Grail indicators - page 10

 
yosuf:
That bigwig has surely left his mark in P(c) and/or N(c), but if he is a newcomer, the latter N(c) will notice his arrival. Generally speaking, if he arrives in the future, we are doomed to an inaccurate prediction. C'est la vie, such is life (c), nothing more can be done. The search for criminals also starts with looking at file cabinets.


Remember what happens when a car with flashing lights squeezes through a traffic jam, speeding everyone in different directions? All the smartest ones immediately lurk behind it. What about carts in crowded markets... same effect. If you spot that trolley and follow it in time, you can make a decent profit.

 
yosuf:

No, just scrolling through the stages of its beginnings again in a popular version. Nostalgia. Still, we'll come to it. But, what the hell, perhaps it can be modernised or improved. The main thing is to finally get its essence to the masses. If even one follower appears, it is already good.

So far, its results since 2009 are the following: at М15, with TP=SL=700 points, the fixed lot is 0.1, without waiting for it; the profit or loss is registered at once at the signal change:

97645 bars in history
Modelled ticks 194264
Modeling quality n/a
Chart mismatch errors 0
Initial deposit 200000.00
Net profit 1398210.59
Total profit 3209397.02
Total loss -1811186.43
Profitability 1.77
Expected payoff 15.01
Absolute drawdown 22480.77
Maximum drawdown 151532.38 (9.41%)
Relative drawdown 20.09% (47048.74)
Total trades 93169
Short positions (% win) 48910 (69.32%)
Long positions (% win) 44259 (71.81%)
Profitable trades (% of all) 65685 (70.50%)
Loss trades (% of all) 27484 (29.50%)
Largest
profitable trade 500.00
Deal Deal loss -699.96
Average
48.86 Profit trade
losing trade -65.90
Maximum number
Continuous wins (profit) 423 (31339.78)
Continuous Losses (Loss) 270 (-48504.02)
Maximum
Continuous Profit (number of wins) 67397.05 (244)
Continuous loss (number of losses) -61605.61 (226)
Average
continuous winnings 25
Continuous loss 11

Why the initial deposit of 200,000 again? Are you willing to lay it out? Start with 200 and see the results!
 
borilunad:
Why the initial deposit of 200,000 again? Are you willing to lay it out? Start with 200 and see the results!
Exactly. It's better to make the initial deposit comparable to the real money that will be pumped into the account. There will be fewer illusions.
 
Mathemat:
Exactly. It is better to make the initial deposit comparable to the real money that will be pumped into the account. There will be fewer illusions.

But a big deposit will sober you up! ))
 
What's the problem ? 200000 = 0.1lot 20000 = 0.01lot 200$ in a cent account.
 
BeerGod:
What is the problem? 200000 = 0.1 lot 20000 = 0.01 lot 200$ in a cent account.

No problem - apart from the fact that Yusuf has already written that the lot size should be constant and about 0.1.

Well, with a 200,000 deposit and a very small for that deposit (0.1) the relative drawdown is 20%. It is too much. The drawdown at the same 0.1 lot is more or less acceptable at a hundred times less deposit.

P.S. I understand why such a 0.1 lot is needed here. Yusuf says that sometimes up to 85 positions are opened. So they pop up, two orders of magnitude...

 
BeerGod:
What's the problem ? 200000 = 0.1lot 20000 = 0.01lot 200$ in a cent account.
No! 200 with 0.01 in a normal account! Leverage 500.
 
yosuf:


B(c) = 1- E

E = Integral(from 0 to t) (t/τ)^(n-1)/G(n)*exp(-t/τ)dt - introduced, by me, function, so that E=H(in)+P(in) .

H(in)= (t/τ)^n/G(n+1)*exp(-t/τ)

P(B) =Integral (0 to t)(t/τ)^(n)/G(n+1)*exp(-t/τ)dt

G(n+1) =Integral(0 to infinite) x^n*exp(-x)dx -Hamma Euler function

G(n+1) = 1*2*3*....*n = n! -for integer values of n;

Let's go further. Let's see what is the nature of the change in the function P(c)

.

The influence of the parameter n on the development of the processes H(in) and P(in) in time tau:

 
yosuf:


B(c) = 1- E

E = Integral(from 0 to t) (t/τ)^(n-1)/G(n)*exp(-t/τ)dt - introduced, by me, function, so that E=H(in)+P(in) .

H(c)= (t/τ)^n/G(n+1)*exp(-t/τ)

P(B) =Integral (0 to t)(t/τ)^(n)/G(n+1)*exp(-t/τ)dt

G(n+1) =Integral(0 to infinite) x^n*exp(-x)dx -Hamma Euler function

G(n+1) = 1*2*3*....*n = n! - for integer values of n;


One more clarification, Yusuf.

Am I writing down the E function correctly? Is there any mistake?

 
borilunad:
Why the initial deposit of 200,000 again? Are you willing to lay it out? Start with 200 and see the result!
It's a cent, that much now and in a real account, = 2K$.
Reason: