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The similarity system

  • #11,261
  • Oct 29, 2014 11:04am Oct 29, 2014 11:04am
Quoting burnssss
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?
You wrote
-) 5,3,2 "-------" X1> X2> X3

isn't it 4,5,1? In this case X1>X2>X3 is false... (according to your state = {4,5,1,3,2})
the 3rd wave is the crowd's wave
  • #11,262
  • Oct 29, 2014 11:15am Oct 29, 2014 11:15am
Quoting Eurusdd
{quote}Yes Burnsss. powerful
Hi, Eurusdd, I have been still following your thoughts of this thread as I am trading probability myself, but the different way.
I have many questions and some contradictions upon these thoughts, BUT I am keeping silence though.
There is a real seed of knowledge in it, I now think.

Anyway, I would like you to give us the original papers of authors, where you take those brillliant formulas you publish.

Best, D.P.
Hunting High and Low
  • #11,263
  • Oct 29, 2014 11:16am Oct 29, 2014 11:16am
Quoting Eurusdd
{image}
Hm, is the theorem really only including those 4(or 8 depending on how you look at it) possibilities?
Are these 3 not included intentionally?
Attached Image
"Mrrraggglhlhghghlgh"
  • #11,264
  • Oct 29, 2014 11:22am Oct 29, 2014 11:22am
Quoting juriFX
{quote} You wrote -) 5,3,2 "-------" X1> X2> X3 isn't it 4,5,1? In this case X1>X2>X3 is false... (according to your state = {4,5,1,3,2})
I try to have subsequences of sequence "state" and i study if subsequence is increasly or no. But i have some doubts, for example the proof study the probability that subsequence n+1 is in sequence n2+1. In this example n2+1 can not be "state" or i dont understand how i can create n2+1 of state.
  • #11,265
  • Oct 29, 2014 11:26am Oct 29, 2014 11:26am
Quoting Eurusdd
{image}
I have coded up my interpretation of this "Decreasing-increasing sequence in set theorem", problem 1, theorem 1 in R. I am not getting the same results (93.75%). I'm getting ~83% depending on how many trials I use. Here is my code. Can anyone see any obvious mistake on my part in how this is implemented? As shown, I tried a few different random number methods in R.

Inserted Code
trials = 100000
success = 0
for (i in 1:trials) {
    x = rnorm(5) #sample(5) #runif(5) #rnorm(5)
    if (((x[1]<x[2] && x[2]<x[3]) || (x[1]>x[2] && x[2]>x[3])) ||
    ((x[1]<x[3] && x[3]<x[5]) || (x[1]>x[3]&& x[3]>x[5])) ||
    ((x[2]<x[3]&& x[3]<x[4]) || (x[2]>x[3] && x[3]>x[4])) ||
    ((x[3]<x[4] && x[4]<x[5]) || (x[3]>x[4] && x[4]>x[5]))) {
        success = success + 1    
    }
}
success / trials
[1] 0.83486

Thanks in advance,
  • #11,266
  • Oct 29, 2014 11:27am Oct 29, 2014 11:27am
Quoting Dr.Phoenix
{quote} Hi, Eurusdd, I have been still following your thoughts of this thread as I am trading probability myself, but the different way. I have many questions and some contradictions upon these thoughts, BUT I am keeping silence though. There is a real seed of knowledge in it, I now think. Anyway, I would like you to give us the original papers of authors, where you take those brillliant formulas you publish. Best, D.P.
Maybe this Math Lady can help us learn the basic...
https://www.youtube.com/watch?v=npboNfgQff8
  • #11,267
  • Oct 29, 2014 11:33am Oct 29, 2014 11:33am
Quoting LiquidGenius
{quote} Hm, is the theorem really only including those 4(or 8 depending on how you look at it) possibilities? Are these 3 not included intentionally? {image}
We are interested in sequences linearly dependent on n. the sequence x(1)>x(2)>x(4) is not valid because 1-2 is not equal to 2-4.

