Hey guys, so the other day I was explaining to a close friend of mine how RnR and Rene's GridMaster with its Dancers work. As I was explaining basic concepts such as hedging, offsetting, etc. an idea occurred to me which prompted me to research a bit and get some numbers. Before I continue I make abstraction of spread, commission and swap for convenience.
The idea involves symmetrical hedges. These come in multiples of 2. N = 4 would be a symmetrical hedge of 4 trades composed of 2 buys and 2 sells. Buys are placed higher than sells at all times in this case. When a symmetrical hedge is activated, a loss is locked-in. This locked-in loss I denote with π. Further I denote P = pip-distance between levels of a grid. Also S = N/2, which stands for the amount of orders of a single side of a symmetrical hedge. π(N;P) is then the locked-in loss of a N-sized symmetrical hedge in a grid with pip-distance P.
Question 1: How is locked-in loss related to N (or S) and P?
Next step is to allow 1 trade to break the symmetry and have a chance at offsetting the locked-in loss π. Take N = 4, P = 5 pips and lets say price activates a 3rd asymmetry-creating sell. Visually:
When price gets at that 3rd sell's level;
- Buy 2= -20 pips
Buy 1 = -15 pips
Sell 1 = +10 pips
Sell 2 = +5 pips
Sell 3 = 0 pips
We impose the rule to only take action at every grid level. In order for that "sell 3" to offset the entire position and come out with the first minimum profit, it has to travel [π(4;5) + P] = 20 + 5 = 25 pips. This happens 5 levels lower.
Question 2: How are these minimum levels of liquidation related to N (or S) and P?
Now let's say we allow a "sell 4" to be triggered when price falls a level from "sell 3". Situation effectively becomes:
Again using "Sell 3" as reference point 0, the first level at which first profit is reached is 3 levels lower. This is 2 levels faster relative to the first case (where asymmetry difference was 1). The profit we obtain here is also equal to P = 5 pips. Another example is when N=6 and P=5. This time we allow Diff = 3. Thus at 3 consecutive levels, sell or buy orders are activated. This is an example:
Locked-in loss π(6;5) = 45 pips (calculated in a similar way as previously shown). Using "Sell 4" as reference point 0; first positive TP is reached after a fall of 5 levels. Net pips won = 15 when immediately capped.
Question 3: What are the dynamics of these minimum levels of liquidation, N, P and profit relative to asymmetry difference (= Diff)?
To answer these, I worked out an Excel spreadsheet (which you can find below). First tab answers Question 1.
As suspected, locked-in loss is not a lineair function of S (remember S = N/2), but a convex function. The exact function is given by:
- π(S;P) = P + [∑ (from i=0 to S-1) 3 + 2i]
I also added a column 'probability of occurrence'. Since - in somewhat normal conditions and considering a certain time constraint - the occurrence of N = 2 symmetrical hedges is much more likely than a hedge of N = 20. Furthermore, if you increase the distance between grid levels, i.e. P, the probabilities again change. E.g. a symmetrical hedge of N=4 and P=5 is more likely to occur than a symmetrical hedge of N=4 and P=25. The 2 graphs are dynamic and are meant to be used with P between 5 and 25 (I arbitrarily configured the probabilities by assuming P between 5 and 25 and a maximum N of 30). P on the first tab is the only input you'll have to change to see how things work. These graphs' purpose is merely visual in order to grasp the concept better. The exact weights of the probabilities in no way should be considered applicable to real life.
Things get really interesting on the 2nd tab "Asymmetry Modes". I analyse up to 5 asymmetry differences. To reiterate; an asymmetry difference is defined as the maximum net difference between the amount of buys and the amount of sells. Diff = 5 allows 5 sells or 5 buys to be activated once price breaks through the boundaries of a symmetrical hedge. Reference point 0 still remains the first buy or sell that breaks the symmetrical hedge. The only input you should play with here is "pips to cover". P is linked with the P on the first tab. So do not manually change it.
Everything should be rather self-explanatory (do not be afraid to ask for clarification!). The graphs at the bottom-right are what's interesting. For P=5:
This graph plots the minimum levels (relative to reference point 0) price has to cross over before a S-sized hedge's locked-in loss can be offset for a profit. The green vertical line considers a hedge of size S=7. In the case of Diff=1; a single 8th buy/sell will have to travel about 50 levels before it can close out the hedge with a profit! Whereas Diff = 5 (with 5 buys or 5 sells) will only have to travel 10 levels from reference point 0 (i.e. the level where the first asymmetry-creating trade of the 5 trades got activated). So Diff = 5 with its 5 trades will cut the to-travel distance by a factor of 5. As you can see Diff = 3 and Diff = 4 are pretty close to Diff = 5 too. In general the higher we go in Diffs, the less impact in terms of lesser-to-travel distance we see at a given S. If we advance in S, the wedge between Diff = 1 and the rest of the Diffs grows exponentially. A downside however is that a higher Diff will quicken the growth of a bigger symmetrical hedges. But put in perspective, the minimum to-travel levels grow much slower. This suggests that Diff=1 is absolutely not effective for this strategy. Too high levels of Diff neither since their impact decreases with higher Diffs + the adverse effect of contributing to the growth of larger symmetrical hedges increases. However, this last adverse effect may be largely mitigated because the probability of occurrence decreases with rising S (cf. Question 1). Also this particular graph does not change with P since it is linked to strict rules and mathematics.
Next comes the last graph that plots the amount of pips won after price traveled the required minimum levels w.r.t. S-sized hedges. This per Diff mode.
This one actually surprised me. It seems Diff=3 is at all times superior to Diff=4 for any size of hedge! When it closes out the entire position, those 3 trades will always bag at least as much or more pips than 4 trades will. Diff=1 always bags 5 pips when it closes out any hedge; again very inefficient and not really worth it if you put into perspective the amount of levels a single trade will have to travel over. Diff=5 shows inferiority to Diff=3 at S=3, S=7, S=8, S=12, S=13, etc. This graph of course will change with the parameter P, but only in absolute values. Relatively speaking it remains the same.
All things considered; using Diff=3 and Diff=5 in accordance with the last graph's sizes of symmetrical hedges should be optimal. The bigger a symmetrical hedge grows the smaller the chances get for it to grow even bigger symmetrically. This because the hedge width grows too which in turn will make it more difficult for price to travel back to the other end of the hedge's boundary. Even after a severe whiplash which could potentially activate e.g. an extreme N=20 symmetrical hedge; if and only if you are well-capitalized, only small lot sizes are used and a decent P is implemented you should normally be fine. It is after all only a matter of time before the market moves hard in one way and covers that locked-in loss. In this sense you'd never see your balance go down. Your DD will increase of course, but I'm wondering whether it would make me feel more at ease if I manage to wholeheartedly rely on the arguments I just wrote in this last paragraph. Also when applied to a few pairs, the interaction should flow rather organically I'd suspect. Some pairs should cater for ones that are stuck in a symmetrical hedge for a certain amount of time by growing the balance, until they get released and start to cater for the pairs that were doing the catering. Of course it will all depend on carefully choosing P for each pair.
In any case, I enjoyed this exercise and thought sharing could be helpful. If anyone should feel like making an EA out of it, I'm always ready and delighted to help
Cheers, Aram
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