Hearst index - page 15
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I dare to assure you that "by eye", you precisely cannot determine with certainty, where is the M1, and where is the week (for example for a series of EURUSD). But the use of the SPX, will show exactly the difference between the various TFs for this quote.
What's that got to do with statistically. People have decided that you can't tell "by sight", so it's fractal. And then they start theorising. Malnenbrot and all the other fractalists.
By the way, even the same Hurst shows different values for different timeframes. Even if these values are not very different, the trends are usually visible.
Who knows about the ISC options?
For example the following variant. After the first calculation, we determine the outliers and assign weights to the data points based on them. After that we repeat the calculation taking the weights into account.
The question is, where is this competently described, so as not to reinvent the wheel?
Who knows about the ISC options?
For example the following variant. After the first calculation, we determine the outliers and assign weights to the data points based on them. After that we repeat the calculation taking the weights into account.
The question where it is competently described, that not to re-invent a bicycle?
Then it is better to calculate the standard deviation, and when counting again, remove points whose standard deviation is 1.5 times greater than the average.
Then it's better to calculate standard deviation and then remove points whose standard deviation is 1.5 times of the mean.
This is an extreme case of what I was asking about. What you suggest means to assign weight to these points =0
The question is the same, where it is competently described?
Who knows about the ISC options?
For example the following variant. After the first calculation, determine the outliers and use them to assign weights to the data points. After that we repeat the calculation taking the weights into account.
The question is where it is competently described, so as not to re-invent the wheel?
Why? There are values A, B, and there are their confidence intervals.
Why? There are A, B values, and there are their confidence intervals.
I assume that by setting the weights you can get a smoother variation index curve. I want to check it. I can of course just impose MA, but it's not so interesting, though maybe we shouldn't look for too complicated ways :)
This is an extreme case of what I was asking about. What you propose is to assign weight to these points =0
The question is the same, where is this intelligently described?
I don't know, based on the assumption that a certain percentage of sample points drop out of the sample and have a noticeable effect on the results.
You can of course look for the right percentage of the most distant points, but it's easier with RMS.
In general it is the opposite of what you said. The correct way is not to take the squared deviation as a weight, but the inverse of it.
This is where the problem of dividing by 0 arises.
Then the coefficient can be thought of as -- 1/(1 + KO) .
Then the repeated target function would be this:
Only the derivatives will have to be recalculated by hand )I don't know, based on the assumption that a certain percentage of sample points drop out of the sample and have a noticeable effect on the results.
You can of course look for the right percentage of the most distant points, but it's easier through RMS.
In general it is the opposite of what you said. The correct way is not to take the squared deviation as a weight, but the inverse of it.
This is where the problem of dividing by 0 arises.
Then the coefficient can be thought of as -- 1/(1 + KO) .
Then the repeated objective function would be this:
Only the derivatives will have to be recalculated by hand )Your version implies a sum of coefficients not equal to 1. Is that correct? Probably correct to normalize them by their own sum.
(1/(1+KOi))/Summ(1/(1+KOi))
Your option implies an amount of coefficient not equal to 1. Is this correct? It is probably correct to normalise them by their own sum.
(1/(1+KOi))/Summ(1/(1+KOi))
It's OK, they are used in the target function, so normalising them will not change the result.
You can check if you want.
I hope you can derive the derivatives?
It's OK, they are used in the target function, so rationing will not change the result.
You can check if you want to.
I hope you can derive the derivatives?
sure :)