If we knew exactly how the price was moving... - page 3
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Are you talking about the distribution of first price differences or something else ?
Too your claim that sl and tp allow such a dashing distribution seems unfounded. To put it mildly. :-)
i don't see the difficulty. alsu basically described in point 1 as well.
Strange reasoning for a person who probably heard something about martingales and the famous theorem of Dub that it is impossible to build a profitable system on a martingale.
Alexey, what can you say if the "known stationary VPW" is ordinary white noise (the integral of it is a Wiener process, a martingale)?
Mathemat, if the noise distribution is HP with mo=0, of course you can't. If mo<>0, then there is a trend component in the cumulative sum series and the way to make money is simple - get into the demolition side. For martingale the best prediction is the last value of the series. If there is a drift, then the best prediction is the last value + mo PRV.
If the distribution is asymmetric, you can select an area using trading rules where the mo will be different from 0. Hypothetical example: we have learned to recognize the gap at the opening of the day. It will be up 10 pips with probability 0.9 or down 90 pips with probability 0.1. MO=10*0.9-90*0.1=0. But there is a good-hearted broker that executes stops and profits at stated prices. We set a long with stop=profit=10 points. Thus, we have mo=10*0.9-10*0.1=8. Of course, this example is far-fetched and has a discrete distribution for simplicity. Although similar scheme was used when there were significant gaps after the weekend and a guaranteed intra-gaap stop-loss was a margin call.
Strange reasoning for a man, who probably heard something about martingales and the famous theorem of Doub that it's impossible to build a profitable system on a martingale.
Alexei, what can you say if the "known stationary PRW" is just plain white noise (the integral of it is a Wiener process, a martingale)?
As Mr. Avals has already mentioned, if a random process is a white noise, its PDF must be symmetrical, for example it can be Gaussian with zero expectation (otherwise the noise will "drift" to one side or another and will not be white anymore). In such a case the necessary profitability condition stated in my previous post is not satisfied, since the area under the graph on the right and on the left is the same and equal to 0.5
I don't see the difficulty. alsu basically described it in point 1 as well.
As I thought, it's about the PRV of the price change on the bar. Something like Close(i)-Open(i) or Open(i+1)-Open(i). However, sl is not triggered by Close or Open, but by the price inside the bar determined by High and Low. Therefore, imho, the above-mentioned PRV is not suitable for determining sl.
There is another point. Asymmetry provided by drift (i.e. trend) has |MO|>0, so it is not a problem to use it. In fact, it is not asymmetry, but non-zero value of MO that is exploited: open to the MO and you will be happy.
Another thing is when MO=0. In this case asymmetry is of quite a different nature - areas under RWP curve to the right and to the left of ordinate axis are equal. Try to take an asymmetrical model distribution that meets this condition (MO=0), generate a sequence of ticks of known density on it and run this strategy on it. I think you will be disappointed.
By the way, regarding point 1 in alsu' s post.
1. Let's select an area on the VPP chart, located on one side of the Y+-spread axis, the area under which is greater than 50%+eps (eps-trading_expenses+planned winning) - this area will be equal to the probability of winning P. Accordingly, the probability of losing Q= 50%-eps.
A skewed near-normal distribution is too lazy to generate.
But for example, let the distribution be
incremental probability
-5
0,02
-0,1
Histogram:
put sl=tp=2p If not triggered exit by market.
мо(лонга)=0*0,4+1*0,1-1*0,02+2*0,2-2*0,28=0,1 -0,02+0,4-0,56=-0,08
if I've counted correctly of course :)
Z.U. And of course this distribution is not like a tick. It is different than HP as far as I remember. At least in zero, there will be no such probability. But it will also be symmetrical. Of course we can analyze it in a real series and "mow" to keep mo equal to zero, but I'm lazy.
... You could of course analyse it from the real series and "mow" it down to zero, but I'm lazy...
...It would be interesting to see what exactly would come out of the real series, but as it does not have any evidentiary power...
Also, it's important to define the purpose of the simulation. If it is a question of sl=tp=2p, then it is not worth the trouble.
And "practice is a criterion of truth" :)))
...It would be interesting to see exactly what would come out of the actual series, but as it does not have any probative value...
Besides, it is important to define the purpose of modelling. If it is a question of sl=tp=2p, then it is not worth the trouble.
And "practice is a criterion of truth" :)))
Of course, this is an abstract example.
And Mathemat's distribution of ticks was posted at https://www.mql5.com/ru/forum/103289/page4. Of course the asymmetry there will not allow the spread to be overplayed. But it's not necessarily about tick increments.
of course this is an abstract example
And Mathemat's distribution of ticks was posted at https://www.mql5.com/ru/forum/103289/page4 There, of course, the asymmetry will not allow the spread to be outplayed.
By the way, has anyone calculated what percentage of guessing should be to beat this spread? But I've found out empirically, that a stable 54% of losses on EURUSD is unbelievable.) So, what to strive for? 60%? 85%?
By the way, has anyone calculated what the percentage of guesses should be in order to beat this spread? But empirically I've found, that a stable 54% on EURUSD is losing a lot))) So, what to strive for? 60%? 85%?
Maybe we are talking about cases when sl=tp? Otherwise the % of profit is little informative without their ratio.
If sl=tp, it would depend on their value. A rough calculation without taking some things into account:
mo=p*tp-(1-p)*sl-spread=(2p-1)*tp-spread>0, where p is the probability of winning
p>spread/2tp+0.5
if for example sl=tp=10p and spread is 2p then p>0.6
and if for example sl=tp=100p, it is enough p>0.51
Of course this is without slippage - only for a positive moe. But even with this you can go for broke, that's why it's better to have some reserve and it depends on MM.
I guess we are talking about cases where sl=tp? Otherwise % gain is not very informative without their ratio.
If sl=tp, it would depend on their value. A rough calculation without taking some things into account:
mo=p*tp-(1-p)*sl-spread=(2p-1)*tp-spread>0, where p is the probability of winning
p>spread/2tp+0.5
if for example sl=tp=10p and the spread is 2p then p>0.6
this is wrong in principle!
0<p<1 is probability
tp, sl are "kilos"
we can't put them in the same key