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interesting. I’m in the process of building a proper framework in python (with no coding experience) so it’ll take me a while but once i get up to speed i’ll certainly be curious to explore this
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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-1This reply has been reported for inappropriate content.
You’re good, @flx23. The main assumption of the theorem was that prize is generated by a stochastic process that gives equal chances for any candle to be either an up or down candle (or having higher highs/lows than the previous candle). However, in his post (see the delivered quotation here) Eurusdd presumes that the probability is shifted whenever two out of the four conditions aren’t met. But indeed, the probability for the last of five values (where the first four have been realized) is still 0.5. Thus. prize will do what it wants to do. However, if you go through the prize history and analyze each chunk of 5 candles in regards to OHLC, then you’ll see that about 93.75 % of these 5-chunks fulfill at least one of the four conditions (see image here). That may mean that prize resembles the output of a stochastic process that has been initially assumed. If you are able to trade with the info that there is at least one increasing or decreasing subsequence in a set of 5 candles, then you may find this theorem useful (however, I couldn’t make any use of it). Another way to use the subsequence idea to your advantage is to count the number of fulfilled increasing/decreasing subsequences within each 5-chunk and use it as a trend-following tool. The beauty of the subsequence theory is that you can always study historical (sub)sequences that are anchored in the most recent (the last) candlestick. Nevertheless, the more time I investigated to find anything useful in Eurusdd’s theories, the more I come to the conclusion that they are merely useless. Maybe they may work with a different interpretation. I conclude that the main reason why some are able to trade this profitable lies within hedging and/or money management. But if you don’t agree with me, please feel free to convince me … Otherwise I’ll stop my research here (have a lot of other research to do).
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.… ” ??? b) Regarding the High/Low.Close/Open price , is it Not the mean/average price, right ?
Any clue on these ??
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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
Wow, nice result @Anti.
a) About the R, how do you get the candle information ? Do you get it directly from MT4 platform (some thing like R+MT4 bridge) ? Or, do you import the candles data from MT4 to your database, and then do calculation using R based on that (so to say, it is R+Mysql / MsSql bridge) ?
b) Regarding the result, can you explain more about the chart ? What does it mean actually ? Lets see the last image, for h=50. At X axis, we have number of (i) recurrence from 0 to 100, and on Y axis, we have cumulative density function from 0 to 1. The curve of h=50 will be almost flat (around = 1 for cumulative density function, after the recurrence reach >= 50 , it is same as the h_value itself ). If the value of (i) recurrence = 0, it means that it is transient zone, this is clear. But i do not understand about this “we can now calculate the probability for seeing a (l)-recurrent event becoming at least a (l+1) event ” ….
Can you explain more ? Sorry for asking too much, i am a newbie … (i hope this is not a silly question
)Muchas gracias por adelantado, D.
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You’re good, @flx23. The main assumption of the theorem was that prize is generated by a stochastic process that gives equal chances for any candle to be either an up or down candle (or having higher highs/lows than the previous candle). However, in his post (see the delivered quotation here) Eurusdd presumes that the probability is shifted whenever two out of the four conditions aren’t met. But indeed, the probability for the last of five values (where the first four have been realized) is still 0.5. Thus. prize will do what it wants to do. However, if you go through the prize history and analyze each chunk of 5 candles in regards to OHLC, then you’ll see that about 93.75 % of these 5-chunks fulfill at least one of the four conditions (see image here). That may mean that prize resembles the output of a stochastic process that has been initially assumed. If you are able to trade with the info that there is at least one increasing or decreasing subsequence in a set of 5 candles, then you may find this theorem useful (however, I couldn’t make any use of it). Another way to use the subsequence idea to your advantage is to count the number of fulfilled increasing/decreasing subsequences within each 5-chunk and use it as a trend-following tool. The beauty of the subsequence theory is that you can always study historical (sub)sequences that are anchored in the most recent (the last) candlestick. Nevertheless, the more time I investigated to find anything useful in Eurusdd’s theories, the more I come to the conclusion that they are merely useless. Maybe they may work with a different interpretation. I conclude that the main reason why some are able to trade this profitable lies within hedging and/or money management. But if you don’t agree with me, please feel free to convince me … Otherwise I’ll stop my research here (have a lot of other research to do).
