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The similarity system

  • #11,281
  • Oct 29, 2014 5:00pm Oct 29, 2014 5:00pm
Quoting Eurusdd
{image} 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...
Thanks Eurusdd (http://www.forexfactory.com/showthread.php?p=7833206#post7833206) so much

Quoting Eurusdd
{image} 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...
Dear
here we consider there is X0 for which we mark X1 either up or down
I'm not a programmer and i don't like ! , but only I try to catch my view !
  • #11,282
  • Oct 29, 2014 6:53pm Oct 29, 2014 6:53pm
Quoting Eurusdd
{quote} 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
No, I meant where h is set such that 97% of bars on the chart are recurrent (either left-only, right-only or fully recurrent), even if the chart contains 23 million bars of data. In that siutuation, the probability of a potential TZ bar becoming resolved (i.e. right-only recurrent) can't be 97%, can it? It must be less, since "right-only-recurrent" is a sub-group of the 97%. In fact, I'm speculating that it will be closer to the 79% figure that I obtained from my small sample size.
You reap what you sow.
  • #11,283
  • Edited 9:00pm Oct 29, 2014 8:22pm | Edited 9:00pm
Quoting Eurusdd
{image} 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...
Lol.. Thanks God and thanks EURUSDD, my 'default' renko shifted step is 12.5% in this last 4 years.

Learning that in geometrical way several years ago... idk if this is 'match' or just similar concept, but is this 'similarity' we are talking about ?...

Attached Image (click to enlarge)
Click to Enlarge Name: USDJPY M11 8point brick - 6.25% shift step geometrical sequences.png Size: 276 KB



MTH
  • #11,284
  • Oct 29, 2014 9:43pm Oct 29, 2014 9:43pm
Quoting NorthTrader
{quote} No, I meant where h is set such that 97% of bars on the chart are recurrent (either left-only, right-only or fully recurrent), even if the chart contains 23 million bars of data. In that siutuation, the probability of a potential TZ bar becoming resolved (i.e. right-only recurrent) can't be 97%, can it? It must be less, since &quot;right-only-recurrent&quot; is a sub-group of the 97%. In fact, I'm speculating that it will be closer to the 79% figure that I obtained from my small sample size.
hi north do you know wat post here that has recurent zone it seems I always see transient zone on every post here😒
  • #11,285
  • Oct 29, 2014 10:31pm Oct 29, 2014 10:31pm
Quoting forextarget
{quote} hi north do you know wat post here that has recurent zone it seems I always see transient zone on every post here😒
Sorry forextarget, I don't understand your question.
You reap what you sow.
  • #11,286
  • Oct 29, 2014 11:10pm Oct 29, 2014 11:10pm
Quoting Eurusdd
{image} 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...
I came up with similar proof last night but I got stuck at 14/16 = 87.5%
Attached Image (click to enlarge)
Click to Enlarge Name: IMG_0621.JPG Size: 1.6 MB


So, for two failed cases 6 and 11: x1, x3, x5 can be larger or lesser than each other so we still have 2/4 = 50% probability.
Therefore, the final prob. is 87.5% + 50%(1-87.5%) = 93.75%

Attached Image (click to enlarge)
Click to Enlarge Name: IMG_0623.JPG Size: 1.6 MB


Thanks Eurusdd
  • #11,287
  • Oct 29, 2014 11:30pm Oct 29, 2014 11:30pm
I just noticed that eurusd fell big. The thing is that fall can be explained using the probabilities I discussed. If you look at the daily chart the next two bar were expected to be negative in a big way. Well it happened on the first!
  • #11,288
  • Oct 30, 2014 12:20am Oct 30, 2014 12:20am
Quoting FX-Jay
{quote} No disrespect to you MAM, I totally appreciate your contributions However, if being misleaded allowed me to get the below results, I'll be more than happy to continue living my life as a misleaded naive young noob that thinks he understands something about TZ & probabilities that doesn't exist in reality. When my account starts going underwater, I will come here and apologize to everyone on the behalf of EURUSDD for wasting their time. As far as I can see, 2 months of pure consistency. As for nothing fantastic coming out of this, I believe...
have you start LIVE yet Fxjay?
  • #11,289
  • Oct 30, 2014 12:42am Oct 30, 2014 12:42am
Quoting Eurusdd
I just noticed that eurusd fell big. The thing is that fall can be explained using the probabilities I discussed. If you look at the daily chart the next two bar were expected to be negative in a big way. Well it happened on the first!
Thanks，as the post#184 discussed “Those interested in the STOCH-bollinger system should not wait for a difference to occur in order to trigger a trade. You can trade the similarities you see. Note that the 1EMA and the stoch lines should 95% of the time take the same shape. So, you can trade expecting that they will end up the same with money-management of course!
For example EURUSD just dropped, why???” I think it will fall down，also the GBP/USD，I will take more test and thank you again
  • #11,290
  • Oct 30, 2014 12:58am Oct 30, 2014 12:58am
Quoting Eurusdd
I just noticed that eurusd fell big. The thing is that fall can be explained using the probabilities I discussed. If you look at the daily chart the next two bar were expected to be negative in a big way. Well it happened on the first!
I see it... I see it.. Yay... Thank You Chief....

