a trading problem for mathematicians

Guys, I finally decided to request to have my account closed here at ET.

Most won't care, some will be delighted (whom I may have insulted)...I learned some things, though I need to unfortunately note that the amount of good things taken away relative to the time spent here is not worth it.

I thank those who actually use their own brains and who produce thoughts that are coherent and who invest the energy to put up with a lot of garbage here on this website.

I myself decided that I do not want to spend my valuable time anymore to try to sift through so little valuable contributions. I also tried to pass on some of my knowledge, some may have discounted it, others appreciated it. But there was just too little that I got out of it.

I wish the best and "good luck" to those who can approve of the following comment:

"I find that the harder I work, the more luck I seem to have." Thomas Jefferson
 
Quote from asiaprop:

Guys, I finally decided to request to have my account closed here at ET.

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KC & The Sunshine Band - Please don't go
http://www.youtube.com/watch?v=w-l5FyA3pgo

Hopefully, you can't kick the addiction! :p

But if you do, crap!, I'll miss you asiaprop. I learned some interesting stuff from your previous posts. You're one of the few mathematically inclined people here on ET.
 
Quote from asiaprop:

Guys, I finally decided to request to have my account closed here at ET.
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I can't recount the number of times I quit drinking for good.
Soon or later you will be back, ET is a nsaty vice. We all need our daily dose of trolling.
 
I'll have to go with 60%. There is a 36% chance that both systems will win for the individual trade as well as a 16% chance that both will lose. Where it gets tricky is the 48% chance that the systems will split. I think you have to take this as half a win and half a loss. So 36% +24% = 60% win, while 16% + 24% = 40% loss. So, by this reasoniing, the win rate remains at 60%.

So the win rate is 36% if a win is counted when only both produce winners and 60% if the case of win/loss or loss/win is split between win/loss for the overall system.
 
Quote from asiapoop:

Your posts on the OP's problem did not make sense nor do your comments on Monty Hall make sense. As said before it delights me to know that the number of ignorant and lazy people does not decrease but is safely floored.
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Hey, asshole, I've developed my own advanced time-series smoother that is faster than SMA2 and smoother than SMA4, so you can stick your stupid ASSumptions where the sun doesn't shine.
Quote from asiapoop:

Guys, I finally decided to request to have my account closed here at ET.
More...
Good riddance. But as others have said, you'll probably punk out and slink back, likely under a new alias. Typical troll behavior.
 
Quote from trend2009:

Suppose I have two trading methods with no correlation to each other. each of the two methods, if applied alone to the market, has a winning rate of 60%. the question is: if the two methods applied at the same time to the market (the signal must be agreed by the two methods at the same time), what is the winning rate?

Any mathematician here gives a try? and show how you get the result?
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Two very simplistic premises:

1. "two trading methods with no correlation to each other".

If we consider this to be true, then we can assume that the events ( trading signals) are independent and not mutually exclusive i.e. both signals can be traded at the same time.

The probability of a winning trade is the probability we get a winner from either strategy ( or both ) minus the probability we get a winner from both strategies at the same time ( we do not want to double count the probs, do we?) i.e.

P(winner trade) = P(A) + P(B) - P ( A and B ) = 0.6 + 0.6 - 0.36 = 84%

Note that this is the probability of getting a winning trade with two truly uncorrelated strategies. It does not say anything about the size of the winner/loser. You will get a winning trade i.e. a non negative single position p&l in 84% of the cases, but you can still have a negative p&l at the portfolio level ( i.e. the other trade (loser) is really bleeding out). Thus, to be more specific and assuming the position size is the same for both strategies in dollar amounts, you need to include at least variance data. In that case (still somewhat simplistically) you will multiply the probabilities by the dev to get a closer result.

2. "the signal must be agreed by the two methods at the same time"

Now, here is where it gets tricky. Think about it logically. This 2nd statement invalidates somewhat the first one. If both strategies were 100% uncorrelated, you would never get an overlapping trade. In this case you would get 100% "winning" results ( i.e. no trades, no losers, no tears, and no positive return either). Now, if the strategies did overlap somewhat, you do not get to trade individually all the original trade set (signals) that resulted in the given 60% empirical return, but only a smaller subset.

What is the return of the smaller subset? With no data, it is anyone's guess. No math mumbo jumbo could answer this!

You need empirical data on at least a test result set to work this out. Maybe you are only trading the losers, maybe only the winners, but maybe the trade set is still significantly large enough ( i.e. if the strategies correlation were close to 1 ) that the end result is still close to 60%.

There you have it, my two free cents for what is worth :cool:
 
Quote from trend2009:

Suppose I have two trading methods with no correlation to each other. each of the two methods, if applied alone to the market, has a winning rate of 60%. the question is: if the two methods applied at the same time to the market (the signal must be agreed by the two methods at the same time), what is the winning rate?

Any mathematician here gives a try? and show how you get the result?
More...

I have not read all the posts, but from what I read, I felt that the points below may have not been addressed.

1. No correlation and independence are two different things--- one can have no correlation and yet a perfect dependence! I have an example that show it somewhere in my blog. In fact, since the introduction of the corelation coefficient sometime around (I believe) 1888 this confusion has been taking place.

2. When two systems are not correlated, one has to indicate with respect to what variables. My understanding is that one usually uses the returns, which is not the same thing as probs.

3. A particular case of 2, where probs and returns might be viewed as same, would be when R/R is 1. Is this what you mean? Even if this case, one still has to deal with issues that no correlation does not mean/imply independence.

Hopefully with the points above some good people here would go back and rework the derivations if there is a need to do so.
 
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