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.… ” ??? b) Regarding the High/Low.Close/Open price , is it Not the mean/average price, right ? Any clue on these ??Hi despacito,
i did a simulation of 5 prices, and i got this result : 60% price will be > 1.2222 , 20% price will be < 1.2211, and the rest 20% prize will be between 1.2211 & 1.2222 , and this cause problem . Now, what to do if we got problem ?? And can we bet that the high of current price will be > 1.2222 , although we just have 60% probability from that 5 simulation prices ? After we did SS based on high prices, is it a good take to do the same for the low and close prices ? hmm, i must go back to my lab and do some tests ….
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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 %.
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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 %.
Hi Anti,
a) Thank you for your information about R and how you did it.
b) Regarding the typo (1.2522 => 1.2222), i am still confused as i read this : “Find the highest and lowest value in your set. The number you are interested in should be above the highest or below the lowest value! “. If we define X1=1.2344, X2=1.2566, X3=1.2211, X4=1.2222 , and current candle is the X5. If current bar must not be included in the set, then the lowest value = 1.2211 and highest value = 1.2566 (X2). But if current bar Must be included in the set, we get 2 linear SS ended with X5: X1-X3-X5 and X3-X4-X5. From these 2 subsets, the lowest value is 1.2211 (X3) and the highest value is 1.2344 (X1). So, the highest value in the subset is : 1.2566 (X2) or 1.2344 (X1) depend on if we Must include current bar or not in the subset. This is not the 1.2222 value as he corrected … Or may be he means some thing else ?
cfdtaleong’s post is also interesting for me , still trying to understand the whole …c) I think i get clue why Kiads distinguish between Skull and star. In term of similarity, the failure is defined as DS and the similarity must be restored, and in term of TRZ, 97% of price are recurrent , and we are hoping for the revisit (in the future). There is similarity here in term of SS if i am not wrong … next i must look up to the sky, expecting the bright light of the Stars shines on me ….
Gracias, D.
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the only thing i found useful in the sim thread was that it makes you look at price action around inefficient market moves. this put me on the path of Order Flow. I just want to be sitting in a pile of imbalanced orders regardless of what various stats or indicators say about the next bar because it’s the imbalanced orders that move price not the indicator or stats :) this can end up being the most important thing you can pay attention to.
Hi Lowphat,
Could you please explain more about imbalance ? I still can not find any clue how to get it. i believe it will appear before gap, and some one talked about same strength of supply/demand (lets say buy and sell strength are almost always in balance on EU) or similar to Bull/Bear. But how to explore this ?
Tom.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 …
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@despacito: 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.
…..
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 …
Great information Anti. very clear like a crystal ball … really appreciate it.
Muchas Gracias, D.This reply has been reported for inappropriate content.
Re. hedging:
I would say: Don’t bother with hedging, especially when scalping and hunting for 1 to 4 pips!
Case 1 – single-pair hedge: Say you open a BUY on pair XY, and market acts against your position. Ok, you could open a SELL on XY (equal size). From now on, both trades would even each other out, but you will pay trading cost twice. If you had just closed your first trade, it would also be neutralized immediately, and you only paid trading cost once. So: why hedging?
Case 2 – multi-pair hedge: BUY XY, running against your position. Your idea is to buy equal value YZ to hedge. Now you pay for your trade twice and have to ‘care’ for two trades from now on. You could also close the XY trade and buy XZ instead: costs you for two trades also, but you have to ‘care’ for just one trade (assuming equal trading cost for the different pairs).
I think user ‘Hanover’ @ FF posted some detailed calculations about hedging vs. non-hedging in FX. His résumé in short: single-pair hedging makes no sense in FX.
In my opinion, that ‘Case 2’ hedge is ok if (and only if) you’re trading a multipair strategy and the XY and YZ trade come from independent trade setups for each pair, and your overall strategy acknowledges that your paying multiple cost. When scalping, single trade cost is always to be considered.
Re. ‘pips’:
‘Pips’ are nice for traders’ smalltalk, but don’t use pips for serious calculations like money management. One pip has a different value on different pairs. For MM use universal units instead, like ATR multiples or similar, or at least normalize the original pips.
s.
Oops: now the post I was referring to is gone!
A good trader is a realist who wants to grab a chunk from the body of a trend, leaving top- and bottom-fishing to people on an ego trip. (Dr. Alexander Elder)
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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 …
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I still see TZs everywhere on my charts hehe. Once seen, never to be unseen.
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Some comments in regards to the similarity methods/thread @FF …
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Heheheheh! Everywhere there is similarity and TZ you will find @anti
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@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?
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I have tried everything in the books in the last 3 years. What I am doing now is the only solution and conclusion I found
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@anti did you get any hints from my posts that helped you find it?
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@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 … ;)
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