Sequences linearly defined on n makes the theorem powerful.
  • #11,268
  • Oct 29, 2014 11:36am Oct 29, 2014 11:36am
Quoting FXEZ
{quote} I have coded up my interpretation of this "Decreasing-increasing sequence in set theorem", problem 1, theorem 1 in R. I am not getting the same results (93.75%). I'm getting ~83% depending on how many trials I use. Here is my code. Can anyone see any obvious mistake on my part in how this is implemented? As shown, I tried a few different random number methods in R. trials = 100000 success = 0 for (i in 1:trials) { x = rnorm(5) #sample(5) #runif(5) #rnorm(5) if (((x[1]<x[2] && x[2]<x[3]) || (x[1]>x[2] && x[2]>x[3])) || ((x[1]<x[3] && x[3]<x[5])...
Thank you for your contribution. Nice to see you here. I will explain the full proof and idea of the statement shortly.
Thanks for your effort though.
  • #11,269
  • Oct 29, 2014 11:40am Oct 29, 2014 11:40am
Quoting Dr.Phoenix
{quote} Hi, Eurusdd, I have been still following your thoughts of this thread as I am trading probability myself, but the different way. I have many questions and some contradictions upon these thoughts, BUT I am keeping silence though. There is a real seed of knowledge in it, I now think. Anyway, I would like you to give us the original papers of authors, where you take those brillliant formulas you publish. Best, D.P.
Hi Dr.
My areas of interest include probability and Number theory with focus on prime number distribution and ergodic theory. Well, anything uncertain or random will get my attention.
Now, proposition 1 is my own result and I have a 7 pages paper to support it, unpublished!
The sub-sequence theorem is not mine but a simple result in real-analysis.
  • #11,270
  • Oct 29, 2014 11:48am Oct 29, 2014 11:48am
I think it is futile to use sub sequence theorem for any prediction of next bar as that would clearly be violation of the theorem itself.

As the theorem says, no matter what the sequence is, the theorem holds. this implies price can be anywhere in the next point and theorem still holds. so clearly the theorem can not put any constraints on where price can go.

I think you cant beat the market from pure numbers properties. Any theorem which works on random numbers is basically saying market prices can be random and will still satisfy the theorem so theorem cannot help predict the price at all.

To trade successfully, you need to find set of conditions that give conditional probability of move in one direction higher than the coin toss and then use those conditions to bet on that move. This is what transient zone somewhat do. In some sense transient zones are playing on the mean reversion property of the price i.e. after moving away from past few bars, chances are price will revert back. this is empirically observed in most markets but still it doesnt give big edge to trade and usually drawdown will be quite high when trading just on mean reversion.
Think : Trade : Live Life
  • #11,271
  • Oct 29, 2014 11:51am Oct 29, 2014 11:51am
Quoting Kiads
{quote} Yes My Friend, My base noob experiences came from offline chart especially renko chart. and as we know that renko chart is 'filtered price chart' or simply zigzag with better view, then we only focus on price movement 'length'/wave length and not distract with candles count, formations shape, time frames limitation, etc. not like candlestick chart, we only have 2 formations , reversal or continuation and the result is always clear. At turning point, we know exactly the 'length' needed to turn it completely, and we can also calculate our...
Hi Mr Kiads - You have some great chart formats and was wondering if you would share some of your TPLs with us here
  • #11,272
  • Oct 29, 2014 11:53am Oct 29, 2014 11:53am
Quoting FXEZ
{quote} I have coded up my interpretation of this "Decreasing-increasing sequence in set theorem", problem 1, theorem 1 in R. I am not getting the same results (93.75%). I'm getting ~83% depending on how many trials I use. Here is my code. Can anyone see any obvious mistake on my part in how this is implemented? As shown, I tried a few different random number methods in R. trials = 100000 success = 0 for (i in 1:trials) { x = rnorm(5) #sample(5) #runif(5) #rnorm(5) if (((x[1]<x[2] && x[2]<x[3]) || (x[1]>x[2] && x[2]>x[3])) || ((x[1]<x[3] && x[3]<x[5])...
Hi FXEZ,
I am also R user and I dont see any mistake.
I also got about 0.83.
If we consider all subsequence, then

