning k or it could be the formula when k = 1, I am not really sure even as a guess from the text.

But it pretty much has to be wrong (or I am misunderstanding other parts of the text), because I would expect things to behave the opposite of what seems to be described.

If the value is very high, I would expect price to drop and if the value is very low, I would expect price to rise. This is because while price does tend to follow the current trend (so being slightly above zero it would still go up and slightly below zero it would still go down), price also tends to revert to the mean - and reversion to mean seems to win out when the difference is large.



-Bruce
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kram
Posted : Friday, September 21, 2018 11:25:08 AM
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Joined: 10/7/2004
Posts: 80

thank you as always

Bruce_L
Posted : Friday, September 21, 2018 11:47:19 AM


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Joined: 10/7/2004
Posts: 65,138

You're welcome.



-Bruce
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