The potential yield of the instrument.

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I have made some simple calculations of currency movements (against the US dollar) on different charts (MN, W1, D1) for different periods:

120 months (last 10 years), 50 weeks (year) and 80 days (current year dynamics)

Now you clearly see why e.g. the Australian and the New Zealander are called as high-yielding currencies


script:

//+------------------------------------------------------------------+
int start()
{
   int Limit=50;
   double Potential=0;

   for(int i=0; i< Limit; i++) Potential+=High[ i]/Low[ i]-1;
   Print("Средняя потенциальная доходность за период "+ Limit+" баров: "+DoubleToStr( Potential/ Limit,8));

   return(0);
}
//+------------------------------------------------------------------+
 

It seems more correct to calculate in this way:

- i runs along the extrema of the ZigZag;

- High and Low are extrema themselves.

Or simply the sum of all ZigZag "knees" divided by min. "Knee".

P.S. Considering the volumes, the majors will come out on top in terms of returns...

 
What a clever girl you are!)
 

sab1uk, how did you go from relative prices (like EUR/USD) to absolute values (abscissa axis on the charts), or am I missing something?

In general, the question of the potential profitability of an instrument is interesting, but cannot be determined by the methods suggested. For example, take an integrated NE (Brownian one-dimensional movement). The potential yield of such instrument is zero (by definition), and the volatility can be any. Therefore, it is not a yield characteristic.

 
The potential yield here refers to the hypothetical maximum return on the instrument.
 
mql4com >> :

It seems more correct to calculate in this way:

- i runs along the extrema of the ZigZag;

- High and Low are extrema themselves.

Or simply the sum of all ZigZag "knees" divided by min. or just the sum of all "knees" of the ZigZag divided by the min.

P.S. If you take volumes into account, the majors will come out on top in terms of profitability...

I do not know about volumes (if you mean trading volumes, not ticks)... they are probably more related to liquidity

this is a double-edged sword - liquidity is inversely proportional to risks but so is potential profitability

 

sab1uk, а как ты от относительных цен (типа EUR/USD) перешёл к абсолютным значениям (ось абсцисс на графиках), или я чего-то не догоняю?

In general, the question of the potential profitability of an instrument is interesting, but cannot be determined in the ways suggested. For example, take an integrated NE (Brownian one-dimensional motion). The potential return of such an instrument is zero (by definition) and the volatility can be anything. Therefore, it is not a yield characteristic.

the percentage cluster type, instead of the difference take the division minus one: High[i]/Low[i]-1

I think there is a middle ground between liquidity and volatility

 
mql4com писал(а) >>

It seems more correct to calculate in this way:

- i runs along the extrema of the ZigZag;

- High and Low are extrema themselves.

Or simply the sum of all the "knees" of the ZigZag divided by the min. "Knee".

P.S. If we take into account volumes, the majors will come out on top in profitability...

The sum of all knees reduced by the spread and then all multiplied by the point value.

 
Integer >> :

The sum of all knees reduced by the spread and then all multiplied by the point value.

Then there is the strongest dependence on the size of the min. knees:

The smaller the min. elbow, the higher the amount.

 
mql4com писал(а) >>

Then there is a strong correlation with the size of the min-knee:

The smaller the min. knee, the higher the amount.

We do min. knee = spread + 1 point - we take everything we can get.

 
Integer писал(а) >>

Do minimum knee = spread + 1 point - take all you can take.

Nah. On historical data, the optimal ZZ is when the average knee is equal to double the spread. There is no bigger hypothetical yield in the world! A step to the side will give a smaller yield (proven strictly).

Except, the value of such an estimate is zero...

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