Inserted Code
trials = 100000
success = 0
for (i in 1:trials) {
    x = rnorm(5) #sample(5) #runif(5) #rnorm(5)
   if ( ((x[1]<x[2] && x[2]<x[3]) || (x[1]>x[2] && x[2]>x[3])) ||
        ((x[1]<x[3] && x[3]<x[5]) || (x[1]>x[3] && x[3]>x[5])) ||
        ((x[2]<x[3] && x[3]<x[4]) || (x[2]>x[3] && x[3]>x[4])) ||
        ((x[3]<x[4] && x[4]<x[5]) || (x[3]>x[4] && x[4]>x[5])) ||
        ((x[1]<x[2] && x[2]<x[4]) || (x[1]>x[2] && x[2]>x[4])) ||
        ((x[1]<x[2] && x[2]<x[5]) || (x[1]>x[2] && x[2]>x[5])) ||
        ((x[1]<x[3] && x[3]<x[4]) || (x[1]>x[3] && x[3]>x[4])) ||
        ((x[2]<x[3] && x[3]<x[5]) || (x[2]>x[3] && x[3]>x[5]))  
) {
        success = success + 1    
    }
}
success / trials

We will get about 0.95. still not 0.9375.
Maybe we need further explanation from eurusdd.

Thank you.

edit: oh, eurusdd just answered to your post when I am writing.
  • #11,273
  • Oct 29, 2014 11:58am Oct 29, 2014 11:58am
Quoting FXEZ
{quote} I have coded up my interpretation of this "Decreasing-increasing sequence in set theorem", problem 1, theorem 1 in R. I am not getting the same results (93.75%). I'm getting ~83% depending on how many trials I use. Here is my code. Can anyone see any obvious mistake on my part in how this is implemented? As shown, I tried a few different random number methods in R. trials = 100000 success = 0 for (i in 1:trials) { x = rnorm(5) #sample(5) #runif(5) #rnorm(5) if (((x[1]<x[2] && x[2]<x[3]) || (x[1]>x[2] && x[2]>x[3])) || ((x[1]<x[3] && x[3]<x[5])...
no errors in this that i can see. this can also be solved analytically though would be quite cumbersome.
Think : Trade : Live Life
  • #11,274
  • Edited 12:09pm Oct 29, 2014 11:59am | Edited 12:09pm
Quoting stt
I think it is futile to use sub sequence theorem for any prediction of next bar as that would clearly be violation of the theorem itself. As the theorem says, no matter what the sequence is, the theorem holds. this implies price can be anywhere in the next point and theorem still holds. so clearly the theorem can not put any constraints on where price can go. I think you cant beat the market from pure numbers properties. Any theorem which works on random numbers is basically saying market prices can be random and will still satisfy the theorem so...
Sure, you are right. The sub-sequence theorem itself does not raelly give you an edge against the market, because it includes subsequences that are not defined linearly on n. Where you got it wrong is the argument I put forward. What happens when we restrict ourselves to subsequences defined linearly on n?
for the case of five points, I gave a result! There will be a 3-subsequences defined linearly on n 93.75% of the time.
That allows you to make money! Think deeply about it!


Defined linearly on n means what?
if you have a sequence listed as follows
x(1),x(2),x(3),x(4),x(5),x(6),x(7),x(8),x(9),x(10)
The the following are 4-subsequences of the sequence above
x(1),x(3),x(5),x(7)
and
x(1),x(2),x(7),x(9).


x(1),x(2),x(7),x(9) is not defined linearly on n because 1-2=-1, 2-7=-5, 7-9=-2: -1, -5 and -2 are not equal.

x(1),x(3),x(5),x(7) is defined linearly on n because 1-3=-2, 3-5=-2, 5-7=-2.
  • #11,275
  • Oct 29, 2014 12:35pm Oct 29, 2014 12:35pm
Quoting vlady1974
{quote} 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.
Yes I think your logic makes more sense. And thankfully Eurusdd point out where my flaw was.

