The potential yield of the instrument. - page 5
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Not impossible.
Except that statements in this form do not make you look good. When making such a statement, you need to back it up with facts, for example, an error in a formula or in taking the derivative... Although, in the post above you complained about the very complicated mathematics in those three formulas. Consequently you can't understand what is reflected there and you are making a statement...
What's that called in plain language? That's right - chit-chat, flubbing. Why would you do that?
Oh, what interesting conclusions you draw! - And complaining and not understanding... Original! The problem is elementary, its solution is obvious. Using such deductions for such a problem is ridiculous, and it's called nonsense.
Or maybe move away from the zig-zag. Consider one knee. Let H=100. Excluding the spread. The maximum is if we take all 100 points. Taking the spread into account, we enter the trade, lose the spread, and leave the trade, lose the spread. So we get the maximum we can take H-2*Spread. In this example, with spread of 2 points, we can take the maximum of 96 points.
Now if H=const=constant, then just multiply by the number of these knees.
Is there a mistakein my statement ? or not ? if not. Then with H=Spred we go into minus. If H=2*Sperd we are in zero. If H>4*Spred then we are in the plus.
Oh, what interesting conclusions you draw! - And complaining and not understanding... That's original! The problem is elementary, its solution is obvious. Using such deductions for such a problem is ridiculous, and it's called nonsense.
So far I haven't seen you give any solution or a reasonable proof! Maybe I missed something in my haste? Well, show me where it is. And if there's nothing to show, just show it and be done with it.
Or maybe move away from the zig-zag. Consider one knee. Let H=100. Excluding the spread. The maximum is if we take all 100 points.
If H=100, the average leverage length tends to 2H=200. Therefore, maximum = 200. I don't get it.
Hang on, Prival. Is that how you want to get a solution for the optimum ZZ when introducing a spread problem?
If we consider only points, then yes, it seems to work that way. I do not see the error. (I can be mistaken, find error in my logical reasoning, I do not see it).
But if we look at it not from the point of view of profit points, but from the point of maximum growth of the deposit, it will be more interesting. We should not take 96 points all at once, but for several times, if we enter % of the deposit each time. Suppose 5%, then there will be a clear maximum
If H=100, the average arm length tends to 2H=200. Therefore, maximum = 200. I don't get it.
Well, let it be 200. From the whole length, the maximum we can take is 200-2*Spred. Let's denote by the arm length L, then at L=2*spred we are at zero.
(I forgot that H is not arm's length, sorry)
let it be 200. The maximum we can take from the whole length is 200-2*Spred. Let's denote the length of the arm by L, then at L=2*spred we are at zero.
(in my sleep I forgot that H is not the arm length, sorry).
Why do you take away the spread twice from each knee? You have twice as many spreads as knees! It's got to be a knee spread. Come on, wake up!
If H=100, the average arm length tends to 2H=200. Therefore, maximum = 200. I don't get it.
as you correctly pointed out the average length tends to 2H - that's about as much as Hurst 0.5. For example Pastukhov and Shiryaev considered this measure (h-volatility) as a property of an instrument and the basis for deciding on its trading method.
But it is wrong to take the average knee value when analytically deducing the maximum earnings, because we are essentially allowed to tweak it. I.e. it is not obvious that the maximum amount will be described as a function of ZZ's average knee multiplied by the number of trades.
I agree that the correct solution is ZZ with spread or spread+1 and the difference will be in the form of trades with zero profit
Prival, the Mathcad-file will help you.
Thanks of course, but I personally don't care how you zigzag. There is a directional movement of 100 pips. the maximum we can do is to take all 100, the spread prevents us. Therefore 100-spread (wake up :-))). This is if we consider only pips. If we look at potential returns not in points, but in rubles, then these 100 points should be taken in several stages (if you use % of your deposit in a deal), then the optimum size of the payoff will appear.