The similarity system | Page 563 | Forex Factory

Options

Go to
Permalinks
Print Thread

Similar Threads

Similarity: A Practical Approach 859 replies

Similarity Trading - The ZZBB Method 215 replies

Rainbow similarity 256 replies

Similarity Stuff 17 replies

Directional Similarity 0 replies

Attachments: The similarity system
Exit Attachments

The similarity system

  • #11,241
  • Oct 29, 2014 9:26am Oct 29, 2014 9:26am
Quoting Eurusdd
For example, suppose you chose your reals to be high of bars. The on the daily chart if 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 1.2222 or below 1.2211. Just two options.
hmm..

not sure how you can conclude that the next bar must be above 1.2222 or below 1.2211. The next bar can be any number (thats really what the sub sequence theorem says and no matter what the numbers are, there will be sequence of n+1 that is either increasing or decreasing).

for e.g. next bar's high can be 1.2220 (between 1.2222 and 1.2211). that will still create a sequence of 2+1 bars

1.2344 1.2222 1.2220


I think you are thinking sub sequence has to be continuous. there is no such claim in the theorem. sub sequence will not always be continuous.

it is easy to create sequence of real numbers where no continuous sub sequence of more than 2 length exist no matter what n is

1 2 1.01 2.01 1.001 2.001 1.002 2.002 1.003 2.003 .......
Think : Trade : Live Life
  • #11,242
  • Edited 9:41am Oct 29, 2014 9:29am | Edited 9:41am
Quoting stt
{quote} hmm.. not sure how you can conclude that the next bar must be above 1.2222 or below 1.2211. The next bar can be any number (thats really what the sub sequence theorem says and no matter what the numbers are, there will be sequence of n+1 that is either increasing or decreasing). for e.g. next bar's high can be 1.2220 (between 1.2222 and 1.2211). that will still create a sequence of 2+1 bars 1.2344 1.2222 1.2220 I think you are thinking sub sequence has to be continuous. there is no such claim in the theorem. sub sequence will not always...
Assume that it is.... given that assumption...leads to a probability statement?

Edit:
If you look at the most recent post, he says listed consecutively. So in his original example, the subsequence can't take into account 1.2566 to 1.2211 or 1.2566 to 1.2222 because the 1.2566 is out of sequence. They don't have to be re-ordered or re-sorted.... they can be randomly generated but are in order of appearance.
A pip is only worth it if you know how much you risked to earn it
  • #11,243
  • Oct 29, 2014 9:32am Oct 29, 2014 9:32am
Quoting stt
{quote} hmm.. not sure how you can conclude that the next bar must be above 1.2222 or below 1.2211. The next bar can be any number (thats really what the sub sequence theorem says and no matter what the numbers are, there will be sequence of n+1 that is either increasing or decreasing). for e.g. next bar's high can be 1.2220 (between 1.2222 and 1.2211). that will still create a sequence of 2+1 bars 1.2344 1.2222 1.2220 I think you are thinking sub sequence has to be continuous. there is no such claim in the theorem. sub sequence will not always...
That post was for a consecutive listed sequence. Corrected
  • #11,244
  • Oct 29, 2014 9:36am Oct 29, 2014 9:36am
Quoting Eurusdd
This theorem is not mine. There are about 22 different proofs of the theorem. If it is true - I cannot say till I prove it myself - then it is the most powerful tool apart from TZ that I have presented here. This theorem means MONEY, if true! The truth of the statement depends on what they mean by subsequence. If they mean a sequence defined linearly on n then I will say there are cases where it is not true. If they allow any possible sequence on n then, yes it is true. So, if the latter is the case then there is one more option for the example...
yes you got it right here. sub sequence does not mean continuous sub sequence. its just any sequence and to be honest it will rarely be continuous sub sequence

PS: sorry i replied earlier without reading all the subsequent posts. but i see you are discussing it further so will review all and then reply more. Thanks!!
Think : Trade : Live Life
  • #11,245
  • Oct 29, 2014 9:38am Oct 29, 2014 9:38am
Quoting Eurusdd
{image}
Ok I assigned random number from 0-9 into x1,x2,x3,x4,x5 and see how the theorem works.

So replacing the set of numbers 0-9 instead with the last 5 high prices, one of the conditions should be valid 93.75% of the time.

If we encounter a failure then we can assume a 0.3% chance of the theorem failing for the next 5 high prices.

