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Help wanted on on statistics / testing approach for prediction
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Help wanted on on statistics / testing approach for prediction

  #1 (permalink)
The fun is in the numbers
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Help wanted on on statistics / testing approach for prediction

I am looking for help on statistics / testing approach to search for rule sets for increased probabilities of certain outcomes.

Situation
I have a database of conditions for EOD results from 1 Feb 2012 forward to 27 June 2014. This equals 605 records.
Each condition has a letter associated with it and these go from A to U.
I want to establish a rule set which will give the highest predictive strength of a condition of the next day's, my desired predicted outcome. (e.g M)

Goal
To establish a systematic way of investigating the possible condition sets to the highest probability combinations for a rule set to predict my outcome, which I can then test my searching my database.

Here is where I am at:
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My goal would be something like a set of rules such as:
1. If D^G and G^J and d^~J then M will happen 68% of the time.
2. If pair 1 or 2 and not pair N1b M will happen 50% of the time.

Perhaps Venn diagrams would be helpful in determining the best rule sets?

I am looking for ideas on an approach to find a solution just as much as a solution.

Thanks in advance.

-----------
Clarifying notes:
1. "^" symbol = the AND condition so D^G is "both D ^ G occur"
2. In the table of occurrences of individual conditions D occurred 223 of 390 records or 57.2% of the time M happened the next day. The percentage on the right 12.4% = 223 of 1796 and is just a relative strength %.
3. "~" symbol = NOT


Last edited by aquarian1; June 29th, 2014 at 07:47 PM.
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  #2 (permalink)
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  #3 (permalink)
Market Wizard
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Interesting question. You may want to message @NJAMC as I know he has applied some machine learning to markets problems that probably/may look a lot like this.

Just curious why did you exclude "B" as a 4th high prediction condition?

Also in your table "all three D^J^G" is 112 but "All 3 pairs" [which I assume is (D^J)^(D^G)^(J^G)] is only 83. When you have D^J^G don't you always have all 3 pairs as well hence the number of occurrences should be the same?

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  #4 (permalink)
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No point finding the highest probability without knowing what is the nature of M. I'd rather not have 99% chance of winning a $500M lottery if it also comes with 1% remaining chance of being kidnapped, tortured and incrementally dismembered for 30 days.

But if you want to find the combination with the highest joint probability, this is trivial. Let Xi denote each observable condition for any nonnegative integer i. We have:

P(M & Xj & Xk) = P(M & Xj) * P(M & Xj | Xk)
The improvement of M & Xj & Xk over M & Xj is really simply:

P(M & Xj & Xk) - P (M & Xj) = P(M & Xj) * P(M & Xj | Xk) - P(M & Xj) = P(M & Xj) * ( P(M & Xj | Xk) - 1)

Axiomatically 0 <= P(S) <= 1 for any set S, so we can conclude P(M & Xj & Xk) - P (M & Xj) <= 0.

In other words, introducing any additional condition will never yield a higher joint probability. Hence, you will find that the combination of conditions that yields the highest joint probability is D alone. It's theoretically impossible to perform better than using D alone.

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  #5 (permalink)
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artemiso View Post
No point finding the highest probability without knowing what is the nature of M. I'd rather not have 99% chance of winning a $500M lottery if it also comes with 1% remaining chance of being kidnapped, tortured and incrementally dismembered for 30 days.

But if you want to find the combination with the highest joint probability, this is trivial. Let Xi denote each observable condition for any nonnegative integer i. We have:

P(M & Xj & Xk) = P(M & Xj) * P(M & Xj | Xk)
The improvement of M & Xj & Xk over M & Xj is really simply:

P(M & Xj & Xk) - P (M & Xj) = P(M & Xj) * P(M & Xj | Xk) - P(M & Xj) = P(M & Xj) * ( P(M & Xj | Xk) - 1)

Axiomatically 0 <= P(S) <= 1 for any set S, so we can conclude P(M & Xj & Xk) - P (M & Xj) <= 0.

In other words, introducing any additional condition will never yield a higher joint probability. Hence, you will find that the combination of conditions that yields the highest joint probability is D alone. It's theoretically impossible to perform better than using D alone.

@artemiso,

Great summary and this looks correct for assuming everything is independent. Given the data used to derive the probabilities is based upon a non-linear non-stationary data set, it might not be fair to say all observations are independent (depends upon how you derive your features), as they may be dependent upon the prior days trend. Your proposal is a great way to start as I would hope most of the data is mostly independent, this is a valid hypothesis and worth trying. I just want to put the caution out there that the results might not run forward in time so validate the solution forward and backward if you can.

