a trading problem for mathematicians

you dont understand the original problem. The question posed is different from asking what the probability is that both signals come up at the same time.

Again you do not seem to understand the question. 60% is NOT the probability of a signal occurring out of all observations but 60% is the probability of a winning trade WHEN a signal is triggered.



Quote from kut2k2:

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%.
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this is the worst and most wrong thing I EVER heard. That you find strategies that have an edge in ISOLATION is what every quant desk is looking for. And if you find other strategies that in isolation (and possibly uncorrelated to existing strategies) also provide an edge then you will be on the right path to build a very successful stat arb or quant book.

No such thing as probability in trading? Well, you could look for astrological entry or exist symbols or evaluation your mood when you get up in the morning but I have made money over the past 12 years of trading based on trades that I believed provide a statistical edge. Nothing could be farther from the truth than saying probabilities are not an essential component in successful trading and betting such as playing poker.


Quote from SnakeEYE:

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!
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Let's say both systems (A & B) give long and short signals such that e.g a long signal means the market will hit +X before it hits -X.

It follows that on occasions where both systems gave a signal, we have the following probabilities:

Market moved X:
A won, B won = 0.6*0.6=0.36
A won, B lost = 0.6*0.4=0.24
A lost, B won = 0.4*0.6=0.24
A lost, B lost = 0.6*0.6=0.16

So, 36% win, 16% lose, 48% no trade because the systems disagreed.

This leaves us with a win rate for the combined signal of 0.36/(0.36+0.16) = 0.69. A small improvement in hit rate, but a big reduction in the number of trades.

We can check this with some other figures. e.g:

If both systems gave a hit rate of 50%, then we get 0.25/(0.25+0.25) = 0.50. No improvement because 50% is what you would get at random anyway. Neither system is providing an edge.

If one system gave 100%, then we get 0.6/(0.6+0.0) = 1.0. You can't improve on 100% win rate.
 
the most logical answer so far, imho. I correct my earlier answer of 60%, it should be slightly more than 60% because you add two strategies that each in isolation provides an edge. This is the precise reason why trading desks search for uncorrelated strategies that provide an edge by trading those in isolation.

Anyone who claims the win rate of 2 strategies that in isolation provide an edge (x>50% win rate)
is smaller than x got it completely wrong.

Additionally those who believe the win rate is 1- [P(A-lose)*P(B-lose)] are also wrong. Take 10 strategies that have a win rate of 50%, you basically claim the win rate of all uncorrelated strategies combined is 1-0.000976 = close to 1? Wrong!


Quote from euclid:

Let's say both systems (A & B) give long and short signals such that e.g a long signal means the market will hit +X before it hits -X.

It follows that on occasions where both systems gave a signal, we have the following probabilities:

Market moved X:
A won, B won = 0.6*0.6=0.36
A won, B lost = 0.6*0.4=0.24
A lost, B won = 0.4*0.6=0.24
A lost, B lost = 0.6*0.6=0.16

So, 36% win, 16% lose, 48% no trade because the systems disagreed.

This leaves us with a win rate for the combined signal of 0.36/(0.36+0.16) = 0.69. A small improvement in hit rate, but a big reduction in the number of trades.

We can check this with some other figures. e.g:

If both systems gave a hit rate of 50%, then we get 0.25(0.25+0.25) = 0.50. No improvement because 50% is what you would get at random anyway. Neither system is providing an edge.

If one system gave 100%, then we get 0.6(0.6+0.0) = 1.0. You can't improve on 100% win rate.
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Quote from asiaprop:

No such thing as probability in trading? I have made money over the past 12 years..
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..and don`t even know the defenition of the 'probability'.

:D
 
Made the Monte Carlo simulation in 20 minutes b4 go to work.
Assume both win and lose have the same profit or loss.
Get into a trade if both signals agree based on a uniform distribution from -1.5 to 1.5:
]-1.5,-0.5] Go short
[-0.5,0.5] Go flat
[0.5,1.5[ Go long
If both signals agree at the same time we take a trade.
When we take a trade we have a win based on a uniform distribution [0,1]. If both systems have a uniform random number smaller than 0.6 we have a win, otherwise it is a loss.
Attached is a 10000 periods 50 times MC simul. Average is in thick black.
Win rate gives: 0.3594. (the expected 0.36)
Now if instead of needing both to be winners we add a condition that if only one of the two is a winner we flat trade.
Win rate gives: 0.6948, an improvement.
So the results really depends how you define EXACTLY how your system would work.
 

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