Which means the high we see in that set is likely indeed the high and we can sell?
I want more brain power!
  • #11,246
  • Oct 29, 2014 9:41am Oct 29, 2014 9:41am
Quoting Kiads
{quote} Gee.. Sorry My Friend I miss your post.. here we go btw, by default, the purpose of this marked ZigZag indicator is to reverse 'the classic' ZZ, and mostly you will go with the last ZZ leg, if 'she' pointing down then you go south, and if 'she' pointing up you go north. This because 'mostly' last leg of ZZ is 'update' (i don't want to say 'repaint' because she's not). So, you go with that 'update' direction, and this somehow give you good result in higher TF. Hope it help and best regards. MTH {file}
Kiads I am glad that you brought this up. It is possible to make a living fading the first zz signal, buying the break of the first bearish dot instead of selling. Since the beginning of this thread I have been looking at how these zz updates can be used to detect mis-pricing.
  • #11,247
  • Oct 29, 2014 9:44am Oct 29, 2014 9:44am
Quoting drloyd
{quote} Kiads I am glad that you brought this up. It is possible to make a living fading the first zz signal, buying the break of the first red dot instead of selling. Since the beginning of this thread I have been looking at how these zz updates can be used to detect mis-pricing.
Explained at this post : http://www.forexfactory.com/showthread.php?p=7831756#post7831756
MTH
  • #11,248
  • Oct 29, 2014 9:45am Oct 29, 2014 9:45am
Quoting cfdtaleong
{quote} Ok I assigned random number from 0-9 into x1,x2,x3,x4,x5 and see how the theorem works. So replacing the set of numbers 0-9 instead with the last 5 high prices, one of the conditions should be valid 93.75% of the time. If we encounter a failure then we can assume a 0.3% chance of the theorem failing for the next 5 high prices. Which means the high we see in that set is likely indeed the high and we can sell?
right
  • #11,249
  • Oct 29, 2014 9:47am Oct 29, 2014 9:47am
Quoting Eurusdd
{image}
Let me try to prove this theorem:

From the 5-length sequence above, we could have the following 3-consecutive sub-sequences:
1 2 3 (*)
1 2 4
1 2 5
1 3 4
1 3 5 (*)
1 4 5

2 3 4 (*)
2 3 5
2 4 5

3 4 5 (*)

So, it's 10 ones. And for each one, for ex: 1 2 3, we could have 4 cases:
1 < 2 < 3
1 < 2 > 3
1 > 2 > 3
1 > 2 < 3
There're 2 cases the sequence is increasing or decreasing (1 < 2 < 3 and 1 > 2 > 3) .

So, prob. of the event in your theorem is: 4*2 / 10*4 = 2 / 10 = 20%

or I missed something?

Thank you!
  • #11,250
  • Edited 10:12am Oct 29, 2014 9:54am | Edited 10:12am
Quoting xixi
{quote} Let me try to prove this theorem: From the 5-length sequence above, we could have the following 3-consecutive sub-sequences: 1 2 3 (*) 1 2 4 1 2 5 1 3 4 1 3 5 (*) 1 4 5 2 3 4 (*) 2 3 5 2 4 5 3 4 5 (*) So, it's 10 ones. And for each one, for ex: 1 2 3, we could have 4 cases: 1 < 2 < 3 1 < 2 > 3 1 > 2 > 3 1 > 2 < 3 There're 2 cases the sequence is increasing or decreasing (1 < 2 < 3 and 1 > 2 > 3) . So, prob. of the event in your theorem is: 4*2 / 10*4 = 2 / 10 = 20% or I missed something? Thank you!
Sure your proof is not correct.
Start with a given state and then look at all the branches of the tree coming out of your state. 1 for up and 0 for down. there should be 32 branches. only two will not have those conditions. so probability of 30/32.
One of those 6.25% cases.
Attached Image
  • #11,251
  • Oct 29, 2014 10:20am Oct 29, 2014 10:20am
Quoting LiquidGenius
Hi, I believe A is correct. B feels like it's double implementing the premises to a bar that's in the past or something.. that's probably not a good way to explain it but I think it's using bad logic. ...
Quoting kprsa
one should just simply measure in the past the relative frequency of what is now going on. to do this, you find the number of same situation N in the past, and measure the number of times it was a certain type of outcome, say N1, and calculate the relative frequency of this to be f=N1/N. Than you say that if there is a sufficient number of situations where this outcome has happened, you have your probability p=h. ...
Quoting Eurusdd
{quote}Now to be close to the true probability p, you can add all time frames and even all pairs. The mt4 platform can take about 250000 bars for a given time-frame. If you do your homework well you should be dealing with approximately 23,000,000 bars. That is very close to enough data. The truth is real! How you apply it is another matter.
Thank you for your replies about my probability question. Yes, I understand what you mean, LiquidGenuis. And k's method is exactly what I was doing to obtain probability A (78.5%). You guys seem to be very good with statistics and probability, so unless someone else refutes choice A I'll assume that the correct probability in my example is 78.5%, not 97%.

Eurusdd suggested that I would need to analyse about 23 million bars of data to get close to the true probability. I've only analysed about 20,000 so far. And I think LiquidGenius (who has produced similar results to me on his blog) analyses about 100,000 bars. I wonder if Eurusdd is suggesting that the true probability will rise towards 97% if you analyse a great deal of data?