Attached Image


MTH
  • #11,291
  • Oct 30, 2014 2:02am Oct 30, 2014 2:02am
Quoting Eurusdd
I just noticed that eurusd fell big. The thing is that fall can be explained using the probabilities I discussed. If you look at the daily chart the next two bar were expected to be negative in a big way. Well it happened on the first!
i see what you mean in a way
  • #11,292
  • Oct 30, 2014 4:31am Oct 30, 2014 4:31am
RecurrenceStatisticV3_text with alert for possible problem bars when ‘ShowChartAlerts’ is set to true. I didn’t get the ‘when a possible problem bar becomes a true problem bar alert’ to work, because I don’t know how and when the indicator calculates that exactly, the things I tried gave me compile errors with that alert. I’m not a coder, just some copy/pasting and some trying.

Thus, to be clear; RecurrenceStatisticv3_text with alert when a possible problem bar appears (the ones with the grey arrows) when ‘ShowChartAlerts’ is set to true. As far as I can see seems it to work fine. And the blue arrows are yellow now, it fits my grey background better

Cheers,

GU
Attached File(s)
File Type: mq4 RecurrenceStatisticV3_Text_Alert.mq4   32 KB | 658 downloads
  • #11,293
  • Oct 30, 2014 4:40am Oct 30, 2014 4:40am
Quoting GU-Night
RecurrenceStatisticV3_text with alert for possible problem bars when ‘ShowChartAlerts’ is set to true.
Very nice, thanks for sharing!
k
  • #11,294
  • Oct 30, 2014 5:18am Oct 30, 2014 5:18am
Quoting kprsa
{quote} Very nice, thanks for sharing! k
this one works without dissappearing,
thanks gu-night
We Win Some, We Loose Some, But Let your Win Rates be More to be Successful
  • #11,295
  • Oct 30, 2014 6:54am Oct 30, 2014 6:54am
mmm, Construction of tree understand it, but regarding the current bars do not understand why conditions Diagram "eurusdd" are not met.
Just looking at the first conditions only X1, X2, X3:
x1> x3> x5
x1> x3 <x5,
If the first condition is true, how is it that the second condition is met?
Insurance is simple and I'm doubting.

Thank you very much

EDIT:
okey, This is where DNA comes in. In the database all these conditions are met except 2, hence it is not accepted that there is a linear subsequence in that sequence.
I hope to be right
  • #11,296
  • Oct 30, 2014 7:23am Oct 30, 2014 7:23am
Quoting Kiads
{quote} Maybe this Math Lady can help us learn the basic... https://www.youtube.com/watch?v=npboNfgQff8
Thanks, I have no problems with math (particularly, probability) since I am Ph.D., but I would like to find the original papers of those authors who wrote them.
Hunting High and Low
  • #11,297
  • Oct 30, 2014 7:28am Oct 30, 2014 7:28am
Quoting Eurusdd
{quote} 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.
Understood than. I am operating with Gauss distribution in volumes.
As to your work unpublished - understood as well. I thought you took somebody else's paper.

still watching....
Hunting High and Low
  • #11,298
  • Edited 9:43am Oct 30, 2014 8:36am | Edited 9:43am
Quoting Eurusdd
{image} 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...

Hi EURUSDD,

i think you are missing couple of paths in this proof. I see you are trying to model price as a binary tree (this is used for COX's option pricing framework widely used http://en.wikipedia.org/wiki/Binomial_options_pricing_model) but in your theorem, each number is randon so in your tree you should have cases like

from x1 go up to x2 then go down a lot to x3 such that x3 is below x1

if you enumerate all the cases, you will get the probabilities same as what FXEZ got via R simulation.

Also i think we cant use properties of random numbers for market predictions as again all they will say is market price can be totally random and still satisfy those probabilities. So that wont allow us to predict.

it is like probability of betting on T after 10 H in sequence is not any greater that .5 even though probability of getting 11 H in 11 tosses is pretty small.
Think : Trade : Live Life
  • #11,299
  • Oct 30, 2014 8:48am Oct 30, 2014 8:48am
Quoting NorthTrader
{quote} No, I meant where h is set such that 97% of bars on the chart are recurrent (either left-only, right-only or fully recurrent), even if the chart contains 23 million bars of data. In that siutuation, the probability of a potential TZ bar becoming resolved (i.e. right-only recurrent) can't be 97%, can it? It must be less, since "right-only-recurrent" is a sub-group of the 97%. In fact, I'm speculating that it will be closer to the 79% figure that I obtained from my small sample size.
The 79% figure you have is quite accurate.
It's the 79 that is trad-able, not the 97. This is what most folks around here are not getting.

79% probability is pretty good, but not good enough to be real profitable. That's why the trick here is how to make this 79% increase to 90+
There's more than one way to do that, including increasing the H value, or playing with the K value.
There are some other ways to do so that are not scientifically proven/valid, but are working more than well for me.

​I used to worry to write these out loud, but since no one is reading what is written & everyone is worried whether the account is real or demo Or if the statemnet is photoshoped or not, then there is no worries of letting the cat have a peak out of the bag a little
  • #11,300
  • Oct 30, 2014 8:49am Oct 30, 2014 8:49am
Quoting stt
{quote} Hi EURUSDD, i think you are missing couple of paths in this proof. If you trying to model it as a binary tree (what is used for COX's option pricing framework widely used) but in the theorem, each number was randon so in your tree you should have cases like from x1 go up to x2 then go down a lot to x3 such that x3 is below x1 if you enumerate all the cases, you will get the probabilities same as what FXEZ got via R silumation. Also i think we cant use properties of random numbers for market predictions as again all they will say is market...
MAKES NO difference. I will not go into the details. I rest my case here.
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