Forum Replies Created
- AuthorPosts
This reply has been reported for inappropriate content.
@anti did you get any hints from my posts that helped you find it?
To be honest, I wasn’t in the mood to watch all of your videos yet. I’ve just seen some parts of it … ;)
This reply has been reported for inappropriate content.
@Ephisi Well, I think I already found it. I have signals with a > 97 % probability and they will generate a few pips on average. However, the cases that don’t work will cause a big damage (some hundret pips). Thus, the question is how to deal with these cases?! … I remember that EU always said that you need to be able to hedge in order to make money with almost everything he introduced (similarity, TZs, etc.). But how could it be hedged?
This reply has been reported for inappropriate content.
Some comments in regards to the similarity methods/thread @FF …
July 31, 2018 at 7:05 am in reply to: Towards a quantitative market view: Trading started from Zero #14362This reply has been reported for inappropriate content.
What exactly?
This reply has been reported for inappropriate content.
Hi again,
although I’ve thought I’m fed up with the similarity ideas (I can just repeat that I haven’t find anything working well for me), while walking to my day job I suddenly started to think about the TZ concept again. It was just an idea and I haven’t tested anything so far. But I wondered how we could deal with potential TZ’s if we’d estimate the height k of a TZ by the potential TZ height of the current bar. In that case we could maybe trade towards forming TZ’s without getting too often caught …
This reply has been reported for inappropriate content.
Thanks, you too.
Thanks a lot, guys. It’s quite interesting to see that there’s no direct relationship between the period and the lag. I don’t like to apply it in trading. However, it seems to be part of CrucialPoint’s indicators (Zero-lag MACD) that uses the Zero-lag EMA described in the paper you mentioned.
Hi, I’m not sure if this really fits here. However, I have a question in regards to Ehlers ZeroLag MA. I’ve tried to find out which lag this “zero lag” version produces in comparison to its calculation period. Unfortunately without much success. Any hints?
-
This reply was modified 8 years, 5 months ago by
Anti.
Enough ideas for homework over the weekend???
Definitively

Oh – I found indeed a shift in behaviour – even in my H1. The indicator window shows the H1 excess activity for a typical week in 2014 (left) and in 2017 (right) … The shift seems to happened somewhere in the beginning of Octobre 2014.

And here is the prize path-volume-structure of EURUSD in terms of the intercepts and slopes …

From the slope relation the outstanding M15 behaviour becomes even more visible.
The question is, what we now could do with these numbers …
-
This reply was modified 8 years, 7 months ago by
Anti.
@simplex: This evening I had some time and thus I followed your suggestion and calculated prize paths and regressions between prize path and volume for EURUD. I obtained the results listed below:

I calculated the results beginning with the smallest TF. Thus, I was amazed to see that there’s a tendency up to the H1 that for higher TFs the prize path explains more variance in volume. I expected the opposite interrelation. However, this relationship collapses when it comes to the H4 and D1.
Another finding is that there’s something strange with the M15. If you look at the slope and intercept you’d usually expect that it increases from M1 to D1. And that’s indeed the case. However, M15 deviates from it – at least my broker’s data.
Here are the graphs (right click on it and choose “show image” to see the higher resulted picture):

In the H4 you can see that the point cloud widens and there may be a bi- or trifurcation. The same bifucation can be seen in the D1 much better. What could cause that bifurcation? The D1 data dates back to August 2010, H4 to December 2009. Was there any big intervention?
-
This reply was modified 8 years, 7 months ago by
Anti.
Yes, it does. Even if I ignore these cases, your idea only explains less than 3 % of total variance …
Seems that the quote function doesn’t work again …
Take you time to do whatever you want ;) I know that this forum isn’t as attracting as it was two years ago since most users are very silent.
Regarding 2*(High-Low)/abs(Open-Close): I’ve tested different versions for the prize path. I think I’ve tested a similar formula, too. However, 2*(High-Low)-abs(Open-Close) explained the most amount of variance under these formulas.
I think we’ve discussed my understanding of the prize path earlier. My understanding was that the minimum prize travelling distance is twice the wicks and the open-close range only once. I think @gg53 disputed that way and suggested the formula high-low. Without looking back to my R script I remember that high-low only explained 60 % in the total variance but 2*(high-low)/abs(open-close) explained a 64 %. Nevertheless, both linear regressions (I wonder if a linear regression is really the best function) where highly significant (p<2.2e-16). Thus, I’d prefer that way. But as soon as I’ve finished my day job I’ll test 2*(high-low)/abs(open-close).
I’ve thought it would be a good idea to investigate the relationship between prize movements and volume. Thus, I’ve used my way of calculating the prize path, calculated it for 603,896 M1 candles and looked at the relationship between prize path and volume:

As you can see, there seems to be a linear relation between volume and the prize path. The red regression line explains more than 64 % of the total variation in volume.
However, the dots above the blue line indicate bars with more than average volume in comparison to their prize path, and thus more activity per pip. I think that these candles can be seen as exhaustion candles. It’s quite interesting to see how most of these exhaustion candles have relatively high volume and quite small prize paths (“ranges”). This is exactly what @gg53 mentioned in his formula (see 3 posts above this post).
Nevertheless, there are some low activity bars, too. They appear in the lower right part of the plot. I think they can be explained with an erratic prize movement in one single direction during the lifetime of the corresponding bar. With “erratic” I mean that prize developed with jumps (e.g. a jump from 0 ticks to 0.3 ticks, etc.).
If we now use the formula for the regression line and subtract the expected (average) volume for each bar from the real tick volume, we may get a measure for the excess activity … I’ll attach my indicator with the hope that we can continue our work on this topic. However, since we are most probably using different brokers, the formula I obtained for the relationship between prize path and volume may be different than the optimal solution for your broker. Additionally, the formula only applies for M1 charts.
-
This reply was modified 8 years, 7 months ago by
Anti.
-
This reply was modified 8 years, 7 months ago by
Anti.
-
This reply was modified 8 years, 7 months ago by
Anti.
Attachments:
You must be logged in to view attached files.Yes, surely. It may only give a hint. Thanks!
The function is quite simple (and logical):
Volume > Volume[1] and Range < Range[1] and (Low < Low[1] or High > High[1])
The analogy: more water (Volume) are coming into the “River” but encounter a narrow creek (smaller Range), so they must deflect and find new, different path. Roll up your sleeves. G.
After studying the volume concepts laid down by GG I found this one again. After adapting and filtering with higher TF signals I obtain sometimes quite accurate turning points. However, what me really wonders is this: We can often see turning points with a high volume and a high range. How does it fit in the water flow analogy? Why should the river change its direction after the flood leaves a big river bed?
Additionally I found a problematic behavior of tick volume in MT4 (and probably with other platforms, too) … a tick doesn’t neccessarily mean a chang in 0.1 pips. Sometimes it can mean a change by 0.2, 0.3, … pips. However, the volume only counts these changes as 1 tick. Thus, you can see very efficient candles where the prize path (abs(open-close)+2*wicks) is bigger than the volume itself. Is this a problem? Any recommendations?
This reply has been reported for inappropriate content.
lol

This reply has been reported for inappropriate content.
Eurusdd wrote:
For example, suppose you chose your reals to be high of bars. The on the daily chart of you set n=2 then you need to look in the DNA for 5 bars. Let us suppose the highs of the last 4 bars are 1.2344, 1.2566, 1.2211, 1.2222, ????? What will be ????? Since there must be a decreasing or increasing 3-subsequence, the high of the next bar must be above 1.2522 or below 1.2211. Just two options.
Now just draw four short lines for the four numbers (1.2344, 1.2566, 1.2211 and 1.2222) on a piece of paper. The general statement of Eurusdd’s 5-linear-sequence-theorem was
where x1 = 1.2344, x2 = 1.2566, x3 = 1.2211, and x4 = 1.2222. We don’t know the outcome of the last of the five candles, x5. But since we expect that at least 1 of these 4 (or 8 if you look at < and >) conditions should be met, we can check the 4 available numbers and proof if they already fulfill one of the statements:
- (i) 1.2344 < 1.2566 > 1.2211
- (iii) 1.2566 > 1.2211 < 1.2222
Thus, conditions (i) and (iii) can’t be fulfilled by our sequence. Let’s further check what our numbers already give in regards to the remaining 2 conditions:
- (ii) 1.2344 > 1.2211 and
- (iv) 1.2211 < 1.2222.
And here is my interpretation (although Eurusdd seems to see it different since he said that we just have to look for the highest and the lowest number of our four numbers):
Since in 93.75 % one of the four conditions should be fulfilled we can bet that the 5th number, x5, is either smaller than 1.2211 (this would fulfill the condition (ii)) or that x5 is bigger than 1.2222 (this would fulfill (iv)).
However, if I want to predict a future direction, I’m just looking for the cases where the two numbers in (ii) and (iv) (that are x1 and x3 or x3 and x4) suggest an identical direction (either < < or > > in both cases). Moreover, if I want to predict the high of x5, I would only take the predictions where both partly conditions gave < <, thus that the outcome is higher than max(x3, x4).
But once again, this doesn’t work since the premise of this subsequence theorem is that the numbers (opens, highs, lows, closes, etc.) are random numbers with an equal chance of being higher or lower than the previous numbers …
This reply has been reported for inappropriate content.
Hi Anti, This is great, very interesting … but i am new, so i have some questions if you don’t mind …. a) Regarding your link to FF , i am not sure there is a typo or it is indeed right about the price above 1.2522 (is it Not 1.2566 ??)
“… the high of the next bar must be above 1.2522 or below 1.2211. Just two options.… ” ???yes, it was a typo of EURUSDD, see here.
b) Regarding the High/Low.Close/Open price , is it Not the mean/average price, right ? Any clue on these ??
In theory you could use any prize characteristics. I think it was @MTH2014 (aka Kiads) who found that for OHLC4 the relative frequency is even higher. However, I still can’t see how we can use it in order to predict a prize direction. My stats on dependent probabilities* have shown that there’s no edge. (* What matters is the relative frequency of an correct predicted outcome, where the prediction is based on the condition which fulfills x1>x3>x5 and x3>x4>x5, where x1>x2>x3 and x2>x3>x4 hasn’t been fulfilled).
Regarding your other post:
I’ve exported the EURUSD data from my MT4’s history center. Still didn’t have the time/muse to experiment with MT4-R-Interfaces.
Regarding the image for h = 50 … It says that on average all prizes within a range of 101 candles (50 to the left, 1 mid-bar, 50 to the right) will be hit by about 17 candles. The probability for (0)-recurrent prizes is smaller than 2 %. However, if a prize has been hit more than 40 times (for instance, in 50 candles), then the probability that the prize will be hit again within the remaining 51 candles is less than 5 %.
-
This reply was modified 8 years, 8 months ago by
Anti.
This reply has been reported for inappropriate content.
Ok, here are some of the results. I’ve used the same terminology as in my above post (a (0)-recurrent event represents transient prizes). For each candle in the EURUSD H1 history I’ve counted how often each prize has been hit h bars before or after the mid-candle (problem bar). Here are some results for h in {5, 25, 50}:

