"Miracle", "digital" "group" movement indicator - page 9

 
Mischek:

Once again . You have made an unsubstantiated assertion . To prove it, only the demonstration of your open orders is suitable .

I won't prove it. What's the point? If you doubt it, you can check it yourself.
 
forex-k:

I won't prove it. What's the point? Whoever doubts can check and see for themselves.


There really is no point. The more so that it is not an easy task - to open a demo, open a cross, lock in two majors and give an investment.

There is no point in saying that you need it.

 
forex-k:

100% synchronized!

It is basically impossible to synchronise them as the closing time of the bar (arrival of the last tick) is different for different currency pairs.

In addition, if you initially block/hedge 100% of a natural cross with a synthetic made of majors, the point value will change in time with changes in the price and it will lead to imbalances. I will not argue about the obvious anymore.

 

The perfect timing can be neglected, you can easily switch to open-ended mode, but it won't change the overall picture.

This dynamics is so insignificant, that even if it wasn't taken into account visually there wouldn't be any difference.

 
forex-k:


why prove what is obvious and is a fact.

here's a screenshot of my index from october 1, the pink line is the so called hedge which consists of two majors and what does it coincide with the cross? (synchronization and point values are accounted for!)

If you don't have a match there, post the code and you will get a popular explanation of why and how.

 
Mischek:


open a demo, open a cross, lock in two majors and give an investment .

Alternatively .
 

Sorry to intrude. But, the question came down to a locus of three currency pairs. It makes sense if you want to make money on swaps. But, why would you want to make so little money?

You solve a system of three linear equations, after exposing the expression, equating the coefficients at D(X), D(X1), D(X2) to zero

D(K1*X1/X + K2*X2/X + K3*X1/X2) = 0,

where D is a delta (I couldn't find such a symbol),

K1, K2, K3 - lot sizes for currency pairs,

X, X1, X2 - synthetic exchange rates. When you solve the expression, the system of equations can be written in crosses.

Further, if (K1, K2, K3) is a solution, then for any r the solution is (r*K1, r*K2, r*K3).

You choose a sign of r such that the swap is positive. You open and wait for the swaps to cover the spreads. As rates change, by solving the equation you can make a decision to adjust the loc.

Now about the delta operator, it has properties:

D(X/Y) = (D(X)*Y - X*D(Y))/(Y*Y);

D(X*Y) = D(X)*Y + X *D(Y);

D(X + Y) = D(X) + D(Y);

D(k*X) = k * D(X),

where k is a constant; X, Y are variables.

 
so how do you think the indices should be calculated correctly? ))))))))))))))))
 
trol222:
so how do you think the indices should be calculated correctly? ))))))))))))))))
Indexes should be calculated so that there is no loss of information. Indexes are synthetic majors. Accordingly, they make as much sense as the available majors.
 
hrenfx:
Indices must be calculated so that there is no loss of information. Indexes are synthetic majors. Accordingly, they make as much sense as the available majors.
I'll get into it again. They are not.
Reason: