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The euro today
Long studs these if predictable for someone, and the stops will allow the DC to work out the profit...Then "the noise is only in the heads" (Paukas).
As long as we apply the indicators, then everything seems clear, what we are doing is getting (identifying) some specific quotient features.
And in the case of the filter. What are we filtering? What do we get, where does what does not pass through the filter go?
It is not clear to me what is clear to you. What kind of indicators do we use? What peculiarities are revealed and how?
МА is a low pass filter, MACD is a bandpass filter (with some stretch). Why don't you ask such questions to MA and MACD indicators as to filters?
What is not clear to me is what is clear to you. What kind of indicators do we use? Which features do we identify and how?
MA is a low-pass filter, MACD is a bandpass filter (with some stretch). Why don't you ask such questions to MA and MACD indicators as to filters?
You can shorten the filter, or you can do something else. For example, move it to the past.
From here it should be clear how much it can be shifted and why the overshoot will be minimal and probably not noticeable to the eye at all.
You have to use weighting functions, otherwise you will just get an SMA. I can even say that I use the Blackman-Hann window.
And then there are BIH filters, but for my purposes they are not suitable.
Could you explain in more detail how to shift the filters to get the minimum delay. If you are aiming for minimum phase excursion, the phase response must be linear, which dictates that its coefficients are symmetrical. In this case the group delay of the filter is equal to half of its length. To decrease filter's delay you need to make its phase response non-linear with higher coefficients on the bars on the right as it is done in Linear Weighted MA, for example. But you claim to use Blackman-Hann window which is symmetric. Apparently you impose another weighted window on it, or shift the cosine arguments, or something else.
There is a mountain of literature written about MA and MACD. It is all very illustrative. But what will happen if the filter has a different front or phase? I do not understand the correlation between the filter concepts and the results of their work
I read this rubbish literature before, just for a laugh, but then I got bored and it wasn't funny. It also said that the strongest MACD signal is divergence. Boo-ha-ha!
There is just as much literature on filters, but it is more useful. You can also read, experiment and everything will become clear.
I read this literary garbage earlier, just for fun, but then I got bored and it became unfunny. It also said that the strongest signal of MAKD is divergence. Boo-ha-ha!
There is just as much literature on filters, but it is more useful. You can also read, experiment and everything will become clear
That's what I tried to do. Changed different filter parameters, got a better fit to the sample. But I didn't see any advantages over the TA analogue - purely at the level of retouching. Both TA and filters don't work at all. Hodrick-Prescott filter has weak relation to quoting. I wrote an article about it
Could you please explain in details how to shift filters to achieve minimal delay. If you strive for minimal phase shift, the phase response should be linear, which dictates that its coefficients are symmetric. In this case the group delay of the filter is equal to half of its length. To decrease filter's delay you need to make its phase response non-linear with higher coefficients on the bars on the right as it is done in Linear Weighted MA, for example. But you claim to use Blackman-Hann window which is symmetric. Apparently you impose another weighted window on it, or shift the cosine arguments, or something else.
It's very simple. There is a filter F[i]=K0*C[i]+K1*C[i+1]+K2*C[i+2]+...+Kn*C[i+n].
Simply replace it with F[i+m]=K0*C[i]+K1*C[i+1]+K2*C[i+2]+...+Kn*C[i+n].
The filter will move backwards by m bars. To calculate the last m bars we seem to be missing C[-1] ... C[-m] (the filter should look into the future). Substitute anything instead, e.g. C[0]. The picture posted above says that this can be done in this case for 150 bars or even more, the error will be almost imperceptible.
Very simply. There is a filter F[i]=K0*C[i]+K1*C[i+1]+K2*C[i+2]+...+Kn*C[i+n].
Simply replace it with F[i+m]=K0*C[i]+K1*C[i+1]+K2*C[i+2]+...+Kn*C[i+n].
The filter will move backwards by m bars. To calculate the last m bars we seem to be missing C[-1] ... C[-m] (the filter should look into the future). Substitute anything instead, e.g. C[0]. The picture above says that this can be done in this case for 150 bars or even more, the error will be almost imperceptible.
Thank you. It worked:
The accuracy of the last Per/2 plot depends on the choice of unknown future values. In the past I did things a little differently: I filtered by the usual Per/2 delayed filter, shifted the filter by Per/2 into the past, and fitted a 3rd degree polynomial with continuous 1st and 2nd derivatives to the last Per/2 price interval. Both methods have the same accuracy as they "fantasise" about the future.
Redrawing