a trading problem for mathematicians

asiaprop · Jun 16, 2011

Quote from newguy05:

This is flat out wrong, you are thinking something different, the AND condition - what is the probability of event A AND B both being true given each event has a 0.6 of being true individually. Then it's 0.6*0.6

What OP is asking is what is the win probability of the TRADE when two strategies with a 0.6 win probability BOTH give the trigger to place said trade. Think about it for a second, if each individual trigger has a 0.6 chance of win, how can waiting for both to trigger reduce your win to 0.36?

The question is valid and clear. It sounds simple until you try to quantify what "winning rate" is in OP's question.

For example using the loaded die template:

Strategy 1: win=0.6 lose=0.4
Strategy 2: win=0.6 lose=0.4

All possible outcome=4
win/win, win/lose, lose/win, lose/lose

All possible WIN outcome=3
win/win,win/lose,lose/win

Probability of each of these outcome occurring then sum it up.
P(WIN) = 0.6*0.6 + 0.6*0.4 + 0.4*0.6 = 0.84 -> 84%

But this is not correct because we are making an assumption it's an OR condition where win/lose and lose/win both equals to a win. Need a quant to fit this into a standard probability template then it will become easy.

Anyone else feeling stupid right now, raise your hand.

I think you may also not fully grasp the original problem. You do not have different outcomes. There is only ONE outcome, WIN OR LOSE, the question is what the probability of winning is after taking the combined "signals" and their individual, uncorrelated, probability of winning into account.

asiaprop · Jun 16, 2011

sorry but you now disappoint me because what you said is bollocks, it has nothing whatsoever to do with the problem at hand. This is a mathematical problem, and while I do not claim my answer must be right I insist there is a mathematical exhaustive answer or else the clear answer that it cannot be calculated with the information at hand. But the answer has nothing to do with how wide stops are. Guaranteed!!!

Quote from jhiro:

Because modelling isn't trading, this can't be answered with mathematics, a traders confidence may grow by taking both together and become more aggressive and widen stops for instance, that's just one tiny variable likely to cause a hurricane in China as it spreads down the line

asiaprop · Jun 16, 2011

so you are telling us the combine win prob is 36%? How can this be? I already disproved you. If strat1 wins 100/100 and strat2 60/100, you claim you have a win prob of 60%???

This is completely incorrect as it violates that strat1 wins always.

Any comments?

Quote from kut2k2:

Even accepting your setup, you still have it wrong. Win/lose or lose/win = wash. The only winning outcome is win/win, which has a probability of .6*.6 = .36

This stuff is basic. No wonder 95% of ET traders are losers.

RCG Trader · Jun 16, 2011

Quote from kut2k2:

Am I the only person here who can read this part of the OP? :

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

What, you people think that's irrelevant? BOTH methods have to agree.

If event A has a probability of p(A) and uncorrelated (aka independent) event B has a probability of p(B), then the joint probability A _AND_ B is

p(A and B) = p(A)*p(B)

This is Probability 101. Even the guy who claims he learned it in elementary school (yeah, right) and then proceeded to blow it with p(A or B or C) might have gotten it IF he could read the OP properly.

Precisely. Each strat has a win rate of 60%, however both have to agree. When means that the strat is essentially one strat with two triggers.

Again, there is a reason why one percent control 40 percent of the wealth

RCG Trader · Jun 16, 2011

Quote from jhiro:

There is no formula that will work it out, you'll have to independently test every time they correlate.. the fact they both give 60% in isolation means fuck all

If math could be applied perfectly to the markets everyone would be rich and we'd have a theory of everything explaining all 11 dimensions with a formula 3 inches long, gravity would be married to the electro-magnetic, weak and strong nuclear forces, the standard model would embrace relativity and Einstein and Heisenberg would kiss, fuck and make up

Actually there is, but no one going to say that.

ElectricSavant · Jun 16, 2011

ummmm...thats 42% sir...

Quote from RCG Trader:

Precisely. Each strat has a win rate of 60%, however both have to agree. When means that the strat is essentially one strat with two triggers.

Again, there is a reason why one percent control 40 percent of the wealth

kut2k2 · Jun 16, 2011

Quote from asiaprop:

so you are telling us the combine win prob is 36%? How can this be? I already disproved you. If strat1 wins 100/100 and strat2 60/100, you claim you have a win prob of 60%???

This is completely incorrect as it violates that strat1 wins always.

Any comments?

You have two coins that each come up heads 60% of the time. What are the chances of getting two heads when you flip them? Obviously it's 36%. Any other combination is a wash or a loser.

If somebody subs a two-headed coin for one of the original coins, you still don't have a 100% win rate, it is only 60%.

MathAndLogic · Jun 16, 2011

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?

There is no holy grail. Here is my superficial take:

(1) Define "winning" as the sum of all profit and loss of the 2 methods being > 0;
(2) Define "losing" as the sum being <= 0;
(3) Methods A and B have winning probability of 0.6 each. (Known)
(4) When a method wins, put a "+" sign in front of it, and "-" sign vice versa;
(5) The winning rate of both methods is

[probability of { (+A -B) > 0 }] or [probability of { (-A +B) > 0}] or
[prob { +A } and prob {+B}] = 1 - (the reverse of the above)

Unless there is definitive info on the relative size of the wins and losses of the two methods, the overall probability cannot be computed under the above definitions. But what is funny is [prob (+A-B)>0] and [prob (-A+B)>0] ; you want something and the reverse of something to have the same property (>0). Think about it. Putting more methods together doesn't seem to beat a coin toss.

For technical analysis, here is something to consider:

Evidence-Based Technical Analysis: Applying the Scientific Method and Statistical Inference to Trading Signals [Hardcover]
David Aronson (Author)

A quote form a review: "In this thought-provoking work, David Aronson tests more than 6,400 technical analysis rules and finds that none of them offer statistically significant returns when applied to trading the S&P 500. "

SnakeEYE · Jun 16, 2011

Quote from jhiro:

the fact they both give 60% in isolation means fuck all

That`s true.

There is no such thing as probability in trading.

If you get the entry right you win - If you get the entry wrong you lose!

intradaybill · Jun 16, 2011

Quote from pbj:

For 2 independent systems at 60% win rate each, the probability of both being wrong is (0.4)*(0.4) = 0.16, so the win rate when both agree is 1 - 0.16 = 84%.

For 3 independent systems at 60% win rate each, the probability of all three being wrong is (0.4)*(0.4)*0.4 = 0.064, so the win rate when all three agree is 1 - 0.064 = 93.6%.

pbj, thanks for the correction. What a blunder of mine in the case of 3 systems. Of course the product space will be A x B x C