There is a great package, RapidMiner which is open-source. Loading the data into that tool may help you look at it different ways to help understand if your hypothesis is correct.

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  #6 (permalink)
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NJAMC View Post
@artemiso,

Great summary and this looks correct for assuming everything is independent. Given the data used to derive the probabilities is based upon a non-linear non-stationary data set, it might not be fair to say all observations are independent (depends upon how you derive your features), as they may be dependent upon the prior days trend. Your proposal is a great way to start as I would hope most of the data is mostly independent, this is a valid hypothesis and worth trying. I just want to put the caution out there that the results might not run forward in time so validate the solution forward and backward if you can.

There is a great package, RapidMiner which is open-source. Loading the data into that tool may help you look at it different ways to help understand if your hypothesis is correct.

Thank-you.

The conditions are not assumed to be independent but each letter is exclusive.

Keep your mind in the future, in the now.
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  #7 (permalink)
The fun is in the numbers
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SMCJB View Post
Interesting question. You may want to message @NJAMC as I know he has applied some machine learning to markets problems that probably/may look a lot like this.

Just curious why did you exclude "B" as a 4th high prediction condition?

Also in your table "all three D^J^G" is 112 but "All 3 pairs" [which I assume is (D^J)^(D^G)^(J^G)] is only 83. When you have D^J^G don't you always have all 3 pairs as well hence the number of occurrences should be the same?

Thank-you

RE: Just curious why did you exclude "B" as a 4th high prediction condition?

- You're correct I would test for it. I was only giving the above as an example of how far I had progressed. I thought it best to get input from others, first, before proceeding further. As my probability and stats are limited others might see a flaw in HOW I am approaching the problem.


------------
RE: Also in your table "all three D^J^G" is 112 but "All 3 pairs" [which I assume is (D^J)^(D^G)^(J^G)] is only 83. When you have D^J^G don't you always have all 3 pairs as well hence the number of occurrences should be the same?

- I think you may well be correct. I will have to go back and check over things. Thank you for highlighting this.
I'm in a bit of a brain fog at the moment (more than usual LOL!)

Keep your mind in the future, in the now.
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  #8 (permalink)
The fun is in the numbers
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artemiso View Post
No point finding the highest probability without knowing what is the nature of M. I'd rather not have 99% chance of winning a $500M lottery if it also comes with 1% remaining chance of being kidnapped, tortured and incrementally dismembered for 30 days.

But if you want to find the combination with the highest joint probability, this is trivial. Let Xi denote each observable condition for any nonnegative integer i. We have:

P(M & Xj & Xk) = P(M & Xj) * P(M & Xj | Xk)
The improvement of M & Xj & Xk over M & Xj is really simply:

P(M & Xj & Xk) - P (M & Xj) = P(M & Xj) * P(M & Xj | Xk) - P(M & Xj) = P(M & Xj) * ( P(M & Xj | Xk) - 1)

Axiomatically 0 <= P(S) <= 1 for any set S, so we can conclude P(M & Xj & Xk) - P (M & Xj) <= 0.

In other words, introducing any additional condition will never yield a higher joint probability. Hence, you will find that the combination of conditions that yields the highest joint probability is D alone. It's theoretically impossible to perform better than using D alone.

Thank-you.

No I am not just looking for the set with the highest probability
"Goal
To establish a systematic way of investigating the possible condition sets to the highest probability combinations for a rule set to predict my outcome, which I can then test my searching my database."

If I follow your reply yes D is highest alone. I think in your reply of "joint probabilities" you are referring to "ands".

There will be some days when D does not occur on those days I could have J^G and knowing what their odd of that giving M are would be useful.

Additionally if D^G is lower than D alone and I have a day with a D^G I would like to know that M has become less likely.

I thought that if one has multiple rules forming a set that one could derive a better trading system.

Keep your mind in the future, in the now.

Last edited by aquarian1; July 3rd, 2014 at 08:13 PM.
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  #9 (permalink)
The fun is in the numbers
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Thank-you all for your replies.

Perhaps I need to go back to the drawing board.

Keep your mind in the future, in the now.
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NJAMC View Post
@artemiso,

Great summary...

Thanks.


NJAMC View Post
...and this looks correct for assuming everything is independent.

Actually, my solution is in the general form and true even for dependent variables. In the special case that {Xi, i is a positive integer} is a collection of pairwise independent variables, you can further decompose the conditional probability P(Xj | Xk) = P(Xj) and P(Xi & Xj) = P(Xi)*P(Xj).

I think what you're meaning to say is that it doesn't solve the problem if each of the random variables {Xi, i is a positive integer} is itself a member of some non-stationary stochastic process. I thank you for pointing out. Well, that's an issue with @aquarian1's methodology...

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