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OK, I have the ref now
TASC, V12:5 [pp 218-221]
The c-Test by William Eckhardt
It is more than clear, when you read the text:
"To determine why a coherency test is necessary, first we look at
some charting procedures that we suspect may be
incoherent — for example,
the fallacy of attaching significance to the sizes of angles on a bar
chart or similar price graph.
It is not that trading systems making use of angles are inferior —
there is such a glut of bad systems in general that it would
hardly be worthwhile to identify only a few — but rather, these
angle-dependent trading systems are pseudo-systems,
incoherent procedures disguised as algorithms."
See also http://groups.yahoo.com/group/amibroker/message/3825 and,
IMO, the subject is over.
DT
--- In amibroker@xxxxxxxxxxxxxxx, "DIMITRIS TSOKAKIS" <TSOKAKIS@xxxx>
wrote:
> Franko,
> we will make a lot of rounds without definitions. Do you have,
> perhaps, Eckhardt reference on the subject ?
> It would help to see the exact text and use it as a common base of
> understanding.
> DT
> --- In amibroker@xxxxxxxxxxxxxxx, Franco Gornati <fgornati@xxxx>
> wrote:
> > DIMITRIS TSOKAKIS wrote:
> > > Franko,
> > > A function[transformation] F is defined as linear when
> > > F(x+y)=F(x)+F(y)
> > > F(k*x)=k*f(x), k is a constant
> > > The linearity of a function has nothing to do with coherence.
> > > A simple example: The RSI transformation is obviously non-
linear
> [do
> > > not expect RSI(10*C)=10*RSI(C),
> > > simply because 10*RSI(C) will vary from 0 to 1000.
> > > On the other side, the RSI transformation should pass the c-
test.
> To
> > > avoid boring maths, simply
> > > x=C;
> > > y=10*C;
> > > Plot(RSIA(x),"RSI[C]",4,1);
> > > Plot(RSIA(y),"RSI[10*C]",1,1);
> > > and you will see two identical plots.
> > > BTW, the same definition implies that MACD is not a coherent
> > > transformation, although it is pretty linear.
> > > See the simple
> > > x=C;
> > > y=10*C;
> > > MACDofx=EMA(x,12)-EMA(x,26);
> > > MACDofy=EMA(y,12)-EMA(y,26);
> > > MACDofxplusy=EMA(x+y,12)-EMA(x+y,26);
> > > Plot(MACDofxplusy,"MACD of (x+y)",2,1);
> > > Plot(MACDofx+MACDofy,"MACD of x +MACD of y",9,1);
> > > Plot(MACDofy,"MACD of 10*x",1,1);Plot(10*MACDofx,"10*MACD of
> x",4,1);
> > > to agree that MACD passes the linearity test,
> > > although it does not pass the c-test [do not expect MACDofy and
> > > MACDofy to have the same graph !!]
> > >
> > > In practical terms, is there any limitation when we apply RSI
or
> > > MACD,
> > > based on the fact that RSI is coherent and MACD is not ?
> > > Dimitris Tsokakis
> >
> > Dimytris,
> >
> > to say that the linearity of a function has nothing to do with
> coherence needs
> > at least the exact definition of dimensionally coherence.
> >
> > A dimensionally coherent function y is expressed as
> >
> > y = f(x) = f(c*x)
> >
> > only if the definition "A system can be said to be dimensionally
> coherent if its
> > results do not change even though the units of measure do" refers
> to the exact
> > value of the indicator. This is probably achieved only by
> normalized indicators
> > in form of ratio. (properties of logs will explain this)
> >
> > But if the definition "A system can be said to be dimensionally
> coherent if its
> > results do not change even though the units of measure do" refers
> to a set of
> > results represented by points (buy and sell points) then a liner
> function does
> > not change any of its x points of minimum or maximum
> >
> > y' = f'(x) = f'(c*x)
> >
> > or any other characteristics.
> >
> > Try again plotting your example but this way
> > x=C;
> > y=10*C;
> > MACDofx=EMA(x,12)-EMA(x,26);
> > MACDofy=EMA(y,12)-EMA(y,26);
> > Plot(MACDofy,"MACD of 10*x",colorBlue,1);
> > Plot(MACDofx,"MACD of x",colorRed,1 + styleOwnScale);
> >
> > They perfectly overlap.
> >
> > In trading, *every* linear function is dimensionally coherent.
But
> it may be
> > that not every dimensionally coherent system is a linear one.
> > So, that's why 'Systems based on absolute indicator values will
not
> be
> > coherent, but systems based on indicator crossovers of moving
> averages
> > (of the indicator), for example, will' as Mark put it. But, you
> could even make
> > it coherent transforming accordingly the absolute value of the
> threshold levels.
> >
> > And 'F(x+y)=F(x)+F(y) F(k*x)=k*f(x), k is a constant' is the same
> as f(a*x + ..
> > + n*z) = a*f(x) + .. + n*f(z) in a less compact way.
> >
> > --
> > Franco Gornati <fgornati@xxxx>
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