A probability theory problem - page 5
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3. if your conclusions are supported by statistical studies, it would be useful to read them.
Unfortunately, I have not kept them, as they have no practical value. But in the near future I will repeat a few runs from memory and post them along with the EA.
How could it be? A negative result is also a result :-) If the research is correct but not of practical benefit, it is a benefit to others - "this way is wasted time" and that is so precious to us. So your altruism will be appreciated.
Going back to addictions I gave the example of "cross", but I don't consider it as such. It is quite an independent exchange rate, just the current speed of information exchange does not allow to earn (because no one wants to lose because of their sluggishness) on the possible difference, and historically there has been the primacy of the dollar, so the pair was called synthetic or cross. So not only the cross can be called without looking at it, but also any pair based on the other two, it is of course understandable. But it is worth taking a closer look at the unambiguous relationship between them. As a guide I suggest you to find 8 identical candlesticks in a row for EUR/USD and simultaneously for GBP/USD. Have you considered such a correlation?
SergNF 23.04.2008 12:35
If only we could find the "dependent events" in the formula EURUSD * GOLD * CocaCola * ???, then we could play around, because, imho, no "Consortium"/DC would have had time to level them out. ;);););)There's no difficulty in this, but I haven't considered it as I don't see it as a prospect.
It's not difficult, but I haven't considered it, because I don't see the perspective in it.
It (perspective) is incommensurably higher comparing to searching for patterns in a sequence of candlesticks.
It's not difficult, but I haven't considered it, because I don't see the perspective in it.
It (perspective) is incommensurably higher compared to searching for patterns in a sequence of candlesticks.
I will answer in reverse order, so as not to be called a "verbalist":
"Coding" is dumb, both in terms of code (comparing doubles, etc. etc.) and "flexibility"
iEURUSD = iBarShift("EURUSD",0,CurrTime,true); iGBPUSD = iBarShift("GBPUSD",0,CurrTime,true); CurrTime = CurrTime + Period() * 60; strEURUSD = ""; strGBPUSD = ""; for(int i = Nbar; i > 0; i--) { if(iClose("EURUSD", 0, iEURUSD+i) > iOpen("EURUSD", 0, iEURUSD+i) ) strEURUSD = strEURUSD + "1"; else strEURUSD = strEURUSD + "0"; if(iClose("GBPUSD", 0, iGBPUSD+i) > iOpen("GBPUSD", 0, iGBPUSD+i) ) strGBPUSD = strGBPUSD + "1"; else strGBPUSD = strGBPUSD + "0"; }But, it seems to me that within the scope of this task ... "will do for the countryside". One more thing - the coincidence of GBPUSD and EURUSD Nbar opening times back from the "current point" I haven't checked - only on the current bar.
2007 г. 30 min bars. In total (I have) 12,216 bars.
- For EURUSD, the maximum number of times a combination of 8 bars occurred is 77 times, i.e. ~0.63%
These are the combinations of "01001001", "10010010", "10011001".
- The same for GBPUSD - 84 times, i.e. ~0.69%.
This is a combination of "01010100".
I.e. it's not interesting any more. :(
The combination of "8 identical candlesticks in a row on EUR/USD and simultaneously on GBP/USD" is not the same, because the number will be almost zero.
The maximum 10 times (i.e. 0.08%) we've encountered 8 bar combinations "on EUR/USD and simultaneously on GBP/USD". Moreover, it was
GBPUSD=01001001
Almost on request ("8 identical candles in a row on EUR/USD and simultaneously on GBP/USD."), but only 10 times a year. i.e. no system/repeatability etc. is out of the question. If anything, just to check:
Actually why did I start this response - I personally - a peddler - not interested in "at the same time ".
I need to know the colour of the next ball, i.e. I need to shift GBPUSD to the right by 1 bar minimum, as well as the topikster.
And ... I already wrote that "statistics is a pseudoscience". especially for "fat tails" ;)
3. if your conclusions are supported by statistical studies, it would be useful to read them.
Unfortunately, I have not kept them, as they have no practical value. But in the near future I will repeat a few runs from memory and post them along with the EA.
How could it be? A negative result is also a result :-) If the research is correct but not of practical benefit, it is a benefit to others - "this way is wasted time" and that is so precious to us. So your altruism will be appreciated.
Attached is an EA (tester version) which searches for a given number of consecutive white (rising) bars and opens on/against them. The filter is a cumulative distance in pips which has been travelled by price over these N bars. Closes the position at the close of the next bar, regardless of reached profit/loss. It works with any pair, on any TF, using "open prices" model. Also one of the profitable runs. There have been better ones, but can't find any more. Profit and number of deals are obviously not enough. You may use it if you are interested. The research continues.
Actually, why did I start this answer - personally I am not interested in "at the same time ".
I need to know the colour of the next ball, i.e. I need to shift GBPUSD at least 1 bar to the right in the "currency pair" contex.
And ... I already wrote that "statistics is a pseudo-science". especially for "fat tails" ;)
Totally agree with the conclusions drawn, especially since they were called for even before the experiments were conducted.
I have a probability problem too.
Taleb in Fooled by Randomness has a paragraph like this:
Парадокс дня рождения
The most intuitive way to describe the problem of dredging up data not statistics is through what's called the birthday paradox, although it's not a real paradox, just a quirk of perception. If you meet someone by chance, there is a one in 365.25 chance that your birthday and theirs are the same, and a much smaller chance that their year of birth is the same. So, the same birthday would be a coincidence that you would be discussing at the dinner table. Now let's look at a situation where there are 23 people in a room. What are the chances of there being two people with the same birthday? Approximately 50%. Since we don't define which people have to have the same birthday, any pairing is appropriate.
The highlighted test struck me, so I decided to test it.
I got 69.23%. Now I'm agonizing with the question - who's counting wrong? ;)
I don't think "69%" has been deliberately replaced by "approximately 50%".
The highlighted test struck me, so I decided to test it.
I got 69.23%. Now I'm torturing myself with the question - who's counting wrong? ;)
I don't think "69%" has been deliberately replaced by "approximately 50%".
Somewhere that's how it works out:
Number of combinations of 23 by 2 = 23!/2!(23-2)!=22*23/2=253
253/365.25=69.2676%
Conclusion: Taleb did not do well in school or studied to be a writer.
Somewhere along the line, this works out to be the case:
Number of combinations of 23 by 2 = 23!/2!(23-2)!=22*23/2=253
253/365.25=69.2676%
Conclusion: Taleb did not do well in school or he studied to be a writer.
By the way, a 100% probability is achieved on a company of only 28 people.
By the way, a 100% probability is achieved on a company of only 28 people.
not 32 ?