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lombardovito
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GOOD HIGH FREQUENCY STRATEGIES

November 5th, 2009, 9:35 pm

what is a good Hf or UHF strategies? a lot of word about Hf but no paper with a good strategy
 
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msperlin
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GOOD HIGH FREQUENCY STRATEGIES

November 5th, 2009, 10:48 pm

QuoteOriginally posted by: lombardovitowhat is a good Hf or UHF strategies? a lot of word about Hf but no paper with a good strategyAnd you honestly can't see the reason why? And I bet you won't also be able to see any paper with one the performs extremelly bad. One because it doesn't add anything to the literature and two because any trading strategy can be inverted by switching the trading signs.
Last edited by msperlin on November 4th, 2009, 11:00 pm, edited 1 time in total.
 
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Marine
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GOOD HIGH FREQUENCY STRATEGIES

November 6th, 2009, 11:50 am

Quotebecause any trading strategy can be inverted by switching the trading signs. This is not correct and misleading. You cannot always just switch a sign for a trading strategy and expect the return curve to be inverted. Usually when this happens it's because you were modelling / backtesting incorrectly i.e. a mistake in what you were actually trying to do.
 
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pnrodriguez
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GOOD HIGH FREQUENCY STRATEGIES

November 6th, 2009, 1:05 pm

I would just start playing with UHF database. Now with NYSE's OpenBook you can even backtest limit-order trading strategies.Speaking about inverting the trading signs, 1-minute Momemtum and 1-minute Contrarian...
 
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pb273
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GOOD HIGH FREQUENCY STRATEGIES

November 6th, 2009, 5:12 pm

QuoteOriginally posted by: Marine This is not correct and misleading. You cannot always just switch a sign for a trading strategy and expect the return curve to be inverted. Usually when this happens it's because you were modelling / backtesting incorrectly i.e. a mistake in what you were actually trying to do.Your statement is also partially correct. The two main reasons why inverting may not work are transaction cost and volatility drift. Volatility drift is because geometric mean (rg) is related to the arithmetic mean (ra) and volatility (s) as rg = ra -1/2 s^2. So, in the case when the strategy is inverted, the new geometric return -rg2 = -ra -1/2 s^2. So as long as the volatility is reasonable, one can invert the sign. Transaction cost, particularly in the case of HF can make a difference, particularly if the absolute returns (abs(ra) ) were low. But if the absolute returns is high, but negative, then inverting the sign will certainly overcome transaction cost.
 
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msperlin
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GOOD HIGH FREQUENCY STRATEGIES

November 7th, 2009, 3:25 pm

QuoteOriginally posted by: MarineQuotebecause any trading strategy can be inverted by switching the trading signs. This is not correct and misleading. You cannot always just switch a sign for a trading strategy and expect the return curve to be inverted. Usually when this happens it's because you were modelling / backtesting incorrectly i.e. a mistake in what you were actually trying to do.In the most simplistic setting (like paper trading) yes, it should invert. Obviuslly if you consider price impact and transactions cost, it won't invert perfectly, but partially.Quote Volatility drift is because geometric mean (rg) is related to the arithmetic mean (ra) and volatility (s) as rg = ra -1/2 s^2. So, in the case when the strategy is inverted, the new geometric return -rg2 = -ra -1/2 s^2. So as long as the volatility is reasonable, one can invert the sign. Transaction cost, particularly in the case of HF can make a difference, particularly if the absolute returns (abs(ra) ) were low. But if the absolute returns is high, but negative, then inverting the sign will certainly overcome transaction cost. I can see your argument regarding volatility drift in the case of arithmetic and geometric. But my argument is much simpler:consider prices P(t), P(t+1) and assuming:1) you can trade any unit at those prices2) transaction costs (round trip) in $ (C) are equal for short and long positionsthen, firstly assuming C=0: - buy 1 unit at P(t) and sell P(t+1) you get dP_Long= P(t-1)-P(t) - buy 1 unit at P(t+1) and sell P(t) you get dP_Short=P(t)-P(t+1) therefore:dP_Long=-dP_Shortwhich is a perfect simetry.Now, assuming C~=0- buy 1 unit at P(t) and sell P(t+1) you get dP_Long= P(t-1)-P(t) - C, - buy 1 unit at P(t+1) and sell P(t) you get dP_Short= P(t)-P(t+1) - C, therefore: dP_Long+C=-(dP_Short+C)dP_Long=-dP_Short - 2Cwhich is not perfectly simetric, but if C is small enough, the arguing for simetry is not that far away.It is also straighforward to prove this for total profit (trades iterating over time) as the simetry property will hold in any sum.Therefore, under my assumptions, switching signs in a trading strategy will switch the sign of total trading profit, but t will only be perfect for C=0.