Market phenomena - page 64

 
There are special stationarity tests, you know which ones (DF, for example). I don't know him, I've only heard of him.
 
Mathemat:
There are special stationarity tests, you know which ones (DF, for example). I don't know it, only heard about it.

I gave a picture of KPSS.

 
Mathemat:
It's a bred safe cable altogether.
According to Peters: the distribution has a leptocurrency: sharp tops and thick tails. According to Mandebrot: the distribution is not normal, but Pareto, where the variance is infinite at all.
 
Mathemat:
Show me your post (or at least the topic), I don't want to waste time looking for it. Especially as the topic here is also quite worthy.
Yes, I do not want to, and the question is not difficult. On a thematic encyclopaedic dictionary this concept is defined only for distributions with one mode. Besides, it is known and intuitively clear that even among these many have no tails at all or very small at any values of the "kurtosis coefficient".
 
-Aleksey-:
I don't feel like it either, and the question isn't complicated. According to the Thematic Encyclopaedic Dictionary this notion is defined only for distributions with one mode. Besides, it is known and intuitively clear that even among these many have no tails at all or very small tails for any values of the "kurtosis coefficient".

Well, it is certain for any distribution, it is just that intertrip in the case of multimode is difficult.

faa: According to Peters: the distribution has leptokursois: sharp tops and thick tails. According to Mandebrot: the distribution is not normal, but Pareto, where the variance is infinite at all.

This has nothing to do with its stationarity. Random walk with fully independent returns, distributed symmetrically about zero by Cauchy with a fixed parameter (i.e. a formally stationary distribution of returns), has thick tails, and the second momentum is infinite. (Actually, Cauchy doesn't even have the first momentum defined.)

At the same time, it is easy to generate a value with floating parameters of a normal distribution whose distribution will have thin tails but will be non-stationary.

 
There is a phenomenon in my recipe with potential for practical application. I'll sketch it out in a minute.
 
alexeymosc:
There is a phenomenon in my recipe with the potential for practical application. I will sketch it out in a minute.

We have a stationary series of random numbers, the autocorrelation between neighbouring terms is close to zero. Moreover, these conditions can be met only partially, not strictly... For our purposes, a series of increments of some currency pair is suitable; I took EURUSD M5 - from terminal A-ri open[0]-open[1] from March 8, 2011 to January 20, 2012:

There it is, my dream row, there it is:

The average of the entire series is close to zero - 0 to five decimal places. Now the basis of the phenomenon. If the value at time t = X(t) is greater than the average of the series, then the next value at time t+1 = X(t+1) will be less than the previous one with 75% probability. Conversely, if the value at t is less than the average, then at t+1 the value will be greater than the previous value with a probability of 75%. 75%. (I'll point out the article on the subject on request.)

If open[0]-open[1] is greater than zero, then the expected rise to the next open will not be greater than open[0]-open[1] with probability 75% (there might be a negative rise and price will go down). Price may go up as well, but probably not more than the distance set by the difference of the previous two Open. Nothing practical comes out so far. Just basic heuristics.

Warning: a question for connoisseurs. If price within a bar moved past open + (open[0]-open[1]), provided that open[0]-open[1] is greater than zero, will price return to the range < open + (open[0]-open[1]) with 75% probability?

Answer: please, Alexey. No, globally (across the entire sample) the probability pattern changes. If price has passed the threshold set by previous values, with nearly 50% probability it will return to the place where it should be according to initial hypothesis 0.75.

And now for a bit of perversion. Let's try to play with open[0]-open[1] dimensions. Maybe there is some additional dependence on the price movement range (volatility).

So, the climax:

The figure shows the case only for open[0]-open[1] <0 (although I mentioned the reverse situation, but still, symmetrically) . In the summary table, column K goes the values of open[0]-open[1] modulo and rounded to 4 decimal places, that is, all variants that are in my original series. Column N is the number of cases. In the column M there are probabilities that the price decreasing inside the bar by the value open[0]-open[1] will be higher than the open + open[0]-open[1]at the future open .That is, it will open the possibility of probabilistic forecasting and even ... shh... profit.

In short, it may be confusing to write. It's something to think about.

The chart shows: the blue line is the probability of price returning to the predicted area, the red line is the number of cases, the abscissa row is the spread of open[0]-open[1] from lesser to greater.

So, for open[0]-open[1] with larger values modulo, the price, after breaking the level set by the previous value of open[0]-open[1] ,tends to return (roll back) to the predicted area, though the probability of this rolling back is lower than 75%.

Here are the results of the simulated trade (I took a spread of 10 five-digit points):

One line for selling, one for buying and their amount. On the ordinate axis are the POINTS.

I am answering questions while I have strength.

That's it.

 

alexeymosc:

If the value at time t = X(t) is greater than the series average, then....
Is there a dependence of probability on amplitude Open[0]-Open[1]?
 
Rorschach:
Is there a probability dependence on the amplitude of Open[0]-Open[1]?
Maybe there is. Try it.
 

What else I will add... Other frames and pairs carry the same ability to work predictably. But I have not checked it.

And another thing - it should not be difficult to make an Expert Advisor (only one adjustable condition, closing of a position by a new bar condition). Maybe I will sketch it myself (in my next life), maybe someone will be interested and...

Reason: