If we knew exactly how the price was moving... - page 8
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Does the distribution by bar (timing) of order openings change as they close?
It's not obvious somehow.It's not obvious somehow.
It's pretty obvious. Pure combinatorics.
coin toss: heads=+1, tails=-1. We start from zero, i.e. SB. Player bets on heads until he loses and stop loss (ends play) at -1.
If sl didn't trigger on the first roll, then it's +1 now
if on the second, then with probability 0.5 +2, with probability 0.5 0. On average 0.5*2+0.5*0=+1
if it didn't work on the third, the average would be 0.25*3+0.75*1=+1.5
etc. on the average the deviation will grow in the positive half-axis, if the stop-loss is not triggered, the same as for tp, only the deviation grows in the negative side. Dependence on the number of throws/time and the value of stop/stop
if on the second, with probability 0.5 +2, with probability 0.5 0.
I pull up the stop all the time, i.e. one can make a mistake only once and then have to start the game again. It turns out that if the stop is not triggered on the second roll, then with probability p=1. +2 and p=0. - zero.
For the experiment to be correct, the interdependence of the takedowns has to be eliminated. Obviously this is a problem with the algorithm.
The bribes are by definition independent, as the condition of independence of the price series increments is fulfilled (it is obtained by integration (commutative sum) of the normally distributed SV with MO=0). Therefore, any observed small dependence between bribes, is random and decreases with increasing number of transactions.
The bribes are by definition independent, because the condition of independence of the price series increments is fulfilled (it is obtained by integration (commutative sum) of the normally distributed SV with MO=0)
We need to see the code of the algorithm.
I pull up the stop all the time, i.e. you can only make a mistake once and then you have to start the game again. It turns out that if the stop is not triggered on the second throw, then with probability p=1. +2 and p=0. - zero.
You pull it up on every bar. During a bar you have a fixed stop, as in the coin example.
If you adapt a coin to your task, then it will look as follows: every 50 passes (for example), we pull up the stop at such-and-such a level from the current position. Conventionally 1bar=50 coin tosses (also by the way HLOC can be displayed). But conclusions remain the same.
gip писал(а) >> Нужно посмотреть код алгоритма.
There is nothing to catch. It's a standard HGC with known parameters on the cyclicality of the sequence - random.
you pull it up on every bar
.During a bar you have a fixed stop, like in the example with the coin.
If you adapt the coin to your task, the following will happen: every 50 passes (for example), we pull up the stop on such-and-such a level from the current position
.Conventionally 1bar=50 coin tosses (also by the way HLOC can be mapped). But the conclusions will remain the same.
So what is the final verdict?
What, the growth of "thick" tails behind the stop (sometimes twice as big as the stop itself (see fig. above) when pulling up the take is explainable?
There is nothing to catch. It uses a standard GSF with known parameters on the cyclicality of the sequence - random.
So, what is the final verdict?
What, the growth of "thick" tails behind the stop (sometimes twice as big as the stop itself (see fig. above) when pulling up the take is explainable?
yes imha. Because the shift towards the stop depends on the value of the take profit. When you decrease the take profit, the offset increases in the opposite direction when it is not triggered.
P.S.more precisely you need to look at your algorithm.
There is nothing to catch. It's a standard RNG with known parameters on the cyclicality of the sequence - random.
Not the algorithm for generating the sequence, but the execution algorithm from which you take the bribe data.
P.S.more precisely you need to look at your algorithm
Checking should be like this: if take profit didn't work, then check if stop loss did. Or vice versa. This condition introduces a dependence I wrote about.