1st and 2nd derivatives of the MACD - page 50
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So what's the problem if there's a series? Acceleration is the 2nd derivative. We construct F[i]-2*F[i+1]+F[i+2] and that's it.
Not that, well, I have to think how to express myself clearly in human language. So now I'll confuse you again.
This is why there has to be a third part. It should answer the question of what C(-1), C(-2), etc. would equal in the absence of external influences on the market.
This is the most important thing. Only it is not clear how the filters calculated on the assumption that C(0)=C(-1)=C(-2)=... For a long time I've been struggling with this problem of determination of future price movement in the absence of external excitations. We need to make some assumptions about our system (linear, for example) and the type of excitations (e.g. momentum).
This is the most important thing. It is just not clear how filters calculated on the assumption that C(0)=C(-1)=C(-2)=... For a long time I've been struggling with this problem of determining the future price movement in the absence of external excitations. We need to make some assumptions about our system (linear, for example) and the type of excitations (e.g. momentum).
Don't calculate filters from such an assumption, it's a temporary solution to at least somehow calculate the filters and understand what to do next. C(-1), C(-2)... should be unknowns, and the filters will help find them. I think so...
Oh, how the radio people are getting off! Fourier isn't enough - they want Voltaire :)
Guys, when there's no excitation, there's no reaction. Sorry if I upset you.
As for looking for market levels in the absence of new information - it's not easy, it's very easy :)
Guys, when there's no arousal, there's no reaction. Sorry if I upset you.
The implication is that we analyse the market reaction to previous excitations and predict this reaction in the future, implying that there will be no new excitations. In simple terms, "predicting the tail of the reaction". At least that's how I understood AlexeyFX's post "...the question of what will be equal to C(-1), C(-2), etc. in the absence of external influences on the market". I may have misunderstood though.
Oh, how the radio people are getting off! Fourier isn't enough - they want Voltaire :)
Removed my post with the "pushing back". Good luck to everyone here!
Removed my post with the "pushing back". Good luck to everyone here!
I took it out. No one insisted, by the way :)
There is a big difference. The filter redraws smoothly and the further in the past, the less. The filter redraws abruptly and immediately in the opposite direction. You may also attach a sound signal that says "What, SUKA, you have not been waiting for?!
I cited ZZ as a demonstration of the problem, while you have walked away from the problem and are advertising a smoothness in the past that you simply do not care about. The problem with trading is that we can't tell the difference between a reversal and a correction. In your terminology, you can't tell on every wave is the amplitude on the right edge or the wave is still going to continue and a reversal is ahead.
For example, its expectation is 0.
MO not equal to zero is a constant, a shift. It doesn't make a difference.
The recognized problem is dispersion, which is not a constant and there are huge markets for trading this non constant. As they say, we see the speck but not the log.