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I also tried now to calculate GBPJPY H1 on 1000 bars. Got two linear regression channels on bars 63 and 164.
I calculate the sample by average bar prices (O+H+L+C)/4.
Good luck and hitchhiking trends.
Still, I suspect a dumping into an infinite loop (something like this) with an unsuccessful channel search.
Also - this non-standard behaviour of the pound - perhaps this is some kind of signal when it becomes too good for this algorithm?
2. Price is USD/EUR and Time is seconds, minutes, hours, etc. What you will add to what. The R/S ratio in the Hearst figure is dimensionless because R and S have the same dimensionality. And that is the only reason it makes sense.
3. In the formula H=Log(R/S)/Log(N/2), the value N is not time, as you might think. N is the number of elements in the sample. The fact that we do not take all events, but only a part of them (counting by Close, for example), dividing the process into equal time intervals, is our problem and it has nothing to do with Hirst.
Indeed you are right. Taking time into account in this parameter is probably nonsense!
So far I stopped at the initial calculation of the curve length as a simple price difference between neighboring points of the parabola. Subjectively I like such calculation better than the difference between High-Low samples so far. No final decision has been made yet. We are conducting experiments. I think that it will take some time of observations in order to understand to what extent this way of calculation R has the right to exist.
I think I've managed to replicate the problem on the history and fix it at the same time. The problem has disappeared as soon as I replaced Lowest(); and Highest(); with my own algorithm for searching. By the way, these functions are not suitable as it is not clear what exactly they output - the longest/lowest function value on a given interval (which is not appropriate) or an extremum (which is required), which is not the same thing - there are values at the ends of the interval and they can be more extreme than extrema but are not extrema - then the samples may be incorrect.
Good luck and good luck with trends.
SZZ I think that you will have to write all functions by yourself to be sure that the algorithms work and it will be easier to rewrite them on VCPP afterwards. :).
Good luck and good luck with trends.
I think I will have to write all the functions by myself to be sure the algorithms work and it will be easier to rewrite them on VCPP. :).
"RE Slawa - answer to zigzag :)"
last post
To search for Hai, the condition should be at least
if((H[k-1]<H[k])&&(H[k]>H[k+1])&&(H[k]>curHi)) for minimum the same way.
Good luck and hit the trends.
All that remains is to justify the choice of a sufficient area around the High to take it as the High of the given sample. I take it that this High must be the maximal point within a radius of +-30% of the sample length? In case it is not so, the sample must be increased in order to determine two things together - the extremum and the sample length? What are your thoughts on this?
Vladislav, do you suppose you will correct the code of the Murray indicator in the light of the new information? We are waiting for the new version ;o)!