From theory to practice - page 486
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Date , Time , Reading
2018.08.28 , 18:40 , -0.00011 ,
2018.08.28 , 18:39 , 0.00002 ,
2018.08.28 , 18:38 , -0.00024 ,
2018.08.28 , 18:37 , -0.00004 ,
2018.08.28 , 18:36 , -0.00019 ,
2018.08.28 , 18:35 , -0.00025 ,
2018.08.28 , 18:34 , 0.00001 ,
I am showing you a file with the following data (GMT+3, period 1440, eurusd till 15th day approximately)
Logically the reading goes out of some frames then there is a signal to open the position. How to determine this frame?
Waiting for advice from Alexander.
Date , Time , Reading
2018.08.28 , 18:40 , -0.00011 ,
2018.08.28 , 18:39 , 0.00002 ,
2018.08.28 , 18:38 , -0.00024 ,
2018.08.28 , 18:37 , -0.00004 ,
2018.08.28 , 18:36 , -0.00019 ,
2018.08.28 , 18:35 , -0.00025 ,
2018.08.28 , 18:34 , 0.00001 ,
I am showing you a file with the following data (GMT+3, period 1440, eurusd till 15th day approximately)
Logically the reading goes out of some frames then there is a signal to open the position. How to determine this frame?
Waiting for advice from Alexander.
We look at the amount of increments.
I see only one deal - I think you posted it here...
The deal is the chicest, but 1 trade in 2 weeks is not enough, of course. That's why I'm working on 8 pairs and I'm going to connect 4 more.
Looking at the amount of increments.
I only see one deal - I think you posted it here...
The deal is the best, but 1 deal in 2 weeks is not enough, of course. That's why I'm working on 8 pairs and I'm going to connect 4 more.
Alexander, from which point in the circle in your graph can you determine? How much from this hindsight graph should already be in the past by that point?
On the fact of going beyond the confidence interval =+-quantile*sqrt(c*lambda*t). What we have discussed in this thread and you were the initiator of this line of research on the price dependence of the "root of T" :))). In this case = +-quantile*(SUM(|returns|)/sqrt(1440)) in the sliding window =1440 (day).
On the fact of going beyond the confidence interval =+-quantile*sqrt(c*lambda*t). What we have discussed in this thread and you were the initiator of this line of research on the price dependence of the "root of T" :))). In this case = +-quantile*(SUM(|returns|)/sqrt(1440)) in the sliding window =1440 (day).
So, without lag, the product of c*lambda need not be evaluated, and the returns are taken from the regular minutes?
So, without lag, the product of c*lambda need not be evaluated, and returns are taken from ordinary minutes?
It's Eugene who takes from ordinary minutes:))) And I work with ticks in Erlang's flows.
So, once again:
Process variance.
D=s^2=c*lambda*t,
where:
c=SUM(|returns|)/t - speed
lambda=SUM(|returns|)/N - average increment
N - number real ticks in time window
t - time in seconds.
Don't worry, Vladimir - the Grail (at least 50% per month) will be found very soon and will be with you and Koldun in the first place.
See the sum of the increments.
So you have to plot the increments from these readings, then calculate the sum of these increments over the observation window?
Hm interesting.
I'm getting something wrong here.
I just summed the increments from the 3rd column in the 1440 window and that's it.
I just summed the increments from the 3rd column in the 1440 window and that was it.
I see, I did the increments and then counted the sum. I see something is wrong. Then I calculated the sum of those amounts again.
And if you look at these readings by themselves without summing up anything useful? At a guess there when going beyond +- 0.00060 the price returns.