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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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