a trading problem for mathematicians

nonlinear5 · Jun 16, 2011

Quote from total_keops:

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.

Would you mind posting your source file? I hope it's an Excel spreadsheet.

abattia · Jun 16, 2011

Those saying the OP hasnât provided enough information are right, IMHO! â¦ However, assuming the two strategies are on the same instrument and timeframe, and have same rules for target and stop, and only differ in their entry signal â¦

Quote from euclid:

... This leaves us with a win rate for the combined signal of 0.36/(0.36+0.16) = 0.69...

... Then, +1 to the above.

Or, another route to get to the same answer:
New win rate = (new number of winning trades) / (new number of all trades)

New number of winning trades
= (A gives signal and wins) INTERSECTION (B gives signal and wins)
= P(A gives signal) x P(A wins) x P(B gives signal) x P(B wins)
= P(A gives signal) x P(B gives signal) x 0.6 x 0.6
= 0.36 x P(A gives signal) x P(B gives signal)

New number of all trades
= (A gives signal and wins) INTERSECTION (B gives signal and wins)
+(A gives signal and loses) INTERSECTION (B gives signal and loses)

= P(A gives signal) x P(A wins) x P(B gives signal) x P(B wins)
+ P(A gives signal) x P(A loses) x P(B gives signal) x P(B loses)

= P(A gives signal) x P(B gives signal) X ((0.6 x 0.6) + (0.4 + 0.4))

= 0.52 x P(A gives signal) x P(B gives signal)

SO, New win rate
= (new number of winning trades) / (new number of all trades)

= 0.36 / 0.52

= 0.6923

noddyboy · Jun 16, 2011

Quote from abattia:

Those saying the OP hasnât provided enough information are right, IMHO! â¦ However, assuming the two strategies are on the same instrument and timeframe, and have same rules for target and stop, and only differ in their entry signal â¦

... Then, +1 to the above.

Or, another route to get to the same answer:
New win rate = (new number of winning trades) / (new number of all trades)

New number of winning trades
= (A gives signal and wins) INTERSECTION (B gives signal and wins)
= P(A gives signal) x P(A wins) x P(B gives signal) x P(B wins)
= P(A gives signal) x P(B gives signal) x 0.6 x 0.6
= 0.36 x P(A gives signal) x P(B gives signal)

New number of all trades
= (A gives signal and wins) INTERSECTION (B gives signal and wins)
+(A gives signal and loses) INTERSECTION (B gives signal and loses)

= P(A gives signal) x P(A wins) x P(B gives signal) x P(B wins)
+ P(A gives signal) x P(A loses) x P(B gives signal) x P(B loses)

= P(A gives signal) x P(B gives signal) X ((0.6 x 0.6) + (0.4 + 0.4))

= 0.52 x P(A gives signal) x P(B gives signal)

SO, New win rate
= (new number of winning trades) / (new number of all trades)

= 0.36 / 0.52

= 0.6923

Can't find the mistake, but the answer is 0.639934408.

I proved it by two methods --
1) I ran a Monte Carlo.
2) It is
NORMDIST(NORMINV(0.6) * ROOT(2))

i.e. convert prob of winning to Sharpe ratio, multiply Sharpe by root(2) and convert back to probability.

Both gives the same result.

noddyboy · Jun 16, 2011

Quote from noddyboy:

Can't find the mistake, but the answer is 0.639934408.

I proved it by two methods --
1) I ran a Monte Carlo.
2) It is
NORMDIST(NORMINV(0.6) * ROOT(2))

i.e. convert prob of winning to Sharpe ratio, multiply Sharpe by root(2) and convert back to probability.

Both gives the same result.

Sorry, my answer was for the percentage of days you win...I think if you calculate for the percentage of days you actually trade, it is 70%.

asiaprop · Jun 16, 2011

Maybe the OP should really re-phrase the whole problem.

I disagree with your interpretation and maybe thats why there are such diverse answers.

To me it sounds there is only one trading system. The idea posed was that there are two signals, NOT systems. if the trader acted on each signal independently then the resulting probability to win would be 60%. But the question is, what is the prob of winning if the trader only acted if BOTH signals were flashed at the same time.

This should have nothing to do with entries and exits, stops, risk, or the like because nobody asked whether the system is profitable, the question was on the win ratio.

Quote from nonlinear5:

This looks right. However, I don't think the original poster formulated the problem precisely enough. Specifically, it's important to realize that any trade is a double-event, described by an entry and an exit. Now let's say you have systems A and B, and each one enters and exits independently. Each one has a win rate of 60%. System C enters the trade when both A and B agree on the entry, and it exits when both A and B agree on the exit. It's not that difficult to see that the win rate of system C may actually be less than 60%.

I'd love to see that modeled in Excel or Matlab.

noddyboy · Jun 16, 2011

Quote from noddyboy:

Can't find the mistake, but the answer is 0.639934408.

I proved it by two methods --
1) I ran a Monte Carlo.
2) It is
NORMDIST(NORMINV(0.6) * ROOT(2))

i.e. convert prob of winning to Sharpe ratio, multiply Sharpe by root(2) and convert back to probability.

Both gives the same result.

The answer above is for a continuous signal. Since the OP is taking two discrete signals, he is out of the market 50% of the time when there is disagreement.

So the Sharpe ratio of the new model is
NORMINV(0.6) * 2

NORMDIST(NORMINV(0.6) * 2) gives
0.693815302

asiaprop · Jun 16, 2011

what does it matter if someone put in a small error the concept is correct, the correct value is around 0.69 (did not calculate any specific numbers as the concept looks right).

At least you people with 0.36 are hopelessly off. How can 2 strategies that both show an edge of 60% win/loss end up at a combined 0.36? LOL!!!

Quote from SnakeEYE:

but that was the most logical answer for asiaprop so far

trade4ever2day · Jun 16, 2011

Quote from abattia:

Those saying the OP hasnât provided enough information are right, IMHO! â¦ However, assuming the two strategies are on the same instrument and timeframe, and have same rules for target and stop, and only differ in their entry signal â¦

I'm afraid this is a special case of the problem, a trivial one.

Quote from abattia:

New number of all trades
= (A gives signal and wins) INTERSECTION (B gives signal and wins)
+(A gives signal and loses) INTERSECTION (B gives signal and loses)

What about the other two cases when A wins and B loses and A loses and B wins?

HowardCohodas · Jun 16, 2011

Anyone interest in tackling one or both of the following which are closer to techniques I have actually used?

Two strategies on the same instrument. Correlation < 1. Different probabilities for each strategy. Trade if either triggers.
Three strategies as above, but at least two out of three agree for entry.

A win is a positive outcome without regard to size. Therefore expectancy does not enter. Nor does length of time for trade.

A strategy is a combination of entry rules. management rules and exit rules.

StarDust9182 · Jun 16, 2011

On a related question, how many problems does it take to stump mathematicians?

Seriously, the problem statement is very unclear for me. The answer of how many times will you win (once or twice) with two systems, how many times will you win with two signals and one system, how many monte carlo simlulations will correctly give the answer etc are confusing the question. IMO the op should clearly re-state the problem and what he wants exactly since I am really confused reading his problem statement.

It is an interesting puzzle but there is no such thing as perfectly uncorrelated signals in my view. The thought is meaningless in the real world.

At the very least, price correlates these two somehow since price is used in the calculations and also in the results. Just my two cents worth on an interesting thread.