Your opinion please - page 5
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* Therefore, make a profit once the price rises again towards point A. I don't think so. You make a profit once price reaches Point-B, I have that outlined in the diagram as Thanks to the Sell. Sure you can come up with a Non-Hedge equivalent in Hind-Sight 20/20. But when you're sitting at Point-B or C without your crystal ball, then there's no equal.
It's got nothing to do with crystal balls or hindsight. The hedged and non-hedged scenarios are exactly the same: at point C (edit: C, not B) you decide to switch from being flat to being 2 units long. Things would become very painful very fast - and equally fast in the hedged and non-hedged scenarios - if the price fell from point C rather than rising.
If the price falls a grid position below C then, in the hedging version, the order at B has been banked at $x, the order from A is in an open loss of 3 x $x, and the new buy order from C is in an open loss of $x. Net position is a loss of 3 x $x. Similarly, in the non-hedged version, the order from A would have been banked at a loss of $x and the new order at C is at a loss of 2 x $x, again for a total loss of 3 x $x.
It's got nothing to do with foresight.
Try out the code I posted two years ago where you create virtual hedged orders which get translated into non-hedged market orders with the same effect on account equity.
The more interesting stuff here is that all this started as a discussion of a game strategy where the reverse triangle hits you hard and ended (mostly because the op, ubzen and i impelemented the system in a hedged way) as a hedging dicussion. The math speak for itself, hedging does not give you an edge but it does simplify the coding. (especially when dealing with gamestrategies).
My calculations show no change if we hedge or if we do not hedge and adjust the lots correctly. (Of course swap is against hedging).
So i personally use hedging for two reasons:
1) simpliciity and the ability to fully recover from a restart without file, global variables or similar stuff
2) for closing a bunch of orders and locking the profit during the clousure.
My calculations show no change if we hedge or if we do not hedge and adjust the lots correctly. (Of course swap is against hedging).
Interesting. My calculations clearly show that it costs an extra spread x 1 (lotsize) to hedge the reversing strategy. The maths doesn't lie, so one of us has his maths wrong. If it is me I could tolerate the "loss of face" in order to improve my trading profitability. Ubzen is still married to the idea that hedged and non-hedged are entirely different (and we all know divorce is painful!)
Let's see if we can both agree on the maths, emotionlessly. I took the example of one reversal as it is the simplest. I calculated an extra spread loss for that particular trade. Do you agree on that, and if not what spread costs do you calculate?
It's got nothing to do with crystal balls or hindsight. The hedged and non-hedged scenarios are exactly the same: at point B you decide to switch from being flat to being 2 units long. Things would become very painful very fast - and equally fast in the hedged and non-hedged scenarios - if the price fell from point C rather than rising.
If the price falls a grid position below C then, in the hedging version, the order at B has been banked at $x, the order from A is in an open loss of 3 x $x, and the new buy order from C is in an open loss of $x. Net position is a loss of 3 x $x. Similarly, in the non-hedged version, the order from A would have been banked at a loss of $x and the new order at C is at a loss of 2 x $x, again for a total loss of 3 x $x.
It's got nothing to do with foresight.
Try out the code I posted two years ago where you create virtual hedged orders which get translated into non-hedged market orders with the same effect on account equity.
Lol, this is like nano-particles theory. At Point-C you decided to go 2-Lots. Why would the guy holding the same Equity as you Not decide to go 2-Lots as well?
Interesting. My calculations clearly show that it costs an extra spread x 1 (lotsize) to hedge the reversing strategy. The maths doesn't lie, so one of us has his maths wrong. If it is me I could tolerate the "loss of face" in order to improve my trading profitability. Ubzen is still married to the idea that hedged and non-hedged are entirely different (and we all know divorce is painful!)
Let's see if we can both agree on the maths, emotionlessly. I took the example of one reversal as it is the simplest. I calculated an extra spread loss for that particular trade. Do you agree on that, and if not what spread costs do you calculate?
The only way I would agree is if you can show the math (but I'll settle for a scenario) which makes it profitable compared to the hedge no-matter-which direction price decided to go. Outside of that I'll maintain my position that there's no difference. I've come to the conclusion that there's no right or wrong answer to this question. Hedge or Close-All have their advantages and dis-advantages depending upon where price decides to go.
Usually it coincides with being trend-adverse vs being range-adverse. The hedge is a range-gambit, the stop-loss is a trend-gambit. No matter what one tries to do, there's no way to optimize a system for both.
Let us all remember, price can go up or down or side-to-side. <--- this is not just black and white.
My calculations show no change if we hedge or if we do not hedge and adjust the lots correctly. (Of course swap is against hedging).
I think I have figured out where our calculations differ :-)
I can see that hedging is costing an extra lotsize x spread. You guys are following the OP's rules of continuing with a trade until it produces the correct amount of profit. This just hides the extra spread cost. The price just has to move that little bit more in the favor of that leg of the strategy in order to get that profit! So according to this thinking both methods will produce exactly the same profit. In fact even the unbalanced swap cost will get wrapped up in this catch all requirement. Hence both methods will simulate as giving exactly the same profit. What will happen is that the hedged trades will, on a long average, tend to be slightly longer (producing more open drawdown). And of course there will be more chance of hitting the dreaded account fail (or 30% drawdown limit). So peace is restored to the cosmos :-)
All this is just like trading 20 pip stoploss, 20 pip takeProfit. As the spread widens you just win less often.
Ok, lets consider this scenario and assume we have a cristall ball to predict it. The results are normalized with gain/cost in pips * lotsize, 2 pips spread in this example
The hedged way:
A: sell 1 lot
B: buy 2 lot
C: close buy
D: close sell
Gain:
The not hedged way:
A: sell 1 lot
B: close sell and open buy 1 lot
C: close buy and sell 1 lot
D: close sell
Gain:
Conclusion: the result is the same, with the hedging you have one trade less and probably less slippage at the cost of higher margin required and possible swap costs.
Conclusion: the result is the same, with the hedging you have one trade less and probably less slippage at the cost of higher margin required and possible swap costs.
If you can show me an example where hedging costs more (without swap) i would be kind of surpriced. (assuming the net position size is the same as for the not hedged way)
Here is my effort...
(edit: corrected buy and sell to agree with colours and calculations!)
The hedged way:
A: buy 1 lot (order 1)
B: sell 2 lot (order 2)
C: close buy & sell
Gain:
Order 1: -1lot * 150pips - 1lot*2pips = -1lot * 152pips
Order 2: 2lot * 100 pips - 2lot* 2 pips = 2*98pips = 1lot * 196pips
Gain total=1lot* (196-152)= 1 lot * 44 pips
The not hedged way:
A: buy 1 lot (order 1)
B: close sell and open sell 1 lot (order 2)
C: close sell
Gain:
Order 1: -1lot * 50pips - 1 lot * 2pips = -1lot*52pips
Order 2= 1lot * 100pips - 1 lot * 2pips = 1lot* 98pips
Gain total= 1 lot * (98-52)= 1 lot*46 pips