I'll think about this some more because if A is the correct choice then I can see the possibility of an interesting paradox.
You reap what you sow.
  • #11,252
  • Oct 29, 2014 10:24am Oct 29, 2014 10:24am
Quoting Eurusdd
{quote}right
Hah thanks!
I want more brain power!
  • #11,253
  • Oct 29, 2014 10:25am Oct 29, 2014 10:25am
Quoting NorthTrader
{quote} {quote} {quote} Thank you for your replies about my probability question. Yes, I understand what you mean, LiquidGenuis. And k's method is exactly what I was doing to obtain probability A (78.5%). You guys seem to be very good with statistics and probability, so unless someone else refutes choice A I'll assume that the correct probability in my example is 78.5%, not 97%. Eurusdd suggested that I would need to analyse about 23 million bars of data to get close to the true probability. I've only analysed about 20,000 so far. And...
The true probability is 1, ALMOST SURELY as stated in the proposition. The 97% is for a typical chart platform and the optimal values for k,h
  • #11,254
  • Oct 29, 2014 10:32am Oct 29, 2014 10:32am
According to your graph, 5 different high, I do the following:
(I used integers for hight so):
state = {4,5,1,3,2}
-) 4,5,1
-) 4,5,3
-) 4,5,2
-) 5,1,3
-) 5,3,2 "-------" X1> X2> X3
-) 1,3,2
In the creation of n + 1, only a subsequence is found in state. Is a decreasing sequence.
Now as I build the tree?
  • #11,255
  • Oct 29, 2014 10:34am Oct 29, 2014 10:34am
Quoting Kiads
{quote} Gee.. Sorry My Friend I miss your post.. here we go btw, by default, the purpose of this marked ZigZag indicator is to reverse 'the classic' ZZ, and mostly you will go with the last ZZ leg, if 'she' pointing down then you go south, and if 'she' pointing up you go north. This because 'mostly' last leg of ZZ is 'update' (i don't want to say 'repaint' because she's not). So, you go with that 'update' direction, and this somehow give you good result in higher TF. Hope it help and best regards. MTH {file}

Kiads,

Thank you very much My Friend for the indicator.

I have downloaded on to my trading platform and am trying it on different time frames with different
values to see how I can use it in my trading.

May the PIPS be with you.
  • #11,256
  • Oct 29, 2014 10:34am Oct 29, 2014 10:34am
Quoting cfdtaleong
{quote} Hah thanks!
I did not support your last statement though.
  • #11,257
  • Oct 29, 2014 10:40am Oct 29, 2014 10:40am
Quoting cfdtaleong
{quote} Ok I assigned random number from 0-9 into x1,x2,x3,x4,x5 and see how the theorem works. So replacing the set of numbers 0-9 instead with the last 5 high prices, one of the conditions should be valid 93.75% of the time. If we encounter a failure then we can assume a 0.3% chance of the theorem failing for the next 5 high prices. Which means the high we see in that set is likely indeed the high and we can sell?
Hey CFD,
If I have got my thinking right, since it leads to a probability statement against a linear sequence, we could say that if the trend was an increasing subsequence which then ceases to be true then we can exit. Not necessarily enter. Because the next set of probabilities would be either 0 subsequence or -1 (decreasing) subsequence which we won't know until the next bar that confirms the condition.
A pip is only worth it if you know how much you risked to earn it
  • #11,258
  • Oct 29, 2014 10:42am Oct 29, 2014 10:42am
Quoting Eurusdd
{quote} Sure your proof is not correct. Start with a given state and then look at all the branches of the tree coming out of your state. 1 for up and 0 for down. there should be 32 branches. only two will not have those conditions. so probability of 30/32. One of those 6.25% cases.{image}
Thanks Eurusdd!
Just look a bit more details...
Seems like you're right.
I'm finding another proof
  • #11,259
  • Oct 29, 2014 10:45am Oct 29, 2014 10:45am
Quoting Eurusdd
{quote} I did not support your last statement though.
Ok will think it through harder
I want more brain power!
  • #11,260
  • Edited 11:12am Oct 29, 2014 10:49am | Edited 11:12am
RevealWormhole recently gave a buy signal that one could have taken to grab a few pips in a small amount of time.

The chart shows the state of the market after the price rose from below the price mark (red line).

The distortion level remains in place as the most recent one and I am wondering if the market will return to it from above.

[ UPDATE: Yes, market did return back down to the marked price. ]

Leaning toward taking the first move away from the price mark as the trade trigger.

Attached Image (click to enlarge)
Click to Enlarge Name: EURUSD-Wormhole-29-Oct-2014-01.png Size: 29 KB
Open to new approaches.
4 traders viewing now