I think to begin understanding the applications of the theorem we need to ask what does compliance with it signify and what does non-compliance with it signify. To start I got a mate to roll off 5 numbers randomly and got the following which I assigned.

8=x1 5=x2 7=x3 4=x4 1=x5, applying the sequencing,

x1<x2<x3 = 8<5<7 = False
x1<x3<x5 = 8<7<1 = False
x2<x3<x4 = 5<7<4 = False
x3<x4<x5 = 7<4<1 = False

Surprisingly to me I was "lucky" enough to get the above. In trading the 2 questions we often ask is if the "trend" will continue or reverse. The above example indicates that no sequence of higher highs was set, period. I would be a fallacy to assume continuation or reversal which I've already committed. We now factor in the fact that the next set of values will likely have one of the sequence returning a "True".

"True" equates to a sequence of higher highs being set out of the possible 4 sequencing. Given this I would still not be able to say for sure if continuation or reversal will happen but I think I can logically assume the following with relative safety :

If in a set of 5 highs, no sequence of higher highs was set base on the set of 4 possible sequencing, then one can expect to see a sequence of higher highs being set in the next set of 5 highs.



Edit : I'm usually wrong
I want more brain power!
  • #11,276
  • Edited 2:47pm Oct 29, 2014 2:17pm | Edited 2:47pm
00000
00001
00010
00011
00100 1
00101 2
00110 3
00111
01000
01001 4
01010
01011 5
01100 6
01101 7
01110
01111
10000
10001
10010 8
10011 9
10100 10
10101
10110 11
10111
11000
11001 12
11010 13
11011 14
11100
11101
11110
11111
There is 14 within the 32 doesn't meet the condition
or May i miss a lot
I hope i understand it
Thanks for all here
I'm not a programmer and i don't like ! , but only I try to catch my view !
  • #11,277
  • Oct 29, 2014 3:33pm Oct 29, 2014 3:33pm
Quoting Sasco_me
00000 00001 00010 00011 00100 1 00101 2 00110 3 00111 01000 01001 4 01010 01011 5 01100 6 01101 7 01110 01111 10000 10001 10010 8 10011 9 10100 10 10101 10110 11 10111 11000 11001 12 11010 13 11011 14 11100 11101 11110 11111 There is 14 within the 32 doesn't meet the condition or May i miss a lot I hope i understand it Thanks for all here
Not like that. I will explain the full thing soon. Take care.
  • #11,278
  • Oct 29, 2014 3:38pm Oct 29, 2014 3:38pm
Quoting Eurusdd
{quote}Not like that. I will explain the full thing soon. Take care.
Thanks Eurusdd (http://www.forexfactory.com/eurusdd)
I'm not a programmer and i don't like ! , but only I try to catch my view !
  • #11,279
  • Oct 29, 2014 4:30pm Oct 29, 2014 4:30pm
Attached Image (click to enlarge)
Click to Enlarge Name: AA.jpg Size: 662 KB

Now in the picture above, on any branch of the tree you move up/down depending on whether the next value is higher/lower.
Apart from the two fails, all branches give you one of those conditions. therefore, if we stop here, we should have probability 14/16=87.5%.
But if we look at the two fails, notice that in the top one x(1)<x(3) but x(5) can be lower or higher than x(3). The reverse is the case for the branch below. If x(5) is higher than x(3), we have to split the remaining probability into two.
Now, 1-87.5%=12.5%.
So, we are interested in 87.5%+1/2(12.5%) = 93.75%.
  • #11,280
  • Oct 29, 2014 4:57pm Oct 29, 2014 4:57pm
I understood now.
The fact look "X1, X3, X5" is linear subsequence.
From there look possibilities. I never thought to apply as well.
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