With these empirical cumulative distribution functions we can now calculate the probability for seeing a
(l)-recurrent event becoming at least a(l+1)event. However, this calculation is based on the assumption that the market is completely random. Thus, in the short run the ‘observed’ probabilities can deviate slightly …Here’s the R code used for calculation of empirical probability distributions:
eu <- read.csv(“C:/Users/Daniel/Desktop/EURUSD60.csv”)
colnames(eu) <- c(“date”,”time”,”open”,”high”,”low”,”close”,”volume”)
head(eu)attach(eu)
h = 5
recur <- NULL
for (i in (h+1):(dim(eu)[1]-h)) {
prize <- seq(from=low,to=high,by=0.00001)
for (j in 1:length(prize)) {
recurrence <- 0
for (k in (i-h):(i+h)) {
if (prize[j]>=low[k] && prize[j]<=high[k]) {
recurrence <- recurrence + 1
}
}
recur <- c(recur,recurrence)
}
print(i)
}h5 <- recur-1
h25 <- recur-1
h50 <- recur-1-
This reply was modified 8 years, 8 months ago by
Anti.
This reply has been reported for inappropriate content.
… way to go …

This reply has been reported for inappropriate content.
What it sounds like (and maybe I’m wrong, a picture is always helpful :) ) is that you’re essentially looking at a histogram of market price – a count of how many times a particular price, or bin of prices, has been touched within time span h. While I haven’t tried this particular method because it seems a little messy, but I suppose not without its potential merit. What I mean is that within the lifetime of a particular bar x, price can run up and down many times. That’s just theory of course. In the rough reflexiveness of the market, hourly bars don’t tend to be too range bound. I think it’s worth a look, at least to understand how that market structure behaves.
Yes, that’s exactly what I did. I’ll post some charts in the future. However, my first findings are that most recurrent levels are hit more often than two times …
This reply has been reported for inappropriate content.
Let me introduce a new terminology. Instead of transient prizes I’ll talk about
(0)-recurrent zones. In general, a prize is(l)-recurrent wheneverl >= 1. It means that the prize under consideration has been hit at least oncehbars after a problem bar. In that way we shift the focus from transient states to recurrent states and look how often one recurrent state has been already hit within the interval[i-h, i+h].Why? The question arose “how do I know if a recurrent zone will be recurrent within the lifetime (in regards to the problem bar) again?”. Or: Will the prize reach again a particular (already
(1)-recurrent zone. I just started to analyze some setups. What I found so far is that for EURUSD H1 withh = 50you’ll have a96.6 %chance to see a further hit of a particular prize that is already(1)-recurrent until the candlei+hhas been build.Before I continue I’d like to see some activity and some useful contributions to this topic (ideas, indicators, concepts, trading strategies, etc. …
-
This reply was modified 8 years, 5 months ago by
- AuthorPosts





-recurrence.png)