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LinRegSlope Divergence
Re: LinRegSlope divergence

* To: metastock@xxxxxxxxxxxxx
* Subject: Re: LinRegSlope divergence
* From: Pierre Tremblay <pt2000@xxxxxxxxxxxxxxxx>
* Date: Mon, 23 Jul 2001 14:01:25 -0400
* References: <000401c112f3\$da6cc980\$45b272c3@xxx>
* Sender: owner-metastock@xxxxxxxxxxxxx

> Hello List,
>
> Please have a look at the formulas:

LinRegSlope Divergence

divergence(C,LinRegSlope(C,2),0.01)=-1

Sell short:
divergence(C,LinRegSlope(C,2),0.01)=1

> They are very profitable in tests, just too good to be true. That's because
> it is... Tests results back-adjust themselves (ie, you know you should have
> gone long 3 bars after the fact, just like the ZigZag indicator).
>
> My question is: what's wrong with the formula?? For all I know LinRegSlope
> indicator doesnt change its apperance on the chart going back. MS Help says
> the divergence function uses ZigZag ... is that it?
>
> Thanks,
> Yarroll

Yarroll,

DIVERGENCE function uses ZIG ZAG function. So you have the same behavior.

Pierre Tremblay

This reminds me of a problem I ran into a while back. Reasoning that a peak or trough, once established, ought to remain a peak or trough, I decided to see whether one could make a profit by taking action after the change of direction had been recognized. So I wrote a system that said, in effect, "Take a position after the price has moved X points down from the peak or up from the trough," where X was the size of the move required to establish that a peak or trough had been made. Worked like a charm! Except, of course, that when I looked at it on short intraday bars, about once a week it would decide that a bunch of peaks and troughs had never happened, and the up or down trend had been continuously in effect since almost forever. If I'd actually been trading, significant money would have gone the wrong direction.

So, fine. Now I accept that there is no way around the peak/trough problem. Yet I can't help thinking that a peak (or trough) established by prices X percent lower (or higher) on either side ought to remain a peak or trough, no matter
what else happens later. Can anyone explain what it is about the way the relevant functions are written that invalidates this reasoning? Or direct me to a resource that will explain it to me, in very small words?

Many thanks.
Owen Davies

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